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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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Which part or parts of the light bulb must be touched for the light bulb to light?
uranmaximum [27]

<u>The tip of the bulb touches the </u>negative end<u> of the battery</u> parts of the light bulb must be touched for the light bulb to light.

What is negative end?

The grounded connection of a car battery is called a negative terminal, and it is typically identified by a minus (-) sign. Typically, it is black and color-coded. The other terminal is a positive terminal that is typically color-coded red or orange and denoted by a plus (+) sign. The grounded terminal attaches to an object that is grounded on the car, usually a bolt in the chassis. The starter motor, as well as other switches, relays, and accessories, are connected to the positive terminal. The car won't start if the battery isn't properly grounded. These days, the majority of car batteries are packed with a charged acid solution that is very explosive and gaseous. In order to keep the battery's inner plates immersed, distilled water may occasionally need to be supplied to the fill-caps.

To learn more about negative end visit:

brainly.com/question/17643791

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3 0
2 years ago
A 1,250 W electric motor is connected to a 220 Vrms, 60 Hz source. The power factor is lagging by 0.65. To correct the pf to 0.9
Black_prince [1.1K]

Answer:

C = 46.891 \mu F

Explanation:

Given data:

v = 220 rms

power factor = 0.65

P = 1250 W

New power factor is 0.9 lag

we knwo that

s = \frac{P}{P.F} < COS^{-1} 0.65

S = \frac{1250}{0.65} < 49.45

s = 1923.09 < 49.65^o

s = [1250 + 1461 j] vA

P.F new = cos [tan^{-1} \frac{Q_{new}}{P}]

solving for Q_{new}

Q_{new} = P tan [cos^{-1} P.F new]

Q_{new} = 1250 [tan[cos^{-1}0.9]]

Q_{new} = 605.40 VARS

Q_C = Q - Q_{new}

Q_C = 1461 - 605.4 = 855.6 vars

Q_C = \frac[v_{rms}^2}{xc} =v_{rms}^2 \omega C

C = \frac{Q_C}{ v_{rms}^2 \omega}

C = \frac{855.6}{220^2 \times 2\pi \times 60}

C = 4..689 \times 10^{-5} Faraday

C = 46.891 \mu F

6 0
3 years ago
Unpolarized light with an intensity of (25.0 A) units is passed through two successive polarizing filters, the first with its po
N76 [4]

Answer:

The intensity of the light after passing through the two polarizing filters is 4.00 units

Explanation:

Io = 25 + A

= 25 + 9

= 34

Intensity of light through first polarizer,

I1 = Io/2

= 34/2

= 17 units

angle between axis of frist and second polarizers, theta = 55 + 6

= 61 degrees

Intensity of light passing through second polarizer,

I2 = I1*cos^2(theta)

= 17*cos^2(61)

= 4.00 units

6 0
3 years ago
Why is the use of carbon- 14 dating limited?​
Vera_Pavlovna [14]

Answer:

because carbon 14 has only a short half life, rather than other elements with longer half lives.

Explanation:

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8 0
2 years ago
The table lists the values for two parameters, x and y, of an experiment. What is the estimated value of y for x=4?
Talja [164]
Y = 16 is correct answer
6 0
4 years ago
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