Answer:

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Explanation:

To calculate the values requested, we can use the formulas for potential energy (PE), kinetic energy (KE), and power (P).

Given:

Rate of water falling (water flow rate) = 2000 kJ/s

Height of fall (highest point to turbine) = 100 m

Height of fall (from highest point to 25 m) = 25 m

Efficiency of conversion = 80% = 0.8

Acceleration due to gravity (g) = 10 m/s²

(a) Potential Energy (PE) of water falling every 1 s at the highest point:

PE = mgh

Here, m is the mass of the water falling per second.

Mass of water falling per second = rate of water falling = 2000 kJ/s

1 kJ = 1000 J

Mass of water falling per second = 2000 kJ/s × (1000 J/kJ) = 2,000,000 J/s

PE = (2,000,000 J/s) × (100 m) = 200,000,000 J/s = 200 MW (megawatts)

Therefore, the potential energy of water falling every 1 second at the highest point is 200 MW.

(b) Kinetic Energy (KE) of water falling through a height of 25 m:

KE = (1/2)mv²

Here, m is the mass of the water falling and v is the velocity of the water at the lower point.

To find v, we can use the equation:

v² = u² + 2gh

where u is the initial velocity (which is assumed to be 0 as the water starts from rest).

v² = 0 + 2 × 10 m/s² × 25 m

v² = 500 m²/s²

v ≈ 22.36 m/s

Mass of water falling = rate of water falling = 2000 kJ/s = 2,000,000 J/s

KE = (1/2) × (2,000,000 J/s) × (22.36 m/s)²

KE ≈ 5,000,000 J/s = 5 MW (megawatts)

Therefore, the kinetic energy of the water when it falls through a height of 25 m is 5 MW.

(c) Power output in 1 second:

Power output = Efficiency × PE

PE = 200 MW (as calculated in part a)

Power output = 0.8 × 200 MW = 160 MW

Therefore, the power output in 1 second is 160 MW.

Note: It's important to remember that these calculations are based on the given values and assumptions provided in the question.