In a hydroelectric dam, water falls 33.0 m and then spins a turbine to generate electricity. what is
δu of 1.0 kg of water? suppose the dam is 80% efficient at converting the water's potential energy to electrical energy. how many kilograms of water must pass through the turbines each second to generate 45.0 mw of electricity? this is a typical value for a small hydroelectric dam.

The change in the potential energy of the water is 323.4 J.

The mass of water that must pass through the turbine each second is 173,933.2 kg/s.

The given parameters;

The change in the potential energy of the water is calculated as follows;

ΔU = mgh

ΔU = 1 x 9.8 x 33

ΔU = 323.4 J

The efficiency of the turbine is calculated as follows;

The mass of water that must pass through the turbine each second is calculated as follows;

Learn more here:link

Explanation:

As we know that potential energy is given as

here we know that

m = 1 kg

h = 33 m

now we have

Part b)

Now we need total power given as

so we have

here we know that it is 80% efficient so we can write

wavelength divides in half and speed stays the same.