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3 years ago

There is more than one answer

Physics

2 answers:

Sveta_85 [38]3 years ago

5 0

Explanation:

the answers are the first 3.

Soloha48 [4]3 years ago

3 0

Answer:

4.6

Explanation:

Ans is 4.6 b coz the difference b/w 4.6 and the other values is comparitively the largest.

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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

3 years ago

Explain how artists use monocular cues to depth perception described in the text to create an impression of three dimensions on

lana66690 [7]

Answer:

The relative size of an object serves as an important monocular cue for depth perception. It works like this: If two objects are roughly the same size, the object that looks the largest will be judged as being the closest to the observer. This applies to three-dimensional scenes as well as two-dimensional images.

Explanation:

6 0

3 years ago

The 25 kg wheel has a radius of gyration about its center O of kO = 300 mm. When the wheel is subjected to the couple moment, it

N76 [4]

Answer:

Explanation:

radius of gyration of wheel k then

k² = r²/2

r² = 2 k²

r = √2 k

= 1.414 x .3 m

r = .4242 m

Moment of inertia of wheel

= mass x radius of gyration ²

= 25 x .3 x .3

= 2.25 kg m²

Friction force acting on it ( sliding )

= μmg , μ being coefficient of kinetic friction

This friction force will create linear acceleration in forward direction

Acceleration produced

= μg

= .6 x 9.8

= 5.88 m / s ²

This will also rotate the wheel , angular acceleration being

linear acceleration / radius

= 5.88 /.4242

= 13.86 radian / s²

3 0

3 years ago

You measure its length as it passes. by how many millimeters do you determine the rod has contracted?

LUCKY_DIMON [66]

It sounds like a special relativity question but I need more info for a total answer. But remember it's length in the lab frame is
L•sqrt(1-(v/c)^2) where L is the rest length, v is its velocity magnitude and c is the speed of light. Sqrt is the square root (I'm on a phone so I can't see the math equation editor)

4 0

4 years ago

Một mặt phẳng vô hạn tích điện đều, mật độ σ = 4.10-9 C/cm2, đặt thẳng đứng trong không khí. Một quả cầu nhỏ có khối lượng 8 g,

dusya [7]

Answer:

The angle is 18.3 degree.

Explanation:

A uniformly charged infinite plane, density σ = 4 x 10^-9 C/cm^2, is placed vertically in air. A small ball of mass 8 g, with charge q = 10^-8 C, hangs close to the plane, so that the string is initially parallel to the plane. Take g = 9.8m/s2. When in equilibrium, by what angle is the string hanging the ball to the plane?

surface charge density, σ = 4 x 10^-5 C/m^2

Charge, q = 10^-8 C

mass, m = 0.008 kg

Let the angle is A and the tension in the string is T.

The electric field due to a plane is

$E =\frac{\varepsilon \sigma }{2\varepsilon o}\\\\E =\frac{4\times 10^{-5}}{2\times 8.85\times 10^{-12}}\\\\E = 2.26\times 10^6 V/m \\$

Now equate the forces,

$T sin A = q E.... (1)\\\\T cos A = m g ..... (2)\\\\divide (1) by (2)\\\\tan A = \frac{10^{-8}\times 2.6\times 10^6}{0.008\times 9.8}\\\\tan A = 0.33\\\\ A = 18.3 degree$

5 0

3 years ago