A circular swimming pool?

Question

A circular swimming pool has a diameter of 18 m, the sides are 4 m high, and the depth of the water is 2.5 m. (The acceleration due to gravity is 9.8m/s^2 and the density of water is 1000kg/m^3 .)
How much work (in Joules) is required to:
(a) pump all of the water over the side?
(b) pump all of the water out of an outlet 2 m over the side?

Answer 1

Swimming pool has diameter 18 m, side = 4 m and depth = 2.5 m.

Top surface of water is 1.5 m (4 - 2.5) from the sides
Bottom part of water is 4 m from the sides

a)
Work to pumb water over the side
= ∫ (density water)(gravity)(volume) dx
= ∫ (1000)(9.8)(pi)(18/2)^2 x dx (from x=1.5 to x = 4)
= ∫ 2492532x dx (from x = 1.5 to x = 4)
= 1246266 [(4)^2 - (1.5)^2]
= 1.7136 x 10^7 Joule

b)

Top surface of water is 3.5 m (6 - 2.5) from outlet over the sides
Bottom part of water is 6 m from outlet over the sides

Work to pumb water over of an outlet 2m over the side
= ∫ (density water)(gravity)(volume) dx
= ∫ (1000)(9.8)(pi)(18/2)^2 x dx (from x=3.5 to x = 6)
= ∫ 2492532x dx (from x = 3.5 to x = 6)
= 1246266 [(6)^2 - (3.5)^2]
= 2.96 x 10^7 Joule

Answer 2

The entire mass of the swimming pool may be assumed to be centered at the center of mass(also the center of gravity in this case).
Thus Volume of water in the swimming pool =81*pi*2.5
Mass of water = density * volume
= 636172 kg
The center of mass is originally at a height of 1.25m
Now change in height of center of mass = 4 - 1.25 = 2.75m
Thus change in potential energy = work done = mgh = 2.75*626172*10 = 16875335 J

For taking the water through an outlet 2m above the side,
Change in height = 4.75. Follow similar procedure for the answer

Answer 3

this could be a difficulty that demands calculus 2 information. you are going to have a fundamental from 0 to a million.5 (water top to the top of the pool). you may wish the quantity = ?(r^2)(?x) = 25??x mass will equivalent the density * quantity so one thousand * 25??x =25000?(3-x)?x F=mg =(9.8)25000?(3-x)?x =245000?(3-x)?x (This places it in newtons) in view that W=Fx, Your paintings could be the stress 245000?(3-x)?x The fundamental would be from 0 to a million.5 with F(x)=245000?(3-x)?x So paintings = fundamental {0 to a million.5} (9800)(pi*5^2)(3 - x)dx i'm hoping that's sturdy for you. this is been awhile in view that i've got carried out calc, yet i be attentive to this is on the properly suited track.