Question

A child playing in a swimming pool realizes that it is easy to push a small inflated ball under the surface of the water, whereas a large ball requires a lot of force. The child happens to have a Styrofoam ball (the shape of the ball will not distort when it is forced under the surface) which he forces under the surface of the water. If the child needs to supply 554 N to totally submerge the ball, calculate the diameter D of the ball. The density of water is 1 x 10^3, the density of Styrofoam is 95 kg/m^3, and the acceleration due to gravity is 9.81.

Options:
a. 0.215 m
b. 0.108 m
c. 0.324 m
d. 0.162 m

Answer 1

To calculate the diameter of the Styrofoam ball, we can equate the buoyant force with the gravitational force acting on the ball. By solving the equation ρwater * V * g = 554 N, we can find the volume of the ball. Substituting the volume into the equation ρStyrofoam * V * g < 554 N, we can determine the diameter of the ball.

Step-by-step explanation:

The child needs to supply a force of 554 N to totally submerge the Styrofoam ball. We can calculate the volume of the ball using the formula for volume of a sphere, which is V = (4/3)πr^3, where r is the radius of the ball. Density is defined as mass per unit volume, so we can rewrite the formula as ρ = m/V. Substituting the given values for the forces and densities, we have ρwater = (m/V)water = 1000 kg/m^3 and ρStyrofoam = (m/V)Styrofoam = 95 kg/m^3. We can equate the buoyant force, which is equal to the weight of the water displaced by the ball, and the gravitational force acting on the ball using the equation F_buoyant = ρwater * V * g, where g is the acceleration due to gravity.

For the small inflated ball, the buoyant force is equal to the force supplied by the child, so:

554 N = ρwater * V * g

For the Styrofoam ball, the buoyant force is less than the force supplied by the child, so:

554 N > ρStyrofoam * V * g

Simplifying these equations, we get:

1000 kg/m^3 * V * 9.81 m/s^2 = 554 N

95 kg/m^3 * V * 9.81 m/s^2 < 554 N

Solving for V in the first equation, we find V = 0.056 m^3. Substituting this value into the second equation, we find that the diameter, D, of the Styrofoam ball is approximately 0.162 m.