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3 years ago

11

Technician A says the names of the components inside an automatic transmission are based on their function; namely what gear are

they used to control. Technician B says the names of the clutches, bands, and servos are based on their location in the transmission. Who is correct

1 answer:

topjm [15]3 years ago

3 0

Answer:

Both A and B are correct

Explanation:

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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:

1

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

What happens when a mixture is separated

iris [78.8K]

Answer:

the answer would be c

Explanation:

3 0

2 years ago

A 10 g bullet is fired into, and embeds itself in, a 2 kg block attached to a spring with a force constant of 19.6 n/m and whose

dmitriy555 [2]

Answer:

$x=0.478\ m$ is the compression in the spring

Explanation:

Given:

mass of the bullet, $m=10\ g=0.01\ kg$

mass of block, $M=2\ kg$

stiffness constant of the spring, $k=19.6\ N.m^{-1}$

initial velocity of the spring just before it hits the block, $u=300\ m.s^{-1}$

<u>Now since the bullet-mass gets embed into the block, we apply the conservation of momentum as:</u>

$m.u=(M+m).v$

$0.01\times 300=(2+0.01)\times v$

$v=1.4925\ m.s^{-1}$

Now this kinetic energy of the combined mass gets converted into potential energy of the spring.

$\rm Kinetic\ energy=Spring\ potential\ energy$

$\frac{1}{2} (M+m).v^2=\frac{1}{2} k.x^2$

$\frac{1}{2} \times (2+0.01)\times 1.4925^2=\frac{1}{2} \times 19.6\times x^2$

$x=0.478\ m$ is the compression in the spring

4 0

3 years ago

Read 2 more answers

46 POINTS will mark brainliest if correct

Mazyrski [523]

46 POINTS will mark brainliest if correct

The radius of Mercury's orbit is r = 5.79 x 1010 m and its orbital period is T=88 days. What is the

magnitude of the orbital velocity for the planet around the sun, assuming a circular orbit?

A.7.21 x 103 m/s

B.8.45* 104 m/s

C.4.79 x 104 m/s

D.5.32 x 104 m/s

5 0

4 years ago

Read 2 more answers

Unpolarized light is passed through an optical filter that is oriented in the vertical direction.

VashaNatasha [74]

In order to solve this problem it is necessary to apply the concepts related to intensity and specifically described in Malus's law.

Malus's law warns that

$I = I_0 cos^2\theta$

Where,

$\theta=$ Angle between the analyzer axis and the polarization axis

$I_0 =$ Intensity of the light before passing through the polarizer

The intensity of the beam from the first polarizer is equal to the half of the initial intensity

$I = \frac{I_0}{2}$

Replacing with our the numerical values we get

$I = \frac{46}{2}$

$I = 23W/m^2$

Therefore the intensity of the light that emerges from the filter is $23W/m^2$

5 0

3 years ago