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vodka [1.7K]
3 years ago
7

Which of the following could lead to a potentially violent situation?

Physics
2 answers:
beks73 [17]3 years ago
7 0
Getting this answer wrong
harkovskaia [24]3 years ago
5 0
<span><span><span>A. Abusive language<span>
B. All of the answer choices could lead to a potentially violent situation</span><span>
C. Harassment and intimidation</span><span>
D. Physical assault or threats of assault</span></span> 

The answer should be A</span></span>
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Unpolarized light of intensity Io is incident on a stack of 7 polarizing filters, each with its axis rotated 17°cw with respect
mash [69]

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

5 0
2 years ago
Consider again the objects you ranked by distance. Suppose each object emitted a burst of light right now. Rank the objects from
Vika [28.1K]

Answer:

Following are the solution to this question:

Explanation:

That light takes a very long time to hit the planet, and the object is far off the earth. The light of such an item near to the planet takes less time to enter it. The star is 2,5 million light-years from the Planet on the far side of the Andromeda Galaxy. But on the other hand, the moon is 15 crore miles from the earth, so sunlight is quickly reached on the ground as the other thing.  

That milky way away from the earth is 66,500 light-years far, that distance between Earth and Orion nebula is 1,344 light-years, with such a distance of 4,367 light-years. The earth is 5.2261 trillion km apart from Pluto.

5 0
3 years ago
Olivia put a glass of water in the freezer. She left it there for three hours. When she returned, the water had turned to ice. W
Darya [45]
A. freezing, when water turns to ice the water is turning from a liquid to a solid.
7 0
3 years ago
NO LINKS PLEASE
Trava [24]

Answer:

2

Explanation:

6 0
3 years ago
You are standing a distance of 17.0 meters from the center of a merry-go-round. The merry-go-round takes 9.50 seconds to go comp
GenaCL600 [577]

Answer:

(a)11.24 m/s

(b)7.44 m/s

(c)409 N

(d)539.55\mu

(e) 0

Explanation:

The period for 1 circle 2\pi of the merry go around is 9.5s. It means the angular speed is:

\omega = \theta / t = 2\pi / 9.5 \approx 0.661 rad/s

(a)The speed is

v = \omega * R = 0.661 * 17 = 11.24 m/s

(b) Centripetal acceleration:

a = \frac{v^2}{R} = \frac{11.24^2}{17} = 7.44 m/s^2

(c) Magnitude of the force that keeps you go around at this acceleration

F = ma = 55 * 7.44 = 409 N

(d) let the coefficient of friction by \mu. The frictional force shall be this coefficient multiplied by normal force reverting gravity of the man

F_f = mg\mu = 55*9.81\mu = 539.55\mu

3 0
3 years ago
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