Energy niagra falls problem water

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SUMMARY

The Niagara Falls hydroelectric generating plant converts the gravitational energy of falling water into electricity, with a height of approximately 45 meters. The energy derived from one kilogram of falling water is calculated using the formula E = mgh, where g is the acceleration due to gravity (9.8 m/s²). To produce one megawatt (10^6 watts) of power, approximately 22.2 kilograms of water must flow through the generators every second, as derived from the energy conversion 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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