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Homework Help: Density Of the Object

  1. Nov 28, 2008 #1
    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. jcsd
  3. Nov 29, 2008 #2
    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
  4. Nov 29, 2008 #3
    i really dont understand....
     
  5. Nov 29, 2008 #4
    V_object=(1003)(831)=8.33x10^5

    but then that is the wrong answer.... how i do this exercise?..help plz
     
  6. Nov 29, 2008 #5

    Doc Al

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    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.
     
  7. Nov 29, 2008 #6
    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
     
  8. Nov 29, 2008 #7

    Doc Al

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    Hint: The part that sticks out is the part that doesn't displace water.
     
  9. Nov 29, 2008 #8
    ok, but how i know that?....
     
  10. Nov 29, 2008 #9

    Doc Al

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    You calculated two volumes. One represents the volume under water, the other represents the total volume.
     
  11. Nov 29, 2008 #10
    so that the answer to the problem is the diference between the total volumne and the volume under the water?....
     
  12. Nov 29, 2008 #11

    Doc Al

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    Yes.
     
  13. Nov 29, 2008 #12
    i put that answer and give me the wrong answer
    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...
     
  14. Nov 29, 2008 #13

    Doc Al

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    Oops... this calculation is incorrect.

    Density = Mass/Volume, so:

    Volume = Mass/Density.

    (Redo your values for V_object and V_water.)
     
  15. Nov 29, 2008 #14
    ok ty.... i get my corrects answer... but can u help me with the other exercise i post about density...
     
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