Water flows at a rate of 6.25 × 10^6 kg per minute over a waterfall - Scottish Highers Physics - Question 6 - 2015

To calculate the power delivered by the water, we can use the equation for gravitational potential energy:

P = rac{E}{t}

where:

• $P$ is the power in watts (W)
• $E$ is the energy in joules (J)
• $t$ is the time in seconds (s)

The energy can be calculated as:

$E = mgh$

where:

• $m$ is the mass in kilograms (kg)
• $g$ is the acceleration due to gravity (approximately 9.81 m/s²)
• $h$ is the height in meters (m)

Given that the flow rate is 6.25 × 10^6 kg per minute, we convert this to kg per second:

$6.25 × 10^6 \text{ kg/min} = \frac{6.25 × 10^6}{60} \text{ kg/s} = 1.0417 × 10^5 \text{ kg/s}$

Now we can calculate the energy delivered in one second (since we are using kg/s):

$E = (1.0417 × 10^5 \text{ kg/s})(9.81 \text{ m/s²})(108 \text{ m})$

Calculating this gives:

$E ≈ 1.11 × 10^7 \text{ J}$

Thus, the power delivered by the water is:

P = rac{1.11 × 10^7 \text{ J}}{1 \text{ s}} = 1.11 × 10^7 \text{ W}

Comparing with the options, the closest answer is:

A 1.13 × 10^7 W.