A plastic cube having a side length of 10cm floats in fresh water. If the volume of thr cub3 that's submerged in water is 40% of it's total volume, what is the density of the cube?(density of water=1000kg/m³).

Answer 1

We can start by finding the volume of the submerged portion of the cube. Since we know that it is 40% of the total volume, we can set up the following equation:

0.4 x total volume = submerged volume

Let's call the total volume "V". The volume of a cube is calculated by multiplying the length, width, and height. Since all sides have the same length in a cube, we can simplify this to:

V = (side length)^3

Plugging in the given value of 10cm for the side length, we get:

V = 10cm x 10cm x 10cm = 1000cm³

Now we can use this to find the volume of the submerged portion:

0.4 x 1000cm³ = 400cm³

Since the cube is floating in water, we know that its weight is equal to the weight of the water it displaces. This is known as Archimedes' principle. The weight of the water it displaces can be calculated using the density of water and the volume of the submerged portion:

Weight of water displaced = density x volume

Weight of water displaced = 1000kg/m³ x 0.4 x 1000cm³/1000000 = 0.4kg

Since the cube is not sinking or rising, its weight must be equal to the weight of the water it displaces. Let's call the density of the plastic cube "D":

Weight of cube = D x V x density of water

Weight of cube = D x 1000cm³ x 1000kg/m³ = D x 1kg

Setting the weight of the cube equal to the weight of the water displaced, we get:

D x 1kg = 0.4kg

Solving for D, we get:

D = 0.4kg/1kg = 0.4

Therefore, the density of the cube is 0.4 times the density of water, or:

Density of cube = 0.4 x 1000kg/m³ = 400kg/m³