Energy

• Oct 26th 2009, 02:30 PM
WhoCares357
Energy
Quote:

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

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

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

I plugged in the numbers and got 1056562500W. According to loncapa (online system) this is wrong. Can anyone see something I did incorrectly?
• Oct 27th 2009, 02:13 PM
skeeter
Quote:

Originally Posted by WhoCares357
I know that $\displaystyle E=E_p+E_k=mgd+.5mv^2$. At the end of the fall there is no more potential energy so all I need is $\displaystyle E_k=.75* (.5mv^2)$.

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

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

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