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December 20, 2014

December 20, 2014

Posted by **Luc** on Thursday, March 1, 2012 at 1:53am.

- physics -
**drwls**, Thursday, March 1, 2012 at 11:57amWhat do you mean by

" 800 k/gm^3 than it does in air " ?

The external fluid does not affect the density.

Did you mean to write:

"What is the volume of a metal which weighs 28g less in kerosene does in air, if the density of kerosene is 800 kg/m^3?

That question can be answered.

The metal will certainly sink to the bottom, and there will be an upward buoyancy force of (0.028 kg)*g = 0.274 N. This equals (kerosene density)*g*V, so

V = 0.274/(800*9.8) = 3.5*10^-5 m^3 = 35 cm^3

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