I need help figuring out this problem. I know ice's density is .9167 but I'm lost beyond that..
Problem:
Icebergs float in the ocean with much of their huge volumes hidden below the surface. What fraction is visible above the water?
What portion of an ice cube floats above the surface of a glass of tap water?
Problem:
Icebergs float in the ocean with much of their huge volumes hidden below the surface. What fraction is visible above the water?
What portion of an ice cube floats above the surface of a glass of tap water?
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Sea water density is about Rho = 1040 kg/m^3 over all. And that means for the weight of the berg to equal the weight of the displaced water we must have Rho g v = rho g V; so that the volume of the displaced sea water is v = V (rho/Rho) = V (916.7/1040) = 0.881442308 V = .88 V. rho = 916.7 kg/m^3 is the berg's density. So about 88% of the berg will be below displacing the water. Thus 12% of its volume will be visible. ANS.