a water polo ball is a hollow sphere of outer radius 15.0 cm and mass 750. grams

Question

a water polo ball is a hollow sphere of outer radius 15.0 cm and mass 750. grams. the ball starts by gloating on the surface of a calm swimming pool. ( assume that the density of the pool's water is 1.000 g/cm^3.)

a) while floating, what fraction of the ball's volume is underwater?

b) suppose that a polo player starts pushing downward on the floating ball, gradually increasing his force. what is the magnitude of the athlete's downward force when the ball has exactly 75% of its volume submerged below the surface?

suppose the total (absolute) air pressure inside the ball is 165 kPa. A driver then takes the ball and begins to swim downward with it in an infinitely deep swimming pool. (Assume that the surface of the swiming pool is located at sea level, so that the atmospheric pressure on its surface is 1.00 atm.)

c) at what depth will the absolute pressure on the outside of the ball exceed 165 kPa. causing the ball to start to collapse?

Explanation / Answer

Weight of ball = Weight of water displaced
Then mass of ball = mass of water displaced (g cancels)
.750 = 1000 * Vw        where density of water is 1000 kg/m^3
Vw = 7.5 * 10E-4 m^3     volume of water displaced
Vb = 4/3 * pi * .15^3 = .0141 m^3
.Fraction under water = 7.5 * 10E-4 / .0141 = .0531
(b) 75% of Vb = .0106 m^3
Additional water displaced = .0106 - 7.5 * 10E-4 = .00982
Fb = .00982 * 1000 * 9.8 = 96.3 N   buoyant force
(c)   Atmospheric presure = 101 kPa
165 - 101 = 64 kPa     pressure required to collaps ball
P = p g d     where p is density of water and d the depth
d = 6.4 *10E4 / (1000 * 9.8) = 6.53 m

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