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.
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Answer:
Explanation:
Difference in the measurement by shuttle crew Δ d = 238857 - 226316 miles
= 12541 miles
Actual measurement = 238857 miles
percentage error
= (12541 / 238857 ) x 100
= 5.25 %
Answer:
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Answer:
C. changing nuclear energy to radiant energy
Explanation:
Nuclear energy takes atoms in their potential state, split them (fission) or fuse them (fusion) creating chain reactions of radiant energy. Most nuclear electrical power plants use fission, radiant energy heats water making steam to spin turbines.
Or think of the atom bomb. Definitely potential energy until the fuse starts detonation and chain reactions. The radiant kinetic energy and shock waves were horrendous.