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sasho [114]
4 years ago
7

The flow rate over Niagara falls is 84,760 cfs (cubic feet per second), and the drop from top to bottom is 167 feet. If this flo

w were harnessed to generate electricity, at what rate would power be produced? Assume 100% conversion of potential energy into electrical energy. How does this compare with the power generated by the Three Gorges Dam?
Physics
1 answer:
jekas [21]4 years ago
8 0

To develop this problem we will calculate the volume from the given flow (volume per unit of time). This process will be accompanied by transforming the units given in the International system.

Through the density-volume-mass ratio, we will calculate the mass. Finally we will proceed to calculate the total power generated.

1 feet = 0.3048 m

\dot{V} = 84760ft^3/s

\dot{V} = 84760ft^3/s (\frac{(0.3058m)^3}{(1ft)^3})

\dot{V} = 2400m^3/s \text{and the Volume each second is } V= 2400m^3

Now the mass of water falling down each second would be

\rho = \frac{m}{V} \rightarrow m = \rho V

m = (1000kg/m^3)(2400m^3)

m = 2.4*10^6kg

Now Height in meters is

h = 167 ft (\frac{(0.3058m)}{(1ft)})

h= 50.9 m

The power produced is equivalent to the work done by gravity over a certain time. At this case that can be expressed as,

P = \frac{W}{t}\\P = \frac{F*h}{t}\\P = \frac{mgh}{t}\\P = \frac{2.4*10^6*9.8*50.9}{1}\\P = 1.197*10^9 Watts

The power of Niagara Falls then would be 1.197GW while the power generated by the Three Gorges Dam is  22,500 MW. This indicates that the Niagara Falls, if it were a dam that covered 100% of the fluid it carries, would exceed the production of the dam in China by thousands.

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4 0
3 years ago
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Answer:

Angle with the +x axis is θ = 79.599degree

Then the velocity of owner = 1.235m/s

Explanation:

Given that the mass of dog is m1 =26.2 kg

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momentum of dog is Py = m1 x u1

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3 years ago
Which statement best explains the path light takes as it travels? A. Light takes a curved path through matter and takes a straig
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A 13-kg sled is moving at a speed of 3.0 m/s. At which of the following speeds will the sled have twice as much kinetic energy?
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K.E. = 1/2 mv²
K.E. is directly proportional to v^2
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