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Density of an iceberg

  1. Oct 31, 2008 #1
    1. The problem statement, all variables and given/known data

    a) If the part of an iceberg above sea level is one ninth of the whole, what is the density of ice?

    b)How much of the iceberg would show if it moved into a fresh water region?

    2. Relevant equations
    Density of sea water=1025kgm^-3
    Density of fresh water 1000kgm^-3

    Weight displaced=upthrust
    Force due to gravity = mg


    3. The attempt at a solution

    For part a) I equated the upthrust and the force due to gravity, by applying Newtons first law. However I am confused as to whether I should also take into account the atmospheric pressure pressing the part of the iceberg above sea level down. In all the stuff I've read no one seems to take it into account when finding the density of an iceberg. Is there a reason for this?

    In part b) I think you again apply Newtons first law and use the ice density calculated in part a),

    [tex]\rho_{ice}*g*v_{ice}=\rho_{water}*g*v_{water}[/tex]

    [tex]\frac{\rho_{ice}}{\rho_{water}}=\frac{v_{water}}{v_{ice}}[/tex]

    Which gives the proportion of ice under the water.

    Please could someone help me?
     
  2. jcsd
  3. Oct 31, 2008 #2

    Vanadium 50

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    If you pick up a piece of paper, do you feel the 1400 pounds of force that air pressure exerts on it?
     
  4. Oct 31, 2008 #3

    Office_Shredder

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    I do :(
     
  5. Oct 31, 2008 #4
    I geuss not, but isn't that because as soon as you pick it up air rushes underneath the paper, very quickly, and the air underneath exerts an equal and oppisite force to the air above, thus we don't feel the pressure force. In the sea this situation is clearly impossible.
     
    Last edited: Oct 31, 2008
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