explanation
a=average velocity/average time
average velocity=0.0+1.2+2.4+3.6/4
average velocity=7.2/4
average velocity=1.8 m/s
average time=0.0+3.0+6.0+9.0/4
average time=18/4
average time=4.5 s
a= average velocity/average time
a=1.8/4.5
a=0.4 m/s²
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:
brainly.com/question/12830237
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The electrons in the positive object are attracted to the negatively charged object. Some of the electrons move from the positive object to the negative object.
an example of this is lightning because the positive electrons on/in the earth are attracted to the negative electrons in the clouds and sky so the positive move to the negative charge.
Answer:
5/2π
Explanation:
According to quizlet the answer is 5/2π