- #1

IntegrateMe

- 217

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^{3}.)

The answer is

**24,127 ft-Ibs**, can anyone figure out why?

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In summary, 24,127 foot-pounds of work is required to pump the sea water to the top of the cylindrical tank.

- #1

IntegrateMe

- 217

- 1

The answer is

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- #2

rock.freak667

Homework Helper

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How do you think you need to approach the problem?

- #3

IntegrateMe

- 217

- 1

so r = 2

deltaV = 4pi(deltaY)

F(y) = weight = 64 Ib/ft

..and that's pretty much all i can think of?

It's very confusing, I'm sorry.

- #4

rock.freak667

Homework Helper

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- #5

IntegrateMe

- 217

- 1

Then to get the force i use the work equation?

so,

W = fnint ((64 Ib/ft

Is that it?

- #6

HallsofIvy

Science Advisor

Homework Helper

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No, that's wrong. The water at therock.freak667 said:

[tex]256\pi\int_0^6 (8- x)dx[/itex] "foot-pounds".

Because the density is constant, you

The work required to pump is the amount of energy or force needed to move a fluid from one location to another.

The work required to pump can be calculated by multiplying the force or pressure applied to the fluid by the distance the fluid is pumped.

The factors that affect the work required to pump include the type and viscosity of the fluid being pumped, the distance the fluid needs to be pumped, and the efficiency of the pumping system.

The unit of measurement for work required to pump is joules (J) or foot-pounds (ft-lb).

The work required to pump can be reduced by optimizing the pumping system, using a more efficient pump, and choosing a fluid with lower viscosity. Additionally, reducing the distance the fluid needs to be pumped can also decrease the work required.

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