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son4ous [18]
4 years ago
9

Light passes straight through two polarizing filters in which the axis of polarization of the second filter is rotated 45 degree

s compared to the first filter. How does the light that passes through the second filter compare to the light that passes through the first filter??
Physics
1 answer:
SIZIF [17.4K]4 years ago
4 0

Answer: If the second polarizing filter is rotated by 45°, only the component (parts of the light beam) of the light parallel to the second filter’s axis is passed.

Explanation: Polarization is the process of converting unpolarized light to polarized light.

Polarized light waves are light waves travel in a single plane.

If two polarizing filters are applied to a light wave,

-If the axes of the first and second filters are aligned in a parallel form, it would cause all of the polarized light passed by the first filter to also pass by the second.

- If the second polarizing filter is rotated by an angle of about 45°, only the parts of the light beam parallel to the second filter’s axis is passed.

- When the axes of the polarization filter are perpendicular, no light is passed by the second filter.

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3 years ago
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A force of 500 N is used to slide a box across a smooth surface; the box moved 5 m in 1.2 seconds. What power is used?
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F = 500N
S = 5m
T = 1.2s
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P = ?

We can use formula P = F * V, just because it was a smooth slide we can assume that average speed was V = S/T = 5 / 1.2 = 50/12

So the final answer would be:

P = F * V = 500 * 50/12

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3 years ago
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7. A stream of water strikes a stationary turbine blade horizontally, as the drawing illustrates. The oncoming water stream has
NNADVOKAT [17]

Answer:

The magnitude of the average force exerted on the water by the blade is 960 N.

Explanation:

Given that,

The mass of water per second that strikes the blade is, \dfrac{m}{t}=30\ kg/s

Initial speed of the oncoming stream, u = 16 m/s

Final speed of the outgoing water stream, v = -16 m/s

We need to find the magnitude of the average force exerted on the water by the blade. It can be calculated using second law of motion as :

F=\dfrac{\Delta P}{\Delta t}

F=\dfrac{m(v-u)}{\Delta t}

F=30\ kg/s\times (-16-16)\ m/s

F = -960 N

So, the magnitude of the average force exerted on the water by the blade is 960 N. Hence, this is the required solution.

6 0
4 years ago
Now find the electromotive force E2(t) induced across the entirety of solenoid 2 by the change in current in solenoid 1. Remembe
lions [1.4K]

Answer:

E_{2} (t) = -\pi\mu_o} \rho^{2} n_{1}n_{2}L\frac{d }{dt}I_{1}(t)

Explanation:

Consider two solenoids that are wound on a common cylinder as shown in fig. 1. Let the cylinder have radius 'ρ' and length 'L' .

No. of turns of solenoid 1 = n₁

No. of turns of solenoid 1 = n₂

Assume that length of  solenoid is much longer than its radius, so that its field can be determined from Ampère's law throughout its entire length:

\oint \overrightarrow {B}\overrightarrow {(r)}.\overrightarrow {dl}= \mu_{o}I

We will consider the field that arises from solenoid 1, having n₁  turns per unit length. The magnetic field due to solenoid 1 passes through solenoid 2, which has n₂ turns per unit length.

Any change in magnetic flux from the field generated by solenoid 1 induces an EMF in solenoid 2 through Faraday's law of induction:

\oint \overrightarrow {B}\overrightarrow {(r)}.\overrightarrow {dl}= -\frac{d}{dt} \phi _{M}(t)

Consider B₁(t) magnetic feild generated in solenoid 1 due to current I₁(t)

Using:

                                  B_{1}(t) =\mu _{o} nI(t)\\ --- (2)

                           

Flux generated due to magnetic field B₁

                      \phi _{1}(t) = \oint \overrightarrow {B_{1}}.dA\\ ---(3)

area of solenoid = A = \pi \rho^{2}

substituting (2) in (3)

                       \phi _{1}(t) = \mu_{o} \pi \rho^{2} n_{1}I_{1}(t) ----(4)

We have to find electromotive force E₂(t) induced across the entirety of solenoid 2 by the change in current in solenoid 1, i.e.

                       E_{2} (t) = -n_{2}L\frac{d \phi_{1}}{dt} ---- (5)

substituting (4) in (5)

E_{2} (t) = -n_{2}L\frac{d }{dt}(\mu_o} \pi \rho^{2} n_{1}I_{1}(t))\\E_{2} (t) = -\pi\mu_o} \rho^{2} n_{1}n_{2}L\frac{d }{dt}I_{1}(t)

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Answer:

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Explanation:

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