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3 years ago

Which statement describes steps involved in the production of hydroelectric power?

Physics

2 answers:

Tju [1.3M]3 years ago

8 0

Answer:

D. Falling water turns a turbine that helps generate electricity.

Explanation:

Here is the proof:

viktelen [127]3 years ago

4 0

Answer:

Answer is D.......Falling water turns a turbine that helps generate electricity.

Explanation:

Hydropower plants capture the energy of falling water to generate electricity. A turbine converts the kinetic energy of falling water into mechanical energy. Then a generator converts the mechanical energy from the turbine into electrical energy.

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A parallel-plate capacitor with circular plates of radius 0.19 m is being discharged. A circular loop of radius 0.28 m is concen

aliina [53]

Answer:

$\dfrac{dE}{dt}=2.59\times 10^{12}\ V/m.s$

Explanation:

Given that

R= 0.19 m

r= 0.28 m

I= 2.6 A

We know that

$I_D=\epsilon _oA\dfrac{dE}{dt}$

A= Area of loop

dE/dt= rate of change of electric filed

I=Displacement current

Here r>R

So A=π R²

Now by putting the values

$I=\epsilon _oA\dfrac{dE}{dt}$

$2.6=8.85\times 10^{-12}\times \pi\times 0.19^2\dfrac{dE}{dt}$

$\dfrac{dE}{dt}=2.59\times 10^{12}\ V/m.s$

7 0

4 years ago

What two factors could the students change to investigate how to increase and decrease the magnetic force between the paperclip

scoundrel [369]

..............................................A

4 0

3 years ago

What is capacitance?

Otrada [13]

Answer:

A, the amount of charge stored per volt

Explanation:

The equation for capacitance is

C = Q/V

therefore, it is charge per volt

5 0

4 years ago

An electron (charge −e=−1.60×10−19C−e=−1.60×10−19C) is at rest at a distance 1.00×10−10 m 1.00×10−10m from the center of a nucle

shtirl [24]

Answer:

$-9.45\times10^{-16} \text{ J}$

Explanation:

According to Coulomb's law, the force of attraction between two point charges, $q_1$ and $q_2$ , separated by a distance $d$ is given by

$F = k\dfrac{q_1q_2}{d^2}$

$k$ is a constant with a value of $9\times10^9\text{ F/m}$ .

When we substitute the values from the question,

$F = (9\times10^9\text{ F/m})\dfrac{(-1.60\times10^{-19} \text{ C})\times(+1.31\times10^{-17}\text{ C})}{(1.00\times10^{-10}\text{ m})^2} = -1.89\times10^{-6} \text{ N}$

This value is negative because it is in a direction towards the positive charge.

The work done in moving the electron from the nucleus is

$W = F\times r$

$W = (-1.89\times 10^{-6} \text{ N})\times(5.00\times10^{-10}\text{ m}) = -9.45\times10^{-16} \text{ J}$

This is negative because work is done on the electron, not by it.

3 0

3 years ago

A powerful searchlight shines on a man. The man's cross-sectional area is 0.500m2 perpendicular to the light beam, and the inten

babymother [125]

Answer:

The magnitude of the force the light beam exerts on the man is 5.9 x 10⁻⁵N

(b) the force the light beam exerts is much too small to be felt by the man.

Explanation:

Given;

cross-sectional area of the man, A = 0.500m²

intensity of light, I = 35.5kW/m²

If all the incident light were absorbed, the pressure of the incident light on the man can be calculated as follows;

P = I/c

where;

P is the pressure of the incident light

I is the intensity of the incident light

c is the speed of light

$P = \frac{I}{c} =\frac{35500}{3*10^8} = 1.18*10^{-4} \ N/m^2$

F = PA

where;

F is the force of the incident light on the man

P is the pressure of the incident light on the man

A is the cross-sectional area of the man

F = 1.18 x 10⁻⁴ x 0.5 = 5.9 x 10⁻⁵ N

The magnitude of the force the light beam exerts on the man is 5.9 x 10⁻⁵ N

Therefore, the force the light beam exerts is much too small to be felt by the man.

8 0

4 years ago