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scoundrel [369]
3 years ago
10

Which best describes a similarity between power plants that use water as an energy source and those that use wind as an energy s

ource?
A.Both use heat from within Earth to produce electricity.
B.Both use kinetic energy to produce electricity.
C.Both use resources that are not abundant.
D.Both use resources that produce a lot of air pollution.
Physics
2 answers:
stiks02 [169]3 years ago
6 0

The correct answer to the question is : B) Both use kinetic energy to produce electricity.

EXPLANATION:

Hydroelectricity is produced when the kinetic energy of the falling water rotates the turbine. The generator associated with the turbine is rotated, and electric energy is produced due to electric magnetic induction.

In case of power plants in which wind is used as energy source, the same process happens. The kinetic energy of wind rotates the propellers of wind turbine which in turn produces electric energy.

Hence, the correct similarity between them is statement B.

disa [49]3 years ago
3 0
The hydropower plant and wind turbines both uses kinetic energy to produce mechanical power and convert the mechanical energy using a generator to an electrical energy. They both have the process to produce energy but they differ in the source the hydropower plant uses water to whit the wind turbines power plant uses wind. Therefore the answer is letter B. 
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What 2 planets have the shortest distance between them
grigory [225]
The two plants that have the shortest distance between them are Mercury and Venus which is roughly <span>31,248,757 miles separated from each other.</span>
3 0
4 years ago
What is the rotational inertia of an object that rolls without slipping down a 2.00-m-high incline starting from rest, and has a
krek1111 [17]

We will apply the concepts of energy conservation to solve the problem. We know that gravitational potential energy is equivalent to the sum of translational kinetic energy and rotational kinetic energy. Additionally, the relation of the angular velocity with the tangential velocity will be determined to eliminate the angular term and obtain the expression of the Inertia in terms of the data given, therefore, we know that

PE_{g} = KE_{trans} +KE_{rot}

mgh = \frac{1}{2} mv^2+\frac{1}{2}I\omega^2

The angular velocity in terms of tangential velocity and radius is defined as,

\omega = \frac{v}{R}

Replacing,

mgh = \frac{1}{2} mv^2+\frac{1}{2} I(\frac{v}{R})^2

Multiplying by R^2,

gh(mR^2) = \frac{v^2}{2} (mR^2)+\frac{v^2}{2}I

\frac{v^2}{2}I = (gh-\frac{v^2}{2})mR^2

I = (\frac{2gh}{v^2}-1)mR^2

Replacing with our values we have,

I = (\frac{2(9.8)(2)}{6^2}-1)mR^2

I = 0.089mR^2

6 0
3 years ago
g The electric field in a sinusoidal wave changes as E =125 N&gt;C2cos 311.2 * 1011 rad&gt;s2t +14.2 * 102 rad&gt;m2x] (a) In wh
joja [24]

Answer:

a) the propagation direction is x, b)   E₀ = 125 N / C² , c) B = 41.67 10⁻⁸ T ,

d)  f = 4.95 10¹¹ Hz, e)  λ = 4.42 10⁻³ m, f) the speed of light ,

Explanation:

The equation they give for the sine wave is

      E = 125 cos (14.2 10² x - 311.2 10¹¹ t)

This expression must have the general shape of a traveling wave

      E = Eo cos (kx - wt + Ф)

we can equal each term between the two equations

a) the propagation direction is x, since it is the term that accompanies the vector k

b) the amplitude is the coefficient before the cosine function

          E₀ = 125 N / C²

c) to find the amplitude of the magnetic field we use that the two fields are in phase

          C = E / B

          B = E / c

          B = 125/3 10⁸

          B = 41.67 10⁻⁸ T

d) the angular velocity is

          w = 311.2 10¹¹ rad / s

angular velocity and frequencies are related

          w = 2π f

           f = w / 2π

           f = 311.2 10¹¹ / 2π

           f = 4.95 10¹¹ Hz

e) the wavelength is obtained from the wave number

          k = 2π /λ

          k = 14.2 10² rad / m

          λ = 2π / k

          λ = 2π / 14.2 10²

          λ = 4.42 10⁻³ m

f) the speed of an electromagnetic wave is the speed of light

g) what is a transverse wave

5 0
3 years ago
While traveling along a highway a drivers velocity is 9m/s. In 12 seconds how far has the car gone?
kupik [55]

While travelling along a highway a drivers velocity is 9 m/s. In 12 seconds, the car has gone for 108 m.

<u>Explanation:</u>

The car travelling along the highway has a uniform velocity of 9 m/s. therefore to calculate the distance travelled by the car with the uniform velocity in 12 seconds, simple formula of velocity is to be employed.

Thereby, as we know that

           \text {velocity}=\frac{\text {displacement}}{\text {time}}

The displacement is along the highway with uniform motion, therefore, distance=displacement

Given that,

Velocity = 9 m/s  

Time = 12 s

Putting the values in the formula, we get,

       9 = \frac{d}{12}

       d=12 \times 9=108 \mathrm{m}

Therefore, the car travels for 108 m in the 12 seconds while travelling along highway with 9 m/s of velocity.

5 0
3 years ago
Power Rating of a Resistor. The power rating of a resistor is the maximum power the resistor can safely dissipate without too gr
IgorLugansk [536]

(a) 273.9 V

The power rating of the resistor is given by

P=\frac{V^2}{R}

where

P is the power rating

V is the potential difference across the resistor

R is the resistance

If the maximum power rating is P=5.0 W, and the resistance of the resistor is R=15 k\Omega = 15000 \Omega, then we can find the maximum potential difference across the resistor by re-arranging the previous equation for V:

V=\sqrt{PR}=\sqrt{(5.0 W)(15000 \Omega)}=273.9 V

(b) 1.6 W

In this case, we have:

R=9.0 k\Omega = 9000 \Omega is the resistance of the resistor

V=120 V is the potential difference across the resistor

So we can find the power rating by using the same formula of part (a):

P=\frac{V^2}{R}=\frac{(120 V)^2}{9000 \Omega}=1.6 W

(c) Maximum voltage: 14.1 V; Rate of heat: 2.00 W and 3.00 W

Here we have two resistors of

R_1 = 100 \Omega\\R_2 = 150 \Omega

and each resistor has a power rating of

P = 2.00 W

So the greatest potential difference allowed in the first resistor is

V=\sqrt{PR_1}=\sqrt{(2.00 W)(100 \Omega)}=14.1 V

While the greatest potential difference allowed in the second resistor is

V=\sqrt{PR_2}=\sqrt{(2.00 W)(150 \Omega)}=17.3 V

So the greatest potential difference allowed not to overheat either of the resistor is 14.1 V.

In this condition, the power dissipated on the first resistor is 2.00 W, while the power dissipated on the second resistor is

P_2 = \frac{V^2}{R_2}=\frac{(14.1 V)^2}{150 \Omega}=1.33 W

And this corresponds to the rate of heat generated in the first resistor (2.00 W) and in the second resistor (1.33 W).

4 0
3 years ago
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