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Mariulka [41]
3 years ago
8

When rocks are analyzed over an increasingly greater distance from a mid-ocean ridge, what can be said about the polarities of t

he rocks?
Physics
1 answer:
zubka84 [21]3 years ago
5 0
They are of more, or opposite polaritys
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Describe a technology used in space exploration.
Alex777 [14]

Answer:

High speed optical communication technology

To be able to communicate from the space to the earth and from earth to space is one of the most essential features required during space exploration.

Explanation:

Space exploration involves going into the space, beyond the earth's atmosphere. Landing on other planets and studying their details, going into deeper space beyond the planets to discover new cosmic events or structures is all a part of space exploration.

The key to analyse the studies and observations is being able to communicate the data collected, photos taken etc to the launch centers or space centers on earth. The space centers on earth should also be able to communicate with the persons or the satellites in space.

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Match the indices of refraction with the corresponding effects on the waves.
Yanka [14]

Answer:

B) waves speed up

C) waves bend away from the normal

Explanation:

The index of refraction of a material is the ratio between the  speed of light in a vacuum and the speed of light in that medium:

n=\frac{c}{v}

where

c is the speed of light in a vacuum

v is the speed of light in the medium

We can re-arrange this equation as:

v=\frac{c}{n}

So from this we already see that if the index of refraction is lower, the speed of light in the medium will be higher, so one correct option is

B) waves speed up

Moreover, when light enters a medium bends according to Snell's Law:

n_1 sin \theta_1 = n_2 sin \theta_2

where

n_1 ,n_2 are the index of refraction of the 1st and 2nd medium

\theta_1,\theta_2are the angles made by the incident ray and refracted ray with the normal to the interface

We can rewrite the equation as

sin \theta_2 = \frac{n_1}{n_2}sin \theta_1

So we see that if the index of refraction of the second medium is lower (n_2), then the ratio \frac{n_1}{n_2} is larger than 1, so the angle of refraction is larger than the angle of incidence:

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