A circular swimming pool has a diameter of 18 m. The circular side of the pool is 3 m high, and the depth of the water is 1.5 m. (The acceleration due to gravity is 9.8 m/s^2 and the density of water is 1000 kg/m 3 kg/m^3.)How much work (in Joules) is required to: i. pump all of the water over the side? ii. pump all of the water out of an outlet 2m over the side?

A) 11223450 J

B) 18705750 J

Explanation:

Work done = force x distance moved.

Force of fluid at bottom of the pool = pghA

Where

P = density of water = 1000 kg/m3

g = acceleration due to gravity = 9.8 m/s2

h = depth of water = 1.5 m

A = area of pool of diameter d = 18 m

A = (¶d^2)/4 = (3.142 x 18^2)/4 = 254.5 m2

Total wall height of pool = 3 m

Force on pool bottom due to water is

F = 1000 x 9.8 x 1.5 x 254.5 = 3741150 N

A) work (in Joules) required to pump all of the water over the side

Height of wall H = 3 m

Work = F x H = 3741150 x 3 = 11223450 J

B) work (in Joules) required to pump all of the water out of an outlet 2m over the side?

Total height will be 2 + 3 = 5 m

Work = F x H = 3741150 x 5 =

18705750 J

this could be newton#'s law of cooling. the cooling curve is exponential