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asambeis [7]
2 years ago
8

Unpolarized light with intensity 300 W m2 is incident on three polarizers, P1, P2, and P3 numbered in the order light reaches th

e polarizers. The transmission axis of P1 and P2 make an angle of 45? . The transmission axis of P2 and P3 make an angle of 30? . How much light is transmitted through P3? Select One of the Following: (a) 60 W m2 (b) 100 W m2 (c) 20 W m2 (d) 10 W
Physics
1 answer:
sweet [91]2 years ago
7 0

Answer:

answer is A corresponding to 60 W

Explanation:

The polarization phenomenon is descriptor Malus's law

      I = I₀ cos² θ

Where I and Io are the transmitted and incident intensities, respectively, θ  is the angle between the direction of polarization of the light and the polarizer.

Unpolarized light strikes the first polarizer and half of it is transmitted, which has the polarization direction of the polarizer

      I₁ = ½ I₀

      I₁ = ½ 300

      I₁ = 150 W

The transmission by the second polarizer (P2) is

      I₂ = I₁ cos² θ 1

      I₂ = 150 cos² 45

      I₂ = 75 W

The transmission of the third polarizer (P3)

     I₃ = I₂ cos² θ 2

     I₃ = 75 cos² 30

     I₃ = 56.25 W

This is the light transmitted by the 56 W system, the closest answer is A corresponding to 60 W

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Explanation:

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ω = 2π / 24 Hr

Now, by definition, an angle is the relationship between the arc s, and the radius r, so we can replace these values in the angular velocity expression, as follows:

ω = (Δs / r) . 1/Δt ⇒ ω = (Δs/Δt). 1/r

But, by definition, Δs/At, is just the linear velocity, v, so we can conclude the following;

ω = v/r ⇒ v = ω. r

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v = 2π /24 hr . 6378 Km = 1,670 Km/hr.

4 0
2 years ago
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mamaluj [8]

Answer:

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Explanation:

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T^2\alpha R^3\\T^2=kR^3.......................(1)

where k is a constant.

From equation (1) we can deduce that the ratio of the square of the period of a planet to the cube of its mean distance from the sun is a constant.

\frac{T^2}{R^3}=k.......................(2)

Let the orbital period of the earth be T_e and its mean distance of from the sun be R_e.

Also let the orbital period of the planet be T_p and its mean distance from the sun be R_p.

Equation (2) therefore implies the following;

\frac{T_e^2}{R_e^3}=\frac{T_p^2}{R_p^3}....................(3)

We make the period of the planet T_p the subject of formula as follows;

T_p^2=\frac{T_e^2R_p^3}{R_e^3}\\T_p=\sqrt{\frac{T_e^2R_p^3}{R_e^3}\\}................(4)

But recall that from the problem stated, the mean distance of the planet from the sun is 16 times that of the earth, so therefore

R_p=16R_e...............(5)

Substituting equation (5) into (4), we obtain the following;

T_p=\sqrt{\frac{T_e^2(16R_e)^3}{(R_e^3}\\}\\T_p=\sqrt{\frac{T_e^24096R_e^3}{R_e^3}\\}

R_e^3 cancels out and we are left with the following;

T_p=\sqrt{4096T_e^2}\\T_p=64T_e..............(6)

Recall that the orbital period of the earth is about 365.25 days, hence;

T_p=64*365.25\\T_p=23376days

4 0
3 years ago
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LenaWriter [7]

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v = √2×9.8×2.8 = 7.41 m/s

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Answer:

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the net force applied to the car is zero.

3 0
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