text: # Density Of the Object
  
  1. Nov 28, 2008
  
  ### Naldo6
  
  An object with density of 831 kg/m^3 and *** of 1003 kg is thrown into te ocean. Find the volume that sticks out of the water. ( use density of water= 1024 kg/m^3)
  
  I use F_b= m'g
  m_water*g=m_object*g
  p_water*V_water*g=p_object*V_object*g
  v_object=(p_water*V_water)/p_object
  
  but what is the volume of the water?....
  
  are my formulas ok?....
  
  2. Nov 29, 2008
  
  ### horatio89
  
  Hey Naldo6,
  
  By the Achimedes' Principle, the volume of water displaced = volume of the block submerged. However, since the object is floating, the volume of water displaced is only a fraction of the volume of the whole object. Your equations are correct, however, your problem may have arisen from your definitions of the variables.
  
  V_object=(p_water*V_water)/p_object
  
  Consider this equation - the V_object here is: the volume of the object itself, or the volume of the object submerged? What should be the subject of the equation? (Hint: look at the statement at the beginning of the post)
  
  Regards,
  Horatio
  
  Last edited: Nov 29, 2008
  3. Nov 29, 2008
  
  ### Naldo6
  
  i really dont understand....
  
  4. Nov 29, 2008
  
  ### Naldo6
  
  V_object=(1003)(831)=8.33x10^5
  
  but then that is the wrong answer.... how i do this exercise?..help plz
  
  5. Nov 29, 2008
  
  ### Staff: Mentor
  
  So far, so good. Now find V_object and V_water and compare.
  
  That's not "the answer", it's a step toward the answer. Good. Now figure out V_water (the volume of water displaced) using your equations. Then calculate out how much of the object's volume sticks out of the water.
  
  6. Nov 29, 2008
  
  ### Naldo6
  
  V_water=(V_object*p_object)/p_water
  so V_water = (8.33x10^5)(831)/(1024)= 6.76x10^5 m^3
  
  ok but how i can calcuta how much of the objects's volume sticks out of the water
  
  7. Nov 29, 2008
  
  ### Staff: Mentor
  
  Hint: The part that sticks out is the part that doesn't displace water.
  
  8. Nov 29, 2008
  
  ### Naldo6
  
  ok, but how i know that?....
  
  9. Nov 29, 2008
  
  ### Staff: Mentor
  
  You calculated two volumes. One represents the volume under water, the other represents the total volume.
  
  10. Nov 29, 2008
  
  ### Naldo6
  
  so that the answer to the problem is the diference between the total volumne and the volume under the water?....
  
  11. Nov 29, 2008
  
  ### Staff: Mentor
  
  Yes.
  
  12. Nov 29, 2008
  
  ### Naldo6
  
  i did:
  
  8.33x10^5 6.76x10^5= 1.57x10^5 m^3 and that is not the crrect answer...plz can u cheq me in my calculus aboves...
  
  13. Nov 29, 2008
  
  ### Staff: Mentor
  
  Oops... this calculation is incorrect.
  
  Density = Mass/Volume, so:
  
  Volume = Mass/Density.
  
  (Redo your values for V_object and V_water.)
  
  14. Nov 29, 2008
  
  ### Naldo6
  
  ok ty.... i get my corrects answer... but can u help me with the other exercise i post about density...
explanation: The document is a forum thread discussing a physics problem related to the density of an object and how much of it sticks out of the water when it is placed in the ocean. The user, Naldo6, asks a question about the problem, and other users, including horatio89 and a staff member, provide guidance and hints to help Naldo6 solve the problem. The thread contains a question and an answer, as Naldo6 eventually arrives at the correct solution with the help of the other users.
distilabel_metadata.content: Evaluate the user given text extract and determine if it contains a math question and optionally and answer to this question.
  
