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?

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!

To find the object's density, we can use the equation:

p_object/p_water = v_water/v_object

Given that three-fourths (3/4) of the object's volume is beneath the surface of water, we can write:

v_water/v_object = 3/4

Next, we need to substitute the density of water, which is 1 g/cm^3, into the equation:

p_object/(1 g/cm^3) = 3/4

To isolate p_object, we can cross-multiply:

p_object = (3/4) * (1 g/cm^3)

Simplifying the expression:

p_object = 3/4 g/cm^3

Therefore, the object's density is 3/4 g/cm^3.

To find the density of the object, we need to use the concept of buoyancy. The buoyant force acting on the object is equal to the weight of the water displaced by the object. When the object floats, it displaces an amount of water equal to its own weight.

Let's assume the density of water is 1 g/cm^3.

Given that three-fourths (3/4) of the object's volume is beneath the surface of the water, it means one-fourth (1/4) of the volume is above the surface. This implies that the weight of the water displaced is equal to the weight of the part of the object above the water.

Let's denote the density of the object as p_o (in g/cm^3). Since the weight is directly proportional to the volume, we have:

p_object/p_water = v_water/v_object

Substituting the given values, we have:

p_o/1 = 1/4

Cross-multiplying, we get:

p_o = 1/4

Therefore, the density of the object is 1/4 g/cm^3.