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).
Well, it sounds like you've got a real waterworks situation going on there! Let's calculate the work required to pump out that water.
First, we need to find the volume of water in the tank. Since we're dealing with a spherical shape, we can use the formula for the volume of a sphere: V = (4/3)πr^3.
V = (4/3) * 3.14 * (9m)^3
V ≈ 3053.63 m^3
Next, we can find the weight of the water in the tank using the formula: weight = density * volume * gravity.
Weight = 1000 kg/m^3 * 3053.63 m^3 * 9.8 m/s^2
Weight ≈ 29,974,236 N
Finally, we can calculate the work required to pump the water out, which is equal to the weight of the water multiplied by the height it needs to be raised: work = weight * height.
Work = 29,974,236 N * 3 m
Work ≈ 89,922,708 Joules
So, approximately 89,922,708 Joules of work would be required to pump all the water out of the tank. Just make sure you don't make any waves while pumping, or else you might end up in a splashy situation!
To find the work required to pump the water out of the spout, we need to calculate the potential energy of the water in the tank.
The potential energy of an object is given by the formula: PE = mgh, where PE is the potential energy, m is the mass of the object, g is the acceleration due to gravity, and h is the height of the object.
In this case, we need to calculate the potential energy of the water in the tank. The mass of the water can be found using its density and volume. The volume of the water in the tank can be calculated as the volume of a spherical cap, which is given by the formula: V = πh^2(3r - h)/3.
Substituting the values, we can calculate the volume of water:
V = π(3^2)(3(9) - 3)/3 = 243π m^3
Next, we can calculate the mass of the water:
m = density × volume = 1000 kg/m^3 × 243π m^3 = 243000π kg
Finally, we can calculate the potential energy:
PE = mgh = 243000π kg × 9.8 m/s^2 × 3 m = 7109160π J
Therefore, the work required to pump the water out of the spout is approximately 7109160π Joules.
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 :)
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.