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PIT_PIT [208]
4 years ago
10

Why do you think the rate of weathering tends to increase at hotter temperatures and higher amounts of rainfall?

Physics
1 answer:
lesya692 [45]4 years ago
7 0

Answer:

Explanation:

Both increase in temperature and availability of water drives physical and chemical weathering and increases their rate.

Increase in temperature causes rocks to expand  and molecules are more free to move about. Constantly heating and cooling of a rock can lead to mechanical fracturing or cracks developing.

A high amount of rainfall facilitates chemical weathering. Chemical weathering is the decomposition of rocks.

The main agent that is important for natural chemical processes to occur is the presence water.

Water contains some other dissolved substances in it.

This is evident in tropical regions of the world

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A proton having an initial velvocity of 20.0i Mm/s enters a uniform magnetic field of magnitude 0.300 T with a direction perpend
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The time interval for which the proton remains in the field is -

Δt = $\frac{\pi R}{40}.

We have a proton entering a uniform magnetic field which is in a direction perpendicular to the proton's velocity.

We have to determine time interval during which the proton is in the field.

<h3>What is the magnitude of force on the charged particle moving in a uniform magnetic field?</h3>

The magnitude of force on the charged particle moving in a uniform magnetic field is given by -

F = qvB sinθ



According to the question, we have -

Entering Velocity (v) = 20 i  m/s

Magnetic field intensity (B) = 0.3 T

Leaving velocity (u) = - 20 j  m/s

Now -

The entering and leaving velocity vectors have 90 degrees difference between them. Therefore, only a quarter of distance of the complete circular path of radius 'R' is traced by the proton. Therefore -

d = $\frac{2\pi r}{4} = $\frac{\pi R}{2}

Since, the radius of circular path is not given, we will assume it R.

Therefore, time for which proton remained in the field is -

t = $\frac{\pi R}{2v} = \frac{\pi R}{40}

Hence, the time interval for which the proton remains in the field is -

Δt = $\frac{\pi R}{40}

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