The weight of a column of air with cross-sectional area 4. 5 m^2 extending from earth's surface to the top of the atmosphere is, 4.56*10^5N.
To find the answer, we have to know about the pressure.
<h3>How to find the weight of a column of air?</h3>
- As we know that the expression of pressure as,

where; F is the force, here it is equal to the weight of the air column, and A is the area of cross section.
- It is given that, the air column is extending from earth's surface to the top of the atmosphere, thus, the pressure will be atmospheric pressure,

- From this, the value of weight will be,

Thus, we can conclude that, the weight of a column of air with cross-sectional area 4. 5 m^2 extending from earth's surface to the top of the atmosphere is, 4.56*10^5N.
Learn more about the pressure here:
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The minimum amount of force needed to move the ball is 15N
Same time
..................
Answer: Induced emf is given by:
ε= Bvl
On putting the values we get
=5×10
−5
×1.50×2
=0.15mV
Explanation:
Hope these helped have a wonderful Christmas break ❄️
The answer is D.nitrogen. the atmosphere is mostly made up of nitrogen.(95% is nitrogen) and (2% <span>oxygen</span>) and and (3% of other gases)