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how come if you jump in a train you will land in the same spot but if you jump on top of the train you won't? i tried to make it

soldier1979 [14.2K]

Answer:

Well, it has to do something with the force of the train. It's pulling you, a force of gravity

However jumping off results in a different sort of energy, the train know is moving farther away as you jump down. So you end up in a different place

Explanation:

3 0

3 years ago

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

The drawing shows four sheets of polarizing material, each with its transmission axis oriented differently. Light that is polari

storchak [24]

Answer:

Explanation:

At the point when light is vertically polarized is incident on the polarizer whose axes are situated at angle points $\theta _1 , \theta _2 , \theta _3$ the intensity power in the wake of going through all the polarisers is given by the Malus law, applied threefold for every one of the three axes.

$I = I_o \ cos \theta _1 \ cos \theta _2 \ cos \theta _ 3$

The heading of the direction of the polarization is equivalent to the pivoted axes of the polarizer provided that light is an electromagnetic wave, its course of polarization is therefore controlled by the electric field part.

∴

a)

When sheet A is removed, the transmitted light goes through B, at 30°.

$I = I_o \ cos ^2 \theta _1 \ cos ^2 \theta _2 \ cos^2 \theta _3$

$I = 20 \ cos ^2 30 \ cos ^2 60 \ cos ^2 30$

$I = 2.81 \ W/m^2$

b)

When B is removed, No light passes since the axis of A and the axis of C are perpendicular to each other.

c)

When C is removed, the intensity is indeed zero since the axes are aligned and adjusted at 90° to one another.

$\mathbf{d) \ I - I_o cos^2 (0) cos^2 (30) \ cos^2 (60)}$

$\mathbf{d = 3.75 \ W/m^2}$

3 0

3 years ago

If a force of 12 N is applied to an object with a mass of 2 kg, the object will accelerate at

klasskru [66]

F- force on the object,
m - mass of the object:
F = 12 N = 12 kgm/s², m = 2 kg
F = m · a
a = F : m = 12 N : 2 kg = 6 m/s²
Answer: C)

4 0

3 years ago

Why does it take 2 days longer for the Moon to go from New Moon to New Moon when viewed form the Earth than it takes for the the

Vesnalui [34]

The time between two new Moon phases or two full Moon phases is 29.5 days. ... The difference of 29.5 and 27.3 is that while the Moon is orbiting the Earth, the Earth is moving along in its orbit so it takes longer for the Moon to reach the same position relative to the Sun.

5 0

3 years ago

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