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anonymous

  • 5 years ago

A circular pool has a radius of 12 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? (water weighs 62.5 lb/ft^3)

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  1. anonymous
    • 5 years ago
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    A = pi * r^2 = 144 * pi Say the thickness is dz So dV = A dz = 144 * pi dz dF = (rho) dV = 144 * pi * (rho) dz dW = d F z = 144 * pi * (rho) z dz The upper layer of water must be lifted 1 ft over the top rim of the pool, whereas the layer at the bottom must be lifted 5 ft. Therefore, the work required to empty the swimming pool of water is \[W = \int\limits_{1}^{5} dW = \int\limits_{1}^{5} 144(\rho)(\pi)zdz\] this is rho ---------^ the symbol that looks like a p Solve integral and you'll get 1728(rho)(pi) plug in density (62.5) for rho and you'll get the answer around 339,292 ft-lb.

  2. anonymous
    • 5 years ago
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    cheers ^_^

  3. anonymous
    • 5 years ago
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    Thanks. Wish I would have done that with the exam. I had all the components I just couldnt put it back together... fun and thanks!

  4. anonymous
    • 5 years ago
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    what is rho?

  5. anonymous
    • 5 years ago
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    rho is this : \[\rho\] it's a greek letter usually meaning density

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