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 527 N to totally submerge the ball, calculate the diameter D of the ball. The density of water is rhow = 1.00 x 10^3 kg/m^3, the density of styrofoam is rhofoam = 95.0 kg/m^3, and the acceleration due to gravity is g = 9.81 m/s^2.

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

Answer 1

The diameter D of the styrofoam ball can be calculated using Archimedes' principle and the provided force needed to submerge the ball. The buoyant force is equal to the weight of the water displaced, and by rearranging the formula for the volume of a sphere, the diameter can be solved.

Step-by-step explanation:

To calculate the diameter D of the styrofoam ball that requires a force of 527 N to totally submerge it in water, we need to apply Archimedes' principle, which states that the buoyant force on an object submerged in a fluid is equal to the weight of the fluid displaced by the object. Here, the weight of the water displaced is equal to the force needed to submerge the ball (527 N).

The buoyant force (F_b) is given by the product of the density of water (rhow), the volume of water displaced (which is the volume of the ball), the acceleration due to gravity (g), and equals the force needed to submerge the ball:
F_b = rhow * V_ball * g = 527 N

The volume of a sphere (V_ball) is (4/3)\(\pi)r^3, and we can express the radius in terms of the diameter (D = 2r). Substituting and solving for the diameter we get:
(4/3)\(\pi)(D/2)^3 * rhow *g = 527 N

Plugging in the known values for rhow (1.00 x 10^3 kg/m^3) and g (9.81 m/s^2) and solving for the diameter D, we can find the correct value which matches one of the provided options.