url
string | fetch_time
int64 | content_mime_type
string | warc_filename
string | warc_record_offset
int64 | warc_record_length
int64 | text
string | token_count
int64 | char_count
int64 | metadata
string | score
float64 | int_score
int64 | crawl
string | snapshot_type
string | language
string | language_score
float64 |
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
http://scubageek.com/articles/wwwrefr.html
| 1,508,246,342,000,000,000 |
text/html
|
crawl-data/CC-MAIN-2017-43/segments/1508187821189.10/warc/CC-MAIN-20171017125144-20171017145144-00312.warc.gz
| 312,529,980 | 3,086 |
# Refraction of Light by Water
When a light beam strikes a plane interface between two media having different optical properties part of it is reflected and part is transmitted. The transmitted beam is refracted, or bent away from the direction of the incident beam. To explain this phenomenon one must resort to the wave theory of light, wherein light is an electromagnetic disturbance consisting of mutually perpendicular oscillating electric and magnetic fields, both of which are perpendicular to the direction of propagation. The theory predicts that the incident, reflected, and refracted beams, as well as the normal to the interface, are all coplanar.
In which direction do the wavefronts travel in the water? In the figure, point A lies at the intersection of a line of incident wave crests and the surface of the water. It must also lie on a line of refracted wave crests. Otherwise, the electrons at A would be required to simultaneously perform two different kinds of oscillations, which, of course, they do not. Lines of crests must therefore be continuous at the surface of the water. Accordingly, as the wavefront in air advances by one wavelength v(a)T (speed times period) and the intersection point A moves to point O, the corresponding wavefront in water must also move one wavelength v(w)T in a manner that maintains a match of wave crests at points A and O, as shown.
sin(i) = v(a)T / AO
sin(r) = v(w)T / BO
sin(i)/v(a) = sin(r)/v(w)
n(a)sin(i) = n(w)sin(r)
sin(r) = 0.75*sin(i)
| 343 | 1,506 |
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 3 | 3 |
CC-MAIN-2017-43
|
latest
|
en
| 0.941547 |
https://cboard.cprogramming.com/c-programming/109427-help-searching-array-longest-sequence-element-printable-thread.html
| 1,508,719,792,000,000,000 |
text/html
|
crawl-data/CC-MAIN-2017-43/segments/1508187825497.18/warc/CC-MAIN-20171023001732-20171023021732-00747.warc.gz
| 638,149,925 | 5,592 |
# Help with Searching Array for Longest Sequence of an Element
• 11-19-2008
w2look
Help with Searching Array for Longest Sequence of an Element
Hi,
I'm new to the forum and I need some help.
My task is to write a program which defines a function to search through a fixed size
array of integers and count the longest sequence (consecutive repetition) of a specific
element in the array. The array and the search element are to be passed as a parameter
to the function and the longest repetition is the result.
For example:
If the following array of 10 integers is input by the user:
1 1 2 2 3 1 1 1 2 3
and the element that you want to find the longest sequence of is 1,
then the program should return the number 3 as the answer.
(because the most consecutive repetitions of 1 in the array at any given point is 3)
I have finished the code to greet, prompt and receive the input from the user, store the
input into an array and execute a print statement for the answer. All of the for loops and
while loops are written including an exit value for when the user would like to quit. I have
created the function, declared the function prototypes and have the proper variables
being passed between the function and main.
However, I am having trouble getting my function to work. I am not sure how to correctly
go about writing the code for the actual searching of the array elements.
I know I need to look at each element separately.
I don't know how to put that into code.
I know I need to then determine if it is equal to the following element in the array.
I don't know how to put that into code.
I know that if it is equal to the next element then I should increment a counter by one to
count how many times it repeats and when the program hits a new number, the value in
the counter should be stored into a new array that will contain the values of the length
of each consecutive repeated number.
I don't know how to put that into code.
Once I have the array containing the lengths of the repetitions, I know how to write the
code to find the max value in that array and print it, but getting to that point is my problem.
If anyone has any suggestions as to how I should begin to code this function,
I would greatly appreciate your help.
.
• 11-19-2008
master5001
There is a book called Algorithms by Kneuth (i think that is the title of the book and the author... idk... to be honest I am just trying to shirk my spanish homework at the moment). It has some really good methodology to use.
• 11-19-2008
blurx
This reminds me of maximum sum subvector problem I had to do. For this you can just do, if(A[i] == 1 && A[i -1] == 1), the count would go up. Then that count would be initialized to some variable, I am calling it highest. Then check to see if the count goes over the highest, and if it does it will initialize it to highest. So then your function would return the variable highest.
This link is for that problem. See the similarities?
• 11-19-2008
w2look
Quote:
Originally Posted by blurx
For this you can just do, if(A[i] == 1 && A[i -1] == 1), the count would go up. Then that count would be initialized to some variable, I am calling it highest. Then check to see if the count goes over the highest, and if it does it will initialize it to highest. So then your function would return the variable highest.
Thanks blurx, but I'm not sure that will work for this problem.
I don't need to check if A[i] == 1
I think what I need to check is if A[i] == " number to be checked" then ++counter
I'm still not sure how to store the value of the counter and then reset the counter to 0,
so I can check to see if it is larger than the next value I create by checking the next sequence.
for instance, in the array I mentioned earlier, the program would check if the first element
of the array was equal to the number to be checked (1). Since it is, it would return a "true"
result and add 1 to the counter.
Then, the program should check to see if the 3rd element is equal to the number to be checked.
In this case, the 3rd element is 2 so it would return a "false" result.
At this point I need to store the value of the counter
(which is 2 for the length of the repeating sequence of 1's) and reset the counter.
Then I need to check if the 4th element of the array is equal to the number to be checked (1).
If it is, add one to the counter and check to see if the 5th is the same and so on.
Whenever the sequence terminates (the next element in the array is not the number being checked)
the program should check the value in the counter against it's previous value. If the new counter value
is larger than the previously stored value, it should be overwritten.
Once it has checked the entire array, it can then return the last value stored which should be the largest.
from my example, given the array
1 1 2 2 3 1 1 1 2 3
the program should check this array one element at a time until it finds the number to be checked (1)
when it finds the number(1), it should add one to the counter. It should then check to see if the next element is (1).
If so, it should add one to the counter. If not, the value of the counter at this point should be stored in a variable
and the counter should be reset to zero. The program should then continue checking the array from the spot it left
off, until it finds the number (1) again. When it does, it should add one to the counter, and so on.
Once it has checked the entire array, then it should return the final value stored in the variable.
I know exactly what I want to do, I just don't know the syntax or proper coding
structure to get it done.
If you or anyone else has more suggestions, I'm all ears.
• 11-19-2008
P4R4N01D
This should work (code tags to keep indentation):
Code:
```1. Read the number in to be searched for 2. Iterate through the list - If number to be checked is found, increment counter - If number found is not the number being looked for and counter > 0: - Store the counter in a seperate array of ints - Reset counter 3. If the array of ints is empty return 0 4. Perform a linear search on the seperate array to find largest element 5. Return largest value```
Hope this helps, there might be quicker ways but you need something to get started before thinking of optimisation.
• 11-19-2008
The standard way to iterate through an array is:
Code:
``` for(i = 0; i < sizeOfArray; i++) { //whatever code you need here. if(Array[i] == numberYouWant) count++; } /* then maybe something like this */ if(count > longestSequence) longestSequence = count;```
Post up your code so we can see what you're trying. If it's pseudo code, that's fine, also.
• 11-24-2008
w2look
Thanks Everyone
I just wanted to thank everyone for posting their suggestions.
I have figured it out with your help and working through the problem.
• 11-25-2008
iMalc
Ya know, a simplification of the Boyer-Moore string matching algorithm, but operating on an array of integers instead of characters would seem like an efficient idea for this.
Okay, what I mean is, after finding a sequence of length n, as you continue through the rest of the array, you may as well step by n each time, unless you hit the number you are looking for. I.e Once you've found a sequence of length n, there's no point in looking for anything but a sequence of length n+1 or greater.
That is all of course if your array is long enough to warrant something smarter than the obvious simple method.
| 1,813 | 7,486 |
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 2.734375 | 3 |
CC-MAIN-2017-43
|
latest
|
en
| 0.940289 |
https://origin.geeksforgeeks.org/how-to-calculate-quantiles-by-group-in-r/
| 1,670,593,153,000,000,000 |
text/html
|
crawl-data/CC-MAIN-2022-49/segments/1669446711396.19/warc/CC-MAIN-20221209112528-20221209142528-00282.warc.gz
| 459,378,637 | 27,396 |
# How to Calculate Quantiles by Group in R?
• Last Updated : 23 Dec, 2021
In this article, we will discuss how to calculate quantiles by the group in R programming language.
To obtain the required quartiles, quantile() function is used.
Syntax:
quantile( data, probs)
Parameters:
• data: data whose percentiles are to be calculated
• probs: percentile value
To group data, we use dplyr module. This module contains a function called group_by() in which the column to be grouped by has to be passed.
Syntax:
group_by(column_name)
To find quantiles of the grouped data we will call summarize method with quantiles() function.
Syntax:
summarize( function )
Example 1: Calculate Quantiles by group by summarizing one quartile with probability 0.5
## R
`# import library ` `library``(dplyr)` `# create dataframe` `df<-``data.frame``(x=``c``(2,13,5,36,12,50),` ` ``y=``c``(``'a'``,``'b'``,``'c'``,``'c'``,``'c'``,``'b'``))` `# create groups` `# calculate quantiles by group` `df %>% ``group_by``(y) %>%` ` ``summarize``(res=``quantile``(x,probs=0.5))`
Output:
Example 2: Calculate quantiles by group by summarizing three quartiles with probability 0.25, 0.5, and 0.75.
## R
`# import library ` `library``(dplyr)` `# create dataframe` `df<-``data.frame``(x=``c``(2,13,5,36,12,50),` ` ``y=``c``(``'a'``,``'b'``,``'c'``,``'c'``,``'c'``,``'b'``))` `# create groups` `# find quantiles` `df %>% ``group_by``(y) %>%` ` ``summarize``(first=``quantile``(x,probs=0.25),` ` ``second=``quantile``(x,probs=0.5),` ` ``third=``quantile``(x,probs=0.75))`
Output:
My Personal Notes arrow_drop_up
Related Articles
| 554 | 1,673 |
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 3.390625 | 3 |
CC-MAIN-2022-49
|
latest
|
en
| 0.599595 |
https://www.speedsolving.com/threads/l5c-sl5c-and-more-new-techniques-and-an-often-dismissed-3x3-method-now-fully-developed.87166/
| 1,656,793,306,000,000,000 |
text/html
|
crawl-data/CC-MAIN-2022-27/segments/1656104204514.62/warc/CC-MAIN-20220702192528-20220702222528-00201.warc.gz
| 1,034,115,461 | 18,276 |
# L5C, SL5C, and more: New techniques and an often dismissed 3x3 method now fully developed
#### Athefre
##### Member
I have developed full L5C and an alternate L5C that I’m calling SL5C (Shortened L5C). The algorithms are developed to fit the following method:
• First block on the left.
• 1x2x2 on the right.
• L5C
• L7E
This is a method that has been proposed a few times over the years. However, it had yet to be fully developed. This is because each time it had been proposed, the community suggested to the proposer that it isn’t good due to the final two steps. Well I think it is time to show that this method could be good. The advantages and disadvantages are obvious and can be debated. Over the past year I have spent a lot of time developing algorithm sets, recognition methods, sub-methods, and techniques for both L5C and L7E. It started with development for 42 and Waterman then I extended it further with development of full L5C, SL5C, and additional ideas. I have developed full L5C (two versions) and an L7E method. So this 3x3 method is now complete and fully useable.
L5C: This is the complete set of 614 algorithms to solve all five corners in a single step.
SL5C: The other I’m calling SL5C (Shortened L5C). In SL5C the algorithms are reduced by 1-2 moves by leaving out the last U* and R’ move for many cases. This takes advantage of one of the strengths of L7E which is that the two remaining R layer edge positions can be either FR + UR or UR + BR and the corners don’t need to be perfectly solved at the start of L7E. So the corners can be in any of five states (solved in any AUF, R U* offset, U R U* offset, U' R U* offset, or U2 R U* offset) and L7E still works as normal. This leaves the cube in the same pseudo L7E state that is in the 42 method but accomplished in fewer moves. The opposite of this could be developed as well, where the square is at dfR and the SL5C algs will often end with the two unsolved edge positions at FR and UR.
I’m calling the algorithm cutting technique truncation. This concept likely best applies to algorithm use. But it should work for some intuitive steps as well. In truncation, a certain number of moves are cut off of the end of a sequence of turns. Depending on the step on which this is used, it can reduce the movecount of the solve and allow the user to have better, more ergonomic solutions versus performing the full length of the step. At first it is being applied to L5C and corner methods in general. However we may be able to find other applications besides just corners. With Zipper, it would be F2L-1 > SL5C > L5E or L5EP if used with ZZ. Though the L5E and L5EP angles need to be accounted for. With Waterman and other corners first related methods, it would be used during CLL, EG, L5C, or any other corner method.
L5C and SL5C movecount and MCC:
SL5C cases that benefited from the L5C truncation:
CLL: 27/42 (64%).
TCLL+: 34/43 (79%).
TCLL-: 34/43 (79%).
UO: 121/162 (75%).
U+: 128/162 (79%).
U-: 134/162 (83%).
Total 478/614 (78%)
I spent a lot of time testing every algorithm so that great ones would be provided in the document. I am just one person though. There are likely better algorithms for a lot of cases and the kinds of algorithms that people use change over time due to new ones being found, new fingertricks, and new hardware. I'm already looking back at the sets that I started with and wanting to regenerate. So through a community effort, we will be able to find even better algorithms.
Recognition:
I have also developed a recognition method which follows the below steps.
• If the D layer corner is on the D layer, use CLL or TCLL recognition. If the D layer corner is on the U layer, determine the orientation of that corner.
• Find the orientation of three of the U layer corners that belong on the U layer. You can choose to always check the three that are on the U layer or any two on the U layer and the one that is currently on the D layer. Choose the positions that you want to check every solve.
• Recognize the pattern in those three corners from step 2.
I think this recognition method is similar to a recognition method that is in one of the Zipper documents. Though I don’t see a description of how it works. The difference here is that there is a full description of the recognition, the idea of the user having free choice of any three U layer corners, and the cases are completely organized by the sub-orientations.
L5C: This is the full 614 algorithms. It requires learning 614 recognition patterns and the full length algorithms.
SL5C: This is the same as L5C but with shortened algorithms.
CCMLL: This is used in the 42 method. A U layer corner is positioned above the right side 1x2x2 and an R turn is performed to make a pseudo CLL. The U layer corners are then solved using CLL or TCLL algorithms. This reduces the number of cases to 42 if an oriented corner is used or 168 if the corner is in any orientation.
My overall opinion: If you want the lowest movecount and the easiest recognition, SL5C is the best. If you want to learn as few algorithms as possible, CCMLL and the 42 method is the way to go.
• SSL5C: There is something about SL5C that I noticed. It seems to be that each case has a shorter or longer compliment version which is the same algorithm but with fewer or additional moves at the end. Example: The SL5C version of (U2) F U R U' R2 F' R U' R U2 R', (U2) F U R U' R2 F' R U' R done in reverse leads to a different looking L5C case. That case, after recognizing using NML5C, can be solved using (U') F U R U' R2 F' R, which is already a full L5C algorithm. This algorithm is the same as the original SL5C algorithm minus the last two moves. This might could be used as a memory aid to learn the two related algs in tandem. So the user would learn it as something like (U2) (U’) F U R U' R2 F' R U' R. This could cut down on the memorization effort. But it may also make recall difficult. This needs more research.
• Union L5C (UL5C): This is an L5C reduction method that I might finish development on sometime. The corners are reduced to the fully solved position, R’ away, U R’ away, U’ R’ away, or U2 R’ away from solved. It works by finding groups of four cases which can all be solved to those pseudo states using the exact same algorithm. The number of algorithms required to learn compared to full L5C is reduced (divided by 4) since the same algorithm will be used for four different cases. This reduction method is pretty much the inverse of CCMLL.
• I went down a long thought path of combining the truncation technique of SL5C with UL5C and CCMLL. These various combinations lead to interesting variants on the number of algorithms required to be learned, the number of recognition patterns, movecount, and recognition style. All of these combinations don't go to any extreme outside of SL5C or 42. None of them have a lower movecount than SL5C and none have fewer required algorithms or patterns than 42.
• A long term plan is to incorporate pseudo L5C recognition into ACRM. This means the use of the L/R stickers for recognition when any right side 1x2x2 is allowed to be built.
First block: 7 moves.
Square: 8 moves.
L5C: 10.47 moves + .75 AUF for standard L5C. 9.00 moves + .75 AUF for SL5C.
L7E: 17-18 moves. More advanced L7E methods or extensions to current L7E methods may be able to reach ~15 moves.
Total:
• ~43-44 moves with standard L5C and a standard L7E method.
• ~41-42 moves with SL5C and a standard L7E method.
• ~39-40 moves with SL5C and a more advanced L7E method.
• ~37-38 moves with SL5C, a more advanced L7E method, and pseudo blockbuilding (any right side 1x2x2).
This is based on full speedsolve reconstructions and the existing algorithm sets. The advanced L7E method number of 15 moves is an estimate right now. The estimate for pseudo blockbuilding is a reduction of 2 moves which is typical based on my experience over the years developing and using pseudo in various methods.
#### GodCubing
##### Member
Excellent work. L7E still needs some work; It is both very promising and very difficult. I think adapting ATCRM to TCMLL would make advanced 42 more manageable. A final note is that the CCMLL algs used in 42 are not optimized for the 7th unsolved edge or the R U* symmetry. Anyone who wants to help develop this method should consider joining the discord (https://discord.gg/MQzYg4Ptff).
| 2,125 | 8,406 |
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 2.8125 | 3 |
CC-MAIN-2022-27
|
latest
|
en
| 0.963401 |
https://www.monroecc.edu/etsdbs/MCCatPub.nsf/OnlineCatalog-CoursePreReq/223DE61427D2151F8525842E00479B92?OpenDocument&Catalog
| 1,719,328,431,000,000,000 |
text/html
|
crawl-data/CC-MAIN-2024-26/segments/1718198866143.18/warc/CC-MAIN-20240625135622-20240625165622-00789.warc.gz
| 778,877,497 | 11,025 |
# Course Descriptions
## MTH 156 - Mathematics for Elementary Teachers II
3 Credits
A continuation of the concepts of MTH 155, which develop the mathematical competency of the teacher or prospective teacher at the elementary level. Students will develop a comprehensive understanding of the mathematical curriculum recommended by the National Council of Teachers of Mathematics (NCTM) Standards using a problem solving approach with appropriate technology. Topics include probability, statistics, measurement, 2 and 3 dimensional geometry, transformational geometry, coordinate geometry, constructions, congruence and similarity. MTH 156 is a special interest course; check for availability.
Prerequisite: MTH 155 with a grade of C or better.
Learning Attributes: WR
New SUNY General Education: SUNY - Mathematics (and Quantitative Reasoning)
Retiring SUNY General Education: SUNY-M - Mathematics (SMAT)
MCC General Education: MCC-CT - Critical Thinking (MCT), MCC-QL - Quantitative Literacy (MQL)
Course Learning Outcomes
1. Compute various descriptive statistics which may include measures of central tendency, measures of spread, or measures of position.
2. Construct, interpret, or analyze the meaning of a variety of graphs.
3. Predict the probability of outcomes of simple or two-stage experiments or events.
4. Analyze geometric shapes which may include classification or distinguishing between attributes of various shapes.
5. Analyze geometric properties or relationships using principles of coordinate geometry.
6. Apply translations (slides), reflections (flips), rotations (turns), or dilations with 2-dimensional figures.
7. Convert within or between Metric and English systems of measurement.
8. Compute or analyze perimeter, area, or volume for a variety of geometric shapes.
9. Apply principles of congruence, similarity, or proportional reasoning to physical or mathematical situations in application problems.
10. Write explanations to problems at an appropriate elementary grade level.
Course Offered Fall, Spring
Use links below to see if this course is offered:
Fall Semester 2024
Summer Session 2024
| 421 | 2,132 |
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 2.8125 | 3 |
CC-MAIN-2024-26
|
latest
|
en
| 0.86748 |
https://lists.evolt.org/archive/Week-of-Mon-20090202/127273.html
| 1,722,870,543,000,000,000 |
text/html
|
crawl-data/CC-MAIN-2024-33/segments/1722640451031.21/warc/CC-MAIN-20240805144950-20240805174950-00220.warc.gz
| 297,015,054 | 2,460 |
# [thelist] Query Design Question
Luther, Ron Ron.Luther at hp.com
Tue Feb 3 09:33:54 CST 2009
```Hi Folks,
Need some input here.
Lets say I have two tables, a simplified Orders table:
Ord_Nbr Part_Nbr Open_Qty
111 AAA 10
222 AAA 5
333 CCC 25
And a simplified Inventory table:
Part_Nbr Inv_Qty
AAA 25
BBB 42
Now, lets say I have two clients who want to see two different reports:
Lucy wants to see all open orders, along with any inventory that might be available to fulfill those orders. Not getting fancy with promising inventory or decrementing on-hand quantities just yet, just looking for visibility on this report. E.g.
