What is 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 atmosp
1 answer:
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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Option c) 1.5 V
Explanation:
<em>As the circuit is build in series first we will find the current passing through the complete circuit. Current stays the same in each element is the series cirucuit, however, the voltage is different.</em>
Voltage is given by the following formula:
V = IR
<em>Because we have to find current through whole circuit, we will first find resistance of the whole circuit.</em>
Equivalent Resistance R(eq): R1 + R2 = 60 + 60 = 120 ohm
Current passing through whole circuit be:
= 0.025
Now we will find out the voltage between C and D:
Current stays the same in series circuit: I = 0.025 c
Resistance between C and D is, R = 60 ohm
Voltage becomes, V = IR = 0.025 * 60 = 1.5 V