1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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

    User Avatar

    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.
     
  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

    User Avatar

    Staff: Mentor

    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

    User Avatar

    Staff: Mentor

    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

    User Avatar

    Staff: Mentor

    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

    User Avatar

    Staff: Mentor

    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...
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Density Of the Object
  1. Density of an Object (Replies: 13)

Loading...