Asked by anonymous
An ice cube floats with 1/9 of its volume below the surface of water, calculate the density of ice.
Answers
Answered by
Henry
Vb = Di/Dw * Vi = Vi/9.
Di/1000 * Vi = Vi/9,
Divide both sides by Vi:
Di/1000 = 1/9, Di = 1000/9 = 111.11 kg/m^3 = Density of the ice cube.
Di/1000 * Vi = Vi/9,
Divide both sides by Vi:
Di/1000 = 1/9, Di = 1000/9 = 111.11 kg/m^3 = Density of the ice cube.
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