Ord_Nbr Part_Nbr Open_Qty Inv_Qty
111 AAA 10 25
222 AAA 5 25
333 CCC 25
Ricky, on the other hand, is an inventory manager. He wants to see all of his inventory, along with any open orders against that stock. (For this report the details are fine. We'll use the sum of the order_qty in a separate dashboard report.) E.g.
Part_Nbr Inv_Qty Ord_Nbr Open_Qty
AAA 25 111 10
AAA 25 222 5
BBB 42
Now, I would normally do this by building two data models; one with the tables left-joined and one with the tables right-joined and run one report off each data model.
I think other prople are planning to propose an alternative design where they would build ONE data model using an outer join to connect the two tables and then construct Lucy's report by adding a 'where Open_Qty > 0' filter and construct Ricky's report by using a 'where Inv_Qty > 0' filter.
Is that really a better approach? (That would be okay. I don't mind being wrong. I'm just not seeing it.)
Any theoretical performance advantages for one approach over the other? (I would think the outer-join would put me in full table scan mode for all the reports and the left-join / right-join would be more efficient. Or am I out in left field again?)
Thanks!
RonL.
```
| 530 | 1,938 |
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 2.5625 | 3 |
CC-MAIN-2024-33
|
latest
|
en
| 0.922772 |
http://metamath.tirix.org/fin23lem7.html
| 1,669,578,898,000,000,000 |
text/html
|
crawl-data/CC-MAIN-2022-49/segments/1669446710417.25/warc/CC-MAIN-20221127173917-20221127203917-00631.warc.gz
| 35,675,723 | 20,262 |
Metamath Proof Explorer < Previous Next > Nearby theorems Mirrors > Home > MPE Home > Th. List > fin23lem7 Unicode version
Theorem fin23lem7 8717
Description: Lemma for isfin2-2 8720. The componentwise complement of a nonempty collection of sets is nonempty. (Contributed by Stefan O'Rear, 31-Oct-2014.) (Revised by Mario Carneiro, 16-May-2015.)
Assertion
Ref Expression
fin23lem7
Distinct variable groups: , ,
Proof of Theorem fin23lem7
Dummy variable is distinct from all other variables.
StepHypRef Expression
1 n0 3794 . . . 4
2 difss 3630 . . . . . . . 8
3 elpw2g 4615 . . . . . . . . 9
43ad2antrr 725 . . . . . . . 8
52, 4mpbiri 233 . . . . . . 7
6 simpr 461 . . . . . . . . . . 11
76sselda 3503 . . . . . . . . . 10
87elpwid 4022 . . . . . . . . 9
9 dfss4 3731 . . . . . . . . 9
108, 9sylib 196 . . . . . . . 8
11 simpr 461 . . . . . . . 8
1210, 11eqeltrd 2545 . . . . . . 7
13 difeq2 3615 . . . . . . . . 9
1413eleq1d 2526 . . . . . . . 8
1514rspcev 3210 . . . . . . 7
165, 12, 15syl2anc 661 . . . . . 6
1716ex 434 . . . . 5
1817exlimdv 1724 . . . 4
191, 18syl5bi 217 . . 3
20193impia 1193 . 2
21 rabn0 3805 . 2
2220, 21sylibr 212 1
Colors of variables: wff setvar class Syntax hints: ->wi 4 <->wb 184 /\wa 369 /\w3a 973 =wceq 1395 E.wex 1612 e.wcel 1818 =/=wne 2652 E.wrex 2808 {crab 2811 \cdif 3472 C_wss 3475 c0 3784 ~Pcpw 4012 This theorem is referenced by: fin2i2 8719 isfin2-2 8720 This theorem was proved from axioms: ax-mp 5 ax-1 6 ax-2 7 ax-3 8 ax-gen 1618 ax-4 1631 ax-5 1704 ax-6 1747 ax-7 1790 ax-10 1837 ax-11 1842 ax-12 1854 ax-13 1999 ax-ext 2435 ax-sep 4573 This theorem depends on definitions: df-bi 185 df-or 370 df-an 371 df-3an 975 df-tru 1398 df-ex 1613 df-nf 1617 df-sb 1740 df-clab 2443 df-cleq 2449 df-clel 2452 df-nfc 2607 df-ne 2654 df-ral 2812 df-rex 2813 df-rab 2816 df-v 3111 df-dif 3478 df-in 3482 df-ss 3489 df-nul 3785 df-pw 4014
Copyright terms: Public domain W3C validator
| 974 | 1,989 |
{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 2.859375 | 3 |
CC-MAIN-2022-49
|
latest
|
en
| 0.109674 |
https://citizenmaths.com/frequency/degrees-per-second-to-degrees-per-hour
| 1,628,196,792,000,000,000 |
text/html
|
crawl-data/CC-MAIN-2021-31/segments/1627046157039.99/warc/CC-MAIN-20210805193327-20210805223327-00101.warc.gz
| 194,571,310 | 12,723 |
# Degree per Second to Degree per Hour Conversions
From
Degree per Second
• Action per Minute
• Attohertz
• Centihertz
• Cycle per Day
• Cycle per Hour
• Cycle per Microsecond
• Cycle per Millisecond
• Cycle per Minute
• Cycle per Month
• Cycle per Nanosecond
• Cycle per Picosecond
• Cycle per Second
• Cycle per Year
• Decahertz
• Decihertz
• Degree per Hour
• Degree per Millisecond
• Degree per Minute
• Degree per Second
• Exahertz
• Femtohertz
• Frame per Second
• Fresnel
• Gigahertz
• Hectohertz
• Hertz
• Kilohertz
• Megahertz
• Microhertz
• Millihertz
• Nanohertz
• Petahertz
• Picohertz
• Revolution per Minute
• Terahertz
• Yoctohertz
• Yottahertz
• Zeptohertz
• Zettahertz
To
Degree per Hour
• Action per Minute
• Attohertz
• Centihertz
• Cycle per Day
• Cycle per Hour
• Cycle per Microsecond
• Cycle per Millisecond
• Cycle per Minute
• Cycle per Month
• Cycle per Nanosecond
• Cycle per Picosecond
• Cycle per Second
• Cycle per Year
• Decahertz
• Decihertz
• Degree per Hour
• Degree per Millisecond
• Degree per Minute
• Degree per Second
• Exahertz
• Femtohertz
• Frame per Second
• Fresnel
• Gigahertz
• Hectohertz
• Hertz
• Kilohertz
• Megahertz
• Microhertz
• Millihertz
• Nanohertz
• Petahertz
• Picohertz
• Revolution per Minute
• Terahertz
• Yoctohertz
• Yottahertz
• Zeptohertz
• Zettahertz
Formula 7,690 deg/s = 7690 x 3600 deg/h = 27,684,000 deg/h
## How To Convert From Degree per Second to Degree per Hour
1 Degree per Second is equivalent to 3,600 Degrees per Hour:
1 deg/s = 3,600 deg/h
For example, if the Degree per Second number is (6.6), then its equivalent Degree per Hour number would be (23,760).
Formula:
6.6 deg/s = 6.6 x 3600 deg/h = 23,760 deg/h
## Degree per Second to Degree per Hour conversion table
Degree per Second (deg/s) Degree per Hour (deg/h)
0.1 deg/s 360 deg/h
0.2 deg/s 720 deg/h
0.3 deg/s 1,080 deg/h
0.4 deg/s 1,440 deg/h
0.5 deg/s 1,800 deg/h
0.6 deg/s 2,160 deg/h
0.7 deg/s 2,520 deg/h
0.8 deg/s 2,880 deg/h
0.9 deg/s 3,240 deg/h
1 deg/s 3,600 deg/h
1.1 deg/s 3,960.0 deg/h
1.2 deg/s 4,320 deg/h
1.3 deg/s 4,680 deg/h
1.4 deg/s 5,040 deg/h
1.5 deg/s 5,400 deg/h
1.6 deg/s 5,760 deg/h
1.7 deg/s 6,120 deg/h
1.8 deg/s 6,480 deg/h
1.9 deg/s 6,840 deg/h
2 deg/s 7,200 deg/h
2.1 deg/s 7,560 deg/h
2.2 deg/s 7,920.0 deg/h
2.3 deg/s 8,280 deg/h
2.4 deg/s 8,640 deg/h
2.5 deg/s 9,000 deg/h
2.6 deg/s 9,360 deg/h
2.7 deg/s 9,720 deg/h
2.8 deg/s 10,080 deg/h
2.9 deg/s 10,440 deg/h
3 deg/s 10,800 deg/h
3.1 deg/s 11,160 deg/h
3.2 deg/s 11,520 deg/h
3.3 deg/s 11,880 deg/h
3.4 deg/s 12,240 deg/h
3.5 deg/s 12,600 deg/h
3.6 deg/s 12,960 deg/h
3.7 deg/s 13,320 deg/h
3.8 deg/s 13,680 deg/h
3.9 deg/s 14,040 deg/h
4 deg/s 14,400 deg/h
4.1 deg/s 14,760.0 deg/h
4.2 deg/s 15,120 deg/h
4.3 deg/s 15,480 deg/h
4.4 deg/s 15,840.0 deg/h
4.5 deg/s 16,200 deg/h
4.6 deg/s 16,560 deg/h
4.7 deg/s 16,920 deg/h
4.8 deg/s 17,280 deg/h
4.9 deg/s 17,640 deg/h
5 deg/s 18,000 deg/h
5.1 deg/s 18,360 deg/h
5.2 deg/s 18,720 deg/h
5.3 deg/s 19,080 deg/h
5.4 deg/s 19,440 deg/h
5.5 deg/s 19,800 deg/h
5.6 deg/s 20,160 deg/h
5.7 deg/s 20,520 deg/h
5.8 deg/s 20,880 deg/h
5.9 deg/s 21,240 deg/h
6 deg/s 21,600 deg/h
6.1 deg/s 21,960 deg/h
6.2 deg/s 22,320 deg/h
6.3 deg/s 22,680 deg/h
6.4 deg/s 23,040 deg/h
6.5 deg/s 23,400 deg/h
6.6 deg/s 23,760 deg/h
6.7 deg/s 24,120 deg/h
6.8 deg/s 24,480 deg/h
6.9 deg/s 24,840 deg/h
7 deg/s 25,200 deg/h
7.1 deg/s 25,560 deg/h
7.2 deg/s 25,920 deg/h
7.3 deg/s 26,280 deg/h
7.4 deg/s 26,640 deg/h
7.5 deg/s 27,000 deg/h
7.6 deg/s 27,360 deg/h
7.7 deg/s 27,720 deg/h
7.8 deg/s 28,080 deg/h
7.9 deg/s 28,440 deg/h
8 deg/s 28,800 deg/h
8.1 deg/s 29,160 deg/h
8.2 deg/s 29,520.0 deg/h
8.3 deg/s 29,880.0 deg/h
8.4 deg/s 30,240 deg/h
8.5 deg/s 30,600 deg/h
8.6 deg/s 30,960 deg/h
8.7 deg/s 31,320.0 deg/h
8.8 deg/s 31,680.0 deg/h
8.9 deg/s 32,040 deg/h
9 deg/s 32,400 deg/h
9.1 deg/s 32,760 deg/h
9.2 deg/s 33,120 deg/h
9.3 deg/s 33,480 deg/h
9.4 deg/s 33,840 deg/h
9.5 deg/s 34,200 deg/h
9.6 deg/s 34,560 deg/h
9.7 deg/s 34,920 deg/h
9.8 deg/s 35,280 deg/h
9.9 deg/s 35,640 deg/h
10 deg/s 36,000 deg/h
20 deg/s 72,000 deg/h
30 deg/s 108,000 deg/h
40 deg/s 144,000 deg/h
50 deg/s 180,000 deg/h
60 deg/s 216,000 deg/h
70 deg/s 252,000 deg/h
80 deg/s 288,000 deg/h
90 deg/s 324,000 deg/h
100 deg/s 360,000 deg/h
110 deg/s 396,000 deg/h
| 1,875 | 4,301 |
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 3.125 | 3 |
CC-MAIN-2021-31
|
latest
|
en
| 0.335197 |
https://www.queryhome.com/puzzle/27788/how-many-squares-are-on-a-chessboard-more-than-64-or-less
| 1,545,006,869,000,000,000 |
text/html
|
crawl-data/CC-MAIN-2018-51/segments/1544376828018.77/warc/CC-MAIN-20181216234902-20181217020902-00388.warc.gz
| 1,025,509,337 | 31,465 |
# How many squares are on a chessboard? More than 64! or less?
35 views
How many squares are on a chessboard? More than 64! or less?
posted Jul 11
64 + 49 + 36 + 25 + 16 + 9 + 4 + 1 = 204 squares which is less than 64!
1, 8x8 square
4, 7x7 squares
9, 6x6 squares
16, 5x5 squares
25, 4x4 squares
36, 3x3 squares
49, 2x2 squares
64, 1x1 squares
| 140 | 345 |
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 2.890625 | 3 |
CC-MAIN-2018-51
|
latest
|
en
| 0.911339 |
https://wordpress.howeverythingworks.org/tag/electric-power-distribution/
| 1,568,775,620,000,000,000 |
text/html
|
crawl-data/CC-MAIN-2019-39/segments/1568514573176.51/warc/CC-MAIN-20190918024332-20190918050332-00349.warc.gz
| 753,867,010 | 21,793 |
## How big is the electric circuit that powers a lamp in my house?
#### When electricity comes out of the wall and through a lamp, where does the circuit loop complete? Does the circuit go all the way back to the power plant? — J, Florida
The electric circuit that powers your lamp extends only as far as a nearby transformer. That transformer is located somewhere near your house, probably as a cylindrical object on a telephone pole down the street or as a green box on a side lawn a few houses away.
A transformer conveys electric power from one electric circuit to another. It performs this feat using several electromagnetic effects associated with changing electric currents—changes present in the alternating current of our power grid. In this case, the transformer is moving power from a high-voltage neighborhood circuit to a low-voltage household circuit.
For safety, household electric power uses relatively low voltages, typically 120 volt in the US. But to deliver significant amounts of power at such low voltages, you need large currents. It’s analogous to delivering hydraulic power at low pressures; low pressures are nice and safe, but you need large amounts of hydraulic fluid to carry much power. There is a problem, however, with sending low voltage electric power long distances: it’s inefficient because wires waste power as heat in proportion to the square of the electric current they carry. Using our analogy again, sending hydraulic power long distances as a large flow of hydraulic fluid at low pressure is wasteful; the fluid will rub against the pipes and waste power as heat.
To send electric power long distances, you do better to use high voltages and small currents (think high pressure and small flows of hydraulic fluid). That requires being careful with the wires because high voltages are dangerous, but it is exactly how electric power travels cross-country in the power grid: very high voltages on transmission lines that are safely out of reach.
Finally, to move power from the long-distance high-voltage transmission wires to the short-distance low-voltage household wires, they use transformers. The long-distance circuit that carries power to your neighborhood closes on one side of the transformer and the short-distance circuit that carries power to your lamp closes on the other side of the transformer. No electric charges pass between those two circuits; they are electrically insulated from one another inside the transformer. The electric charges that are flowing through your lamp go round and round that little local circuit, shuttling from the transformer to your lamp and back again.
## Is it safe to locate my dog’s bed near the power adapter for the telephone?
#### My dog’s bed is on the floor just to the left and below the transformer plug for our house phone. She has been sleeping there for years. She has been experiencing problems lately and I would like to know if the transformer could be emitting some type of harmful waves that could be making her not feel well. — SH, Florida
While I’m sorry to hear that your dog isn’t well, I doubt that electromagnetic fields are responsible for her infirmities. The fields from the telephone adapter are too weak to have any significant effect and 60-Hz electromagnetic fields don’t appear to be dangerous even at considerably stronger levels.
To begin with, plug-in power adapters are designed to keep their electromagnetic fields relatively well contained. They’re engineered that way not because of safety concerns but because their overall energy efficiencies would diminish if they accidentally conveyed power to their surroundings. Keeping their fields inside keeps their energy inside, where it belongs. Moreover, any electric and magnetic fields emerging from an adapter probably don’t propagate as waves and instead fall off exponentially with distance. As a result, it should be fairly difficult to detect electric or magnetic fields more than a few inches from the adapter.
Even if the adapter did project significant electric and magnetic fields all the way to where your dog sleeps, it’s still unlikely that they would cause any harm. For years, researchers have been looking for a correlation between high-voltage electric power lines and a variety of human illnesses, notably childhood cancers such as leukemia. As far as I know, no such correlation has ever been demonstrated. In all likelihood, if there are any risks to being near 60-Hz electric or magnetic fields, those risks aren’t large enough to be easily recognized.
In contrast to power adapters, cell phones deliberate emit electric and magnetic fields in order to communicate with distant receivers on cell phone towers. Those fields are woven together to form electromagnetic waves that propagate long distances and definitely don’t vanish inches from a cell phone. Any electromagnetic hazard due to a power adapter pales in comparison to the same for cell phones.
Furthermore, cell phone operate at much higher frequencies than the alternating current power line. A typical cell phone frequency is approximately 1 GHz (1,000,000,000 Hz), while ordinary alternating current electric power operates at 60 Hz (50 Hz in Europe). Higher frequencies carry more energy per quanta or “photon” and are presumably more dangerous. But even though cell phones are held right against heads and radiate microwaves directly into brain tissue, they still doen’t appear to be significantly dangerous. As unfond as I am of cell phones, I can’t condemn them because of any proven radiation hazard. Their biggest danger appears to be driving with them; I don’t understand why they haven’t been banned from the hands of drivers.
Lastly, there are no obvious physical mechanisms whereby weak to moderate electric and magnetic fields at 60-Hz would cause damage to human or canine tissue. We’re essentially non-magnetic, so magnetic fields have almost no effect on us. And electric fields just push charges around in us but that alone doesn’t cause any obvious trouble. Research continues into the safety of electromagnetic fields at all frequencies, but this low-frequency stuff (power lines and cell phones) doesn’t seem to be unsafe.
## Is it safe to turn off computer equipment by turning off the power strip?
#### Does it matter how I turn off electronic devices? I have installed a power surge strip and it’s easiest for me to simply turn off that strip. Is it better for the devices to turn them off individually first? For the computer itself, I perform the shutdown procedure first. — A, Seattle, Washington
As long you shutdown the computer first, turning off the power strip is fine. Essentially all modern household computer devices are designed to shut themselves down gracefully when they lose electrical power and that’s exactly what they down when you turn off the power strip.
In fact, turning off the power strip is likely to save energy as well. Many computer devices have two different “off” switches: one that stops them from doing their normal functions and one that actually cuts off all electrical power. Computers in particular don’t really turn off until you reach around back and flip the real power switch on the computer’s power supply. The same is true of television monitors and home theater equipment.
In general, any device that has a remote control or that can wake itself up to respond to a pretty button or to some other piece of equipment is never truly off until you shut off its electrical power. Our homes are now filled with electronic gadgets that are always on, waiting for instructions. Keeping them powered up even at a low level consumes a small amount of electrical power and it adds up. Last I heard, this always-on behavior of our gadgets consumes something on the order of 1% of our electrical power. Whatever it is, it’s too much. So by turning off your power strip and completely stopping the flow of power to your computer, your speakers, your monitor, etc., you are saving energy. You lose the convenience of being able to turn everything on from your couch with a remote, but who cares. Energy is too precious to waste for such nonessential conveniences.
## Is DC power transmission possible?
#### In your response to Question 891, you wrote of the advantages of alternating current power transmission. Hasn’t lately there been some discussion of going to DC power transmission? I believe it is supposed to have superior operating properties when transmitting power over large distances. I have tried to find the reference, I think I came across the comment either in New Scientist or Scientific American. — JM, United Kingdom
You’re right that DC (direct current) power transmission has some important advantages of AC (alternating current) power transmission. In alternating current power transmission, the current reverses directions many times per second and during each reversal there is very little power being transmitted. With its power surging up and down rhythmically, our AC power distribution system is wasting about half of its capacity. It’s only using the full capacity of its transmission lines about half of each second. Direct current power, in contrast, doesn’t reverse and can use the full capacity of the transmission lines all the time.
DC power also avoids the phase issues that make the AC power grid so complicated and fragile. It’s not enough to ensure that all of the generators on the AC grid are producing the correct amounts of electrical power; those generators also have to be synchronized properly or power will flow between the generators instead of to the customers. Keeping the AC power grid running smoothly is a tour-de-force effort that keeps lots of people up at night worrying about the details. With DC power, there is no synchronization problem and each generating plant can concentrate on making sure that their generators are producing the correct amounts of power at the correct voltages.
Lastly, alternating currents tend to flow on the outsides of conductors due to a self-interaction between the alternating current and its own electromagnetic fields. For 60-cycle AC, this "skin effect" is about 1 cm for copper and aluminum wires. That means that as the radius of a transmission line increases beyond about 1 cm, its current capacity stops increasing in proportion to the cross section of the wire and begins increasing in proportion to the surface area of the wire. For very thick wires, the interior metal is wasted as far as power delivery is concerned. It’s just added weight and cost. Since direct current has no skin effect, however, the entire conductor can be carry current and there is no wasted metal. That’s a big plus for DC power distribution.
