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neonofarm [45]
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
Physics
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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3 years ago
what happens to the water after it rains? a. precipitation b. runoff c. condensation user: what do organisms from the ocean use
Vedmedyk [2.9K]
The rain gets evaporated in to water vapor and is returned to the clouds where they go through condensation and then they poud down as rain or A.K.A,  Precipitation.
4 0
3 years ago
Two identical conducting spheres, A and B, carry equal charge. They are separated by a distance much larger than their diameters
Vesnalui [34]

Answer:

C. \frac{3F}{8}

Explanation:

Let initial charges on both spheres be,q

F=\frac{Kq^2}{d^2}   \ \ \  \ \ \  \ \ \  \ \_i

When the sphere C is touched by A, the final charges on both will be,\frac{q}{2}

#Now, when C is touched by B, the final charges on both of them will be:

q_c=q_d=\frac{q/2+q}{2}\\\\=\frac{3q}{4}\\

Now the force between A and B is calculated as:

F\prime=\frac{k\times\frac{q}{2}\times \frac{3q}{4}}{d^2}\\F\prime=\frac{3F}{8}

Hence the electrostatic force becomes 3F/8

4 0
3 years ago
Which of these is NOT a chemical property of matter?
marshall27 [118]

C. Magnetism

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7 0
3 years ago
g calculate the effectiveness radiation dosage in sieverts for a 79 kg person who is exposed 6.8x10^9
Elza [17]

Answer:

The answer is "\bold{dosage = 0.031 rem}"

Explanation:

please find the complete question in the attached file.

Given value:

m = 79\  kg  \\\\n = 3.4 \times  10^9 \\\\E = 5.5  \times  10^{-13} \\\\ RBE = 15

\to E = n E\\

        = 3.4  \times  10^9  \times  5.5  \times  10^{-13} \\\\      = 1.87 \times  10^{-3}

\to E(absorbed) = 1.87  \times 10^{-3}  \times  0.87 = 1.63  \times  10^{-3}

calculating the radiation absorbed per kg:

= \frac{1.63  \times  10^{-3}}{79}  \\\\ = 2.06  \times  10^{-5} \\\\ = 0.00206 \  rad

\to Dosage = 0.00206  \times  15 \\

                 = 0.031 \ \ rem

4 0
3 years ago
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