Asked by Maria
A floating cube is 10 cm on a side and has a density of 800 kg/m^3 . IT is floating in fresh water that has a density of 1000 kg/m^3. what percent of the cube is above the surface of the water?
Again, my professor says it is 20%, but how??
Again, my professor says it is 20%, but how??
Answers
Answered by
Writeacher
Maria/Andy/Ana/jessica
Please keep the same name in all your posts.
Please keep the same name in all your posts.
Answered by
bobpursley
the mass of the cube is density*volume=.8g/cm^3 * 1000cm^3=.8kg
the volume of the water displacing this (mass of water=8kg also)
volume=8kg/1000kg/m^3= (.8/1000)10^6 cm^3
= 800cm^3
So the volume above water is 200 cm^3 or 20 percent
the volume of the water displacing this (mass of water=8kg also)
volume=8kg/1000kg/m^3= (.8/1000)10^6 cm^3
= 800cm^3
So the volume above water is 200 cm^3 or 20 percent
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