Acircular swimming pool has a diameter of 24 ft, the sides are 5 ft high, and the depth of the water is 4 ft. how much work is required to pump all of the water out over the side. density of water= 62.5 lb/a circular swimming pool has a diameter of 24 ft,. show all steps. the answer is [w= 108,000a circular swimming pool has a diameter of 24 ft, ft-lb]

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 F  ×  d

we know w  = 62.5  

so

W = 62.5 πr²   (x+1) dx

W = 62.5 π(12)²  

W = 62.5 π(144)  (12)

W = 108000 π

hence W = 108000 π

in simple terms, the principle states that the buoyancy force on an object is equal to the weight of the fluid displaced by the object, or the density of the fluid multiplied by the submerged volume times the gravitational acceleration, g. this is also known as upthrust.