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.
Solution:
Pressure = Force/Area
P = F/A
P = 1 atm = 1.013*10^5 pascals
From this, the value of weight will be,
F = mg = P * A
= 1.013*10^5 * 4.5
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.
The force perpendicularly applied to an object's surface divided by the area over which it is dispersed is known as pressure. The pressure as compared to the surrounding air pressure is known as gauge pressure. To express pressure, a variety of units are employed.
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