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bixtya [17]
3 years ago
12

A beam of unpolarized light with a power of 1.0 mW passes through two ideal linear polarizers in series. An ideal polarizerabsor

bs100% of the light perpendicular to itspolarization axis, and none of the light parallel to itspolarization axis. What should the angle between the polarization axes of the two polarizers be if you want the power emerging from the second polarize?
Physics
1 answer:
Mashutka [201]3 years ago
3 0

Answer:

the axes of the two polarizers are aligned, the angle between this axis must be zero

Explanation:

Polarizers are anisotropic systems where absorption is selective in each axis, in this case they indicate that on an axis they let 100% pass and perpendicular to this axis no light passes.

  The first polarizer defines a direction of light that of its axis, it is said that the light is polarized, when this light reaches the second polarized, so that it can pass the axis must be in the direction of the polarized beam.

In summary, the axes of the two polarizers are aligned, the angle between this axis must be zero

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A charge of 25 nC is uniformly distributed along a straight rod of length 3.0 m that is bent into a circular arc with a radius o
Greeley [361]

Answer:

E = 31.329 N/C.

Explanation:

The differential electric field dE at the center of curvature of the arc is

dE = k\dfrac{dQ}{r^2}cos(\theta ) <em>(we have a cosine because vertical components cancel, leaving only horizontal cosine components of E. )</em>

where r is the radius of curvature.

Now

dQ = \lambda rd\theta,

where \lambda is the charge per unit length, and it has the value

\lambda = \dfrac{25*10^{-9}C}{3.0m} = 8.3*10^{-9}C/m.

Thus, the electric field at the center of the curvature of the arc is:

E = \int_{\theta_1}^{\theta_2} k\dfrac{\lambda rd\theta  }{r^2} cos(\theta)

E = \dfrac{\lambda k}{r} \int_{\theta_1}^{\theta_2}cos(\theta) d\theta.

Now, we find \theta_1 and \theta_2. To do this we ask ourselves what fraction is the arc length  3.0 of the circumference of the circle:

fraction = \dfrac{3.0m}{2\pi (2.3m)}  = 0.2076

and this is  

0.2076*2\pi =1.304 radians.

Therefore,

E = \dfrac{\lambda k}{r} \int_{\theta_1}^{\theta_2} cos(\theta)d\theta= \dfrac{\lambda k}{r} \int_{0}^{1.304}cos(\theta) d\theta.

evaluating the integral, and putting in the numerical values  we get:

E = \dfrac{8.3*10^{-9} *9*10^9}{2.3} *(sin(1.304)-sin(0))\\

\boxed{ E = 31.329N/C.}

4 0
3 years ago
A college dorm room measures 14 ft wide by 13 ft long by 6 ft high. What is the air in it under normal conditions?
kirza4 [7]

Complete question:

A college dormitory room measures 14 ft wide by 13 ft long by 6 ft high. Weight density of air is 0.07 lbs/ft3. What is the weight of air in it under normal conditions?

Answer:

the weight of the air is 76.44 lbs

Explanation:

Given;

dimension of the dormitory, = 14 ft by 13 ft by 6 ft

density of the air, = 0.07 lbs/ft³

The volume of the air in the dormitory room = 14 ft x 13 ft x 6 ft

                                                                          = 1092 ft³

The weight of the air = density  x  volume

                                   = 0.07 lbs/ft³  x  1092 ft³

                                   = 76.44 lbs

Therefore, the weight of the air is 76.44 lbs

6 0
3 years ago
To turn a coil of wire into a magnet, run a(n) ____ through it.
evablogger [386]

A.) Electromagnetic Current

please mark me as the brainliest

7 0
4 years ago
Read 2 more answers
Which of the following is an example of acceleration? I. A car speeds up. II. A car slows down. III. A car travels in a straight
svetlana [45]
I., II., and IV. are examples of acceleration. III. isn't.
6 0
4 years ago
On Earth, plasma exists in the ionosphere, in flames, and in chemical and nuclearexplosions. Matter in a controlled thermonuclea
nasty-shy [4]

The cost of developing thermonuclear power with plasmabe defended because D. It can provide an inexpensive power source.

<h3>How did the cost of developing thermonuclear power defended?</h3>

The  cost of developing thermonuclear power defended becvause we can see in the paragraph how it was told that the generation of ths power can be donee through the understanding of  the occurrence of plasmain nature,

It should be noted that this  thermonuclear power with plasmabe  posses the characteristics which make it to exist in the ionosphere, and it can be felt in the flames as well; as in the chemical and nuclearexplosions.

In conclusion the power can be seen as an inexpensive source power because the p[roduction of this power cn be found in most of the thing that can be found around us as discused above.

Therefore, option D is correct.

Read more about cost at:

brainly.com/question/25109150

#SPJ1

3 0
1 year ago
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