Energy niagra falls problem water

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The discussion focuses on calculating the energy obtained from one kilogram of falling water at Niagara Falls, which is approximately 45 meters high. The gravitational energy formula, E = mgh, is used, where g is the acceleration due to gravity (9.8 m/s²). The incorrect initial calculation of 441 joules for 1 kg of water is noted, and clarification is provided that 1 megawatt equals 10^6 watts or joules per second. The conversation emphasizes the need to determine the mass flow rate of water required to generate one megawatt of power. The participants seek assistance in correctly relating mass, energy, and time in their calculations.
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Homework Statement



In the Niagara Falls hydroelectric generating plant, the energy of falling water is converted into electricity. The height of the falls is about 45 meters. Assuming that the energy conversion is highly efficient, approximately how much energy is obtained from one kilogram of falling water? Therefore, approximately how many kilograms of water must go through the generators every second to produce a megawatt of power (106 watts)?

____kg/s


Homework Equations



Gravitational energy = (mass) × g × (height).


The Attempt at a Solution



I tried to find energy based on the fact one drop of water is one gram.
I ended up pluging in the values and got (45)(9.8) = 441
I found this to be incorrect.
Not sure how to relate mass to seconds to energy to height...
=/
 
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For 1kg of water, the energy derived=mgh joules.
For mega watts, its not 106, but 10^6 watts of power. This is equal to 10^6joules/second.

Can you work it out now?
 

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