physics

A solid object floats with three-fourths of its volume beneath the surface of eater. What is the object's density? Give answers in g/cm^3.

p_o=?
p_object/p_water = v_water/v_object
p_o/(1 g/cm^3) = ?
for the volume side would it be 3/4
so that p_o/(1 g/cm^3)=3/4?

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asked by Tammy
  1. I assume you mean water, not eater. I also assume it is fresh water with a density of 1.000 g/cm^3

    If the object density is "rho1", the water desnity is "rho2", and the volume is V,
    (3/4) V *rho2 = V*rho1
    rho1 = 3/4

    Your answer is correct. Good job!

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    posted by drwls

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