The great advantage of AC power transmission has always been that it can use transformers to convey power between electrical circuits. Transformers make it easy to move AC power from a medium-voltage generating circuit to an ultrahigh-voltage transmission line circuit to a medium-voltage city circuit to a low-voltage neighborhood circuit. DC power transmission can’t use transformers directly because transformers need alternating currents to move power from circuit to circuit. But modern switching electronics has made it possible to convert electrical power from DC to AC and from AC to DC easily and efficiently. So it is now possible to move DC power between circuits by converting it temporarily into AC power, sending it through a transformer, and returning it to DC power. They can even use higher frequency AC currents and consequently smaller transformers to move that power between circuits. It’s a big win on all ends. While I haven’t followed the developments in this arena closely, I would not be surprised if DC power transmission started to take hold in the United State as we transition from fossil fuel power plants to renewable energy sources. Using those renewable sources effectively will require that we handle long distance transmission better than we do now and we’ll have to develop lots of new transmission infrastructure. It might well be DC transmission.
## Are there high voltages around fluorescent lamps?
#### Do ballasts of fluorescent light fixtures produce a high voltage arc that ionizes gases in the tube during start up? If so what sort of voltages are we talking about? — SC, Australia
A traditional fluorescent lamp needs a ballast to limit the current flowing through its gas discharge. That’s because gas discharges have strange electrical characteristics, most notably a regime of “negative” electrical resistance: the voltage drop across the discharge actually decreases as the current in the discharge increases. If you connect a gas discharge lamp to a voltage source without anything to limit the current and start the discharge, the current flowing through the lamp will rise essentially without limit and the lamp will quickly destroy itself. As a kid, I blew up several small neon lamps by connecting them directly to the power line without any current limiter. That was not a clever or safe idea, so don’t try it!
The standard current limiter for fluorescent lamps and other discharge lamps that are powered from 60-cycle (or 50-cycle) alternating current has been an electromagnetic coil known as a ballast. When that coil is in series with the discharge, the coil’s self-inductance limits how quickly the current flowing through the lamp can rise and therefore how much power the lamp can consume before the alternating current reverses direction. The discharge winks on and off with each current reversal and never draws more current than it can tolerate. Unfortunately, the lamp’s light also winks on and off and some people can see that flicker, especially with their peripheral vision.
Actually, the ballast usually has another job to do in a traditional fluorescent lamp: it acts as a transformer to provide the current needed to heat the electrode filaments at the ends of the lamp. Heating those electrodes helps drive electrons out of the metal and into the lamp’s gas so that the gas becomes electrically conducting. In total then, the ballast receives alternating current electric power from the power line and prepares it so that all the lamp filaments are heated properly and a limited current flows through the lamp from one electrode to the other.
In modern fluorescent lamps with heated electrodes, however, the role of the ballast has been usurped by a more sophisticated electronic power conditioning device. That device converts 60-cycle alternating current electric power into a series of electrical energy pulses, typically at about 40,000 pulses per second, and delivers them to the lamp. The lamp’s flicker is almost undetectable because it is so fast and the limited energy in each pulse prevents the discharge from consuming too much power. It’s a much better system. Compact fluorescent lamps use it exclusively.
So where might high voltage fit into this story? Well, there are some fluorescent lamps that don’t heat their electrodes with filaments. They rely on the discharge itself to drive electrons out of the electrodes and into the gas to sustain the discharge. But that begs the question: “how does such a lamp start its discharge?” It uses high voltage. Because of cosmic rays and natural radioactivity, gases always have some electric charges in them: ions and electrons. When the voltage difference between the two ends of the lamp becomes very large, the electric field in the lamp propels those naturally occurring ions and electrons into the constituents of the lamp violently enough to start the lamp’s discharge. The voltages needed to start these “cold cathode” lamps are typically in the low thousands of volts. For example, the cold cathode fluorescent lamps used in laptop computer displays start at about 2000 volts and then operate at much lower voltages.
## Is it important to ground a microwave oven?
#### A co-worker who is an intelligent electrical engineer said an ungrounded microwave is dangerous because microwaves can then escape through the holes in the door. Aside from the electrical dangers, I disagreed because I think it is just the size of the holes vs. the wavelength of the microwaves. Does lack of a ground allow some microwaves to escape through the holes in the microwave door? — LG, Maine
You’re right. Whether the microwave oven is grounded or not makes no difference on its screen’s ability to prevent microwave leakage. In fact, the whole idea of grounding something is nearly meaningless at such high frequencies. Since electrical influences can’t travel faster than the speed of light and light only travels 12.4 cm during one cycle of the oven’s microwaves, the oven can’t tell if it’s grounded at microwave frequencies; its power cord is just too long and there just isn’t time for charge to flow all the way through that cord during a microwave cycle.
When you ground an appliance, you’re are making it possible for electric charge to equilibrate between that appliance and the earth. The earth is approximately neutral, so a grounded appliance can’t retain large amounts of either positive or negative charge. That’s a nice safety feature because it means that you won’t get a shock when you touch the appliance, even if one of its power wires comes loose and touches the case. Any charge that the power wire tries to deposit on the case will quickly flow to the earth as the appliance and earth equilibrate.
But charge can’t escape from the appliance through the grounding wire instantly. Light takes about 1 nanosecond to travel 1 foot and electricity takes a little longer than that. For charge to leave your appliance for the earth might well require 50 nanoseconds or more. That’s not a problem for ordinary power distribution, so grounding is generally a great idea. Each cycle of the 60-Hz AC power in the U.S. takes 18 milliseconds to complete, so the appliance and earth have plenty of time to equilibrate with one another. But a cycle of the microwave power in the oven takes less about 0.4 nanoseconds to complete and there’s just no time for the appliance and earth to equilibrate. At microwave frequencies, the electric current flowing through a long wire is wavelike, meaning that at one instant in time the wire has both positive and negative patches, spaced half a wavelength apart along its length. It’s carrying an electromagnetic ripple.
The metal screen on the oven’s door has to reflect the microwaves all by itself. It does this without a problem because the holes are so much smaller than 12.4 centimeters that currents easily flow around them during a cycle of the microwaves. Those currents are able to compensate for the holes in the screens and cause the microwaves to reflect perfectly.
## If a bird lands on a high-voltage wire, will it be injured?
#### A bird lands on an uninsulated 10,000 volt power line. Will it become extra crispy? — RKS, Texas
No. Birds do this all the time. What protects the bird is the fact that it doesn’t complete a circuit. It touches only one wire and nothing else. Although there is a substantial charge on the power line and some of that charge flows onto the bird when it lands, the charge movement is self-limiting. Once the bird has enough charge on it to have the same voltage as the power line, charge stops flowing. And even though the power line’s voltage rises and falls 60 times a second (or 50 times a second in some parts of the world), the overall charge movement at 10,000 volts just isn’t enough to bother the bird much. At 100,000 volts or more, the charge movement is uncomfortable enough to keep birds away, so you don’t see them landing on the extremely high-voltage transmission lines that travel across vast stretches of countryside.
The story wouldn’t be the same if the bird made the mistake of spanning the gap from one wire to another. In that case, current could flow through the bird from one wire to the other and the bird would run the serious risk of becoming a flashbulb. Squirrels occasionally do this trick when they accidentally bridge a pair of wires. Some of the unexpected power flickers that occur in places where the power lines run overhead are caused by squirrels and occasionally birds vaporizing when they let current flow between power lines.
## Lightbulbs and Power
#### What does it mean if a light bulb uses 60 watts? — B, Los Angeles
The watt is a unit of power, equivalent to the joule-per-second. One joule is about the amount of energy it takes to raise a 12 ounce can of soda 1 foot. A 60 watt lightbulb uses 60 joules-per-second, so the power it consumes could raise a 24-can case of soda 2.5 feet each second. Most tables are about 2.5 feet above the floor. Next time you leave a 60-watt lightbulb burning while you’re not in the room, imagine how tired you’d get lifting one case of soda onto a table every second for an hour or two. That’s the mechanical effort required at the generating plant to provide the 60-watts of power you’re wasting. If don’t need the light, turn off lightbulb!
## Do brownouts or other power outages damage appliances?
#### If a home looses some of its power during a power outage and the lights shine dim, will it burn up the motor in the refrigerator? Will it damage other appliances (TV, VCR. stereo. etc)? Should the main disconnect be shut off? — J, Ohio
Power outages come in a variety of types, one of which involves a substantial decrease in the voltage supplied to your home. The most obvious effect of this voltage decrease is the dimming of the incandescent lights, which is why it’s called a “brownout.” The filament of a lightbulb is poor conductor of electricity, so keeping an electric charge moving through it steadily requires a forward force. That forward force is provided by the voltage difference between the two wires: the one that delivers charges to the filament and the one that collects them back from the filament. As the household voltage decreases, so does the force on each charge in the filament. The current passing through the filament decreases and the filament receives less electric power. It glows dimly.
At the risk of telling you more than you ever want to know, I’ll point out that the filament behaves approximately according to Ohm’s law: the current that flows through it is proportional to the voltage difference between its two ends. The larger that voltage difference, the bigger the forces and the more current that flows. This ohmic behavior allows incandescent lightbulbs to survive decreases in voltage unscathed. They don’t, however, do well with increases in voltage, since they’ll then carry too much current and receive so much power that they’ll overheat and break. Voltage surges, not voltage decreases, are what kill lightbulbs.
The other appliances you mention are not ohmic devices and the currents that flow through them are not simply proportional to the voltage supplied to your home. Motors are a particularly interesting case; the average current a motor carries is related in a complicated way to how fast and how easily it’s spinning. A motor that’s turning effortlessly carries little average current and receives little electric power. But a motor that is struggling to turn, either because it has a heavy burden or because it can’t obtain enough electric power to overcome starting effects, will carry a great deal of average current. An overburdened or non-starting motor can become very hot because it’s wiring deals inefficiently with the large average current, and it can burn out. While I’ve never heard of a refrigerator motor dying during a brownout, it wouldn’t surprise me. I suspect that most appliance motors are protected by thermal sensors that turn them off temporarily whenever they overheat.
Modern electronic devices are also interesting with respect to voltage supply issues. Electronic devices operate on specific internal voltage differences, all of which are DC — direct current. Your home is supplied with AC — alternating current. The power adapters that transfer electric power from the home’s AC power to the device’s DC circuitry have evolved over the years. During a brownout, the older types of power adapters simply provide less voltage to the electronic devices, which misbehave in various ways, most of which are benign. You just want to turn them off because they’re not working properly. It’s just as if their batteries are worn out.
But the most modern and sophisticated adapters are nearly oblivious to the supply voltage. Many of them can tolerate brownouts without a hitch and they’ll keep the electronics working anyway. The power units for laptops are a case in point: they can take a whole range of input AC voltages because they prepare their DC output voltages using switching circuitry that adjusts for input voltage. They make few assumptions about what they’ll be plugged into and do their best to produce the DC power required by the laptop.
In short, the motors in your home won’t like the brownout, but they’re probably protected against the potential overheating problem. The electronic appliances will either misbehave benignly or ride out the brownout unperturbed. Once in a while, something will fail during a brownout. But I think that most of the damage is down during the return to normal after the brownout. The voltages bounce around wildly for a second or so as power is restored and those fluctuations can be pretty hard some devices. It’s probably worth turning off sensitive electronics once the brownout is underway because you don’t know what will happen on the way back to normal.
## I understand how a transformer changes voltage, but how does it regulate the amp…
#### I understand how a transformer changes voltage, but how does it regulate the amperage? – DE
A transformer’s current regulation involves a beautiful natural feedback process. To begin with, a transformer consists of two coils of wire that share a common magnetic core. When an alternating current flows through the primary coil (the one bringing power to the transformer), that current produces an alternating magnetic field around both coils and this alternating magnetic field is accompanied by an alternating electric field (recall that changing magnetic fields produce electric fields). This electric field pushes forward on any current passing through the secondary coil (the one taking power out of the transformer) and pushes backward on the current passing through the primary coil. The net result is that power is drawn out of the primary coil current and put into the secondary coil current.
But you are wondering what controls the currents flowing in the two coils. The circuit it is connected to determines the current in the secondary coil. If that circuit is open, then no current will flow. If it is connected to a light bulb, then the light bulb will determine the current. What is remarkable about a transformer is that once the load on the secondary coil establishes the secondary current, the primary current is also determined.
Remember that the current flowing in the secondary coil is itself magnetic and because it is an alternating current, it is accompanied by its own electric field. The more current that is allowed to flow through the secondary coil, the stronger its electric field becomes. The secondary coil’s electric field opposes the primary coil’s electric field, in accordance with a famous rule of electromagnetism known as Lenz’s law. The primary coil’s electric field was pushing backward on current passing through the primary coil, so the secondary coil’s electric field must be pushing forward on that current. Since the backward push is being partially negated, more current flows through the primary coil.
The current in the primary coil increases until the two electric fields, one from the primary current and one from the secondary current, work together so that they extract all of the primary current’s electrostatic energy during its trip through the coil. This natural feedback process ensures that when more current is allowed to flow through the transformer’s secondary coil, more current will flow through the primary coil to match.
| 5,520 | 28,249 |
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 2.890625 | 3 |
CC-MAIN-2019-39
|
longest
|
en
| 0.947084 |
https://www.nickzom.org/blog/tag/depth-of-available-water/
| 1,603,682,396,000,000,000 |
text/html
|
crawl-data/CC-MAIN-2020-45/segments/1603107890273.42/warc/CC-MAIN-20201026031408-20201026061408-00065.warc.gz
| 829,602,098 | 27,193 |
How to Calculate and Solve for Depth of Available Water | Irrigation Water Requirement
The image above represents depth of available water.
To compute for depth of available water, three essential parameters are needed and these parameters are Apparent Specific Gravity of the Soil (s), Depth of the Root Zone of the Plant (d) and Water Content (m).
The formula for calculating depth of available water:
dw = s . d . m
Where:
dw = Depth of Available Water
s = Apparent Specific Gravity of the Soil
d = Depth of the Root Zone of the Plant
m = Water Content
Let’s solve an example;
Find the depth of available water when the apparent sepcific gravity of the soil is 10, the depth of the root zone of the plant is 8 and the water content is 4.
This implies that;
s = Apparent Specific Gravity of the Soil = 10
d = Depth of the Root Zone of the Plant = 8
m = Water Content = 4
dw = s . d . m
dw = (10) . (8) . (4)
dw = 320
Therefore, the depth of available water is 320.
Calculating for the Apparent Specific Gravity of the Soil when the Depth of Available Water, the Depth of the Root Zone of the Plant and the Water content is Given.
s =dw / d . m
Where;
s = Apparent Specific Gravity of the Soil
dw = Depth of Available water
d = Depth of the Root Zone of the Plant
m = Water Content
Let’s Solve an example;
Find the apparent specific gravity of the soil when the depth of available water is 60, the depth of the root zone of the plant is 10 and the water content is 3.
This implies that;
dw = Depth of Available water = 60
d = Depth of the Root Zone of the Plant = 10
m = Water Content = 3
s =dw / d . m
s =60 / 10 . 3
s =60 / 30
s = 2
Therefore, the apparent specific gravity of the soil is 2.
| 455 | 1,714 |
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 4.1875 | 4 |
CC-MAIN-2020-45
|
latest
|
en
| 0.812455 |
https://file.scirp.org/Html/2-2310772_79635.htm
| 1,686,151,298,000,000,000 |
text/html
|
crawl-data/CC-MAIN-2023-23/segments/1685224653930.47/warc/CC-MAIN-20230607143116-20230607173116-00609.warc.gz
| 299,013,178 | 14,645 |
Research on Model and Algorithm of Task Allocation and Path Planning for Multi-Robot
Open Journal of Applied Sciences
Vol.07 No.10(2017), Article ID:79635,9 pages
10.4236/ojapps.2017.710037
Research on Model and Algorithm of Task Allocation and Path Planning for Multi-Robot
Zhenping Li, Xueting Li*
School of Information, Beijing Wuzi University, Beijing, China
Received: September 15, 2017; Accepted: October 13, 2017; Published: October 16, 2017
ABSTRACT
Based on the modeling of robot working environment, the shortest distance matrix between points is solved by Floyd algorithm. With the objective of minimizing the sum of the fixed cost of robot and the cost of robot operation, an integer programming model is established and a genetic algorithm for solving the model is designed. In order to make coordination to accomplish their respective tasks for each robot with high efficiency, this paper uses natural number encoding way. The objective function is based on penalty term constructed with the total number of collisions in the running path of robots. The fitness function is constructed by using the objective function with penalty term. Based on elitist retention strategy, a genetic algorithm with collision detection is designed. Using this algorithm for task allocation and path planning of multi-robot, it can effectively avoid or reduce the number of collisions in the process of multi-robot performing tasks. Finally, an example is used to validate the method.
Keywords:
Path Planning, Task Allocation, Collision Detection, Mathematical Model, Genetic Algorithm
1. Introduction
With the development of industrial robot technology [1] , a single robot cannot meet the needs of society, and many robots cooperate to accomplish complex and dangerous tasks [2] . In the process of multi-robot collaborative performing tasks, if the collision occurs between robots, it will cause losses such as robot damage, prolong the working time, and even make the system paralyzed, so planning the path for robots should reduce the occurrence of collisions as far as possible.
For the problem of robot’s path planning, many scholars have studied it. Particle swarm optimization algorithm or navigation algorithm are used to solve the problem [3] [4] [5] [6] [7] . In the aspect of robot’s task allocation, task allocation strategy [8] and ant colony algorithm are adopted [9] [10] . In the part of collision avoidance, improved artificial potential field obstacle avoidance method [11] and robot’s collision avoidance algorithm based on RRT are adopted [12] .
These scholars have proposed different algorithms to solve the problems of robot’s path planning, task allocation and collision avoidance respectively. But few people establish integer programming models and corresponding algorithms to solve them. Considering the maximum running time of a single robot, this paper aims at minimizing the sum of the robot’s fixed cost and the robot’s running cost. An integer programming model for task allocation and path planning of multi-robot with collision detection is proposed. Introducing collision penalty term, a genetic algorithm with collision detection is designed to calculate the optimal number of robots, the tasks completed by each robot and the collision free path of each robot, making the total cost of the detection tasks lowest.
2. Problem Description and Mathematical Model
2.1. Problem Description
A factory uses m robots to complete the monitoring of n monitoring points. Each robot filled with power can travel limited distance continuously and each monitoring point is monitored with equal time. When the task is executed, each robot starts from the platform at the same time, and tries to avoid collisions among them during the process of performing tasks, and finally returns to the departure platform. In the process of completing tasks, the robot performs only one task, each task requires only one robot to complete, each task is only performed once, and all tasks will be executed. The call cost and the operating cost of per unit distance are known for each robot, calculating the optimal number of robots to complete tasks, and planning the path of each robot, making the total cost of monitoring tasks completed lowest.
2.2. Mathematical Model
In order to establish the model, the following symbols are defined:
0: Represents departure platform;
$M=\left\{1,2,\cdots ,m\right\}$ : Represents a set of available robots;
L: Represents a set of feasible walking points for all robots；
$D=\left\{1,2,\cdots ,n,n+1\right\}$ : Represents a set of points to be monitored;
n + 1: Represents return platform can be considered as a replicate departure platform;
G: Represents a set of generally walk able grid points, and $L=D\cup G$ ;
$T=\left\{1,2,\cdots ,t\right\}$ : Represents the running time set of robots;
${d}_{ij}$ : Represents the distance from the i point to the j point of robots $i,j=0,1,2,\cdots ,n,n+1;$
${t}_{ij}$ : Represents the driving time from the i point to the j point of robots, $i,j=0,1,2,\cdots ,n,n+1;$
s: Represents the monitoring time of each robot at each point;
g: Represents the fixed cost of using a robot;
f: Represents the per unit distance cost of each robot;
$A=\left[\begin{array}{cccc}{a}_{11}& {a}_{12}& \cdots & {a}_{1r}\\ {a}_{21}& {a}_{22}& \cdots & {a}_{2r}\\ ⋮& ⋮& & ⋮\\ {a}_{r1}& {a}_{r2}& \cdots & {a}_{rr}\end{array}\right]$ : Represents the adjacency distance matrix between
feasible grids; when ${a}_{ij}=1$ represents the point i adjacent to the point of j, i and j are feasible grid points, otherwise ${a}_{ij}=0$ .
