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Ilya [14]
3 years ago
9

A vertically polarized beam of light of intensity 100 W/m2 passes through two ideal polarizers. The transmission axis of the fir

st polarizer makes an angle of 20.0° with the vertical, and the transmission axis of the second one makes an angle of 40.0° with the vertical. What is the intensity of the light after it has passes through both polarizers?
Physics
1 answer:
TEA [102]3 years ago
7 0

To solve the problem it is necessary to apply the Malus Law. Malus's law indicates that the intensity of a linearly polarized beam of light, which passes through a perfect analyzer with a vertical optical axis is equivalent to:

I=I_0 cos^2\theta

Where,

I_ {0} indicates the intensity of the light before passing through the polarizer,

I is the resulting intensity, and

\theta indicates the angle between the axis of the analyzer and the polarization axis of the incident light.

Since we have two objects the law would be,

I=I_0cos^2\theta_1*cos^2(\theta_2-\theta_1)

Replacing the values,

I=100*cos^2(20)*cos^2(40-20)

I=100*cos^4(20)

I=77.91W/m^2

Therefore the intesity of the light after it has passes through both polarizers is 77.91W/m^2

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algol13

Answer:

g₂ = 11 m/s²

Explanation:

The value of free-fall acceleration on the surface of a planet is given by the following formula:

g = \frac{Gm}{r^2}

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g = free-fall acceleration

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FOR PLANET 1:

g_1 = \frac{Gm_1}{r_1^2}\\\\\frac{Gm_1}{r_1^2} = 22 m/s^2 --------------------- equation (1)

FOR PLANET 2:

g_2 = \frac{Gm_2}{r_2^2}\\\\g_2 = \frac{G(2m_1)}{(2r_1)^2}\\\\g_2 = \frac{1}{2}\frac{Gm_1}{r_1^2}\\\\

using equation (1):

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Answer:

v = 11.0 m/s at 198.6° (18.6° south of west)

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Explanation:

I assume you want to find the final velocity and the change in kinetic energy.

Take east to be +x and north to be +y.

Momentum is conserved in the x direction:

(1050 kg) (0 m/s) + (750 kg) (-25.0 m/s) = (1050 kg + 750 kg) vₓ

vₓ = -10.4 m/s

Momentum is conserved in the y direction:

(1050 kg) (-6.00 m/s) + (750 kg) (0 m/s) = (1050 kg + 750 kg) vᵧ

vᵧ = -3.50 m/s

The magnitude of the final velocity is:

v² = (-10.4 m/s)² + (-3.50 m/s)²

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The direction of the final velocity is:

θ = atan(-3.50 m/s / -10.4 m/s)

θ = 198.6°

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KE₀ = 253,275 J

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KE = 108,682 J

The change in kinetic energy is:

ΔKE = 108,682 J − 253,275 J

ΔKE ≈ -145,000 J

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