Approximately 5.5 X10^6 kg of water fall 50m over Niagara Falls each second.

a) What is the decrease in the gravitational potential energy of the water-Earth system each second?
b) If all this energy could be converted to electrical energy (can’t happen) at what rate would electrical energy be supplied? (the mass of 1 m^3 of water is 1000kg)
c) If the electrical energy were sold at 1 cent/kW *h what would be the yearly cost?

a)would I use 1/2mv^2=mgh? But the thing is velocity isn't given.
For b) and c) confuses me could you please guide me into the right direction?

jgfj

To calculate the decrease in gravitational potential energy of the water-Earth system each second (a), you are correct that you can use the equation: ΔPE = mgh, where m is the mass, g is the acceleration due to gravity (9.8 m/s^2), and h is the height of the fall.

a) In this case, you're given that the mass of water falling per second is 5.5 x 10^6 kg and the height is 50 m. So, you can calculate the decrease in gravitational potential energy per second as follows:
ΔPE = (5.5 x 10^6 kg) x (9.8 m/s^2) x (50 m) = 2.7 x 10^9 J

For b) and c), we need to consider the conversion of gravitational potential energy to electrical energy. Since not all energy can be converted due to inefficiencies, it is essential to consider the conversion efficiency. Without the efficiency, we can calculate the maximum electrical energy supplied if all gravitational potential energy is converted.

b) The formula for electrical energy (E) is power (P) multiplied by time (t). Mathematically, it is expressed as E = Pt. Since power is the rate at which energy is transferred or work is done, we can relate it to the decrease in gravitational potential energy per second (ΔPE) calculated in part a). Therefore, we can write P = ΔPE/t.

To find the electrical energy rate, we need to know the time it takes for the water to fall 50 m. Assuming it falls in 1 second, we can calculate the electrical energy rate as:
P = ΔPE/t = (2.7 x 10^9 J) / (1 s) = 2.7 x 10^9 W

c) To determine the yearly cost, we need to convert the electrical energy supplied (in watts) to kilowatt-hours (kWh) and multiply it by the cost per kilowatt-hour.

Since 1 W = 1 J/s and there are 3600 seconds in an hour, we can calculate the electrical energy supplied in kilowatt-hours as follows:
Energy (in kWh) = (2.7 x 10^9 W) x (3600 s) / (1000 W/kW) / (1000 kWh/MWh) = 9720 kWh

Now, assuming the cost is 1 cent/kWh, we can calculate the yearly cost as:
Yearly cost = (9720 kWh) x ($0.01/kWh) x (365 days) = $3546

Please note that the actual electrical energy production and costs at Niagara Falls are more complex and depend on various factors, including the efficiency of energy conversion, infrastructure, and market rates. This calculation is a simplified theoretical scenario for understanding the concepts involved.