# Problem №3 (10 points)

An electric field is created in space, the intensity of which depends on the coordinate as shown in the figure.

![](https://cdn.mathpix.com/cropped/2024_05_06_9e5a3306654225a310b5g-07.jpg?height=477&width=964&top_left_y=107&top_left_x=89)

This field moved a particle with a mass of 1 mg and a charge of 1 mC from a point with a coordinate of 5 cm to a point with a coordinate of 15 cm. Determine the speed of the charge at the final point, if it was at rest at the initial point.
# Solution:

The intensity at the point with a coordinate of 10 cm is $200 \mathrm{~B} / \mathrm{M}$

The potential difference between the final and initial points is equal to the area under the graph:

$\Delta \varphi=\left(\frac{100+200}{2} \cdot 0.05\right)+(200 \cdot 0.05)=17.5 B$

## b) points

The work done by the electric field is equal to the kinetic energy acquired by the charge:

$q \Delta \varphi=\frac{m v^{2}}{2}$

## From which the speed of the charge

$v=\sqrt{\frac{2 q \Delta \varphi}{m}}=\sqrt{\frac{2 \cdot 10^{-3} \cdot 17.5}{10^{-6}}} \approx 187 \mathrm{M} / \mathrm{c}$

## Problem №4 (15 points)

An object was placed in front of a biconvex lens. As a result, an image was obtained that had the same size as the object. After replacing the lens with a biconcave one, the radii of curvature of the surfaces of which were equal to the radii of the original lens, the image of the object shifted by $240 \mathrm{~cm}$. Determine the optical power of the original lens. The materials of the lenses are the same. The object was not moved.

## Solution:

The original lens is a converging lens, and the object is located at its double focal distance.

(3 points)

The second lens is a diverging lens, and from the formula $D=(n-1)\left(\frac{1}{R_{1}}+\frac{1}{R_{2}}\right)$, it follows that it has the same optical power in magnitude.

(3 points)

The thin lens formula for the second case is:

$-\frac{1}{F}=\frac{1}{2 F}-\frac{1}{f}$

Additionally, from the condition, it follows:

$2 F+f=2.4$

(3 points)

Solving this system, we get: $F=0.9$ m

## The optical power of the first lens:

$D=\frac{1}{F}=\frac{10}{9}$ D $n m p$

# Multidisciplinary Engineering Olympiad "Star"  NATURAL SCIENCES  2016 

March 29, 2016

PROTOCOL № 1 of the meeting of the jury

PRESENT: Keller A.V., Zalyapin V.I., Zamishlyaeva A.A., Vorontsov A.G., Kuts D.A., Gusev A.V.

HEARD: about the distribution of points among the winners and prizewinners of the olympiad

## DECIDED:

## 11th grade

- to consider the winners of the olympiad and award diplomas of 1st degree to participants who scored 100 - 90 points;
- to consider the prizewinners of the olympiad and award diplomas of 2nd degree to participants who scored 89 - 70 points;
- to consider the prizewinners of the olympiad and award diplomas of 3rd degree to participants who scored 69 - 45 points.

## 10th grade

- to consider the winners of the olympiad and award diplomas of 1st degree to participants who scored 100 - 95 points;
- to consider the prizewinners of the olympiad and award diplomas of 2nd degree to participants who scored 94 - 70 points;
- to consider the prizewinners of the olympiad and award diplomas of 3rd degree to participants who scored 69 - 40 points.

## 9th grade

- to consider the winners of the olympiad and award diplomas of 1st degree to participants who scored 100 - 90 points;
- to consider the prizewinners of the olympiad and award diplomas of 2nd degree to participants who scored 89 - 70 points;
- to consider the prizewinners of the olympiad and award diplomas of 3rd degree to participants who scored 69 - 40 points.

## 8th grade

- to consider the winners of the olympiad and award diplomas of 1st degree to participants who scored 100 - 75 points;
- to consider the prizewinners of the olympiad and award diplomas of 2nd degree to participants who scored 74 - 65 points;
- to consider the prizewinners of the olympiad and award diplomas of 3rd degree to participants who scored 64 - 40 points.

## 7th grade

- to consider the winners of the olympiad and award diplomas of 1st degree to participants who scored 100 - 90 points;
- to consider the prizewinners of the olympiad and award diplomas of 2nd degree to participants who scored 89 - 70 points;
- to consider the prizewinners of the olympiad and award diplomas of 3rd degree to participants who scored 69 - 40 points.

## 6th grade

- to consider the winners of the olympiad and award diplomas of 1st degree to participants who scored 100 - 90 points;
- to consider the prizewinners of the olympiad and award diplomas of 2nd degree to participants who scored 89 - 80 points;
- to consider the prizewinners of the olympiad and award diplomas of 3rd degree to participants who scored 79 - 40 points.

## Chairmen of the jury:

![](https://cdn.mathpix.com/cropped/2024_05_06_9e5a3306654225a310b5g-10.jpg?height=257&width=582&top_left_y=1096&top_left_x=180)

Keller A.V.

Vorontsov A.G.