Assuming 100% efficient energy conversion, how much water stored behind a 50 centimeter high hydroelectric dam would be required to charge the battery?

Complete question is;

Assuming 100% efficient energy conversion how much water stored behind a 50 centimeter high hydroelectric dam would be required to charge the battery with power rating, 12 V, 50 Ampere-minutes.

Amount of water required to charge the battery = 7.35 m³

Explanation:

The formula for Potential energy of the water at that height = mgh

Where;

m = mass of the water

g = acceleration due to gravity = 9.8 m/s²

h = height of water = 50 cm = 0.5 m

We know that in density, m = ρV

Where;

ρ = density of water = 1000 kg/m³

V = volume of water

So, potential energy is now given as;

Potential energy = ρVgH = 1000 × V × 9.8 × 0.5 = (4900V) J

Now, formula for energy of the battery is given as;

E = qV

We are given;

q = 50 A.min = 50 × 60 = 3,000 C

V = 12 V

Thus;

qV = 3,000 × 12 = 36,000 J

E = 36,000 J

At a 100% conversion rate, the energy of the water totally powers the battery.

Thus;

(4900V) = (36,000)

4900V = 36,000

V = 36,000/4900

V = 7.35 m³

the first, second, and fifth options

acceleration = (change in speed) / (time for the change)

change in speed = (20-120)= -100 m/s.

1.5 min = 90 sec

acceleration = -100/90 = -1.11 m/s^2.

we don't care about the 150kg or the 2,500 meters. they don't make any difference.

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