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3 years ago

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"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."

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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