Question

A full water tank in the shape of a hemispherical bowl of radius 8 m.

How much work is required to pump all the water to a height of 5 m above the tank? (The density of water is 1000 kg/m3).

Answer 1

The work required to pump all the water to a height of 5 m above the tank is approximately 105,529,460 Joules.

Step-by-step explanation:

To calculate the work required to pump all the water to a height of 5 m above the tank, we need to determine the mass of the water and the height it is being lifted.

The volume of the hemispherical tank can be calculated using the formula for the volume of a hemisphere:

V = (2/3)πr³

where r is the radius of the tank. Plugging in the given radius of 8 m, we find that the volume of the tank is approximately 2144.66 m³.

Since the tank is full, the volume of water it contains is equal to the volume of the tank itself. Thus, the mass of the water is the volume of the water multiplied by its density.

m = V x density = 2144.66 m³ x 1000 kg/m³ = 2,144,660 kg

The work required to lift the water to a height of 5 m is given by the formula:

W = mgh

where m is the mass of the water, g is the acceleration due to gravity (approximately 9.8 m/s²), and h is the height. Plugging in the values, we get:

W = 2,144,660 kg x 9.8 m/s² x 5 m = 105,529,460 J

Therefore, the work required to pump all the water to a height of 5 m above the tank is approximately 105,529,460 Joules.

Answer 2

The total work required to pump all the water to a height of 5 meters above the tank is approximately
`\( 84,070,694.93 \)` joules (since work is measured in joules in the SI system).

To calculate the work required to pump all the water to a height of 5 meters above the tank, we need to use the concept of work from physics, where work is the product of force and distance. The force, in this case, is the weight of the water, which is the product of the volume of the water, the density of water, and the acceleration due to gravity.

The work done to lift a thin slice of water at a certain height to the top of the tank is given by the formula:

`\[ \text{Work} = \text{Force} * \text{Distance} \]`

`\[ \text{Work} = (\text{Volume of water slice}) * (\text{Density of water}) * (g) * (\text{Distance to top of tank}) \]`

Since the tank is hemispherical, we'll integrate over the height of the hemisphere to find the total work done.

First, let's consider a thin slice of water at height
`\( h \)` with a thickness
`\( dh \)`. The volume of this slice,
`\( dV \)`, is the area of the circle at height
`\( h \)` times
`\( dh \)`, and the area
`\( A \)` is
`\( \pi r^2 \)`, where
`\( r \)` is the radius of the circle at height
`\( h \)`. Using the Pythagorean theorem, we know that
`\( r^2 = R^2 - (R - h)^2 \)`, where \( R \) is the radius of the hemisphere (8 m).

1. Calculate the radius of the circular slice at height \( h \):

`\[ r(h) = √(R^2 - (R - h)^2) \]`

2. Calculate the volume of the thin slice:

`\[ dV = A \cdot dh = \pi r(h)^2 \cdot dh \]`

3. Calculate the force exerted by the water slice, which is its weight:

`\[ dF = dV \cdot \text{density} \cdot g \]`

The density of water is
`\( 1000 \text{ kg/m}^3 \)` and
`\( g \)` (acceleration due to gravity) is
`\( 9.8 \text{ m/s}^2 \)`.

4. The distance that each slice must be lifted is \( 5 \text{ m} - h \), since the tank itself is 8 meters high and we need to lift the water 5 meters above the tank, the total lifting height is \( 13 - h \) meters.

5. Calculate the work for the thin slice:

`\[ dW = dF \cdot (5 \text{ m} - h) \]`

6. Integrate this over the height of the hemisphere, from \( 0 \) to \( R \):

`\[ W = \int_0^R dW \]`

`\[ W = \int_0^R \pi (R^2 - (R - h)^2) \cdot 1000 \cdot 9.8 \cdot (5 - h) \, dh \]`

Let's perform this integration to find the total work done.

7. Integrate and simplify:

`\[ W = 9800\pi \int_0^8 (64 - (8 - h)^2)(5 - h) \, dh \]`

The total work required to pump all the water to a height of 5 meters above the tank is approximately
`\( 84,070,694.93 \)` joules (since work is measured in joules in the SI system).