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3 years ago

7

The most common type of dam is an impoundment hydroelectric plant, which consists of a dam across a river, and a reservoir (lake

) behind the dam. How does this type of dam work

1 answer:

Naya [18.7K]3 years ago

6 0

Answer:

A turbine spins the water to generate electricity.

Explanation:

This is impoundment facility is called a hydro electric power dam, it is basically a big hydropower system which uses the dam to store river water in a reservoir. The water released from the reservoir flows directly through a turbine, the turbine spins it, which afterwards activates a generator to generate electricity.

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A typical wall outlet voltage in the United States is 120 volts. Personal MP3 players require much smaller voltages, typically 4

alexgriva [62]

Answer:

Number of turns on the secondary coil of the adapter transformer is 10.

Explanation:

For a transformer,

$\frac{V_{s} }{V_{p} } = \frac{N_{s} }{N_{p} }$

where $V_{s}$ is the voltage induced in the secondary coil

$V_{p}$ is the voltage in the primary coil

$N_{s}$ is the number of turns of secondary coil

$N_{p}$ is the number of turns of primary coil

From the given question,

$\frac{487*10^{-3} }{120}$ = $\frac{N_{s} }{2464}$

⇒ $N_{s}$ = $\frac{2462*487*10^{-3} }{120}$

= 9.999733

∴ $N_{s}$ = 10 turns

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3 years ago

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The two types of glaciers are the U type which is found in the Rocky Mountains, and the L type which can be found in Switzerland

Arturiano [62]

THE ANSWER IS TRUE
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4 0

2 years ago

A mass of 100 g stretches a spring 5 cm. If the mass is set in motion from its equilibrium position with a downward velocity of

Rudiy27

Answer:

Explanation:

Given

mass of spring $m=100\ gm$

extension in spring $x=5\ cm$

downward velocity $v=70\ cm/s$

Position in undamped free vibration is given by

$u(t)=A\cos \omega _0t+B\sin \omega _0t$

where $\omega _0^2=\frac{k}{m}$

also $\frac{k}{m}=\frac{g}{L}$

$\omega _0^2=\frac{k}{m}=\frac{9.8}{0.05}$

$\omega _0=14$

$u(t)=A\cos(14t)+B\sin(14t)$

it is given

$u(0)=0$

$u'(0)=70\ cm/s$

substituting values we get

$A=0$

$u(t)=B\sin (14t)$

$u'(t)=14B\cos (14t)$

$70=14B$

$B=\frac{10}{2}$

$B=5$

$u(t)=5\sin (14t)$

3 0

3 years ago

El movimiento de los cuerpos y la formación de imágenes son ejemplos de esta clase de fenómeno

77julia77 [94]

Answer:

depende de que fenómenos nos referimos de acuerdo al los cuerpos de formación puede aver movimiento contante

7 0

3 years ago

A 500kg car skids to a stop at a traffic light, leaving behind a 18.25m skid mark as it comes to a rest. Assuming that the car i

Nastasia [14]

Answer:

Coefficient of friction will be 0.587

Explanation:

We have given mass of the car m = 500 kg

Distance s = 18.25 m

Initial velocity of the car u = 14.5 m/sec

As the car finally stops so final velocity v = 0 m/sec

From second equation of motion

$v^2=u^2+2as$

$0^2=14.5^2+2\times a\times 18.25$

$a=-5.76m/sec^2$

We know that acceleration is given by

$a=\mu g$

$5.76=\mu\times 9.81$

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So coefficient of friction will be 0.587

6 0

3 years ago