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Anna [14]
2 years ago
8

This model shows an example of a fog bank formation. This can happen in the Great Lakes area as warm summer air moves across the

water and cools down quickly. Which statement explains how this happens?
A) The cooled air becomes warm again after passing over the lake.

B) The moisture in the cooled air condenses, forming a thick fog.

C) The cooled air forms thick snow clouds that resemble dense fog.

D) The air surrounding the lake rises, making clouds break into fog.
Physics
1 answer:
Vadim26 [7]2 years ago
5 0

Fog bank formation occurs when the moisture in the cooled air condenses, forming a thick fog.

<h3>How Fog banks formed?</h3>

Fog banks form at sea where cool air moves quickly over the surface of the ocean that is warm. The cool incoming air lowers the temperature of the air just above the water surface and water vapor condenses into fog.

So we can conclude that Fog bank formation occurs when the moisture in the cooled air condenses, forming a thick fog.

Learn more about fog here: brainly.com/question/18943608

#SPJ1

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b) See the proof below

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Explanation:

Part a

For this case we have the following differential equation:

\frac{dA}{dt}= kA

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We can rewrite the differential equation like this:

\frac{dA}{A} =k dt

And if we integrate both sides we got:

ln |A|= kt + c_1

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A_o = C

So then our solution for the differential equation is given by:

A(t) = A_o e^{kt}

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\frac{1}{2} A_o = A_o e^{kt}

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Applying natural log we have this:

ln (\frac{1}{2}) = kt

And then the value of t would be:

t = \frac{ln (1/2)}{k}

And using the fact that ln(1/2) = -ln(2) we have this:

t = -\frac{ln(2)}{k}

Part b

For this case we need to show that the solution on part a can be written as:

A(t) = A_o 2^{-t/T}

For this case we have the following model:

A(t) = A_o e^{kt}

If we replace the value of k obtained from part a we got:

k = -\frac{ln(2)}{T}

A(t) = A_o e^{-\frac{ln(2)}{T} t}

And we can rewrite this expression like this:

A(t) = A_o e^{ln(2) (-\frac{t}{T})}

And we can cancel the exponential with the natural log and we have this:

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Part c

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\frac{A_o}{8}= A_o 2^{-\frac{t}{T}}

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t = T \frac{ln(8)}{ln(2)}

We can rewrite this expression like this:

t = T \frac{ln(2^3)}{ln(2)}

Using properties of natural logs we got:

t = 3T \frac{ln(2)}{ln(2)}= 3T

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Hopes this helps!

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