Define the following decision variables:
${x}_{ijkt}=\left\{\begin{array}{l}1,\text{Robot}k\text{fromthepointof}i\text{arrivesatthepointof}j\text{directly}\text{\hspace{0.17em}}\text{at}\text{\hspace{0.17em}}\text{time}t\\ 0,\text{otherwise};k\in M;t\in T;i,j\in L\end{array}$
${y}_{ikt}=\left\{\begin{array}{l}1,\text{robot}k\text{servicesforthepointof}i\text{\hspace{0.17em}}\text{at}\text{\hspace{0.17em}}\text{time}\text{\hspace{0.17em}}t\\ 0,\text{otherwise};k\in M,t\in T,i\in D\end{array}$
$Cjkpt=\left\{\begin{array}{l}1,{x}_{ijkt}={x}_{ujpt}=1,\text{\hspace{0.17em}}\text{Robot}\text{\hspace{0.17em}}k\text{\hspace{0.17em}}\text{and}\text{\hspace{0.17em}}\text{Robot}\text{\hspace{0.17em}}p\text{\hspace{0.17em}}\text{occur}\text{\hspace{0.17em}}\text{collision}\text{\hspace{0.17em}}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{at}\text{\hspace{0.17em}}\text{the}\text{\hspace{0.17em}}\text{point}\text{\hspace{0.17em}}\text{of}\text{\hspace{0.17em}}j\text{\hspace{0.17em}}\text{at}\text{\hspace{0.17em}}\text{time}\text{\hspace{0.17em}}t\\ 0,\text{otherwise};t\in T;i,u\in L;j\in L;k\ne p\in M\end{array}$
In order to reduce collisions between robots’ paths, a penalty term is constructed based on the number of collisions between paths. By introducing the penalty term into the objective function, an integer programming model for the task allocation and path planning of multi-robot with collision penalty is established.
$\mathrm{min}z=g\sum _{k\in M}\sum _{j\in D}\sum _{t\in T}{x}_{0jkt}+f\sum _{k\in M}\sum _{i\in L}\sum _{j\in L}\sum _{t\in T}{d}_{ij}i{x}_{ijkt}+\lambda \sum _{j\in L}\sum _{k\in M}\sum _{p\ne k\in M}\sum _{t\in T}{C}_{ikpt}$ (1)
$\sum _{j\in L}{x}_{0jkt}=1,k\in M$ (2)
$\sum _{i\in L}\sum _{t\in T}{x}_{in+1kt}=1,k\in M\text{\hspace{0.17em}}$ (3)
$\sum _{i\in L}{x}_{i0kt}=0,k\in M;t\in T\text{ }$ (4)
$\sum _{j\in L}{x}_{n+1jkt}=0,k\in M ;t\in T$ (5)
$\sum _{k\in M}\sum _{t\in T}{y}_{ikt}=1,i\in D\text{}$ (6)
$\sum _{i\in L}{x}_{ijkt}=\sum _{u\in L}{x}_{juk\left(t+s+1\right)}\text{\hspace{0.17em}}j\in D;k\in M;t\in T$ (7)
$\sum _{i\in L}{x}_{ijkt}=\sum _{u\in L}{x}_{juk\left(t+1\right)}\text{\hspace{0.17em}}j\in G;k\in M;t\in T$ (8)
$\sum _{i\in L}{x}_{ijkt}={y}_{jkt}\text{\hspace{0.17em}}j\in D;k\in M;t\in T$ (9)
$2{C}_{jkpt}\le {y}_{jkt}+{y}_{jpt}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}j\in L;k,p\in M;t\in T$ (10)
${y}_{jkt}+{y}_{jpt}\le 1+{C}_{jkpt}\text{\hspace{0.17em}}\text{\hspace{0.17em}}j\in L;k,p\in M;t\in T$ (11)
${x}_{ijkt}\le {a}_{ij}\text{\hspace{0.17em}}i,j\in L;k\in M;t\in T\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}$ (12)
${x}_{ijkt}=0,1\text{}i,j\in L;k\in M;t\in T$ (13)
${C}_{jkpt}=0,1\text{\hspace{0.17em}}j\in L;k,p\in M;t\in T$ (14)
${y}_{jkt}=0,1\text{\hspace{0.17em}}j\in L;k\in M;t\in T$ (15)
(1) Represents the minimum total cost with a penalty item; (2) Represents each robot must leave from platform 0; (3) Represents each robot must return to the platform point after completing the inspection task; (4) Represents each robot cannot return to the departure platform 0; (5) Represents each robot cannot leave the platform $n+1$ ; (6) Represents each monitoring point is also being serviced by one robot; (7) Represents each robot arrived at the monitoring point of j must leave from the monitoring point of j; (8) Represents each robot arriving at the general grid point J must leave from the first J point; (9) Represented a robot serviced for a monitoring point must have reached the monitoring point; (10), (11) Represent robot k and robot p occur collision at time t; (12) Represents robots can only walk into an adjacent grid within a unit time; (13), (14), (15) Represent a decision variable value constraint;
Since it will take too long to solve the integer programming model directly, we will design a genetic algorithm with collision detection to solve the task allocation planning problem for large scale robots.
3. Genetic Algorithm with Collision Detection
The steps of the genetic algorithm will not be described, and then the collision detection method proposed in this paper will be described in detail [13] .
In order to avoid the robot’s collision loss, the collision penalty term is added to the calculation of the objective function and the fitness value. The collision penalty term is defined according to the number of collisions between the robot paths. The following examples are combined with specific examples to describe the collision number detection algorithm.
According to the layout of a warehouse, the working environment after modeling by using the serial number grid method is shown in Figure 1. The gray squares represent the infeasible areas, the red squares represent the departure platforms, and the blue squares represent the points to be monitored. There are 10 monitoring points in the warehouse, which are indicated by serial numbers $\left\{18,30,112,135,157,196,223,235,245,267\right\}$ , and the red departure platform is indicated by serial number 264.
1) First, Figure 1 is shown with the adjacency distance matrix A, and the adjacency matrix is used to store the graph. Floyd algorithm [14] [15] is used to solve the shortest path between any two points in the network. In this paper, the Floyd algorithm is implemented by MATLAB, and the shortest distance matrix and the routing matrix between any two points in the graph are calculated by the adjacency distance matrix, and can query the shortest distance between any two points and routing.
2) Suppose the robots walk within one grid per unit time, and the distances traveled by robots are quantified by the number of grids. Two units of time are
Figure 1. Working environment of robots.
needed to be monitored for the grid of the monitoring points, and the monitoring points are quantized into two walk able grids. For example, with the maximum travel time of the robot as a restriction, decoding a feasible solution is 264-245-196-235-267-264 264-157-18-223-264 264-135-30-112-264, in which the path of the ${R}_{1}$ can be simply described as $264\to 246\to 245\to 229\to 213$ $\to 196\to 197\to 215\to 233\to 234\to 235\to 251\to 267\to 266\to 265\to 264$ . Similarly, the path of ${R}_{2}$ and the path of ${R}_{3}$ can be obtained.
3) The paths for each robot to complete their monitoring tasks can be traced by the routing matrix calculated by the Floyd algorithm. Then the shortest distance matrix is solved according to the Floyd algorithm, and the driving distance of each robot is obtained. Each robot starts at the same time from the starting platform and runs at the same speed, In the process of performing tasks, the robot will collide when and when the two robots arrive at the same grid point at the same time. It can be described as follows:
$\left\{\begin{array}{l}2{C}_{jkpt}\le {y}_{jkt}+{y}_{jpt}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}j\in L;k,p\in M;t\in T\\ {y}_{jkt}+{y}_{jpt}\le 1+{C}_{jkpt}\text{ }\text{\hspace{0.17em}}j=L;k,p\in M;t\in T\end{array}$ (16)
4) In order to reduce the collision, a penalty term is added to the individual objective function according to the total number of collisions of each robot path in the individual, so that the population is continuously optimized. First, a collision algorithm is used to solve the total number of collisions $CN$ between all paths of two robots corresponding to a chromosome, which is used as a measure of punishment. Take $\lambda CN$ as the penalty term, among which $\lambda \left(\lambda =0.6\right)$ is coefficient.
$CN=\sum _{j\in L}\sum _{k\in M}\sum _{p\ne k\in M}\sum _{t\in T}{C}_{jkpt}$ (17)
$\text{penalty}=\lambda CN$ (18)
${f}_{new}\left({x}_{i}\right)={f}_{old}\left({x}_{i}\right)+\text{penalty}$ (19)
$CN$ is the total number of collisions on a single chromosome, $\lambda \left(0<\lambda <1\right)$ is the Correction parameter of penalty. ${f}_{old}\left({x}_{i}\right)$ is the corresponding objective function value of individual ${x}_{i}$ , ${f}_{new}\left({x}_{i}\right)$ is the sum of corresponding objective function and collision penalties of individual ${x}_{i}$ .
5) Selection, crossover, mutation and other operations are performed according to genetic algorithms until the iteration goes to the maximum iteration path. The approximate optimal solution or optimal solution obtained is the optimal path.
4. Case Solving
In the warehouse layout shown in Figure 1, the gray grid represents the infeasible area, the red grid represents the departure platform and the blue grid represents the monitoring point. At the same time, each robot starts from the departure platform and completes the monitoring task of ten points to be monitored, which is $\left\{18,30,112,135,157,196,223,235,245,267\right\}$ .The warehouse has a total of 10 robots available, each robot filled with electricity can work continuously for 60 units of time and the speed of each robot is equal. The fixed cost of call a robot is 60 yuan and the cost of each robot driving unit distance is 2 yuan.
The program code is written by MATLAB, and the total number of collisions of each chromosome is calculated by the collision detection method, and the fitness of the penalty term is further calculated. Set the population size is 50 and the maximum evolution algebra is 100. In windows 7 (g4, 2G, 32 operating system) environment run the algorithm program 30 times [13] .
According to the working environment of the robot in Figure 1, the adjacency matrix is used to store Figure 1. Based on the advantages of the Floyd algorithm, the distance matrix and the routing matrix between any two points in the adjacency distance matrix can be calculated. The optimum number of robots is 3 based on a genetic algorithm with collision detection. The program with collision detection algorithm runs 30 times, and in the 30 run, the corresponding statistical results of the running distance of 3 robots are: the average value is 86.13, the maximum value is 90, the minimum value is 80, and the standard deviation is 2.36. The total cost of 3 robots completed tasks are: the average total cost is 357.62, the maximum is 364, the minimum is 340, and the standard deviation is 5.51. The statistical results of running the program for 30 times are: the average time is 33.65, the maximum is 37.86, the minimum is 29.96, and the standard deviation is 2.39.
Figure 2 is an iterative graph of an optimal solution. It can be seen that when the genetic algorithm program with collision detection runs, it converges to the
Figure 2. Convergence process of genetic algorithm.
optimal solution about 26 iterations, and the optimal solution does not change after 26 generations.
Figure 3 is a collision free path corresponding to 3 robots solved in this paper, which can be simply described as:
V1: 264--245--112--18--264;
V2: 264--223--157--196--267--264;
V3: 264--135--30--235--264.
It can be seen from Figure 3 that the optimal running path of 3 robots can be solved by using genetic algorithm with collision detection. Although the paths of V1, V2 and V3 shown in Figure 3 having overlapping parts, 3 robots leave the platform at the same time and the robots run at the same speed, so the time to run to the same path point is different and there will be no collisions actually.
5. Concluding Remarks
This paper studies the problem of multi-robot path planning based on genetic algorithm. Firstly, the adjacency matrix is used to store the example data. The shortest distance matrix and the routing matrix between the detected points are solved by using the Floyd algorithm according to the adjacency distance matrix [16] . The shortest time matrix is obtained from the shortest distance matrix, and the initial population is generated randomly according to the genetic algorithm. The task is assigned to robots with the longest running time of each robot as the constraint. Using the backtracking routing function of Floyd algorithm, the path of each robot is obtained according to the tasks assigned to each robot. The genetic algorithm of collision detection is used to solve the task allocation and path planning model of multi-robot, getting the optimum number of robots. The tasks each robot needs to accomplish, and the least cost collision free path is planned at the same time.
On the one hand, a collision detection algorithm is designed to avoid collisions in the running process of multi-robots; on the other hand, with the longest running time of each robot as the constraint, using the least robot completes
Figure 3. The running path of robots this paper.
the monitoring tasks making the total cost lowest. The genetic algorithm with collision detection designed in this paper can solve the problem of task allocation and path planning of multi-robot simultaneously.
In this paper, assuming that all robots start from the same platform, the collision between multi-robot can be further reduced by adjusting the starting platform of each robot actually; this paper does not consider the priority of robots to avoid collisions, so the robot can be based on a certain priority rules for collision avoidance. In the following research, we will consider these factors and further study the problem.
Acknowledgements
This work was supported by National Natural Science Foundation of China (71540028), Teaching Master of Beijing High Tech Project (G02040011); the Funding Project of High Level Cultivation of Beijing Wuzi University (GJB20164005).
Cite this paper
Li, Z.P. and Li, X.T. (2017) Research on Model and Algorithm of Task Allocation and Path Planning for Multi-Robot. Open Journal of Applied Sciences, 7, 511-519. https://doi.org/10.4236/ojapps.2017.710037
References
1. 1. Sun, F. (2012) Research on the Development of Industrial Robots in China. Science Technology and Engineering, 12, 2912-2918.
2. 2. Zhang, F. (2005) Improved Market Approach for Multi Robot Collaborative Exploration. Control and Decision, 20, 516-524.
3. 3. Zeng, N. (2016) Path Planning for Intelligent Robot Based on Switching Local Evolu-tionary PSO Algorithm. Emerald Insight, 36, 120-126.
4. 4. Dong, Y. (2016) Disordered and Multiple Destinations Path Planning Methods for Mobile Robot in Dynamic Environment. Journal of Electrical and Computer Engineering, 10, 1-10. https://doi.org/10.1155/2016/3620895
5. 5. Lu, Y. and Zhao, Y, (2015) Path Planning of Narrow Space Based on Improved Genetic Algorithm. Application Research of Computers, 32, 414-418.
6. 6. Lie, X. (2011) A Tabu Search Autonomous Navigation Algorithm for Mobile Robot. Control and Decision, 26, 1310-1314.
7. 7. Tang, W. (2012) The Improved Optimization Algorithm Is Used for Path Planning of Robots. Science Technology and Engineering, 12, 7599-7601.
8. 8. Chen, X. (2013) Multi Robot Task Allocation Based on Partitioning. Journal of Southern Yangtze University (Natural Science Edition), 12, 412-414.
9. 9. Cao, Z. and Wu, B. (2013) Assignment of Multi Robot Tasks Based on Improved Ant Colony Algorithm. Modular Machine Tools and Automatic Machining Technology, 2, 34-37.
10. 10. Luo, X. and Liu, H. (2017) Application of Optimized Fuzzy Decision Algorithm in Multi Homing Vehicle Scheduling Problem. Science Technology and Engineering, 17, 85-90.
11. 11. Yin, X. and Hu, Y. (2016) Research on Path Planning of Robot Obstacle Avoidance in Unknown Environment. Science Technology and Engineering, 33, 221-225.
12. 12. Li, H. and Liang, Y. (2012) Research on Robot Collision Avoidance Motion Planning Algorithm Based on RRT. Journal of Shenzhen Institute of Information Technology, 10, 21-26.
13. 13. Li, Z. and Li, X. (2016) Genetic Algorithm for Task Allocation And Path Planning of Multi-Robot System. Journal of Mathematical Sciences and Application, 11, 1-7.
14. 14. Wang C. and Li, X. (2017) Leg Deformation Strategy of Antiriot Robot Based on Floyd Algorithm. Science Technology and Engineering, 17, 70-73.
15. 15. Zhu, H. and Zhang, Y. (2011) Find All the Shortest Paths among Nodes Based on Im-proved Floyd Algorithm. Network and Multimedia, 35, 65-67.
16. 16. Zuo, X. and Shen, W. (2017) Improved Algorithm for Multiple Shortest Path Problem Based on Floyd Algorithm. Computer Science, 44, 222-227.
| 5,792 | 21,946 |
{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 50, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 2.59375 | 3 |
CC-MAIN-2023-23
|
latest
|
en
| 0.918571 |
https://mathoverflow.net/questions/391501/can-z-ranks-successor-cardinals-ordinal-inaccessibility-be-equal-to-zf
| 1,709,134,895,000,000,000 |
text/html
|
crawl-data/CC-MAIN-2024-10/segments/1707947474737.17/warc/CC-MAIN-20240228143955-20240228173955-00314.warc.gz
| 392,417,961 | 28,785 |
# Can Z + Ranks + Successor cardinals + Ordinal inaccessibility be equal to ZF?
[EDIT: The axiom of successor cardinals was found by an answer by Greg Kirmayer, not to be capturing the intended meaning of it, which is simply reflected by its name, i.e. the existence of a successor cardinal for every cardinal. A corrective note had been inserted below that axiom.
Is Z + Rank + Successor cardinals + Ordinal inaccessibility = ZF?
Where:
Ranks: $$\forall x \exists \alpha: x \in V_\alpha$$
where $$V_\alpha$$ is the $$\alpha^{th}$$ stage of Von Neumann's universe (the cumulative hierarchy).
Successor cardinals: if $$f$$ is a definable function, then: $$\forall \text{ordinal } \alpha \ \exists \text{ ordinal }\beta \, (\neg \forall \gamma < \beta \exists \lambda < \alpha : f(\lambda)=\gamma)$$ By $$\text{ordinal}$$ it means a von Neumann ordinal defined in the usual manner.
For every ordinal there is an ordinal such that no surjective function (definable in the language of set theory) from the former to the latter can exist.
[NOTE]: The above formulation suffers a flaw, therefore the correct formulation is: $$\forall \text{ordinal } \alpha \, \exists \text{ordinal } \beta: \not \exists f (f: \beta \hookrightarrow \alpha)$$, where $$\hookrightarrow$$ signify "injective function".
Ordinal inaccessibility: if $$f$$ is a definable function, then: $$\forall \gamma \, (\neg \forall \text{ ordinal } \alpha \,\exists \beta < \gamma : \alpha \leq f(\beta))$$ Any function [definable in the language of set theory] coming from an ordinal cannot have every ordinal being smaller than or equal to some ordinal in its range. That is, its range cannot be a cofinal subclass of the class of all ordinals.
The idea is that every uncountable cardinal in ZF is either Regular and therefore a successor cardinal, or otherwise a singular limit cardinal. Here, both of those kinds would be constructed from below, and the axiom of Rank assures that all sets are built successively within those ordinally indexed stages. Ordinal inaccessibility is I think equivalent to Ordinal Replacement which by itself is actually weak, it can only build stages up to $$V_{\omega_1}$$, and of course successor cardinals can only build the next stages, but together it seems that they can act to build up the whole of Von Neumann's universe. So, I thought that the above would prove full Replacement. Its easy to prove the result with Choice (in the form of every set is bijective to some von Neumann ordinal); but without it the proof is elluding me?
• Can you specify exactly what you mean by ordinal in this theory? And I think there is a typo in the definition of ordinal inaccessibility; is $\lambda=\gamma$? Apr 29, 2021 at 21:37
• @FarmerS, thanks for spotting the typo. About ordinals, those are the usual von Neumann ordinals, i.e. transitive sets of transitive sets, that are $\in$-well founded. Apr 29, 2021 at 21:54
• Do you mean "transitive set whose elements are strictly linearly ordered by $\in$"? Apr 29, 2021 at 21:57
• @FarmerS, I mean a transitive set whose elements are strictly well ordered by $\in$. An this is the usual official definition of von Neumann ordinals, and it is also equivalent to the one I gave in my prior comment. Apr 29, 2021 at 21:59
• Hmm, under ZF (in particular, Foundation), all sets are $\in$-wellfounded, but not in general $\in$-wellordered. Apr 29, 2021 at 22:05
This theory doesn't prove Replacement (assuming the consistency of an inaccessible, at least).
Assume ZFC + $$\kappa$$ is inaccessible and force over $$V$$ to add $$\kappa$$-many Cohen reals (i.e. the forcing is finite support product $$\Pi_{\alpha<\kappa}\mathbb{C}_\alpha$$ where each $$\mathbb{C}_\alpha$$ is just Cohen forcing). This forcing is ccc, so preserves all cardinals and cofinalities. So $$\kappa$$ is still a weakly inaccessible cardinal in $$V[G]$$. Now let $$M=V_\kappa^{V[G]}$$. Note that $$M$$ models the axioms you mention (in particular as $$\kappa$$ is regular in $$V[G]$$). But it does not satisfy Replacement: Working in $$V[G]$$, let $$X$$ be the set of all wellorders of $$V_{\omega+1}$$ of ordertype $${<\kappa}$$. Then $$X\in V_\kappa$$. Consider the function sending $$W\in X$$ to the ordertype of $$W$$. This is definable over $$V_\kappa^{V[G]}$$ (actually without parameters), and is a surjection $$X\to\kappa$$ there.
Note that $$M$$ does satisfy Choice, formulated as "for every function there is a choice function". However, it does not satisfy the statement that "every wellorderable set is bijectable with an ordinal". (The set $$X'$$ of equivalence classes of wellorders in $$X$$, where they are equivalent if they have the same ordertype, is such a set.) Although ordinal inaccessibility holds in $$M$$, $$X'$$ is wellorderable in ordertype that of the class of ordinals.
Note that "ordinal" can also be defined as (i) the equivalence class of all wellordered sets of a given ordertype. Under ZF, this is equivalent to (ii) the von Neumann ordinals "transitive sets whose elements are strictly linearly ordered by $$\in$$" (which under ZF is equivalent to (ii') "transitive sets whose elements are well ordered by $$\in$$"). But in the model $$M$$, (i) and (ii) do not coincide (but (ii) and (ii') do).
