*problem: 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 *formulas: Gravitational energy = (mass) × g × (height). *attempts: 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.... =/
that's the second part of the question... the first part is: " how much energy is obtained from one kilogram of falling water?"
huh well, theres only one answer to submit i think they were simply restating the question to better explain so are u saying that there are multiple steps to the problem or is it fine to say 441 J = 441 kg/s ?
Oh, I think I see now. So 1 kg produces 441J. You need to produce 106 mega watts = 106*10^6 J/s... so in 1s you need to produce 106*10^6J ? If 1kg produces 441J, how many kg produces 106*10^6J
I'm sorry. I got confused with the power... for some reason I thought it was 106 megawatts... the power is 1 megawatt = 10^6 watts = 10^6 J/s so 10^6/441 = 2267.57 2267.57 kg/s. check yourself also to make sure this makes sense and there aren't any mistakes...