Energy of a random atomic and molecular is called molecules energy
The correct answer is C) the focus
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:
I don't really understand the wording of this question but I'd totally cuddle a Koala! If they weren't so mean
Explanation:
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Answer:
See Explanation Below
Explanation:
This question is incomplete.
I'll answer this question on general terms. You'll get your result if you apply the steps I'll highlight below.
To start with; what's Ampere law;
It states that for any closed loop path, the sum of the length elements times the magnetic field in the direction of the length element is equal to the permeability times the electric current enclosed in the loop.
The simple translation of this law is that, the sum of current in the close loop gives us the desired result.
Rephrasing your question;
Three currents, (I1 = +3A, I2 = +4A and I3 = -5A) are passing through a surface bounded by a closed path. The currents have different values and directions. According to Ampere’s law, what is the value of I on the right side of this equation?
First, we take note of the signs in front of the given currents.
The negative sign in front of I3 means that; it is moving in opposite direction of I1 and I2.
To calculate the value of I.
The value of I is the sum of the three currents:
i.e. 3A + 4A - 5A
I = 2 A