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IrinaK [193]
3 years ago
13

Which of the following weathering processes involves the constant freezing and thawing of water?

Physics
2 answers:
muminat3 years ago
7 0

Answer: frost wedging

Explanation:

Weathering is a process of disintegration of rocks into parts or pieces due to the affect of physical and chemical agents. Physical agent includes wind, pressure, gravity and temperature. Chemical agents include water, carbon dioxide, acid rain and living organisms.

Frost wedging is a process of physical weathering. In this the rapid heating and freezing of the rock takes place due to the fluctuations in the weather conditions of a region. On cooling the water present in the rocks get solidified as a result of this the ice being solid in nature occupies more space in the rocks. This produces disturbance in the internal structure of the rocks. This leads to the disintegration of the rocks.

katrin2010 [14]3 years ago
6 0
I think it is spheroidal weathering
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Which of the following describes one of Galileo Galilei's contributions to the development of the concept of gravity?
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Answer:

The correct option is;

c. He conducted experiments with balls and inclined planes

Explanation:

The inclined plane experiment conducted by Galileo Galilei was one of his most important experiments.

The experiment consisted of a grooved board with which a small metal ball is rolled down in, so as to analyze the held Aristotelian ideas about motion.

In the experiment, the focus was on acceleration, a concept that the previously held ideas about motion

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I need help with this.
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A.The new discovery might cause the theory to be dismissed since it contradicts it.

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D.The new discovery might further deny the theory since it presents evidence that contradicts it.

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3 years ago
The temperature of water in a beaker is 45°C. What does this measurement represent?
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A certain car traveling 33.0mph skids to a stop in 39m from the point where the brakes were applied. In approximately what dista
Greeley [361]

Answer: 156.02 metre.

Explanation:

Give that a certain car traveling 33.0mph skids to a stop in 39m from the point where the brakes were applied.

Let us use third equation of motion,

V^2 = U^2 + 2as

Since the car is decelerating, V = 0

And acceleration a will be negative.

U = 33 mph

S = 39 m

Substitute both into the formula

0 = 33^2 - 2 × a × 39

0 = 1089 - 78a

78a = 1089

a = 1089 / 78

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If we assume that the car decelerate at the same rate.

the distance the car will stop had it been going 66.0mph will be achieved by using the same formula

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Therefore, the car would stop at

156.02 m

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3 years ago
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