The density of a solid cube is 678.2 kg/m^3. (Note that each edge is 4.8 cm in length and the mass is 75g.) When the cube is placed in a container of water, it is observed to float (it reaches equilibrium when part of the cube is below he surface or water.) What is the magnitude of the buoyancy force acting on the cube in N, and what percentage of the cube's volume is underwater while in equilibrium?

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aristocles aristocles · Answer 1 · 2019-11-25T02:24:38+01:00

Answer:

Part a)

F = 0.735 N

Part b)

[tex]percentage = 67.9%[/tex]

Explanation:

As we know that mass of the cube is given as

[tex]M = \rho V[/tex]

[tex]M = \rho a^3[/tex]

[tex]M = (678.2) (4.8 \times 10^{-2})^3[/tex]

[tex]M = 0.075 kg[/tex]

Now we know that cube is floating in the water

So net force due to weight of the cube must be counter balanced by buoyancy force on the liquid

so we have

[tex]F_b = mg[/tex]

[tex]F_b = 0.075 \times 9.81[/tex]

[tex]F_b = 0.735 N[/tex]

Part b)

Percentage of volume submerged into the liquid is given as

[tex]F_b = \rho V g[/tex]

[tex]0.735 = 1000 V \times 9.8[/tex]

[tex]V = 7.71 \times 10^{-5} m^3[/tex]

now percentage of submerged liquid is given as

[tex]percentage = \frac{7.71 \times 10^{-5}}{(4.8 \times 10^{-2})^3}\times 100[/tex]

[tex]percentage = 67.9%[/tex]