Is drifting around in space
The ratio of the intensity between light intensity that emerges from the last filter and unpolarized light of intensity, I₀ is It/I₀ = 0.2925
To answer the question, we need to know what polarization of light is.
<h3>What is polarization of light?</h3>
This is when the electric field vector of light is oscillating in one plane.
- Now for light of intensity I' which is initially unpolarized, its intensity after polarization is I = 1/2I'.
- Also, for light initially polarized, its intensity after polarization is I"' = I"cos²Ф where Ф is the angle between the initial direction and the direction of polarization.
<h3>Intensity of light through each polarized filter</h3>
Given that we have 7 polarizing filters, each rotated 17° cw with respect to the previous filter.
So, since the light is initially unpolarized,
- The intensity through the first polarizing filter is I₁ = 1/2I₀ where I₀ is the initial intensity.
- The intensity through the second polarizing filter is I₂ = I₁cos²17°= 1/2I₀cos²17°
- The intensity through the third polarizing filter is I₃ = I₂cos²17° = 1/2I₀cos⁴17°
- The intensity through the fourth polarizing filter is I₄ = I₃cos²17° = 1/2I₀cos⁶17°
- The intensity through the fifth polarizing filter is I₅ = I₄cos²17° = 1/2I₀cos⁸17°
- The intensity through the sixth polarizing filter is I₆ = I₅cos²17° = 1/2I₀cos¹⁰17°
- The intensity through the seventh polarizing filter is I₇ = I₆cos²17° = 1/2I₀cos¹²17°.
<h3>The ratio of the intensity between light intensity that emerges from the last filter and unpolarized light of intensity</h3>
Since I₇ is the last intensity I₇ = It = 1/2I₀cos¹²17°.
So, It/I₀ = 1/2cos¹²17°
= 1/2(0.9563)¹²
= 1/2 × 0.5850
= 0.2925
So, the ratio of the intensity between light intensity that emerges from the last filter and unpolarized light of intensity, I₀ is It/I₀ = 0.2925
Learn more about intensity of polarized light here:
brainly.com/question/25402491
Answer:
37.5 N Hard
Explanation:
Hook's law: The force applied to an elastic material is directly proportional to the extension provided the elastic limit of the material is not exceeded.
Using the expression for hook's law,
F = ke.............. Equation 1
F = Force of the athlete, k = force constant of the spring, e = extension/compression of the spring.
Given: k = 750 N/m, e = 5.0 cm = 0.05 m
Substitute into equation 1
F = 750(0.05)
F = 37.5 N
Hence the athlete is pushing 37.5 N hard
Answer:
28.7%
Explanation:
efficiency = work output /work input × 100
<span>95 km/h = 26.39 m/s (95000m/3600 secs)
55 km/h = 15.28 m/s (55000m/3600 secs)
75 revolutions = 75 x 2pi = 471.23 radians
radius = 0.80/2 = 0.40m
v/r = omega (rad/s)
26.39/0.40 = 65.97 rad/s
15.28/0.40 = 38.20 rad/s
s/((vi + vf)/2) = t
471.23 /((65.97 + 38.20)/2) = 9.04 secs
(vf - vi)/t = a
(38.20 - 65.97)/9.04 = -3.0719
The angular acceleration of the tires = -3.0719 rad/s^2
Time is required for it to stop
(0 - 38.20)/ -3.0719 = 12.43 secs
How far does it go?
65.97 - 38.20 = 27.77 M</span>