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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Thrust is a reaction force described quantitatively by Newton's third law. When a system expels or accelerates mass in one direction, the accelerated mass will cause a force of equal magnitude but opposite direction on that system. Mathematically can be written as,

Here,
v = speed of the exhaust gases measured relative to the rocket.
= Rate of change of mass with respect to time
Our values are given as,


Replacing we have that


If we let
p as the directed multigraph that has no isolated vertices and has an Euler circuit
q as the graph that is weakly connected with the in-degree and out-degree of each vertex equal
The statement we have to prove is
p ←→q (for biconditional)
Since
p → q (assuming that p is strongly connected to q)
q ← p (since p is strongly connected to q)
Therefore, the bicondition is satisfied