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a circular swimming pool has a diameter of 16 ft, the sides are 6 ft high, and the depth of the water is 5 ft. how much work (in ft-lb) is required to pump all of the water out over the side? (use the fact that water weighs 62.5 lb/ft3. round your answer to the nearest integer.)

User Benjist
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1 Answer

17 votes
17 votes

Final answer:

The work required to pump all the water out of the circular swimming pool is approximately 607152.71 lb-ft.

Step-by-step explanation:

To calculate the work required to pump all the water out of the circular swimming pool, we need to find the volume of the pool. The volume of the pool can be calculated using the formula for the volume of a cylinder: V = πr^2h, where r is the radius of the pool and h is the height.

In this case, the radius is half the diameter, so r = 8 ft. The volume of the pool is V = π(8^2)(6) = 301.5928 ft^3.

Next, we need to convert the volume to mass using the density of water, which is 62.5 lb/ft^3.

The mass of the water is m = V * density = 301.5928 ft^3 * 62.5 lb/ft^3 = 18849.55 lb.

Finally, we can calculate the work using the formula work = force * distance.

The force required to lift the water is equal to its weight, so force = mass * gravity = 18849.55 lb * 32.2 ft/s^2 = 607152.71 lb-ft/s^2.

The distance is the height of the pool, which is 6 ft. Therefore, the work required to pump all of the water out over the side is approximately 607152.71 lb-ft.

User Jowayne
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