- #1
Ryker
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Homework Statement
[tex]10m^{3}[/tex] of water flows over a 50m high dam every second. What's the maximum power of the falling water at the bottom of the dam?
Homework Equations
[tex]W = mgh[/tex]
[tex]P = \frac{\Delta W}{\Delta t}[/tex]
[tex]P = Fv[/tex] - apparently not
The Attempt at a Solution
I used the equation
[tex]P = \frac{mgh}{\Delta t} = \frac{10kg\times 9.8\frac{m}{s^{2}}\times 50m}{1s} = 4,9 MJ[/tex]
But why isn't it possible to use [tex]P = Fv = mgv[/tex] with [tex]m = 10000kg[/tex] and [tex]v = \sqrt{2gh}[/tex]. I mean, say you were to drop a block, holding 10000kgs of water from the same dam. It would hit the bottom with [tex]v = 31,3\frac{m}{s}[/tex]. But the power calculated this way is smaller than power calculated above? Why is that? The difference can't be the speed, since the water is hitting the bottom with the speed mentioned and in one second 10000kgs of it do that.
What am I not seeing here and why the difference?
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