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goblinko [34]
3 years ago
15

"an open tank has the shape of a right circular cone. the tank is 6 feet across the top and 5 feet high. how much work is done i

n emptying the tank by pumping the water over the top edge? note: the density of water is 62.4 lbs per cubic foot."
Physics
1 answer:
Nonamiya [84]3 years ago
3 0

Answer:

The amount of workdone by in emptying the tank by pumping the water over the top edge is \mathbf{ 1169.99 \pi \ ft-lbs}

Explanation:

Given that the tank is 6 feet across the top and 5 feet high.

Using the similar triangles.

\dfrac{3}{5}  = \dfrac{r}{y}

5r = 3y

r = \dfrac{3}{5} y

Thus; each disc is a circle with area

A = \pi ( \frac{3}{5} y)^2

The weight of each disc is ;

m = \rho_{water}A

= 62.4 × \pi ( \frac{3}{5} y)^2

= \dfrac{561.6}{25} \pi y^2

The distance pumped is 5-y

Thus; the workdone in pumping the tank by pumping the water over the top edge is :

W = \dfrac{561.6 \pi}{25} \int\limits^5_0 {(5-y)} \, y^2dy

W = \dfrac{561.6 \pi}{25} \int\limits^5_0 {(5y^2-y^3)} dy

W = \dfrac{561.6 \pi}{25}[52.083]

\mathbf{W = 1169.99 \pi \ ft-lbs}

The amount of workdone by in emptying the tank by pumping the water over the top edge is \mathbf{ 1169.99 \pi \ ft-lbs}

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