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Artemon [7]
3 years ago
9

Alternate freezing and thawing often leads to ______.

Physics
2 answers:
Evgen [1.6K]3 years ago
8 0
Alternate freezing and thawing often leads to :
A. Creep
this occurrence usually occurs when the soils in that specific region is subjected to the temperature below 0 Celcius

hope this helps
WITCHER [35]3 years ago
7 0

The answer from the given options is "A. Creep".


Alternate freezing and thawing often leads to "creep".


Creep refers to the low mass development of soil and soil material down slants, driven fundamentally by gravity yet facilitated by immersion with water and by substitute freezing and thawing. Generally not distinguishable with the exception of through expanded perception.

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Answer:

25.632 m²

Explanation:

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4 0
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dangina [55]

1) Medium "b" has more optical density

2) Light must hit the interface between the two mediums perpendicularly

Explanation:

1)

Refraction occurs when light propagates from a medium into a second medium.

The optical density of a medium is given by its index of refraction, which is defined as:

n=\frac{c}{v}

where

c is the speed of light in a vacuum

v is the speed of light in a medium

Higher index of refraction means higher optical density, and light propagater slower into a medium with higher optical density.

In this problem, light propagates faster through medium "a" than medium "b": this means that medium "a" has lower refractive index of medium "b", and so "b" has more optical density.

2)

We can answer this part by referring to Snell's law, which gives the relationship between the direction of the incident ray and of the refracted ray when light passes through the interface between two media:

n_1 sin \theta_1 = n_2 sin \theta_2

where

n_1, n_2 are the index of refraction of the two mediums

\theta_1, \theta_2 are the angle of incidence and of refraction (the angle that light makes with the normal to the surface in medium 1 and medium 2)

Here we want the direction of propagation of the light ray not to change: this means that it must be

sin \theta_1 = sin \theta_2 (1)

However, here we have two mediums "a" and "b" with different index of refraction, so

n_1\neq n_2

Therefore the only angle that can satisfy eq.(1) is

\theta_1 = \theta_2 = 0

So, the light must hit the surface perpendicular to the interface between the two mediums.

Learn more about refraction:

brainly.com/question/3183125

brainly.com/question/12370040

#LearnwithBrainly

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Explanation:

I just did this exact unit in bio last week I hope this helps ;)

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