How high is the top of the spout from the center of gravity of the tank ???
radius + 3 = 9 + 3 = 12 meters
So you have to lift the weight of the water 12 meters.
mass of water in tank = m = (4/3) pi r^3 * 1000 Kg
weight = w = m g = 9.8 * m
work = weight * distance lifted = (4/3) pi 9^3 * 1000 Kg * 9.8 * 12 Joules
Note, You will need more because the water has velocity when it exits but we do not have data for that
(1/2) m v^2 :)
A tank is full of water. Find the work required to pump the water out of the spout. (Use 9.8 m/s^2 for g. Use 1000 kg/m^3 as the density of water. Assume r = 9 m and h = 3 m.)
The tank is a spherical shape with r as the radius(9m). On top of the tank is a spigot with h as the height (3m).
2 answers
For now, assuming that the weight of water is 1 kg/m^2 (so we just have to worry about figuring the volume * distance)
moving the center of mass to the height of the top of the sphere (forget the spout -- assume the water just runs out the top) requires
4/3 π * 9^3 * 9 = 8748π J
Doing the calculus, and figuring that the cross-section area of a slice of water at a distance of y units above the bottom of the sphere is π√(81-(9-y)^2) (this is correct, because even though 9-y goes from 9..-9, it is squared)
∫[0..18] πr^2h dy = ∫[0..18] π(81-(9-y)^2)(18-y) dy = 8748π J
Now just multiply that by the correct density (N/m^3) and increase the height by the length of the spout, and you should get Damon's answer.
moving the center of mass to the height of the top of the sphere (forget the spout -- assume the water just runs out the top) requires
4/3 π * 9^3 * 9 = 8748π J
Doing the calculus, and figuring that the cross-section area of a slice of water at a distance of y units above the bottom of the sphere is π√(81-(9-y)^2) (this is correct, because even though 9-y goes from 9..-9, it is squared)
∫[0..18] πr^2h dy = ∫[0..18] π(81-(9-y)^2)(18-y) dy = 8748π J
Now just multiply that by the correct density (N/m^3) and increase the height by the length of the spout, and you should get Damon's answer.