A circular swimming pool has a diameter of 16 meters, its sides are 4 meters high, and the depth of the water is 3 meters. How much work (in Joules) is required to pump the water over the side? Assume the acceleration due to gravity is 9.8 m/s2 and the density of water is 1,000 kg/m3. Round the result to three decimal places.

To calculate the work required to pump the water over the side of the pool, we need to find the mass of the water first, then use the formula for work done against gravity.

Find the volume of water in the pool:

The pool is a cylinder, so we'll find the volume of the cylinder filled with water.

Volume of cylinder = πr²h

where r is the radius of the pool and h is the depth of the water.

Given:

Diameter (d) = 16 meters

Radius (r) = d/2 = 16/2 = 8 meters

Depth of water (h) = 3 meters

Substituting these values:

Volume = π * (8)^2 * 3 = 192π cubic meters

Find the mass of the water:

Mass = Volume * Density of water

Density of water = 1000 kg/m³

Mass = 192π * 1000 = 192000π kg

Find the force required to lift the water:

Force = Mass * Acceleration due to gravity

Acceleration due to gravity = 9.8 m/s²

Force = 192000π * 9.8 = 1881600π N

Find the work done:

Work = Force * Distance

Since the water is being pumped over the side, the distance is the height of the pool.

Distance = 4 meters

Work = 1881600π * 4 Joules

Now, we calculate the value:

Work = 1881600π * 4 ≈ 23615898.36 Joules

Rounded to three decimal places, the work required to pump the water over the side is approximately 23,615,898.36 Joules.