Potential energy of this water were converted in to electrical energy

AI Thread Summary
The discussion centers on calculating the potential energy converted to electrical energy from water flowing over Niagara Falls, which is 128 meters high at a rate of 1.67e6 kg/s. The formula used is PE/s = m * g * h, resulting in a power output of 2,096,985,600 J/s. The calculation suggests that if half of this energy is converted, the power produced would be approximately 1,048,492,800 watts. The main concern raised is about the correctness of the calculation and the clarity of the question posed. The thread emphasizes the importance of understanding energy conversion in hydroelectric power generation.
sundeepsingh
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q. water flows over niagra falls which is 128m high at an average rate of 1.67e6 Kg/s. The acceleration due to gravity is 9.81. If half of the potential energy of this water were converted into electrical energy how much power will be produced by these falls.

Please help asap

MY work: PE/s = m. g. h
1.67e6 Kg/s . 9.81 . 128
2096985600 J/s
Ans/2
 
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Thread 'Variable mass system : water sprayed into a moving container'

Thread 'Variable mass system : water sprayed into a moving container'

Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
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