Energy

**WhoCares357** · October 26th 2009, 02:30 PM

Each second, 1250 m3 of water passes over a waterfall 115 m high. Three-fourths of the kinetic energy gained by the water in falling is transferred to electrical energy by a hydroelectric generator. At what rate does the generator produce electrical energy? (The mass of 1 m3 of water is 1000 kg.)

I know that $E=E_p+E_k=mgd+.5mv^2$ . At the end of the fall there is no more potential energy so all I need is $E_k=.75* (.5mv^2)$ .

To get v I used $v=\sqrt{2gd}$ .

So the final equation is $E_k=.75*(mgd)$ .

I plugged in the numbers and got 1056562500W. According to loncapa (online system) this is wrong. Can anyone see something I did incorrectly?

**skeeter** · October 27th 2009, 02:13 PM

Originally Posted by WhoCares357

I know that $E=E_p+E_k=mgd+.5mv^2$ . At the end of the fall there is no more potential energy so all I need is $E_k=.75* (.5mv^2)$ .

To get v I used $v=\sqrt{2gd}$ .

So the final equation is $E_k=.75*(mgd)$ .

I plugged in the numbers and got 1056562500W. According to loncapa (online system) this is wrong. Can anyone see something I did incorrectly?

significant figures, maybe ?

1.06 GW

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