Question

Water flows over the edge of a waterfall at a rate of 1.2 x 10^6 kg/s. There are 50.0 m between the top and bottom of the waterfall. How much power is generated by the falling water by the time it reaches bottom?

Answer 1

5.88×10⁸ W

Step-by-step explanation:

Power = change in energy / time

P = mgh / t

P = (m/t) gh

P = (1.2×10⁶ kg/s) (9.8 m/s²) (50.0 m)

P = 5.88×10⁸ W

Answer 2

The power generated is about
`588 * 10^(8)`Watt when the waterfall from the waterfall and reaches the bottom.

Step-by-step explanation:

As per given question, the rate of water is
`1.2 * 10^(6) \mathrm{kg} / \mathrm{s}`

From height (h) of 50m and acceleration due to gravity is 9.8
`\mathrm{m} / \mathrm{s}^(2)`

we know that Potential Energy,
`\mathrm{PE}=\mathrm{m} * \mathrm{g} * \mathrm{h}`

The potential energy of
`1.2 * 10^(6) \mathrm{kg}` (m) water for one second is written as

`\mathrm{PE}=\mathrm{m} * \mathrm{g} * \mathrm{h}`

`\mathrm{PE}=\left(1.2 * 10^(6)\right) * 9.8 * 50`

`\mathrm{PE}=588 * 10^(6) \text { Joule }`

But power output of 1 Watt = 1 Joule / second. So the power generated in the waterfall is
`588 * 10^(6) \mathrm{Watt}` or we can also write as 588 Mega Watts.