• so what we need to get replacement is to add to the above the axiom that every well ordered set is bijective to an ordinal. I think we won't need sucessor cardinals by then. Apr 30, 2021 at 5:45
• I don't know what you mean by "coincide" in your last remark. But it appears to me that should I've used Scott's ordinals instead of von Neumanns in the formulation of the above axioms, then the resulting system would prove full replacement, since every well ordered set would have a Scott ordinal and even an initial segement of Scott ordinals that is order isomorphic to it, then well ordered replacement would follow and this would prove full Replacement over Z + Foundation. Apr 30, 2021 at 7:23
• If every wellordered set is bijective with a von Neumann ordinal (hence in fact has the ordertype some von Neumann ordinal), then yes, I agree you get replacement. By "coincide" I just mean "are equivalent". If you use Scott's ordinals instead of von Neumann's, I agree you get replacement, but I think to first get well ordered replacement, it seems to require a small argument to observe that there is no Scott ordinal with ordertype that of the class of all von Neumann ordinals... May 2, 2021 at 19:56
We are assuming that a formula π(x,y) with 2 free variables defines a function means that for every x there is a unique y such that π(x,y).
We assume that the axiom schema of Successor cardinals and the axiom schema of Ordinal inaccessibility are the result of replacing "f is a definable function" by "π(x,y)
defines a function", replacing "π(π)=πΎ" by π(π,πΎ), and replacing πΌβ€π(π½) by "there is a b such that π(π½,b) and πΌβ€b".
We note that the axiom schema of Successor cardinals, is an immediate consequence of the axiom schema of Ordinal inaccessibility.(Suppose that π(x,y) defines a function and c ia an ordinal.
By Ordinal inaccessibility, there is an ordinal πΌ with the property that βπ½<c:π(π½,b)-->("b is not an ordinal" or b<πΌ). Then Β¬βπΎ<(πΌ+1)βπ<c:π(π,πΎ), since Β¬π(π,πΌ) for all π<c.)
Z + Ranks + Ordinal inaccessibility has the same consequences as Z + Ranks + Ordinal Replacement.
Proof: Suppose Ordinal Replacement holds, π(x,y) defines a function, πΎ is an ordinal, and for every ordinal πΌ (βπ½<πΎβb:πΌβ€bβ§π(π½,b)). Let π(x,y) be the formula ("x is an ordinal" and "y is an ordinal" and π(x,y)) or ("x is an ordinal" and (βt(π(x,y)-->"t is not an ordinal") and y=0) or ("x is not an ordinal" and y=x). By Ordinal Replacement βπ΅βπ¦(π¦βπ΅ββπ₯βπΎπ(π₯,π¦)). But then every ordinal is in UB. Now suppose Ordinal inaccessibility holds, π(x,y) is a formula in two free variables x,y, βπ₯[πππππππ(π₯)ββ!π¦(πππππππ(π¦)β§π(π₯,π¦))], and A is a set of ordinals. Let π(x,y) be the formula ("x is an ordinal" and "y is an ordinal" and π(x,y)) or ("x is not an ordinal" and x=y). Then π(x,y) defines a function. By Ordinal inaccessibility, there is an ordinal πΌ with the property βπ½<(UA)βb(π(π½,b)-->b<πΌ). Then there is a set B such that
bβB<-->bβπΌβ§βπ₯βπ΄(π(x,b)). Therefore Ordinal Replacement holds.
If ZF is consistent then Z + Ranks + Ordinal inaccessibility + Choice does not prove Replacement.
Proof: See the answer of Joel David Hamkins to "Is full Replacement provable in Z + Ordinal Replacement?". (His argument shows that in a model of ZFC, β¨ππ1,ββ© satisfies Z + Ranks + Ordinal Replacement + Choice but not all instancess of Replacement.)
• the form of axiom of choice I meant is that every set is bijective to some ordinal. In that form, it'll prove the result. I'll edit it to that effect. May 1, 2021 at 7:48
• Thanks for this answer, I see that the ordinal $\beta$ occurs "after" each function, while it should be prior to all of them, which cannot be done easily. I've corrected the answer to the conventional way of expressing it. May 1, 2021 at 9:11
| 2,807 | 9,335 |
{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 39, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 3.0625 | 3 |
CC-MAIN-2024-10
|
latest
|
en
| 0.846122 |
https://ncarkyslosa.web.app/1386.html
| 1,652,941,018,000,000,000 |
text/html
|
crawl-data/CC-MAIN-2022-21/segments/1652662525507.54/warc/CC-MAIN-20220519042059-20220519072059-00195.warc.gz
| 489,716,913 | 4,174 |
# Compact set in metric space pdf
A metric space which is sequentially compact is totally bounded and complete. Jan 02, 2017 a video explaining the idea of compactness in r with an example of a compact set and a noncompact set in r. Intro real analysis, lec 33, euclidean metric, triangle. Every continuous function on a compact set is uniformly continuous. An open sets family of a metric space is defined next and it has been shown that the metric space with any open sets family is a topological space. In fact, every compact metric space is a continuous image of the cantor set. In euclidean space, a set being compact is equivalent to. A metric space is just a set x equipped with a function d of two variables which measures the distance between points. A subset a of x is compact with respect to the subspace topology on a if. A metric space is said to be locally compact if every point has a compact neighborhood. Schep in this note we shall present a proof that in a metric space x. Metric spaces a metric space is a set x that has a notion of the distance dx,y between every pair of points x,y.
We then have the following fundamental theorem characterizing compact metric spaces. X with x 6 y there exist open sets u containing x and v containing y such that u t v 3. It is stronger then usual continuity at every point because here depends only on the and not on the point nonexample. Compact sets in metric spaces are complete mathonline. The sequential compactness is equivalent to socalled countable compactness. Denition theclosureof a, denoted a, is the smallest closed set containing a alternatively, the intersection of all closed sets containing a. Denition theinteriorof a, denoted inta, is the largest open set contained in a alternatively, the union of all open sets contained in a.
A set k in a metric space x, d is said to be compact if any open cover u. Let f n be a decreasing sequence of closed nonempty subsets of x, and let g n fc n. Theorem in metric space, a subset kis compact if and only if it is sequentially compact. Say a metric space xis sequentially compact if every sequence in xhas a subsequence that converges in x. So unlike with closed and open sets, a set is \compact relative a subset y if and only if it is compact relative to the whole space.
A sequentially compact subset of a metric space is bounded and closed. A metric space x is compact if every open cover of x has a finite subcover. Spaces of continuous functions if the underlying space x is compact, pointwise continuity and uniform continuity is the same. A pathconnected space is a stronger notion of connectedness, requiring the structure of a path. A metric space is called sequentially compact if every sequence in x has a convergent subsequence. A of open sets is called an open cover of x if every x. Pdf in this article, we mainly formalize in mizar 2 the equivalence among a. A metric space which is totally bounded and complete is also sequentially compact. When we discuss probability theory of random processes, the underlying sample spaces and eld structures become quite complex. The answer is yes, and the theory is called the theory of metric spaces. The equivalence between closed and boundedness and compactness is valid in nite dimensional euclidean spaces and some special in nite dimensional space such as c1k. Theorem each compact set k in a metric space is closed and bounded. In other words, the compact sets in rn are characterized by the bolzanoweierstrass property.
We do not develop their theory in detail, and we leave the veri. The definition below imposes certain natural conditions on the distance between the points. Informally, 3 and 4 say, respectively, that cis closed under. So, lets look at an example of a subset of a metric space that is not compact. Compact spaces connected sets relative compactness theorem suppose x. We will also be interested in the space of continuous rvalued functions cx. Suppose that x is a sequentially compact metric space. Compact sets in metric spaces uc davis mathematics. If you do not wish to use the heineborel theorem for metric spaces as suggested in the answer by igor rivin then here is another way of proving that a compact metric space is complete. Note that in metric spaces the notions of compactness and sequential compactness coincide. The following properties of a metric space are equivalent. Xis compact, and x j2kis a cauchy sequence, then there exists x2ksuch that lim j. A metric space is compact if and only if it is sequentially compact. Proof if k is a closed set in x, then k is compact lemma 5.
If uis an open cover of k, then there is a 0 such that for each x2kthere is a. A path from a point x to a point y in a topological space x is a continuous function. Some of this material is contained in optional sections of the book, but i will assume none of that and start from scratch. Compact metric space yongheng zhang when we say a metric space xis compact, usually we mean any open covering of xhas a nite subcovering. A metric space x is compact if every open cover of x has a. A metric space is a set in which we can talk of the distance between any two of its elements. Every compact metric space is second countable, and is a continuous image of the cantor set. A metric space is complete if every cauchy sequence converges.
A metric space x is sequentially compact if every sequence of points in x has a convergent subsequence converging to a point in x. A video explaining the idea of compactness in r with an example of a compact set and a noncompact set in r. In general, it may be more difficult to show that a subset of a metric space is compact than to show a subset of a metric space is not compact. Cauchy sequence in x has a convergent subsequence, so, by lemma 6 below. Note that every compact space is locally compact, since the whole space xsatis es the necessary condition. Intro real analysis, lec 33, euclidean metric, triangle inequality, metric spaces, compact sets. Proposition each closed subset of a compact set is also. R 0,1 from the real number line to the closed unit interval, and define a topology on k so that a sequence in k converges towards if and only if converges towards f x. On the other hand, the closed unit ball of the dual of a normed space is compact for the weak topology. Turns out, these three definitions are essentially equivalent. Compact spaces connected sets open covers and compactness.
Pdf some notes on compact sets in soft metric spaces. A space is locally compact if it is locally compact at each point. Roughly speaking, a metric on the set xis just a rule to measure the distance between any two elements of x. Notes on metric spaces these notes introduce the concept of a metric space, which will be an essential notion throughout this course and in others that follow. A metric space is sequentially compact if every sequence has a convergent subsequence. Ais a family of sets in cindexed by some index set a,then a o c. X, there exists an open neighborhood u of x with closure u. A metric or topological space is compact if every open cover of the space has a nite subcover. A subset, k, of m is said to be compact if and only if every open cover of k by open sets in m. Remarks in the theory of point set topology, the compactness implies the sequential compactness, but not vice versa. Suppose kis a subset of a metric space xand k is sequentially compact. Euclidean spaces are locally compact, but infinitedimensional banach.
1378 794 681 922 13 101 1451 254 832 1275 1095 240 1080 1409 1289 1351 1055 525 1499 1139 723 1420 472 955 541 459 318 6 209 1463 57
| 1,715 | 7,636 |
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 3.5 | 4 |
CC-MAIN-2022-21
|
latest
|
en
| 0.921348 |
https://www.unitsconverters.com/en/Squaremicrometre-To-Squaremil/Unittounit-306-332
| 1,627,260,897,000,000,000 |
text/html
|
crawl-data/CC-MAIN-2021-31/segments/1627046151972.40/warc/CC-MAIN-20210726000859-20210726030859-00306.warc.gz
| 1,119,309,211 | 37,148 |
Formula Used
1 Square Micrometer = 1E-12 Square Meter
1 Square Meter = 1550003099.9938 Square Mil
1 Square Micrometer = 0.0015500030999938 Square Mil
## Square Micrometers to Square Mils Conversion
µm² stands for square micrometers and mi² stands for square mils. The formula used in square micrometers to square mils conversion is 1 Square Micrometer = 0.0015500030999938 Square Mil. In other words, 1 square micrometer is 646 times smaller than a square mil. To convert all types of measurement units, you can used this tool which is able to provide you conversions on a scale.
## Convert Square Micrometer to Square Mil
How to convert square micrometer to square mil? In the area measurement, first choose square micrometer from the left dropdown and square mil from the right dropdown, enter the value you want to convert and click on 'convert'. Want a reverse calculation from square mil to square micrometer? You can check our square mil to square micrometer converter.
How many Square Meter is 1 Square Micrometer?
1 Square Micrometer is equal to 1E-12 Square Meter. 1 Square Micrometer is 1000000000000 times Smaller than 1 Square Meter.
How many Square Kilometer is 1 Square Micrometer?
1 Square Micrometer is equal to 1E-18 Square Kilometer. 1 Square Micrometer is 1E+18 times Smaller than 1 Square Kilometer.
How many Square Centimeter is 1 Square Micrometer?
1 Square Micrometer is equal to 1E-08 Square Centimeter. 1 Square Micrometer is 100000000 times Smaller than 1 Square Centimeter.
How many Square Millimeter is 1 Square Micrometer?
1 Square Micrometer is equal to 1E-06 Square Millimeter. 1 Square Micrometer is 1000000 times Smaller than 1 Square Millimeter.
## Square Micrometers to Square Mils Converter
Units of measurement use the International System of Units, better known as SI units, which provide a standard for measuring the physical properties of matter. Measurement like area finds its use in a number of places right from education to industrial usage. Be it buying grocery or cooking, units play a vital role in our daily life; and hence their conversions. unitsconverters.com helps in the conversion of different units of measurement like µm² to mi² through multiplicative conversion factors. When you are converting area, you need a Square Micrometers to Square Mils converter that is elaborate and still easy to use. Converting Square Micrometer to Square Mil is easy, for you only have to select the units first and the value you want to convert. If you encounter any issues to convert, this tool is the answer that gives you the exact conversion of units. You can also get the formula used in Square Micrometer to Square Mil conversion along with a table representing the entire conversion.
Let Others Know
| 632 | 2,754 |
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 2.578125 | 3 |
CC-MAIN-2021-31
|
latest
|
en
| 0.820872 |
http://slideplayer.com/slide/3621586/
| 1,529,557,893,000,000,000 |
text/html
|
crawl-data/CC-MAIN-2018-26/segments/1529267864022.18/warc/CC-MAIN-20180621040124-20180621060124-00183.warc.gz
| 286,966,747 | 28,650 |
# MFMcGrawChap 13d-Liquids - Revised 4-3-101 Chapter 13 LIQUIDS.
## Presentation on theme: "MFMcGrawChap 13d-Liquids - Revised 4-3-101 Chapter 13 LIQUIDS."— Presentation transcript:
MFMcGrawChap 13d-Liquids - Revised 4-3-101 Chapter 13 LIQUIDS
MFMcGrawChap 13d-Liquids - Revised 4-3-102 This lecture will help you understand: Pressure Pressure in a Liquid Buoyancy in a Liquid Archimedes’ Principle What Makes an Object Sink or Float Flotation Pascal’s Principle Surface Tension & Capillarity
MFMcGrawChap 13d-Liquids - Revised 4-3-103 The force per unit area that one object exerts on another In equation form: Pressure depends on area over which force is distributed - a force “density.” Units: N/m 2, lb/ft 2, or Pa (Pascals) Pressure force area Pressure
MFMcGrawChap 13d-Liquids - Revised 4-3-104 Pressure Pressure depends upon the surface area over which the force is applied. The vertical block exerts a greater force on the table than the horizontal block.
MFMcGrawChap 13d-Liquids - Revised 4-3-105 Pressure Example: The teacher between nails is unharmed because force is applied over many nails. Combined surface area of the nails results in a tolerable pressure that does not puncture the skin.
MFMcGrawChap 13d-Liquids - Revised 4-3-106 When you stand on one foot instead of two, the force you exert on the floor is less. B.the same. C.more. D.None of the above. Pressure CHECK YOUR NEIGHBOR-Trick Question
MFMcGrawChap 13d-Liquids - Revised 4-3-107 When you stand on one foot instead of two, the force you exert on the floor is less. B.the same. C.more. D.None of the above. Comment: Distinguish between force and pressure! Pressure CHECK YOUR ANSWER - Did you get caught?
MFMcGrawChap 13d-Liquids - Revised 4-3-108 When you stand on one foot instead of two, the pressure you exert on the floor is less. B.the same. C.more. D.None of the above. Pressure CHECK YOUR NEIGHBOR
MFMcGrawChap 13d-Liquids - Revised 4-3-109 When you stand on one foot instead of two, the pressure you exert on the floor is less. B.the same. C.more. D.None of the above. Explanation: Twice as much, in fact! Pressure CHECK YOUR ANSWER
MFMcGrawChap 13d-Liquids - Revised 4-3-1010 Characteristics of a Liquid Liquids can flow to fill up an arbitrary shape. Liquids can not support a shear stress. Liquids (for our purposes) are in compessive We can’t squeeze a given amount of liquid into a smaller volume. Liquids are not like nerf balls In general - fluid means a liquid or a gas.
MFMcGrawChap 13d-Liquids - Revised 4-3-1011 Pressure in a Liquid Force per unit area that a liquid exerts on an object Depth dependent and not volume dependent Example: Swim twice as deep, then twice as much weight of water above you produces twice as much pressure on you.
MFMcGrawChap 13d-Liquids - Revised 4-3-1012 Pressure in a Liquid Pressure acts equally in all directions Example: Your ears feel the same amount of pressure under water no matter how you tip your head. Bottom of a boat is pushed upward by water pressure. Pressure acts upward when pushing a beach ball under water. A shrunken styrofoam head bearing geologist Mike Rowe's signature and the scrawls and artwork of other science party members demonstates the effects of the deep-sea pressures that bear down on Alvin. Passengers inside the sub are protected by the vehicle's thick, titanium hull. This head began as a standard-sized wig form. Photograph: Sonya Senkowsky.
MFMcGrawChap 13d-Liquids - Revised 4-3-1013 What's the deal with the styrofoam heads? The styrofoam heads are something we do for fun on the research cruises when all of the work is done. We buy them from local stores in La Paz or a wholesaler in the US. They are the same size as a normal human head when we get them. We draw on them with Sharpies and then tie them in a dive bag on the side of the ROV so that when the ROV is at depth (10 meters of water depth = 1 atmosphere of pressure = 15 pounds per square inch) of say, 3000 meters, the head will be compressed to about 1/10 of its size! It's a bit like a souvenir from the deep. I've put pictures of the three that I've done at MBARI. http://www.mbari.org/staff/jones/shrunken.htmhttp://www.mbari.org/staff/jones/shrunken.htm A fluid can not support a shear stress.
MFMcGrawChap 13d-Liquids - Revised 4-3-1014 Pressure in a Liquid Independent of shape of container: Whatever the shape of a container, pressure at any particular depth is the same. In equation form: Liquid pressure weight density depth
MFMcGrawChap 13d-Liquids - Revised 4-3-1015 Water pressure provided by a water tower is greater if the tower is taller. B.holds more water. C.Both A and B. D.None of the above. Pressure in a Liquid CHECK YOUR NEIGHBOR
MFMcGrawChap 13d-Liquids - Revised 4-3-1016 Water pressure provided by a water tower is greater if the tower is taller. B.holds more water. C.Both A and B. D.None of the above. Explanation: Only depth, not amount of water, contributes to pressure. Pressure in a Liquid CHECK YOUR ANSWER
MFMcGrawChap 13d-Liquids - Revised 4-3-1017 Pressure Depends on Depth Independent of the volume of water behind the dam the size of the dam is determined by the depth of the water behind the dam.
MFMcGrawChap 13d-Liquids - Revised 4-3-1018 Pressure in a Liquid Effects of water pressure Acts perpendicular to surfaces of a container Liquid spurts at right angles from a hole in the surface. –The greater the depth, the greater the exiting speed.
MFMcGrawChap 13d-Liquids - Revised 4-3-1019 Pressure in a Liquid What happens to the water streaming out of the cup if you drop it?
MFMcGrawChap 13d-Liquids - Revised 4-3-1020 Fountain Height
MFMcGrawChap 13d-Liquids - Revised 4-3-1021 Buoyancy in a Liquid Buoyancy Apparent loss of weight of a submerged object Amount equals the weight of water displaced
MFMcGrawChap 13d-Liquids - Revised 4-3-1022 Buoyancy in a Liquid Displacement rule: A completely submerged object always displaces a volume of liquid equal to its own volume. Example: Place a stone in a container that is brimful of water, and the amount of water overflow equals the volume of the stone.
MFMcGrawChap 13d-Liquids - Revised 4-3-1023 Buoyant Force and Overflow Weight
MFMcGrawChap 13d-Liquids - Revised 4-3-1024 A cook who measures a specific amount of butter by placing it in a measuring cup with water in it is using the principle of buoyancy. B.displacement rule. C.concept of density. D.All of the above. Buoyancy in a Liquid CHECK YOUR NEIGHBOR
MFMcGrawChap 13d-Liquids - Revised 4-3-1025 A cook who measures a specific amount of butter by placing it in a measuring cup with water in it is using the principle of buoyancy. B.displacement rule. C.concept of density. D.All of the above. Buoyancy in a Liquid CHECK YOUR ANSWER
MFMcGrawChap 13d-Liquids - Revised 4-3-1026 Buoyancy in a Liquid Buoyant force Net upward force that a fluid exerts on an immersed object = weight of water displaced Example: The difference in the upward and downward forces acting on the submerged block is the same at any depth
MFMcGrawChap 13d-Liquids - Revised 4-3-1027 Buoyant Force
MFMcGrawChap 13d-Liquids - Revised 4-3-1028 How many forces act on a submerged body at rest in a fluid? One—buoyancy B.Two—buoyancy and the force due to gravity C.None—in accord with the equilibrium rule, F = 0 D.None of the above. Buoyancy in a Liquid CHECK YOUR NEIGHBOR
MFMcGrawChap 13d-Liquids - Revised 4-3-1029 How many forces act on a submerged body at rest in a fluid? One—buoyancy B.Two—buoyancy and the force due to gravity C.None—in accord with the equilibrium rule, F = 0 D.None of the above. Buoyancy in a Liquid CHECK YOUR ANSWER
MFMcGrawChap 13d-Liquids - Revised 4-3-1030 Buoyancy in a Liquid Sink or float? Sink when weight of submerged object is greater than the buoyant force. Neither sink nor float when weight of a submerged object is equal to buoyant force—object will remain at any level. Float when weight of submerged object is less than the buoyant force it would have when submerged — when floating, buoyant force = weight of floating object.
