The answer's 25,000 joules
half m v squared
half 2000, x 5 squared
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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Answer:
The mass of Uranium present in a 1.2mg sample is 
Explanation:
The ration between Uranium mass and total sample mass is:
For a sample of mass 1.2 mg, the amount of uranium is:

Answer:
B
Explanation:
Friction always eventually stops whatever is in motion.
higher temp = higher energy = higher frequency = shorter wavelength