Asked by Tammy
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?
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?
Answers
Answered by
drwls
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!
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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