MFMcGrawChap 13d-Liquids - Revised 4-3-1031 Archimedes’ Principle Archimedes’ principle: Discovered by Greek scientist Archimedes. Relates buoyancy to displaced liquid. States that an immersed body (completely or partially) is buoyed up by a force equal to the weight of the fluid it displaces. Applies to gases and liquids.
MFMcGrawChap 13d-Liquids - Revised 4-3-1032 Archimedes’ Principle Apparent weight of a submerged object Weight out of water — buoyant force Example: If a 3-kg block submerged in water apparently “weighs” 2 kg, then the buoyant force or weight of water displaced is 1 kg.
MFMcGrawChap 13d-Liquids - Revised 4-3-1033 On which of these blocks submerged in water is the buoyant force greatest? 1 kg of lead B.1 kg of aluminum C.1 kg of uranium D.All the same. Archimedes’ Principle CHECK YOUR NEIGHBOR
MFMcGrawChap 13d-Liquids - Revised 4-3-1034 On which of these blocks submerged in water is the buoyant force greatest? 1 kg of lead B.1 kg of aluminum C.1 kg of uranium D.All the same. Explanation: The largest block is the aluminum one. It displaces more water and therefore experiences the greatest buoyant force. All blocks are 1 kg, therefore the material with the smallest density has the largest volume. Archimedes’ Principle CHECK YOUR ANSWER
MFMcGrawChap 13d-Liquids - Revised 4-3-1035 When a fish expands its air bladder, the density of the fish decreases. B.increases. C.remains the same. D.None of the above. Archimedes’ Principle CHECK YOUR NEIGHBOR
MFMcGrawChap 13d-Liquids - Revised 4-3-1036 When a fish expands its air bladder, the density of the fish decreases. B.increases. C.remains the same. D.None of the above. Archimedes’ Principle CHECK YOUR ANSWER
MFMcGrawChap 13d-Liquids - Revised 4-3-1037 When a fish decreases the size of its air bladder, the density of the fish decreases. B.increases. C.remains the same. D.None of the above. Archimedes’ Principle CHECK YOUR NEIGHBOR
MFMcGrawChap 13d-Liquids - Revised 4-3-1038 When a fish decreases the size of its air bladder, the density of the fish decreases. B.increases. C.remains the same. D.None of the above. The bladder is normally empty, if the it is decreased then it is replaced with more dense tissue. Therefore the density of the fish increases. Archimedes’ Principle CHECK YOUR ANSWER
MFMcGrawChap 13d-Liquids - Revised 4-3-1039 When a submarine takes water into its ballast tanks, its density decreases. B.increases. C.remains the same. D.None of the above. Archimedes’ Principle CHECK YOUR NEIGHBOR
MFMcGrawChap 13d-Liquids - Revised 4-3-1040 When a submarine takes water into its ballast tanks, its density decreases. B.increases. C.remains the same. D.None of the above. Archimedes’ Principle CHECK YOUR ANSWER
MFMcGrawChap 13d-Liquids - Revised 4-3-1041 When a submerged submarine expels water from its ballast tanks, its density decreases. B.increases. C.remains the same. D.None of the above. Archimedes’ Principle CHECK YOUR NEIGHBOR
MFMcGrawChap 13d-Liquids - Revised 4-3-1042 When a submerged submarine expels water from its ballast tanks, its density decreases. B.increases. C.remains the same. D.None of the above. Explanation: This is how a submerged submarine is able to surface. Archimedes’ Principle CHECK YOUR ANSWER
MFMcGrawChap 13d-Liquids - Revised 4-3-1043 Flotation Principle of flotation: –A floating object displaces a weight of fluid equal to its own weight. Example: A solid iron 1-ton block may displace 1/8 ton of water and sink. The same 1 ton of iron in a bowl shape displaces a greater volume of water—the greater buoyant force allows it to float. Archimedes’ Principle
MFMcGrawChap 13d-Liquids - Revised 4-3-1044 The reason a person finds it easier to float in saltwater compared with freshwater is that in saltwater the buoyant force is greater. B.a person feels less heavy. C.Neither of these. D.None of the above. Archimedes’ Principle CHECK YOUR NEIGHBOR
MFMcGrawChap 13d-Liquids - Revised 4-3-1045 The reason a person finds it easier to float in saltwater compared with freshwater is that in saltwater the buoyant force is greater. B.a person feels less heavy. C.Neither of these. D.None of the above. Explanation: A floating person has the same buoyant force whatever the density of water. A person floats higher because a smaller volume of the denser saltwater is displaced. Archimedes’ Principle CHECK YOUR ANSWER
MFMcGrawChap 13d-Liquids - Revised 4-3-1046 On a boat ride, the skipper gives you a life preserver filled with lead pellets. When he sees the skeptical look on your face, he says that you’ll experience a greater buoyant force if you fall overboard than your friends who wear Styrofoam-filled preservers. He apparently doesn’t know his physics. B.He is correct. Archimedes’ Principle Hewitt Turns Sadistic
MFMcGrawChap 13d-Liquids - Revised 4-3-1047 On a boat ride, the skipper gives you a life preserver filled with lead pellets. When he sees the skeptical look on your face, he says that you’ll experience a greater buoyant force if you fall overboard than your friends who wear Styrofoam-filled preservers. He apparently doesn’t know his physics. B.He is correct. Explanation: He’s correct, but what he doesn’t tell you is you’ll drown! Your life preserver will submerge and displace more water than those of your friends who float at the surface. Although the buoyant force on you will be greater, the net force downward is greater still! Archimedes’ Principle Hewitt Turns Sadistic
MFMcGrawChap 13d-Liquids - Revised 4-3-1048 You place an object in a container that is full to the brim with water on a scale. The object floats, but some water spills out. How does the weight of the object compare with the weight of the water displaced? Weight of object is greater than weight of water displaced. B.Weight of object is less than weight of water displaced. C.Weight of object is equal to weight of water displaced. D.There is not enough information to decide. Flotation CHECK YOUR NEIGHBOR
MFMcGrawChap 13d-Liquids - Revised 4-3-1049 You place an object in a container that is full to the brim with water on a scale. The object floats, but the water spills out. How does the weight of the object compare with the weight of the water displaced? Weight of object is greater than weight of water displaced. B.Weight of object is less than weight of water displaced. C.Weight of object is equal to weight of water displaced. D.There is not enough information to decide. Flotation CHECK YOUR ANSWER Explanation: This principle is wonderfully illustrated with Scotland’s Falkirk Wheel.
MFMcGrawChap 13d-Liquids - Revised 4-3-1050 The Falkirk Wheel’s two caisson are brimful of water and the same weight, regardless of whether there are boats in them. This makes rotation and lifting almost effortless. Falkkirk Wheel
MFMcGrawChap 13d-Liquids - Revised 4-3-1051 Falkkirk Wheel
MFMcGrawChap 13d-Liquids - Revised 4-3-1052 Archimedes’ Principle Denser fluids will exert a greater buoyant force on a body than less dense fluids of the same volume. Example: Ship will float higher in saltwater (density = 1.03 g/cm 3 ) than in freshwater (density = 1.00 g/cm 3 ).
MFMcGrawChap 13d-Liquids - Revised 4-3-1053 Archimedes’ Principle Applies in air –The more air an object displaces, the greater the buoyant force on it. –If an object displaces its weight, it hovers at a constant altitude. –If an object displaces less air, it descends.
MFMcGrawChap 13d-Liquids - Revised 4-3-1054 As you sit in class, is there a buoyant force acting on you? No, as evidenced by an absence of lift B.Yes, due to displacement of air Archimedes’ Principle CHECK YOUR NEIGHBOR
MFMcGrawChap 13d-Liquids - Revised 4-3-1055 As you sit in class, is there a buoyant force acting on you? No, as evidenced by an absence of lift B.Yes, due to displacement of air Explanation: There is a buoyant force on you due to air displacement, but much less than your weight. Archimedes’ Principle CHECK YOUR ANSWER
MFMcGrawChap 13d-Liquids - Revised 4-3-1056 What Makes an Object Float or Sink? Whether an object floats or sinks depends upon the volume of the object. volume of the fluid displaced. For an object to float: Weight of object is less than buoyant force of the liquid, i.e., less than the weight of the liquid it displaces. Odd fact - A chuck of iron (7.86 gm/cm 3 ) will float in mercury (13.6 gm/cm 3 ).
MFMcGrawChap 13d-Liquids - Revised 4-3-1057 What Makes an Object Float or Sink? Three rules: 1.An object more dense than the fluid in which it is immersed will sink. 2.An object less dense than the fluid in which it is immersed will float. 3.An object having a density equal to the density of the fluid in which it is immersed will neither sink nor float.
MFMcGrawChap 13d-Liquids - Revised 4-3-1058 Artesian Diver
MFMcGrawChap 13d-Liquids - Revised 4-3-1059 Two solid blocks of identical size are submerged in water. One block is lead and the other is aluminum. Upon which is the buoyant force greater? On the lead block B.On the aluminum block C.Same on both blocks D.There is not enough information to decide. What Makes an Object Float or Sink? CHECK YOUR NEIGHBOR
MFMcGrawChap 13d-Liquids - Revised 4-3-1060 Two solid blocks of identical size are submerged in water. One block is lead and the other is aluminum. Upon which is the buoyant force greater? On the lead block B.On the aluminum block C.Same on both blocks D.There is not enough information to decide. What Makes an Object Float or Sink? CHECK YOUR ANSWER Explanation: The buoyant force depends upon the volume of the block that is submerged. Since both submerged blocks are the same size, they displace the same volume of water. So they have the same buoyant force.
MFMcGrawChap 13d-Liquids - Revised 4-3-1061 Pascal’s Principle A change in pressure at any point in a confined fluid is transmitted everywhere throughout the fluid. (This is useful in making a hydraulic lift.) The pressure transmitted throughout a confined liquid is pressure over and above that already in the liquid.
MFMcGrawChap 13d-Liquids - Revised 4-3-1062 Pascal’s Principle Pressure applied to one portion of a confined liquid is transferred through out the liquid.
MFMcGrawChap 13d-Liquids - Revised 4-3-1063 Pascal’s Principle Pressing on the “wrong” side of the hydraulic lift doesn’t take advantage of Pascal’s Principle In a real hydraulic lift thesmaller force is applied using an air compressor.
MFMcGrawChap 13d-Liquids - Revised 4-3-1064 Apply a force F 1 here to a piston of cross-sectional area A 1. The applied force is transmitted to the piston of cross-sectional area A 2 here. Pascal’s Principle
MFMcGrawChap 13d-Liquids - Revised 4-3-1065 Mathematically, Here we simplify the problem by ignoring the contribution of the difference in the fluid column heights. Pascal’s Principle
MFMcGrawChap 13d-Liquids - Revised 4-3-1066 Example: Assume that a force of 500 N (about 110 lbs) is applied to the smaller piston in the previous figure. For each case, compute the force on the larger piston if the ratio of the piston areas (A 2 /A 1 ) are 1, 10, and 100. 50,000 N100 5000 N10 500 N1 F2F2 Using Pascal’s Principle: Pascal’s Principle
MFMcGrawChap 13d-Liquids - Revised 4-3-1067 Surface Tension & Capillarity In the next chapter on gases we’ll think of atoms as rigid balls that ricochet off one another. Here we think of atoms as sticky balls which demonstrate the property of capillarity. An interesting example of capillarity (not in the text) involves the tallness of trees. The cohesive forces of water explains the transport of water from roots to the top of tall trees. When a single water molecule evaporates from the cell membrane inside a leaf, it is replaced by the one immediately next to it due to the cohesive forces between water molecules. A pull is created on the column of water which is continuous from leaves to roots. Water can be lifted far higher than the 10.2 m that atmospheric pressure would serve—even to 100 m in this way, the height of the largest trees.
Quick Questions
MFMcGrawChap 13d-Liquids - Revised 4-3-1069 Consider a boat loaded with scrap iron in a swimming pool. If the iron is thrown overboard into the pool, will the water level at the edge of the pool rise, fall, or remain unchanged? 1. Rise 2. Fall 3. Remain unchanged
MFMcGrawChap 13d-Liquids - Revised 4-3-1070 1. Rise 2. Fall 3. Remain unchanged Consider a boat loaded with scrap iron in a swimming pool. If the iron is thrown overboard into the pool, will the water level at the edge of the pool rise, fall, or remain unchanged?
MFMcGrawChap 13d-Liquids - Revised 4-3-1071
MFMcGrawChap 13d-Liquids - Revised 4-3-1072 In the presence of air, the small iron ball and large plastic ball balance each other. When air is evacuated from the container, the larger ball 1. rises. 2. falls. 3. remains in place.
MFMcGrawChap 13d-Liquids - Revised 4-3-1073
MFMcGrawChap 13d-Liquids - Revised 4-3-1074 The weight of the stand and suspended solid iron ball is equal to the weight of the container of water as shown above. When the ball is lowered into the water the balance is upset. The amount of weight that must be added to the left side to restore balance, compared to the weight of water displaced by the ball, would be 1. half.2. the same. 3. twice.4. more than twice.
MFMcGrawChap 13d-Liquids - Revised 4-3-1075
MFMcGrawChap 13d-Liquids - Revised 4-3-1076 A pair of identical balloons are inflated with air and suspended on the ends of a stick that is horizontally balanced. When the balloon on the left is punctured, the balance of the stick is 1. upset and the stick rotates clockwise. 2. upset and the stick rotates counter-clockwise. 3. unchanged.
MFMcGrawChap 13d-Liquids - Revised 4-3-1077
MFMcGrawChap 13d-Liquids - Revised 4-3-1078 Consider an air-filled balloon weighted so that it is on the verge of sinking—that is, its overall density just equals that of water. Now if you push it beneath the surface, it will 1. sink. 2. return to the surface. 3. stay at the depth to which it is pushed.
MFMcGrawChap 13d-Liquids - Revised 4-3-1079
MFMcGrawChap 13d-Liquids - Revised 4-3-1080 Summary Pressure Pressure in a Liquid Buoyancy in a Liquid Archimedes’ Principle What Makes an Object Sink or Float Flotation Pascal’s Principle Surface Tension & Capillarity
Extras
MFMcGrawChap 13d-Liquids - Revised 4-3-1082 Densities of Common Substances
MFMcGrawChap 13d-Liquids - Revised 4-3-1083 The density of the block of wood floating in water is 1. greater than the density of water. 2. equal to the density of water. 3. less than half that of water. 4. more than half the density of water. 5. … not enough information is given.
| 6,121 | 22,721 |
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 4.1875 | 4 |
CC-MAIN-2018-26
|
latest
|
en
| 0.797689 |
https://plarium.com/forum/en/soldiers-inc/entertainment-community/31281_there-are-3-000-diamonds-hidden-in-one-of-these-pumpkins-/
| 1,531,876,713,000,000,000 |
text/html
|
crawl-data/CC-MAIN-2018-30/segments/1531676589980.6/warc/CC-MAIN-20180718002426-20180718022426-00310.warc.gz
| 723,854,088 | 11,424 |
This topic is closed
## There are 3,000 Diamonds hidden in one of these pumpkins!
14 Replies
Character
1 November, 2016, 10:09 AM UTC
There are 3,000 Diamonds hidden in one of these pumpkins. Try to guess in which one for a chance to win them!
Struck with fear, ZGH agents ran away into the night, screaming and squealing... But they had just enough time to hide some Diamonds inside one of the pumpkins. In which one?
One of the Commanders to give the correct answer will get 3,000 Diamonds!
Trick or treat! Happy Halloween!
UTC +3:00
0
User
1 November, 2016, 1:03 PM UTC
pumpkin... pumpkin on the right side!
miracles in life...
UTC +7:00
0
User
1 November, 2016, 6:19 PM UTC
In the middle pumpkin .
UTC +2:00
0
User
1 November, 2016, 6:20 PM UTC
The pumpkin in the middle
UTC +1:00
0
User
1 November, 2016, 6:28 PM UTC
the largest pumpkin (in middle),
lighted in blue.
UTC +1:00
0
User
1 November, 2016, 10:39 PM UTC
The pumpkin on the left.
UTC -5:00
0
User
2 November, 2016, 2:29 AM UTC
3000 diamonds are a huge amount...therefore I guess that they are hidden in the middle pumpkin!
UTC +0:00
0
User
2 November, 2016, 4:56 PM UTC
black pumpkin, at extreme left! oops...my mistake, at extreme left is avatar of Mr.Black, non a pumpkin! (though, the resemblance is remarkable!) AHAHAHAHA!
"Glory to the Bomb and his Divine destructive Cloud"
UTC +0:00
1
Moderator
3 November, 2016, 5:15 PM UTC
The pumpkin of the right !
A lost battle is a battle that is believed to be lost
UTC +1:00
0
User
5 November, 2016, 9:22 AM UTC
when do we know who won?
UTC +1:00
0
User
5 November, 2016, 4:59 PM UTC
each time you close a contest, could you please also explain to us the correct solution of the riddle?
... Otherwise it is absurd to declare a winner without that others understand why!!
UTC +1:00
0
User
8 November, 2016, 3:01 PM UTC
Here is a random puzzle!
miracles in life...
UTC +7:00
0
Moderator
10 November, 2016, 4:44 PM UTC
Pumpkin of the middle
A lost battle is a battle that is believed to be lost
UTC +1:00
0
User
12 November, 2016, 3:45 AM UTC
The middle as diamonds will give off the light
UTC +2:00
0
| 677 | 2,134 |
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 2.6875 | 3 |
CC-MAIN-2018-30
|
latest
|
en
| 0.864074 |
https://stat.ethz.ch/pipermail/r-sig-meta-analysis/2022-December/004285.html
| 1,701,534,858,000,000,000 |
text/html
|
crawl-data/CC-MAIN-2023-50/segments/1700679100427.59/warc/CC-MAIN-20231202140407-20231202170407-00082.warc.gz
| 611,927,462 | 3,635 |
# [R-meta] R-sig-meta-analysis Digest, Vol 67, Issue 2
James Pustejovsky jepu@to @end|ng |rom gm@||@com
Thu Dec 15 16:53:39 CET 2022
```Hi Yuhang,
For the sort of probability calculations you are trying to do, it is best
to use the transformed scale (i.e., the Fisher z scale). These sorts of
calculations rely on ALL of the model's distributional assumptions, so it
is critical to use the scale that best conforms to those assumptions. That
said, you can still interpret the results of these calculations in the
original metric.
Say that zi = transf.rtoz(ri) and that zi is an approximately unbiased
estimate of the parameter zeta-i. We can interpret zeta-i as the r-to-z
transformation of the correlation parameter rho-i. Assume that zeta-i
follows a normal distribution (possibly with a hierarchical structure, like
in your example, but I'll ignore that for now). Then this assumption
implies that rho-i also follows some distribution, but a non-normal one.
However, you can still do percentile calculations on the zeta-i scale. For
instance, based on the model, we can calculate
Pr(zeta-i > 0)
Since transf.rtoz(0) = 0, then
Pr(zeta-i > 0) = Pr(rho-i > 0)
More generally, suppose you want to find Pr(rho-i > c_r). Then you can take
c_z = transf.rtoz(c_r) and find
Pr(zeta-i > c_z),
which will correspond to Pr(rho-i > c_r). For small values of c_r, the
cutoffs are very similar, e.g.
c_r = 0.2 implies c_z = 0.2027
c_r = 0.3 implies c_z = 0.3095
c_z = 0.4 implies c_z = 0.4236
etc.
All of these calculations work more generally with the sort of multi-level
random effects model that you described in your example.
Cheers,
James
On Wed, Dec 14, 2022 at 7:33 PM Yuhang Hu <yh342 using nau.edu> wrote:
> Dear Wolfgang,
>
> Many thanks for your response. To make sure, as a general principle, do you
> recommend always computing those probabilities based off of the
> r2z-transformed effect size estimates?
>
> My intuition regarding transforming the variance components back to their
> original scale came from the fact that the emmeans package does a similar
> thing (for standard errors) via the delta method (
>
> https://cran.r-project.org/web/packages/emmeans/vignettes/transformations.html#after
> ).
>
> And my desire to accurately back transform the variance components was
> solely because I thought it could improve interpretability given the ease
> of working with correlations than r2z transformed correlations.