  Output rules:
  
  1. Briefly describe the document and mention if it contains a question or a question and answer.
  2. Finalize the output with: Label: no-question,question,question-and-answer,multiple-question,multiple-question-and-answer
distilabel_metadata.role: system
distilabel_metadata.content: # Density Of the Object
  
  1. Nov 28, 2008
  
  ### Naldo6
  
  An object with density of 831 kg/m^3 and *** of 1003 kg is thrown into te ocean. Find the volume that sticks out of the water. ( use density of water= 1024 kg/m^3)
  
  I use F_b= m'g
  m_water*g=m_object*g
  p_water*V_water*g=p_object*V_object*g
  v_object=(p_water*V_water)/p_object
  
  but what is the volume of the water?....
  
  are my formulas ok?....
  
  2. Nov 29, 2008
  
  ### horatio89
  
  Hey Naldo6,
  
  By the Achimedes' Principle, the volume of water displaced = volume of the block submerged. However, since the object is floating, the volume of water displaced is only a fraction of the volume of the whole object. Your equations are correct, however, your problem may have arisen from your definitions of the variables.
  
  V_object=(p_water*V_water)/p_object
  
  Consider this equation - the V_object here is: the volume of the object itself, or the volume of the object submerged? What should be the subject of the equation? (Hint: look at the statement at the beginning of the post)
  
  Regards,
  Horatio
  
  Last edited: Nov 29, 2008
  3. Nov 29, 2008
  
  ### Naldo6
  
  i really dont understand....
  
  4. Nov 29, 2008
  
  ### Naldo6
  
  V_object=(1003)(831)=8.33x10^5
  
  but then that is the wrong answer.... how i do this exercise?..help plz
  
  5. Nov 29, 2008
  
  ### Staff: Mentor
  
  So far, so good. Now find V_object and V_water and compare.
  
  That's not "the answer", it's a step toward the answer. Good. Now figure out V_water (the volume of water displaced) using your equations. Then calculate out how much of the object's volume sticks out of the water.
  
  6. Nov 29, 2008
  
  ### Naldo6
  
  V_water=(V_object*p_object)/p_water
  so V_water = (8.33x10^5)(831)/(1024)= 6.76x10^5 m^3
  
  ok but how i can calcuta how much of the objects's volume sticks out of the water
  
  7. Nov 29, 2008
  
  ### Staff: Mentor
  
  Hint: The part that sticks out is the part that doesn't displace water.
  
  8. Nov 29, 2008
  
  ### Naldo6
  
  ok, but how i know that?....
  
  9. Nov 29, 2008
  
  ### Staff: Mentor
  
  You calculated two volumes. One represents the volume under water, the other represents the total volume.
  
  10. Nov 29, 2008
  
  ### Naldo6
  
  so that the answer to the problem is the diference between the total volumne and the volume under the water?....
  
  11. Nov 29, 2008
  
  ### Staff: Mentor
  
  Yes.
  
  12. Nov 29, 2008
  
  ### Naldo6
  
  i did:
  
  8.33x10^5 6.76x10^5= 1.57x10^5 m^3 and that is not the crrect answer...plz can u cheq me in my calculus aboves...
  
  13. Nov 29, 2008
  
  ### Staff: Mentor
  
  Oops... this calculation is incorrect.
  
  Density = Mass/Volume, so:
  
  Volume = Mass/Density.
  
  (Redo your values for V_object and V_water.)
  
  14. Nov 29, 2008
  
  ### Naldo6
  
  ok ty.... i get my corrects answer... but can u help me with the other exercise i post about density...
distilabel_metadata.role: user
distilabel_metadata.raw_output_q_a_detector_0: The document is a forum thread discussing a physics problem related to the density of an object and how much of it sticks out of the water when it is placed in the ocean. The user, Naldo6, asks a question about the problem, and other users, including horatio89 and a staff member, provide guidance and hints to help Naldo6 solve the problem. The thread contains a question and an answer, as Naldo6 eventually arrives at the correct solution with the help of the other users.
  
  Label: question-and-answer