A plastic cube having a side of length 10cm floats on fresh water. If the volume of the cube that is submerged into water is 40 percent of its total volume ,what is the density of the cube?

Answer 1

Let's start by finding the volume of the cube. Since the cube has a side length of 10cm, its volume is:

V = s^3 = 10^3 = 1000 cm^3

Next, we can find the volume of the cube that is submerged in water. We know that this volume is 40% of the total volume, so we can write:

V_submerged = 0.4V = 0.4(1000) = 400 cm^3

To find the density of the cube, we need to know its mass. We can use the fact that the cube is floating to find its mass. When an object is floating, the buoyant force (which is equal to the weight of the water displaced by the object) is equal to the weight of the object. Since the cube is not sinking or floating to the top, we know that its weight (which is equal to its mass times gravity) is equal to the buoyant force. In other words:

weight = buoyant force
m*g = ρ_water*V_submerged*g

where m is the mass of the cube, g is the acceleration due to gravity, ρ_water is the density of fresh water, and V_submerged is the volume of water displaced by the cube (which we found earlier).

Simplifying this equation, we can solve for the mass of the cube:

m = ρ_water*V_submerged

Plugging in the values we know, we get:

m = 1000 kg/m^3 * 0.4 * 10^-3 m^3 = 0.4 kg

Finally, we can find the density of the cube by dividing its mass by its volume:

ρ = m/V = 0.4 kg / 0.001 m^3 = 400 kg/m^3

Therefore, the density of the cube is 400 kg/m^3.