Answer: A is Compression and B is Rarefaction.
Explanation:
i think it's right. hope it helps.
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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<span>c the pattern of the magnetic fields lines</span>
Stamens are the <span>part of the flower is the male reproductive organ.</span>
Answer:
0.11 m/s
Explanation:
From the question given above, the following data were obtained:
Initial displacement (d1) = 1.09 m
Final displacement (d2) = 2.55 m
Time (t) = 12.8 s
Average velocity =?
Next, we shall determine the total displacement (i.e change in displacement). This can be obtained as follow:
Initial displacement (d1) = 1.09 m
Final displacement (d2) = 2.55 m
Total displacement = d2 – d1
Total displacement = 2.55 – 1.09
Total displacement = 1.46 m
Finally, we shall determine the average velocity of the beetle. This can be obtained as follow:
Total Displacement = 1.46 m
Total time (t) = 12.8 s
Average velocity =?
Average velocity = Total Displacement / Total time
Average velocity = 1.46/12.8
Average velocity = 0.11 m/s
Thus, the average velocity of the beetle is 0.11 m/s