Coastal mountains force cool, moist air from oceans to rise as it moves toward land. Clouds form, and precipitation falls on

10 answers
Question:

Coastal mountains force cool, moist air from oceans to rise as it moves toward land. Clouds form, and precipitation falls on the side of the mountain. On the side of the mountain, very little precipitation falls.

Answers

C. Ice sheets make it hard for people to meet their needs.

Explanation: Antarctica is what one could compare to a frozen desert. It is dead and barren and because of the ice sheets that make up nearly all of Antarctica, almost nothing can grow.

A permanent settlement wouldn't even be worth setting up, other than studying the wildlife.

The answers that apply;

They increase precipitation on the windward sides of mountain rangesThey force cool, moist air from oceans to rise as they move toward land

The air above the oceans in coastal areas become humid as the sun heats up the oceans in the day. The oceans, however, remain cooler than the land because the water has a high heat capacity. Therefore, as air on land rises (due to lower density), it is replaced by the cooler air mass from the oceans  (this is called a sea breeze). When the cool humid air encounters mountains inland, it is forced to rise and it gets even cooler. The moisture in the air condenses and precipitates in the windward side of the mountain.

answer:

c

explanation:

b

Explanation:

Coastal mountains force cool, moist air from oceans to rise as it moves toward land. Clouds form, and precipitation falls on the windward side of the mountain. On the leeward side of the mountain, very little precipitation falls.

D: They force cool, moist air from oceans to rise as it moves toward land.

Explanation:

It's correct

I would say B

Explanation:

Mountains, Piedmont, and costal plains

Explanation:

Option 4, They force cool, moist air from oceans to rise as it moves toward land.

I'm not sure how to explain, I'm sorry! I hope you got your answer though :)

Have a good day!

-Emma

snow

Explanation:

Since the process undergoes adiabatic expansion, hence q = 0 and ΔU = w.

We can sole this problem using the following derivation:

[tex]ln(\frac{T_2}{T_1} )=-(\gamma -1)ln(\frac{V_f}{V_i} )=-(\gamma -1)ln(\frac{T_2}{T_1}\frac{P_i}{P_f} )\\=-(\gamma -1)ln(\frac{T_2}{T_1})-(\gamma -1)ln(\frac{P_i}{P_f})\\=-(\frac{\gamma -1}{\gamma})ln(\frac{P_i}{P_f})\\=-(\frac{\frac{C_{p,m}}{C_{p,m}-R} -1}{\frac{C_{p,m}}{C_{p,m}-R}})ln(\frac{P_i}{P_f})\\\\ln(\frac{T_2}{T_1} )==-(\frac{\frac{C_{p,m}}{C_{p,m}-R} -1}{\frac{C_{p,m}}{C_{p,m}-R}})ln(\frac{P_i}{P_f})\\\\Substituting\ values:\\\\[/tex]

[tex]ln(\frac{T_2}{T_1} )=-(\frac{\frac{28.86}{28.86-8.314} -1}{\frac{28.86}{28.86-8.314}})ln(\frac{0.802\ atm}{0.602\ atm})=-0.0826\\\\ln(\frac{T_2}{T_1} )=-0.0826\\\\Taking\ exponential\ of\ both \ sides:\\\\\frac{T_2}{T_1} =e^{-0.0826}\\\\T_2=0.9207T_1\\\\T_2=0.9207*288\\\\T_2=265\ K\\[/tex]

Since T2 = 265 K, we should expect a snow

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