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2 years ago

Nonrenewable energy resources do not include which of the following?

Physics

2 answers:

Tems11 [23]2 years ago

8 0

Hydrogen fuel cells is the answer

Lerok [7]2 years ago

8 0

Answer: The correct answer is-

hydrogen fuel cells.

Hydrogen fuel cells are used for production of electricity whenever it is required. It is an electrochemical cell that is used to generate electricity when it is supplied with hydrogen and oxygen, which together drain as water and generate electricity.

Renewable energy sources are those that can be replenished or renewed naturally ( such as wind power, solar energy, geothermal energy). In other words, these sources of energy never run off.

Hydrogen fuel can be produced by means of renewable energy sources such as wind or solar energy.

Thus, hydrogen fuel cells do not come under non renewable energy resources.

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Three ideal polarizing filters are stacked, with the polarizing axis of the second and third filters at 21 degrees and 61 degree

kvv77 [185]

Answer:

When second polarizer is removed the intensity after it passes through the stack is

$I_f_3 = 27.57 W/cm^2$

2 When third polarizer is removed the intensity after it passes through the stack is

$I_f_2 = 102.24 W/cm^2$

Explanation:

From the question we are told that

The angle of the second polarizing to the first is $\theta_2 = 21^o$

The angle of the third polarizing to the first is $\theta_3 = 61^o$

The unpolarized light after it pass through the polarizing stack $I_u = 60 W/cm^2$

Let the initial intensity of the beam of light before polarization be $I_p$

Generally when the unpolarized light passes through the first polarizing filter the intensity of light that emerges is mathematically evaluated as

$I_1 = \frac{I_p}{2}$

Now according to Malus’ law the intensity of light that would emerge from the second polarizing filter is mathematically represented as

$I_2 = I_1 cos^2 \theta_1$

$= \frac{I_p}{2} cos ^2 \theta_1$

The intensity of light that will emerge from the third filter is mathematically represented as

$I_3 = I_2 cos^2(\theta_2 - \theta_1 )$

$I_3= \frac{I_p}{2}(cos^2 \theta_1)[cos^2(\theta_2 - \theta_1)]$

making $I_p$ the subject of the formula

$I_p = \frac{2L_3}{(cos^2 \theta [cos^2 (\theta_2 - \theta_1)])}$

Note that $I_u = I_3$ as $I_3$ is the last emerging intensity of light after it has pass through the polarizing stack

Substituting values

$I_p = \frac{2 * 60 }{(cos^2(21) [cos^2 (61-21)])}$

$I_p = \frac{2 * 60 }{(cos^2(21) [cos^2 (40)])}$

$=234.622W/cm^2$

When the second is removed the third polarizer becomes the second and final polarizer so the intensity of light would be mathematically evaluated as

$I_f_3 = \frac{I_p}{2} cos ^2 \theta_2$

$I_f_3$ is the intensity of the light emerging from the stack

substituting values

$I_f_3 = \frac{234.622}{2} * cos^2(61)$

$I_f_3 = 27.57 W/cm^2$

When the third polarizer is removed the second polarizer becomes the

the final polarizer and the intensity of light emerging from the stack would be

$I_f_2 = \frac{I_p}{2} cos ^2 \theta_1$

$I_f_2$ is the intensity of the light emerging from the stack

Substituting values

$I_f_2 = \frac{234.622}{2} cos^2 (21)$

$I_f_2 = 102.24 W/cm^2$

7 0

2 years ago

Suppose a soccer ball is kicked from the ground at an angle 20.0º above the horizontal at 8.00 m/s. The y-velocity is determined

julia-pushkina [17]

initial speed of the ball by which it is kicked is 8 m/s

now the angle at which it is kicked is 20 degree

here we will have

$v_y = v sin20$

$v_y = 8 sin20 = 2.74 m/s$

now we will have

$\Delta y = v_y t + \frac{1}{2}at^2$

so if the ball again land on the ground at same level then we have

$\Delta y = 0$

$0 = 2.74 t - \frac{1}{2}(9.8) t^2$

$0 = 2.74 - 4.9 t$

$t = 0.56 s$

so total time will be 0.56 s

8 0

3 years ago

How does the energy moves from the suns core to the photosphere?

Minchanka [31]

Answer: Energy from the core travels by radiation through the radiative

zone, then by convection through the convection zone.

Explanation:

4 0

3 years ago

A jet plane flying 600 m/s experiences an acceleration of 4g when pulling out of the dive. What is the radius of curvature of th

Gnoma [55]

Answer:

91.84 m/s²

Explanation:

velocity, v = 600 m/s

acceleration, a = 4 g = 4 x 9.8 = 39.2 m/s^2

Let the radius of the loop is r.

he experiences a centripetal force.

centripetal acceleration,

a = v² / r

39.2 x r = 600 x 600

r = 3600 / 39.2

r = 91.84 m/s²

Thus, the radius of the loop is 91.84 m/s².

8 0

2 years ago

What is an instrument commonly used to measure wind speed?

xeze [42]

The correct answer is letter D. Anemometer. It is a device that is used to measure wind speed. It is a very common weather station instrument and is available to use and to make. Anemos, from the greek word that means wind.

5 0

2 years ago

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