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A circular swimming pool has a diameter of 14 m, the sides are 4 m high, and the depth of the water is 3 m. How much work (in Joules) is required to pump all of the water over the side?

(The acceleration due to gravity is [tex]9.8 \, \text{m/s}^2[/tex] and the density of water is [tex]1000 \, \text{kg/m}^3[/tex].)

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A tank, shaped like a cone, has a height of 9 meters and a base radius of 2 meters. It is placed so that the circular part is upward. It is full of water, and we have to pump it all out by a pipe that is always leveled at the surface of the water. Assume that a cubic meter of water weighs [tex]10000 \, \text{N}[/tex], i.e., the density of water is [tex]10000 \, \text{N/m}^3[/tex].

How much work (in Joules) is required to pump all of the water out of the tank?

Answer :

Final answer:

To pump all of the water over the side of the circular swimming pool, we need to calculate the work required. By finding the mass of the water and using the formula for work, we can determine the amount of work in Joules. Plugging in the given values, the work required is 13,531,200 Joules.

Explanation:

To calculate the work required to pump all of the water over the side of a circular swimming pool, we need to calculate the gravitational potential energy of the water. The mass of the water can be found by multiplying the density of water (1000 kg/m³) by its volume, which is the area of the pool's surface (πr²) multiplied by the height of the water (3 m). The work is then given by the formula: work = mass x acceleration due to gravity x height. Plugging in the values and converting the units, we can calculate the work required.

Using the given values for the diameter of the pool (14 m) and the height of its sides (4 m), we can calculate the radius of the pool (7 m).

  • The area of the pool's surface is then π(7 m)² = 154 m². Multiplying the area by the height of the water (3 m) gives us the volume of the water: 462 m³.
  • Multiplying this volume by the density of water (1000 kg/m³) gives us the mass: 462,000 kg.
  • Finally, using the acceleration due to gravity (9.8 m/s²) and the height of the water (3 m), we can calculate the work required: work = (462,000 kg) x (9.8 m/s²) x (3 m) = 13,531,200 Joules.

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Rewritten by : Barada

Work done in pumping water out of a circular pool Work done in pumping all the water over the side of a swimming pool can be obtained by calculating the gravitational potential energy of water and can be calculated as follows:

Therefore, the amount of work required to pump all the water over the side is 5745600 J.2. Work done in pumping water out of a conical tankLet V be the volume of water in the conical tank and H be the height of water. Since the tank is conical, the volume of water in the tank at height h is given by.

Where R is the base radius of the tank. Therefore, the volume of water in the tank is given by: The mass of water is given by:

m = density of water × volume

of Since the pipe is always leveled at the surface of the water, the work done to lift water from height h to height H is given by: W = mghwhere g is the acceleration due to gravity and h is the height of the water being pumped out.

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