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A 10.0 kg ball weighs 98.0 N in air and weighs 65.0 N when submerged in water. The volume of the ball is

Question

A 10.0 kg ball weighs 98.0 N in air and weighs 65.0 N when submerged in water. The volume of the ball is

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Solution

The weight of an object is the force of gravity acting on it. When the ball is in air, its weight is 98.0 N. When it is submerged in water, it weighs less, 65.0 N, because of the buoyant force exerted by the water.

The buoyant force is equal to the weight of the fluid displaced by the object. In this case, the fluid is water. The weight of the water displaced by the ball is 98.0 N - 65.0 N = 33.0 N.

The weight of water is its mass times the acceleration due to gravity. The mass of water is its volume times its density. Therefore, the weight of water is its volume times its density times the acceleration due to gravity.

We can set up the equation:

33.0 N = Volume * Density of water * 9.8 m/s^2

We know that the density of water is approximately 1000 kg/m^3. We can substitute this value into the equation:

33.0 N = Volume * 1000 kg/m^3 * 9.8 m/s^2

Solving for Volume gives:

Volume = 33.0 N / (1000 kg/m^3 * 9.8 m/s^2) = 0.00337 m^3

So, the volume of the ball is 0.00337 cubic meters.

This problem has been solved

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