A cylindrical reservoir of diameter 4ft ang height

Mark Anthony Ramilo

Mark Anthony Ramilo

Answered question

2022-07-14

A cylindrical reservoir of diameter 4ft ang height 6ft is half-full of water weighing 10lb/ft³. Find the work done emptying the water over the top

Answer & Explanation

star233

star233

Skilled2023-05-29Added 403 answers

To find the work done emptying the water over the top of the cylindrical reservoir, we need to calculate the weight of the water in the reservoir and multiply it by the height from which it is being emptied.
The volume of the cylindrical reservoir can be calculated using the formula:
V=πr2h
where r is the radius and h is the height of the reservoir.
Given that the diameter of the reservoir is 4ft, the radius can be calculated as r=d2=42=2ft.
Substituting the values into the volume formula, we have:
V=π(2)2(6)=24π
Since the reservoir is half-full, the volume of water in the reservoir is half of the total volume:
Vwater=12·24π=12π
The weight of water can be calculated by multiplying the volume of water by the density of water:
Weight=Vwater·density=12π·10lb/ft3=120πlb
To find the work done emptying the water over the top, we need to multiply the weight of water by the height from which it is being emptied. Given that the reservoir is half-full, the height from which the water is being emptied is equal to the height of the reservoir, which is 6ft.
Therefore, the work done emptying the water over the top is:
Work=Weight·Height=120πlb·6ft=720πft-lb
Hence, the work done emptying the water over the top of the cylindrical reservoir is 720πft-lb.

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