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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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A night-light uses 3 W.If the voltage is 120 V, what is the current? A.0.025A B.40 A C.360A D.0.25A
Tomtit [17]
The answer to this question is A. 



6 0
3 years ago
Two vectors are given by a = 8.6i + 5.1 j and b = 931 + 9.5.
skelet666 [1.2K]

Answer:

<u>Assuming b = 9.3i + 9.5j</u> <em>(b = 931 + 9.5 is wrong):</em>

a) a×b = 34.27k

b) a·b = 128.43

c) (a + b)·b = 305.17

d) The component of a along the direction of b = 9.66

Explanation:

<u>Assuming b = 9.3i + 9.5j</u> <em>(b = 931 + 9.5 is wrong)</em> we can proceed as follows:  

a) The vectorial product, a×b  is:

a \times b = (8.6*9.5 - 5.1*9.3)k = 34.27k

b) The escalar product a·b is:

a\cdot b = (8.6*9.3) + (5.1*9.5) = 128.43

c) <u>Asumming (a</u><u> </u><u>+ b)·b</u> <em>instead a+b·b</em> we have:

(a + b)\cdot b = [(8.6 + 9.3)i + (5.1 + 9.5)j]\cdot (9.3i + 9.5j) = (17.9i + 14.6j)\cdot (9.3i + 9.5j) = 305.17

d) The component of a along the direction of b is:

a*cos(\theta) = \frac{a\cdot b}{|b|} = \frac{128.43}{\sqrt{9.3^{2} + 9.5^{2}}} = 9.66

I hope it helps you!                        

5 0
3 years ago
A small ball of mass 2.00 kilograms is moving at a velocity 1.50 meters/second. It hits a larger, stationary ball of mass 5.00 k
rewona [7]

The kinetic energy of the small ball before the collision is

                             KE  =  (1/2) (mass) (speed)²

                                     = (1/2) (2 kg) (1.5 m/s)

                                     =    (1 kg)  (2.25 m²/s²)

                                     =        2.25 joules.

Now is a good time to review the Law of Conservation of Energy:

                     Energy is never created or destroyed. 
                     If it seems that some energy disappeared,
                     it actually had to go somewhere.
                     And if it seems like some energy magically appeared,
                     it actually had to come from somewhere.

The small ball has 2.25 joules of kinetic energy before the collision.
If the small ball doesn't have a jet engine on it or a hamster inside,
and does not stop briefly to eat spinach, then there won't be any
more kinetic energy than that after the collision.  The large ball
and the small ball will just have to share the same 2.25 joules.

3 0
3 years ago
2 decaliters + 800 deciliters = __________ liters?
Stolb23 [73]
Answer is 100 liters
8 0
3 years ago
Read 2 more answers
A tugboat tows a ship with a constant force of magnitude F1. The increase in the ship's speed during a 10 s interval is 3.0 km/h
Yuki888 [10]

Answer:

The magnitude of F₁ is 3.7 times of F₂

Explanation:

Given that,

Time = 10 sec

Speed = 3.0 km/h

Speed of second tugboat = 11 km/h

We need to calculate the speed

v_{1}=\dfrac{3.0\times10^{3}}{3600}

v_{1}=0.833\ m/s

The force F₁is constant acceleration is also a constant.

F_{1}=ma_{1}

We need to calculate the acceleration

Using formula of acceleration

a_{1}=\dfrac{v}{t}

a_{1}=\dfrac{0.833}{10}

a_{1}=0.083\ m/s^2

Similarly,

F_{2}=ma_{2}

For total force,

F_{3}=F_{2}+F_{1}

ma_{3}=ma_{2}+ma_{1}

The speed of second tugboat is

v=\dfrac{11\times10^{3}}{3600}

v=3.05\ m/s

We need to calculate total acceleration

a_{3}=\dfrac{v}{t}

a_{3}=\dfrac{3.05}{10}

a_{3}=0.305\ m/s^2

We need to calculate the acceleration a₂

0.305=a_{2}+0.083

a_{2}=0.305-0.083

a_{2}=0.222\ m/s^2

We need to calculate the factor of F₁ and F₂

Dividing force F₁ by F₂

\dfrac{F_{1}}{F_{2}}=\dfrac{m\times0.83}{m\times0.22}

\dfrac{F_{1}}{F_{2}}=3.7

F_{1}=3.7F_{2}

Hence, The magnitude of F₁ is 3.7 times of F₂

3 0
3 years ago
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