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?

Sorry, I know this is 5 years late, however:

For a) you would use the potential energy equation: mgh

For b) you would use the power equation: Ep(answer from question a))/1s
This answer should be practically the same as answer a), what you've done in question b) is changed the units. (From Joules to Watts)

For c) you would multiply your answer from b) by the total hours per year and divide that product by 1000W (to find kW).
After this you'll have cents per year, however your yearly cost or income should be in dollars per year. In order to find this simply divide your cents total by 100 cents to find dollars.

To solve this problem, we need to calculate the gravitational potential energy and then use that to answer parts a), b), and c).

a) The decrease in gravitational potential energy each second can be calculated using the formula:

ΔPE = mgh

where:
ΔPE = change in potential energy
m = mass of water (5.5 x 10^6 kg)
g = acceleration due to gravity (approximately 9.8 m/s^2)
h = height of the fall (50 m)

By substituting the given values into the formula, we get:

ΔPE = (5.5 x 10^6 kg) x (9.8 m/s^2) x (50 m)

b) To calculate the electrical energy that could be supplied, we can convert the change in gravitational potential energy to electrical energy. Since we know that 1 m^3 of water has a mass of 1000 kg, and the height of the fall is 50 m, we can calculate the volume of water falling in 1 second by dividing the mass of water falling by the mass per unit volume, which gives us:

Volume = (5.5 x 10^6 kg) / (1000 kg/m^3)

Then, we can calculate the electrical energy supplied using the equation:

Electrical energy = Volume x ΔPE

c) Finally, to calculate the yearly cost, we need to know the rate at which electrical energy is being supplied per hour (kWh). We can calculate the electrical energy supplied in one hour by multiplying by 3600 (the number of seconds in an hour). Then, we can calculate the yearly cost using the following equation:

Yearly cost = (Electrical energy supplied per hour) x (cost per kWh) x (Hours in a year)

Given the information in part c), we can substitute the values to calculate the yearly cost.

Please note that in reality, it is not possible to convert all the energy into electrical energy, and there may be losses in the conversion process. This calculation assumes 100% efficiency.

To find the decrease in gravitational potential energy of the water-Earth system each second (part a), we can use the formula for gravitational potential energy:

U = mgh

where U is the gravitational potential energy, m is the mass, g is the acceleration due to gravity, and h is the height. In this case, m = 5.5 x 10^6 kg (mass of water falling per second) and h = 50 m (height of the fall).

a) Decrease in gravitational potential energy per second:
U = (5.5 x 10^6 kg) x (9.8 m/s^2) x (50 m)
U ≈ 2.7 x 10^9 J/s

So, the decrease in gravitational potential energy of the water-Earth system each second is approximately 2.7 x 10^9 Joules per second.

For part b, we need to calculate the rate at which electrical energy would be supplied if all the gravitational potential energy could be converted. Since 1 m^3 of water weighs 1000 kg, we need to determine the volume of water falling per second.

To find the volume of water per second:
Volume = Mass / Density
Volume = (5.5 x 10^6 kg) / (1000 kg/m^3)
Volume ≈ 5500 m^3/s

Since the height is 50 m, the potential energy is equivalent to mgh. If we multiply this by the volume of water per second, we can find the electrical energy supplied per second.

b) Electrical energy supplied per second:
Energy = Volume x Potential energy
Energy = (5500 m^3/s) x (9.8 m/s^2) x (50 m)
Energy ≈ 2.7 x 10^6 J/s

So, if all the gravitational potential energy could be converted into electrical energy, it would be supplied at a rate of approximately 2.7 x 10^6 Joules per second.

For part c, we have the rate of electrical energy supplied per second from part b. To find the yearly cost, we need to convert the energy into kilowatt-hours (kWh) and multiply it by the cost per kWh.

To convert the energy from joules to kilowatt-hours (kWh):
Energy (kWh) = Energy (J) / (3.6 x 10^6 J/kWh)

c) Yearly cost:
Cost = Energy (kWh) x Cost per kWh
Cost = (2.7 x 10^6 J/s) / (3.6 x 10^6 J/kWh) x (1 kWh/3600 s) x (365 days/year) x (24 hours/day) x (1 cent/kWh)
Cost ≈ $707,638.89

So, based on the given cost of 1 cent per kilowatt-hour, the yearly cost would be approximately $707,638.89.

no answer