>
> Thanks again,
> Yuhang
>
> On Wed, Dec 14, 2022 at 4:01 AM <r-sig-meta-analysis-request using r-project.org
> >
> wrote:
>
> > Send R-sig-meta-analysis mailing list submissions to
> > r-sig-meta-analysis using r-project.org
> >
> > To subscribe or unsubscribe via the World Wide Web, visit
> > https://stat.ethz.ch/mailman/listinfo/r-sig-meta-analysis
> > or, via email, send a message with subject or body 'help' to
> > r-sig-meta-analysis-request using r-project.org
> >
> > You can reach the person managing the list at
> > r-sig-meta-analysis-owner using r-project.org
> >
> > than "Re: Contents of R-sig-meta-analysis digest..."
> >
> >
> > Today's Topics:
> >
> > 1. Using variance components with effect size transformation
> > (Yuhang Hu)
> >
> > ----------------------------------------------------------------------
> >
> > Message: 1
> > Date: Tue, 13 Dec 2022 20:28:55 -0700
> > From: Yuhang Hu <yh342 using nau.edu>
> > To: R meta <r-sig-meta-analysis using r-project.org>
> > Subject: [R-meta] Using variance components with effect size
> > transformation
> > Message-ID:
> > <
> > CA+dzWjp7MVtvytKgzK8eAQmGUzRsVex2zdvsi86UGP2QwHcgTA using mail.gmail.com>
> > Content-Type: text/plain; charset="utf-8"
> >
> > Dear Meta Community,
> >
> > I have a 3-level meta-regression model with a categorical moderator (3
> > levels) plus some covariates fit as:
> >
> > m6 = rma.mv(r2z_transformed ~ 0 + cat_mod + covariates, random = ~ 1 |
> > study/effect)
> >
> > I wonder to calculate probabilities from the distribution of effects as
> > explained in this post (
> >
> https://stat.ethz.ch/pipermail/r-sig-meta-analysis/2022-August/004136.html
> > ),
> > whether I should use the transformed variance components (A) or
> > back-transformed variance components (B)?
> >
> > A: confint.rma.mv(m6) # transformed i.e., 'r2z'
> > B: confint.rma.mv(m6, transf = transf.ztor) # back-transformed i.e.,
> 'z2r'
> >
> > Currently, A and B give the same result, and I wonder why?
> >
> > Thanks,
> > Yuhang
> >
> > [[alternative HTML version deleted]]
> >
> >
> >
> >
> > ------------------------------
> >
> > Subject: Digest Footer
> >
> > _______________________________________________
> > R-sig-meta-analysis mailing list @ R-sig-meta-analysis using r-project.org
> > To manage your subscription to this mailing list, go to:
> > https://stat.ethz.ch/mailman/listinfo/r-sig-meta-analysis
> >
> >
> > ------------------------------
> >
> > End of R-sig-meta-analysis Digest, Vol 67, Issue 2
> > **************************************************
> >
>
> [[alternative HTML version deleted]]
>
> _______________________________________________
> R-sig-meta-analysis mailing list @ R-sig-meta-analysis using r-project.org
> To manage your subscription to this mailing list, go to:
> https://stat.ethz.ch/mailman/listinfo/r-sig-meta-analysis
>
[[alternative HTML version deleted]]
```
| 1,463 | 5,395 |
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 3 | 3 |
CC-MAIN-2023-50
|
latest
|
en
| 0.871572 |
http://encyclopedia.kids.net.au/page/ph/Phase_transition
| 1,611,058,214,000,000,000 |
text/html
|
crawl-data/CC-MAIN-2021-04/segments/1610703518240.40/warc/CC-MAIN-20210119103923-20210119133923-00086.warc.gz
| 35,421,239 | 9,205 |
## Encyclopedia > Phase transition
Article Content
# Phase transition
A phase transition is the transformation of a thermodynamic system from one phase to another. The distinguishing characteristic of a phase transition is an abrupt sudden change in one or more physical properties, in particular the heat capacity, with a small change in a thermodynamic variable such as the temperature. Examples of phase transitions are:
As discussed in the article on phases, phase transitions come about when the free energy of a system is non-analytic for some choice of thermodynamic variables. This non-analyticity generally stems from the interactions of an extremely large number of particles in a system, and does not appear in systems that are too small.
### Ehrenfest classification
The first attempt at classifying phase transitions was the Ehrenfest classification scheme, which grouped phase transitions based on the degree of non-analyticity involved. Though useful, Ehrenfest's classification is flawed, as we will discuss in the next section.
Under this scheme, phase transitions were labelled by the lowest derivative of the free energy that is discontinuous at the transition. First-order phase transitions exhibit a discontinuity in the first derivative of the free energy with a thermodynamic variable. The various solid/liquid/gas transitions are classified as first-order transitions, as the pressure, which is the first derivative of the free energy with volume, changes discontinuously across the transitions. Second-order phase transitions have a discontinuity in a second derivative of the free energy. These include the ferromagnetic phase transition in materials such as iron, where the magnetization, which is the first derivative of the free energy with the appplied magnetic field strength, increases continuously from zero as the temperature is lowered below the Curie temperature. The magnetic susceptibility[?], the second derivative of the free energy with the field, changes discontinuously. Under the Ehrenfest classication scheme, there could in principle be third, fourth, and higher-order phase transitions.
### Modern classification of phase transitions
The Ehrenfest scheme is an inaccurate method of classifying phase transitions, for it is based on the mean field theory of phases (to be described in a later section.) Mean field theory is inaccurate in the vicinity of phase transitions, as it neglects the role of thermodynamic fluctuations. For instance, it predicts a finite discontinuity in the heat capacity at the ferromagnetic transition, which is implied by Ehrenfest's definition of "second-order" transitions. In real ferromagnets, the heat capacity diverges to infinity at the transition.
In the modern classification scheme, phase transitions are divided into two broad categories, named similarly to the Ehrenfest classes:
The first-order phase transitions are those that involve a latent heat. During such a transition, a system either absorbs or releases a fixed (and typically large) amount of energy. Because energy cannot be instantaneously transferred between the system and its environment, first-order transitions are associated with "mixed-phase regimes" in which some parts of the system have completed the transition and others have not. This phenomenon is familiar to anyone who has boiled a pot of water: the water does not instantly turn into gas, but forms a turbulent mixture of water and water vapor bubbles. Mixed-phase systems are difficult to study, because their dynamics are violent and hard to control. However, many important phase transitions fall in this category, including the solid/liquid/gas transitions.
The second class of phase transitions are the continuous phase transitions, also called second-order phase transitions. These have no associated latent heat. Examples of second-order phase transitions are the ferromagnetic transition, the superfluid transition, and Bose-Einstein condensation.
### Critical points
In systems containing liquid and gaseous phases, there exist a special combination of pressure and temperature, known as the critical point, at which the transition between liquid and gas becomes a second-order transition. Near the critical point, the fluid is sufficiently hot and compressed that the distinction between the liquid and gaseous phases is almost non-existent.
This is associated with the phenomenon of critical opalescence[?], a milky appearance of the liquid, due to density fluctuations at all possible wavelengths (including those of visible light).
### Symmetry
Phase transitions often (but not always) take place between phases with different symmetry. Consider, for example, the transition between a fluid (i.e. liquid or gas) and a crystalline solid. A fluid, which is composed of atoms arranged in a disordered but homogenous manner, possesses continuous translational symmetry: each point inside the fluid has the same properties as any other point. A crystalline solid, on the other hand, is made up of atoms arranged in a regular lattice. Each point in the solid is not similar to other points, unless those points are displaced by an amount equal to some lattice spacing.
Generally, we may speak of one phase in a phase transition as being more symmetrical than the other. The transition from the more symmetrical phase to the less symmetrical one is a symmetry-breaking process. In the fluid-solid transition, for example, we say that continuous translation symmetry is broken.
The ferromagnetic transition is another example of a symmetry-breaking transition, in this case time-reversal symmetry. The magnetization of a system switches sign under time-reversal (one may think of magnetization as produced by tiny loops of electrical current, which reverse direction when time is reversed.) In the absence of an applied magnetic field, the paramagnetic phase contains no net magnetization, and is symmetrical under time-reversal; whereas the ferromagnetic phase has a net magnetization and is not symmetrical under time-reversal.
The presence of symmetry-breaking (or nonbreaking) is important to the behavior of phase transitions. It was pointed out by Landau that, given any state of a system, one may unequivocally say whether or not it possesses a given symmetry. Therefore, it cannot be possible to analytically deform a state in one phase into a phase possessing a different symmetry. This means, for example, that it is impossible for the solid-liquid phase boundary to end in a critical point like the liquid-gas boundary. However, symmetry-breaking transitions can still be either first or second order.
Typically, the more symmetrical phase is on the high-temperature side of a phase transition, and the less symmetrical phase on the low-temperature side. This is certainly the case for the solid-fluid and ferromagnetic transitions. This happens because the Hamiltonian of a system usually exhibits all the possible symmetries of the system, whereas the low-energy states lack some of these symmetries (this phenomenon is known as spontaneous symmetry breaking[?].) At low temperatures, the system tends to be confined to the low-energy states. At higher temperatures, thermal fluctuations allow the system to access states in a broader range of energy, and thus more of the symmetries of the Hamiltonian.
When symmetry is broken, one needs to introduce one or more extra variables to describe the state of the system. For example, in the ferromagnetic phase one must provide the net magnetization, whose direction was spontaneously chosen when the system cooled below the Curie point. Such variables are instances of order parameters, which will be described later. However, note that order parameters can also be defined for symmetry-nonbreaking transitions.
Symmetry-breaking phase transitions play an important role in cosmology. It has been speculated that, in the hot early universe, the vacuum (i.e. the various quantum fields that fill space) possessed a large number of symmetries. As the universe expanded and cooled, the vacuum underwent a series of symmetry-breaking phase transitions. For example, the electroweak transition broke the SU(2)×U(1) symmetry of the electroweak field into the U(1) symmetry of the present-day electromagnetic field. This transition is important to understanding the asymmetry between the amount of matter and antimatter in the present-day universe (see electroweak baryogenesis[?].)
### Critical exponents and universality classes
Continuous phase transitions are easier to study than first-order transitions due to the absence of latent heat, and they have been discovered to have many interesting properties. The phenomena associated with continuous phase transitions are called critical phenomena, due to their association with critical points.
It turns out that continuous phase transitions can be characterized by parameters known as critical exponents[?]. For instance, let us examine the behavior of the heat capacity near such a transition. We vary the temperature T of the system while keeping all the other thermodynamic variables fixed, and find that the transition occurs at some critical temperature Tc. When T is near Tc, the heat capacity C typically has a power law behaviour:
$C \sim |T_c - T|^{-\alpha}$
The constant α is the critical exponent associated with the heat capacity. It is not difficult to see that it must be less than 1 in order for the transition to have no latent heat. Its actual value depends on the type of phase transition we are considering. For -1 < α < 0, the heat capacity has a "kink" at the transition temperature. This is the behavior of liquid helium at the "lambda transition" from a normal state to the superfluid state, for which experiments have found α = -0.013±0.003. For 0 < α < 1, the heat capacity diverges at the transition temperature (though, since α < 1, the divergence is not strong enough to produce a latent heat.) An example of such behavior is the 3-dimensional ferromagnetic phase transition. In the three-dimensional Ising model[?] for uniaxial magnets, detailed theoretical studies have yielded the exponent α ∼ 0.110.
Some model systems do not obey this power law behavior. For example, mean field theory predicts a finite discontinuity of the heat capacity at the transition temperature, and the two-dimensional Ising model has a logarithmic divergence. However, these systems are an exception to the rule. Real phase transitions exhibit power law behavior.
Several other critical exponents - β, γ, δ, ν, and η - are defined, examining the power law behavior of a measurable physical quantity near the phase transition.
It is a remarkable fact that phase transitions arising in different systems often possess the same set of critical exponents. This phenomenon is known as universality. For example, the critical exponents at the liquid-gas critical point have been found to be independent of the chemical composition of the fluid. More amazingly, they are an exact match for the critical exponents of the ferromagnetic phase transition in uniaxial magnets. Such systems are said to be in the same universality class. Universality is a prediction of the renormalization group theory of phase transitions, which states that the thermodynamic properties of a system near a phase transition depend only on a small number of features, such as dimensionality and symmetry, and is insensitive to the underlying microscopic properties of the system.
References
• Anderson, P.W., Basic Notions of Condensed Matter Physics, Perseus Publishing (1997).
• Goldenfeld, N., Lectures on Phase Transitions and the Renormalization Group, Perseus Publishing (1992).
• Landau, L.D. and Lifshitz, E.M., Statistical Physics Part 1, vol. 5 of Course of Theoretical Physics, Pargamon, 3rd Ed. (1994).
All Wikipedia text is available under the terms of the GNU Free Documentation License
Search Encyclopedia
Search over one million articles, find something about almost anything!
Featured Article
Reformed churches ... Church (USA) (Anglo-Scot Presbyterians and Congregationalists) Protestant Reformed Church[?] (Dutch Reformed - GKN) One of the most conservative Reformed/Calvinist ...
This page was created in 41.4 ms
| 2,438 | 12,279 |
{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 2.734375 | 3 |
CC-MAIN-2021-04
|
longest
|
en
| 0.928562 |
https://astrostatistics.wordpress.com/2013/11/13/statistics-citations-a-conversion-table-for-astronomers/
| 1,529,714,193,000,000,000 |
text/html
|
crawl-data/CC-MAIN-2018-26/segments/1529267864848.47/warc/CC-MAIN-20180623000334-20180623020334-00071.warc.gz
| 548,248,653 | 17,164 |
## Statistics citations: a conversion table for astronomers
While checking up on astro ph today (as distraction from my reading on efficient simulation from huge GP models) I noticed this embarrassingly awful contribution to astronomical model selection by some Italian faculty (at first I thought masters/PhD students): http://arxiv.org/pdf/1311.2736.pdf . The authors aim to compute the marginal likelihoods for two competing models: one with two parameters and the other with three parameters, so it would be easy enough to do these integrals directly via quadrature. But instead they try out and compare the harmonic mean, Laplace approximation, thermodynamic integration, and nested sampling; ostensibly for pedagogical purposes. However, there’s very limited discussion of these methods and crucial missing details of their implementation, but *most worryingly* all return very different Bayes factors (quoted with no uncertainties) … and the authors don’t even investigate why this is!!!! E.g. from their table 5 the four methods for estimating the same Bayes factor give 1418, 145, 586, and 171 (in the same order as above). It’s bizarre. No surprises that the HME is a fail, but what has happened to thermodynamic integration? My suspicion is that they’ve chosen a linear temperature sequence (perhaps also too coarsely spaced) where a logarithmic one (cf. Friel & Pettitt 2008) would have been appropriate. But we’ll never know because they don’t discuss this. They also advocate uniform and Jeffreys priors for model selection …
Amongst various other things that annoyed me about the above paper is a common grievance of mine: that astronomical citations are too often given for standard statistical concepts. With this in mind I thought it might compile (and keep updated with all suggestions welcome) a list of transformations for converting between the neophyte’s astronomical reference system and the old hand’s statistical one.
the Laplace approximation for marginal likelihood estimation:
Gregory (2005); Ntzoufras (2009) -> Tierney & Kadane (1986)
the Harmonic Mean Estimator:
Ntzoufras (2009) -> Newton & Raftery (1994)
parallel tempering:
Gregory (2005); Handberg & Campante (2011) -> Geyer 1994
the thermodynamic integration identity:
Gregory (2005) -> Gelman & Meng (1998); Lefebvre (2010)
thermodynamic integration in practise:
Gregory (2005) -> Lartillot & Phillipe (2006); Friel & Pettitt (2008)
the Bayesian approach to hypothesis testing:
Gregory (2005) -> Jeffreys (1935,1961); Kass & Raftery (1995)
On the plus side today there was a nice paper by Gomez et al. (including Wolpert as guest statistician) on a sensitivity analysis via emulators for galaxy formation models, which I intend to read further when I get a chance: http://arxiv.org/pdf/1311.2587.pdf
| 664 | 2,805 |
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 2.609375 | 3 |
CC-MAIN-2018-26
|
latest
|
en
| 0.895605 |
https://www.airmilescalculator.com/distance/cpr-to-adq/
| 1,621,007,776,000,000,000 |
text/html
|
crawl-data/CC-MAIN-2021-21/segments/1620243991428.43/warc/CC-MAIN-20210514152803-20210514182803-00432.warc.gz
| 660,428,613 | 74,438 |
# Distance between Casper, WY (CPR) and Kodiak, AK (ADQ)
Flight distance from Casper to Kodiak (Casper–Natrona County International Airport – Kodiak Airport) is 2223 miles / 3578 kilometers / 1932 nautical miles. Estimated flight time is 4 hours 42 minutes.
Driving distance from Casper (CPR) to Kodiak (ADQ) is 3308 miles / 5323 kilometers and travel time by car is about 64 hours 52 minutes.
## Map of flight path and driving directions from Casper to Kodiak.
Shortest flight path between Casper–Natrona County International Airport (CPR) and Kodiak Airport (ADQ).
## How far is Kodiak from Casper?
There are several ways to calculate distances between Casper and Kodiak. Here are two common methods:
Vincenty's formula (applied above)
• 2223.252 miles
• 3577.977 kilometers
• 1931.953 nautical miles
Vincenty's formula calculates the distance between latitude/longitude points on the earth’s surface, using an ellipsoidal model of the earth.
Haversine formula
• 2217.709 miles
• 3569.056 kilometers
• 1927.136 nautical miles
The haversine formula calculates the distance between latitude/longitude points assuming a spherical earth (great-circle distance – the shortest distance between two points).
## Airport information
A Casper–Natrona County International Airport
City: Casper, WY
Country: United States
IATA Code: CPR
ICAO Code: KCPR
Coordinates: 42°54′28″N, 106°27′50″W
B Kodiak Airport
City: Kodiak, AK
Country: United States
Coordinates: 57°45′0″N, 152°29′38″W
## Time difference and current local times
The time difference between Casper and Kodiak is 2 hours. Kodiak is 2 hours behind Casper.
MDT
AKDT
## Carbon dioxide emissions
Estimated CO2 emissions per passenger is 243 kg (536 pounds).
## Frequent Flyer Miles Calculator
Distance:
2223
Elite level bonus:
0
Booking class bonus:
0
### In total
Total frequent flyer miles:
2223
Round trip?
| 503 | 1,881 |
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 2.765625 | 3 |
CC-MAIN-2021-21
|
latest
|
en
| 0.807388 |
https://www.lmfdb.org/Variety/Abelian/Fq/2/49/aw_ic
| 1,591,133,637,000,000,000 |
text/html
|
crawl-data/CC-MAIN-2020-24/segments/1590347426801.75/warc/CC-MAIN-20200602193431-20200602223431-00301.warc.gz
| 803,976,763 | 26,073 |
# Properties
Label 2.49.aw_ic Base Field $\F_{7^{2}}$ Dimension $2$ Ordinary No $p$-rank $1$ Principally polarizable Yes Contains a Jacobian No
## Invariants
Base field: $\F_{7^{2}}$ Dimension: $2$ L-polynomial: $( 1 - 7 x )^{2}( 1 - 8 x + 49 x^{2} )$ Frobenius angles: $0$, $0$, $\pm0.306389419005$ Angle rank: $1$ (numerical) Jacobians: 0
This isogeny class is not simple.
## Newton polygon
$p$-rank: $1$ Slopes: $[0, 1/2, 1/2, 1]$
## Point counts
This isogeny class is principally polarizable, but does not contain a Jacobian.
$r$ 1 2 3 4 5 6 7 8 9 10 $A(\F_{q^r})$ 1512 5612544 13838478696 33226260480000 79781820476078952 191575128888463574016 459984415854857021776872 1104427232468159779307520000 2651730809531246043410493784296 6366805760003910324030652455650304
$r$ 1 2 3 4 5 6 7 8 9 10 $C(\F_{q^r})$ 28 2338 117628 5763646 282438268 13840846306 678219946012 33232917276286 1628413575601372 79792266286268578
## Decomposition and endomorphism algebra
Endomorphism algebra over $\F_{7^{2}}$
The isogeny class factors as 1.49.ao $\times$ 1.49.ai and its endomorphism algebra is a direct product of the endomorphism algebras for each isotypic factor. The endomorphism algebra for each factor is: 1.49.ao : the quaternion algebra over $$\Q$$ ramified at $7$ and $\infty$. 1.49.ai : $$\Q(\sqrt{-33})$$.
All geometric endomorphisms are defined over $\F_{7^{2}}$.
## Base change
This is a primitive isogeny class.
## Twists
Below are some of the twists of this isogeny class.
Twist Extension Degree Common base change 2.49.ag_ao $2$ (not in LMFDB) 2.49.g_ao $2$ (not in LMFDB) 2.49.w_ic $2$ (not in LMFDB)
Below is a list of all twists of this isogeny class.
Twist Extension Degree Common base change 2.49.ag_ao $2$ (not in LMFDB) 2.49.g_ao $2$ (not in LMFDB) 2.49.w_ic $2$ (not in LMFDB) 2.49.ai_du $4$ (not in LMFDB) 2.49.i_du $4$ (not in LMFDB)
| 717 | 1,872 |
{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 2.796875 | 3 |
CC-MAIN-2020-24
|
latest
|
en
| 0.574568 |
http://www.jiskha.com/display.cgi?id=1366642284
| 1,493,299,150,000,000,000 |
text/html
|
crawl-data/CC-MAIN-2017-17/segments/1492917122167.63/warc/CC-MAIN-20170423031202-00504-ip-10-145-167-34.ec2.internal.warc.gz
| 594,993,498 | 3,720 |
# Math help
posted by on .
