Respuesta :
Answer:
W = 108000 π
Explanation:
Given data
diameter = 24 ft = so r = 24/2 = 12 ft
high = 5 ft
depth = 4 ft
Density of water= 62.5 lb
to find out
How much work is required
solution
we will apply here weight equation that i s
dW = density × area
artea = πr²
so that Work done wil be [tex]\int_{0}^{4}[/tex] F × d
we know w = 62.5
so
W = 62.5 πr² [tex]\int_{0}^{4}[/tex] (x+1) dx
W = 62.5 π(12)² [tex](x^{2} /2 + x )^{4}_0[/tex]
W = 62.5 π(144) (12)
W = 108000 π
hence W = 108000 π
Work done by circular swimming pool is amount of energy used to pump the water. Work done is required to pump all the water out over the side is 108000 j.
What is work done to pump water?
The amount of energy which is required to pump the water at a depth to the ground is known as the work done of pump the water.
Given information-
The diameter of the circular swimming pool is 24 ft.
The height of the circular swimming pool is 5 ft.
The depth of the circular swimming pool is 4 ft.
The density ([tex]\rho[/tex]) of water is 62.5 lb.
The equation for the work done can be given as,
[tex]dW=\rho\times A \;dx \\dW=\rho\times \pi r^2 dx[/tex]
Let the distance is x. For this the integration limits should be 1 to 5 for 4 ft depth. Thus put the value of in the above equation with integrating it as,
[tex]W=62.5\pi\times (12)^2\int\limits^5_1 {x} \, dx \\W=62.5\times \pi\times 144 [\dfrac{x^2}{2}]^5_1\\W=62.5\times \pi\times 144 [\dfrac{5^2-1^2}{2}]\\W=62.5\times \pi\times 144 [\dfrac{24}{2}]\\W=108000\pi \rm J[/tex]
Hence, the work is required to pump all the water out over the side is 108000 j.
Learn more about the work done of pump the water here;
https://brainly.com/question/13148947
