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Volume of a object by being submerged

  1. Nov 21, 2004 #1
    I have a multiple choice question where a rock weighs 1400 N in air and 900 N apparent weight in fresh water with 998 kg/m^3 density (i think), The question is what is the volume of the rock. 2 of the choices are .14 m^3 and .051 m^3. I have tried
    p = m / v and 1400 - 500 and division by g's and densities, i have come too both of these conclusions many different ways, coming to either answer about the same amount of times. Does anyone know the true formula for this problem.
     
  2. jcsd
  3. Nov 21, 2004 #2
    Well : 900 = 1400 - 500 that is OK. So let's take 10 for g, then the mass of the water that is moved by the submerged rock (you know, Archimedes' law) is 50kg. Divide this by the density of water and you get the volume that was moved, so this needs to be equal to the volume of the rock : answer 0.051...
    marlon
     
  4. Nov 21, 2004 #3
    why 10 for g, isn't g 9.8 and where does the 50 kg's come from
     
  5. Nov 22, 2004 #4
    Well then take 9.8 for g...

    500 N is the upward force and is equal to the weight of the replaced water. So you have that 500 = mg and if we take g to be 10, then m = 50

    regards
    marlon
     
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