A gas station charges \$2.19 per gallon of gas. Use function notation to describe the relationship between the total cost C(g) and the number of gallons purchased g.
C(g) = –2.19g
g = 2.19C(g)
C(g) = g + 2.19
C(g) = 2.19g
Please tell me the explination and the answer.
• Math help - ,
C(g) = 2.19g
• Math help - ,
I think this is correct.
#### Answer This Question
First Name: School Subject: Answer:
#### Related Questions
More Related Questions
Post a New Question
| 145 | 506 |
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 3.28125 | 3 |
CC-MAIN-2017-17
|
latest
|
en
| 0.845905 |
http://blogs.scienceforums.net/ajb/2012/01/26/quantum-field-theory-by-l-h-ryder/
| 1,571,423,115,000,000,000 |
text/html
|
crawl-data/CC-MAIN-2019-43/segments/1570986684425.36/warc/CC-MAIN-20191018181458-20191018204958-00157.warc.gz
| 34,696,929 | 9,296 |
# Quantum Field Theory by L.H. Ryder
Quantum Field Theory Quantum field theory is at the heart of modern physics and forms the backbone of the standard model, which is our current best understanding of the laws of particles and forces. The reputation of QFT is that it is very difficult to learn. Quantum field theory by Lewis H. Ryder is a solid modern pedagogical introduction to the ideas and techniques of QFT. The book assumes some familiarity with quantum mechanics and special relativity.
The readership is graduate students in theoretical physics. This book is clearly of a pedagogical nature and contains very detailed worked examples and proofs of statements.
The book consists of 11 chapters.
Chapter 1 gives a general overview of particle physics. The reader is exposed to the basic idea of QFT here and the 4 forces of nature. This chapter gives one a taste for the standard model though the book is not really a book about particle physics.
Chapter 2 introduces single-particle relativistic wave equations. Here the Klein-Gordon equation the Dirac equation & antiparticles, the Maxwell and Proco equations are covered. The importance of the Lorentz and Poincare groups in physics are outlined. The differential geometry of Maxwell’s equations is also presented.
Classical field theory is the topic of Chapter 3. Here the Lagrangian formulation and variational principles are reviewed. The Bohm-Aharonov effect is presented as is Yang-Mills theory from a geometric perspective.
Chapter 4 deals with the canonical quantisation and particles. Due to technical difficulties the Klein-Gordon and Dirac equations cannot be single particle equations. In this chapter the reader starts to deal with quantum field theory as a theory of “many particles”. The real and complex Klein-Gordon fields, Dirac fields and the electromagnetic field are dealt with via canonical quantisation.
From a modern perspective path-integrals are the root to quantisation of fields. In Chapter 5 the Feynman path-integral formulation of quantum mechanics is presented. Topics covered here include perturbation theory & S-matrices and Coulomb scattering. This chapter lays down the ideas of path-integrals ready for QFT.
Chapter 6 develops the path-integral quantisation and Feynman rules for scalar and Dirac fields. Topics covered include: generating functionals, functional integration, Green’s functions & propagators, interacting fields, fermions & anticommuting variables and S matrix formula.
Chapter 7 moves on to discuss path-integral quantisation of gauge fields. Topics covered include: gauge fixing, Faddeev-Popov ghosts, Feynman rules in the Lorentz gauge, the Ward-Takahashi identity in QED, the BRST transformations and the Slavnov-Taylor identities.
Spontaneous symmetry breaking and the Weinberg-Salam model are the topic of Chapter 8. Topics covered include: the Goldstone theorem, spontaneous breaking of gauge symmetries, superconductivity and the Weinberg-Salam model.
Chapter 9 covers the subject of renormalisation. Topics covered include: divergences in QFT, dimensional analysis, regularisation, loop expansions, counter-terms, the renormalisation group, 1-loop renormalisation of QED, renormalisabilty of QCD, asymptotic freedom, anomalies, and renormalisation of Yang-Mills theories with spontaneous symmetry breaking.
Chapter 10 introduces the notion of topological objects in field theory. Topics covered include: the sine-Gordon kink, the Dirac monopole, instantons and theta-vacua.
N= 1 supersymmety in 4 dimensions is the topic of Chapter 11. The theory is built at first in component form and then the power of superspace methods are exposed. Topics covered include the super-Poincare algebra, superspace & super fields, chiral super fields and the Wess-Zumino model.
Paperback: 507 pages
Publisher: Cambridge University Press; 2 edition (6 Jun 1996)
Language English
ISBN-10: 0521478146
ISBN-13: 978-0521478144
| 848 | 3,944 |
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 2.59375 | 3 |
CC-MAIN-2019-43
|
latest
|
en
| 0.917225 |
http://www.jiskha.com/display.cgi?id=1229577778
| 1,498,190,552,000,000,000 |
text/html
|
crawl-data/CC-MAIN-2017-26/segments/1498128319992.22/warc/CC-MAIN-20170623031127-20170623051127-00131.warc.gz
| 590,748,017 | 4,178 |
# math probability
posted by on .
this is a probability question...
Suppose you are asked to choose a whole number between 1 and 13 inclusive.
(a) what is the probability that it is odd?...7/13
(b) What is the probability that it is even?...6/13
(c) what is the probability that it is a multiple of 3?...4/13
(d) what i the probability that it is odd or a multiple of 3?.. This is where i am stumped. the book says the answer is 9/13 but i cant find it.
• math probability - ,
the first 3 are correct
for d)
the odds are 1,3,5,7,9,11,13
multiples of 3 are 3,6,9,12
now the numbers which are either odd OR a multiple of 3 are 1,3,5,6,7,9,11,12,13
or nine of them
so prob = 9/13
There is a formula which says
Prob(A or B) = Prob(A) + Prob(B) - Prob(A and B)
= 4 + 7 - 2 = 9
notice that the 3 and 9 are in both sets, so if we just add up the count, we counted them twice, so we have to subtract one of the counts.
• math probability - ,
P(b and 2)
• math probability - ,
help me is suck at math and probability is very hard
### Answer This Question
First Name: School Subject: Answer:
### Related Questions
More Related Questions
Post a New Question
| 350 | 1,166 |
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 3.90625 | 4 |
CC-MAIN-2017-26
|
latest
|
en
| 0.946979 |
http://www.ux1.eiu.edu/~cfadd/1360/21KineticTheory/HmwkSol.html
| 1,553,010,369,000,000,000 |
text/html
|
crawl-data/CC-MAIN-2019-13/segments/1552912201996.61/warc/CC-MAIN-20190319143502-20190319165502-00404.warc.gz
| 370,286,231 | 7,905 |
# Homework Solutions
From the FIFTH edition
Questions: 1, 2, 4, 6, 12, 15
Problems: *, 3, **, 8, 13, ****, 25, *****, ******, 33
Be sure and do these; do not just wait and watch me do them in class!
Q21.1 Dalton's law of partial pressures states: The total pressure of a mixture of gases is equal to the sum of the partial pressures of gases making up the mixture. Give a convincing argument of this law based on the kinetic theory of gases.
Pressure is a result of atoms (or molecules) bombarding the sides of
the container. If there are more atoms in the container, there will be
more bombarding the sides of the container and this will increase the pressure. Dalton's law that the total pressure is the sum of the partial pressure is another way of saying the total force or pressure due to the sum of all the atoms hitting the side of the container is the sum of the force or pressure due to the different kinds of atoms hitting the side of the container.
Q21.2 One container is filled with helium gas and another with argon gas. If both containers are at the same temperature, which molecules have the higher rms speed?
At the same temperature, the average value of the translational kinetic energy [(1/2) m <v2>] will be the same for both helium and argon. Argon has a larger mass so it will have a smaller value of <v2> or vrms. Helium has a smaller mass so it will have a larger value of <v2> or vrms.
Q21.4 Although the average speed of gas molecules in thermal equilibrium at some temperature is greater than zero, the average velocity is zero. Explain.
The average velocity must be zero since the center of mass of the gas is not moving.
Q21.6 A liquid partially fills a container. Explain why the temperature of the liquid decreases when the container is partially evacuated. (It is possible to freeze water using this technique.)
The faster-moving atoms are the ones that escape from the liquid's
surface (that is, they are the ones that evaporate). That means the
average speed of the remaining atoms is decreased -- and the
temperature decreases as the average speed of the atoms decreases.
Q21.12 Why does a diatomic gas have a greater thermal energy content per mole than a monatomic gas at the same temperature?
The molar specific heat of a diatomic molecule is CV = (5/2) R, meaning U = (5/2) n R T, while, for a monatomic gas, CV = (3/2) R or U = (3/2) n R T. Since the molar specific heat is greater for a diatomic molecule, there is more internal energy stored inthe motion of the molecules for the same temperature than for that temperature in a monatomic gas.
Q21.13 An ideal gas is contained in a vessel at 300 K. If the temperature is increased to 900 K,
(a) by what factor does the rms speed of each molecule change?
The temperature has increased by a factor of 3. The rms speed goes like the square root of the temperature so we have increased the rms speed by SQRT(3)=1.732
(b) by what factor does the pressure in the vessel change?
The Ideal Gas Law still holds, P V = n R T. If the volume remains constant, an increase in temperature by a factor of 3 means an increase in pressure by the same factor of 3.
SQRT[] means "Square Root of" because it is easier for me to write SQRT than to go to a graphics editor and construct a Square Root sign.
21.*. Find the rms speed of nitrogen molecules under standard conditions, 0.0oC and 1.00 atm pressure. Recall that 1 mole of any gas occupies a volume of 22.4 liters under standard conditions.
Eq 21.7, on page 590, tells us
T = 0oC = 273 K
R = 8.31 J / mol-K
M = 28 g/mole = 28 x 10 - 3 kg/mole
vrms = SQRT[ 3 (8.31 J/mol-K)(273 K)/(28 x 10 - 3 kg/mole)]
vrms = SQRT [2.43 x 105 (J/mol-K)(K)/(kg/mole)]
vrms = SQRT [2.43 x 105 (J/kg)(N-m/J)(kg-m/s2/N)]
vrms = SQRT [2.43 x 105 (m2/s2)]
vrms = 493 m/s
Be careful with the units. They're important! They're not just an afterthought.
21.3 In a 30-s interval, 500 hailstones strike a glass window of area 0.60 m2 at an angle of 45o to the window surface. Each hailstone has a mass of 5.0 g (0.005 kg) and a speed of 8.0 m/s. If the collisions are elastic, find the average force and pressure on the window.
vx = v cos 45o
vx = (8.0 m/s)(0.707) = 5.66 m/s
px = m vx = (0.005 kg)(5.66 m/s) = 2.83 x 10 - 2 kg-m/s
px = 2 px = 5.66 x 10 - 2 kg-m/s
F = pxtot/t
F = (500)(5.66 x 10 - 2 kg-m/s)/30 s
F = 0.943 kg-m/s2
F = 0.943 N
P = F/A
P = 0.943 N / 0.60 m2
P = 1.57 N/m2
P = 1.57 Pa
21.** Calculate the rms speed of an H2 molecule at 250oC.
Of course, this is basically the same as problem 21.1 above. But it is always good to work similiar problems -- just to see that the idea is the same even with different numbers. This time
T = 250oC = 523 K
M = 2 g/mole = 2 x 10 - 3 kg/mole
As with 21.1, we begin with Eq 21.7, from page 590,
R = 8.31 J / mol-K
vrms = SQRT[ 3 (8.31 J/mol-K)(523 K)/(2 x 10 - 3 kg/mole)]
vrms = SQRT [6.52 x 106 (J/mol-K)(K)/(kg/mole)]
vrms = SQRT [6.52 x 106 (J/kg)(N-m/J)(kg-m/s2/N)]
vrms = SQRT [6.52 x 106 (m2/s2)]
vrms = 2 550 m/s
21.8 If the rms speed of a helium atom at room temperature is 1350 m/s, what is the rms speed of an oxygen (O2) molecule at this temperature? (The molar mass of O2 is 32 and the molar mass of He is 4.)
This problem, too, begins with Eq 21.7,
vrmsO/vrmsH = SQRT[3 R T/MO] / SQRT[3 R T/MH]
vrmsO/vrmsH = SQRT[ (3 R T/MO) / (3 R T/MH) ]
vrmsO/vrmsH = SQRT[ MH/MO ]
vrmsO/vrmsH = SQRT[ 4/16 ]
vrmsO/vrmsH = SQRT[ 1/4 ]
vrmsO/vrmsH = 1/2
vrmsO/vrmsH = 0.5
vrmsO = 0.5 vrmsH
vrmsO = 0.5 (1350 m/s)
vrmsO = 675 m/s
21.*** One mole of xenon gas at 20.0oC occupies 0.0224 m3. What is the pressure exerted by the Xe atoms on the walls of a container?
P V = n R T
P = n R T / V
P = (1 mole) (8.31 J/mole-K) (293 K) / 0.0224 m3
P = 1.09 x 105 (J/m3)
Now, what in the world is a J/m3?
P = 1.09 x 105 (J/m3) [ N-m / J ]
P = 1.09 x 105 N/m2
P = 1.09 x 105 Pa
P = 1.09 x 105 Pa [1 atm / 1.013 x 105 Pa]
P = 0.029 atm
21.13 Calculate the change in internal energy of 3.0 moles of helium gas when its temperature is increased by 2.0 K.
For an ideal gas, we would have
U = (3/2) n R T
U = (3/2) n R T
U = (3/2) (3 moles) (8.31 J/mole-K) (2.0 K)
U = 74.8 J
While helium behaves very nearly as an ideal gas, we can do this calculation for the actual behavior of helium.
U = n Cv T
From Table 21.2 on page 647 of the text, we find that, for helium
cv = 12.5 J/mole-K
U = (3 moles)(12.5 J/mole-K)(2.0 K)
U = 75.0 J
21.**** One mole of an ideal monatomic gas is at an initial temperature of 300 K. The gas undergoes an isovolumetric process acquiring 500 J of heat. It then undergoes an isobaric process losing the same amount of heat. Determine (a) the new temperature of the gas and (b) the work done on the gas.
[[ Note: Part of the earlier solution posted on the web was wrong.]]
For the isovolumetric (constant volume) process, from A to B, no work is done,
WAB = 0
Q = U = 500 J
For the isobaric (constant pressure), process, from B to C, the work is
WBC = P V
We know the initial temperature, TA = 300 K. To find WBC we will need the temperatures TB and TC at states B and C so we can find P V for the expression above.
QAB = n CV T = 500 J
(n) ((3/2) R) (T) = 500 J
(1 mole) ((3/2) 8.31 J/mole-K) (T) = 500 J
( 12.5 J / K ) T = 500 J
T = (500/12.5) K
T = 40 K
TB = 300 K + 40 K
TB = 340 K
QBC = n CPT
- 500 J = (n) ((5/2) R) (T)
- 500 J = (1 mole) ((5/2) 8.31 J/mole-K) (T)
- 500 J = ( 20.8 J / K ) T
T = - (500/20.8) K
T = - 24.1 K
TC = 340 K - 24.1 K
TC = 315.9 K
Wtot = WAB + WBC
WAB = 0
WBC = P V
Qtot = Wtot + Utot
Qtot = 500 J + ( - 500 J) = 0
Wtot = - Utot
U = n CV T
U = (n) ((3/2) R) (Ttot)
T = 40.0 K - 24.1 K
T = 15.9 K
U = (n) ((3/2) R) (Ttot)
U = (1 mole)(12.5 J/mole-K)(15.9 K)
U = 199 J
Wtot = - Utot
W = - 199 J
[[ This final answer is the same as was posted earlier but part of the earlier solution was wrong. ]]
21.25 Two moles of an ideal gas ( = 1.40) expands slowly and adiabatically from a pressure of 5.0 atm and a volume of 12.0 liters to a final volume of 30.0 liters.
(a) What is the final pressure of the gas?
or
Pf = (5.0 atm) (12.0 L/30.0 L)1.4
Pf = (5.0 atm) (0.4)1.4
Pf = (5.0 atm) (0.2773)
Pf = 1.386 atm
(b) What are the initial and final temperatures?
Again (or "as usual"?!?!?), we will find the Ideal Gas Law very useful;
P V = n R T
T = P V / n R
Ti = Pi Vi / n R
Ti = (5.0 atm) (12.0 L)/[(2.0 moles) (8.31 J/mole-K)]
Be careful! While that equation is true, it is also clumsy or cumberson since we have such strange or mixed units. Since we have pressure in atmospheres and volume in liters, we will be far better served to use R in units of L-atm instead of joules! Again (and again and again and again!), watch the units! Units are not an "add-on"; they're important -- even vital!
Ti = (5.0 atm) (12.0 L)/[(2.0 moles) (0.0821 L-atm/mole-K)]
Ti = 365.4 K
Tf = Pf Vf / n R
Tf = (1.386 atm) (30.0 L)/[(2.0 moles) (0.0821 L-atm/mole-K)]
Tf = 253.2 K
21.***** Four liters of a diatomic ideal gas ( = 1.40) confined to a cylinder are put through a closed cycle. The gas is initially at 1.0 atm and at 300 K. First, its pressure is tripled under constant volume. It then expands adiabatically to its original pressure and finally is compressed isobarically to its original volume.
(z) What is its initial pressure?
P V = n R T
P = n R T / V
(a) Draw a PV diagram of this cycle.
Here is a rough sketch:
(b) Determine the volume at the end of the adiabatic expansion.
As we move along the adiabat,
or
or
(c) Find the temperature of the gas at the start of the adiabatic expansion
At state A,
PA VA = n R TA
n = PA VA / R TAA
At state B, the start of the adiabatic expansion,
PB VB = n R TBB
TB = PB VB / n R
TB = PB VB /[(PA VA / R TA) R]
TB = TA [ PB VBB /(PA VA) ]
TB = TA (PB/PA) (VB/VA)
[ or, we could have gotten this about as easily by noting
P V = n R T
P V / T = n R
PA VA/TA = PB VB/TB
TB = TA (PB/PA) (VB/VA)
Having done this both ways, this way seems noticeably easier. But, of course, the result is the same.]
TB = TA (PB/PA) (VB/VA)
Of course,
VB = VA
TB = 300 K (3 atm/1 atm) (1)
TB = 300 K (3)
TB = 900 K
(d) Find the temperature at the end of the cycle.
We know the pressure and volume for state C, the end of the adiabatic expansion,
PC = 1 atm
VC = 8.77 L
P V = n R T
P V / T = const
PA VA / TA = PC VC / TC
TC = TA (PC/PA) (VC/VA)
TC = 300 K (1 atm/1 atm) (8.77 L/4 L)
TC = 300 K (1) (2.19)
TC = 657.5 K
(e) What was the net work done for this cycle?
Look at the work for each piece and then sum those;
WAB = 0
WBC = - U
T = Tf - Ti = TC - TB
T = 657.5 K - 900 K
T = - 242.5
U = n CV T
U = n [ (5/2) R ] [ - 242.5 K ]
How many moles do we have? Go back to state A and find out;
P V = n R T
n = [ P V/T ] [1/R]
n = [ 1 atm - 4 L / 300 K] [1/R]
U = [ 1 atm - 4 L / 300 K] [1/R] [ (5/2) R ] [ - 242.5 K ]
U = [ 4 atm - L / 300 ] [ (5/2)] [ - 242.5 ]
U = - 8.08 L-atm
We have done this unit coversion before. We could just go back and get it. But it's worth doing from "scratch", too;
U = - 8.08 L-atm [0.001 m3/L] [1.013 x 105 (N/m2)/atm]
U = - 819 J
WBC = + 819 J
WCA = P V
WCA = (1 atm) (4 L - 8.77 L)
WCA = - 4.77 L-atm
WCA = - 4.77 L-atm [0.001 m3/L] [1.013 x 105 (N/m2)/atm]
WCA = - 483 J
Wtot = WAB + WBC + WCA
Wtot = 0 + 819 J - 483 J
Wtot = 336 J
While I have labeled this Wtot, as in the total work, we could also call this the net work and label it Wnet.
21.****** A 5.0-liter vessel contains 0.125 mole of an ideal gas at 1.50 atm. What is the average translational kinetic energy of a single molecule?
<KE> = (1/2) m <v2>
<KE> = (1/2) m <v2> = (1/2) k T
So, first, we must find the temperature T
P V = n R T
T = P V / n R
T = (1.5 atm) (5.0 L) / [(0.125 mole) (0.0821 L-atm/mole-K)]
T = 731 K
<KE> = (1/2) k T = (1/2) (1.38 x 10 - 23 J/K) (731 K)
<KE> = 5.04 x 10 - 21 J
21.33 Consider 2.0 moles of an ideal diatomic gas. Find the total heat capacity at constant volume and at constant pressure if
(a) the molecules rotate but do not vibrate and
CV = (5/2) R
Ctot = n CV = 5 R
(b) the molecules rotate and vibrate.
CV = (7/2) R
Ctot = n CV = 7 R
| 4,284 | 12,204 |
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 3.5 | 4 |
CC-MAIN-2019-13
|
latest
|
en
| 0.902194 |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.