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

The power generated is about $588 \times 10^{8}$Watt when the waterfall from the waterfall and reaches the bottom.

Explanation:

As per given question, the rate of water is $1.2 \times 10^{6} \mathrm{kg} / \mathrm{s}$

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

we know that Potential Energy, $\mathrm{PE}=\mathrm{m} \times \mathrm{g} \times \mathrm{h}$

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

$\mathrm{PE}=\mathrm{m} \times \mathrm{g} \times \mathrm{h}$

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

$\mathrm{PE}=588 \times 10^{6} \text { Joule }$

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

Answer 2

Answer:

5.88×10⁸ W

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