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Bezzdna [24]
3 years ago
12

Explain how a single piece of polarizing film can be used to adjust the intensity of a beam of light..

Physics
1 answer:
masya89 [10]3 years ago
6 0

Answer:

Sunlight and almost every other form of natural and artificial illumination produces light waves whose electric field vectors vibrate in all planes that are perpendicular with respect to the direction of propagation.

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Why is pure oxygen stored as a liquid under pressure
yKpoI14uk [10]
<h2>Answer: It is highly flammable.</h2>

Explanation:

Liquid oxygen is created from oxygen atoms that have been forced to assume the liquid state due to <u>compression (change of pressure) and temperature modification. </u>

Specifically this is achieved by cooling the oxygen enough to change it to its liquid state. So,<u> as the temperature drops, the atoms move more slowly because they have less energy. </u>

In this sense, in the liquid state it is easier to store and mobilize oxygen, taking into account that it is a highly flammable gas.

3 0
3 years ago
Which theory of the origin of the moon would be most supported if the moon had an irregular shape, like an asteroid? impact theo
Fiesta28 [93]
<span>Its the impact theory.
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3 0
3 years ago
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Which of the following will cause an increase in gas pressure in a closed container?
Shalnov [3]
The answer is A. Or the first option. Pressure is changed by lowering the pressure, not reducing the volume. You would assume its C but its A. 
5 0
2 years ago
The ocean may pull away rom the shore as a tsunami approaches?
ale4655 [162]
Uh if theres a tsunami it probably will.
6 0
2 years ago
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Multiple-Concept Example 7 and Interactive LearningWare 26.1 provide some helpful background for this problem. The drawing shows
Vlada [557]

Answer:

n = 1.4266

Explanation:

Given that:

refractive index of crystalline slab n = 1.665

let refractive index of fluid is n.

angle of incidence θ₁ = 37.0°

Critical angle \theta _c = sin^{-1} (\frac{n}{n_{slab}} )

sin \theta _ c =\frac{n}{n_{slab}}

According to Snell's law of refraction:

n sin \theta _1 = n_{slab}  \ sin \  (90- \theta_c)

At point P ; 90 - \theta _2  \leq \theta _c

\theta _2 = 90 - \theta _c

Therefore:

n \ sin \theta_1 = n_{slab} \sqrt{(1-sin^2 \theta _c)}  \\ \\ n \ sin \theta_1 = n_{slab}   \sqrt{(1- \frac{n}{n_{slab}} )}

Then maximum value of refractive index  n of the fluid is:

n = \frac{n_{slab}}{\sqrt{1+ sin^2 \theta _1 } }

n = \frac{1.665}{\sqrt{1+ sin^2 \  37} }

n = 1.4266

3 0
2 years ago
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