The answer is B increase the resistance of the circuit
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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Answer:

Explanation:
We are given that
Distance between plates=d=2.2 cm=


Using 
We have to find the magnitude of E in the region between the plates.
We know that the electric field for parallel plates





Where 
Substitute the values


Hence, the magnitude of E in the region between the plates=
Newton also understood that white light can be separated into its components because each ray of color is deviated by the glass of the prism by a different amount. He realized, for example, that red light is consistently less deviated than violet light.
Answer:
Explanation:
Given
N0 = 20kg (original substance)
decay constant λ = 0.179/sec
time t = 300s
We are to find N(t)
Using the formula;
n(t) = N0e^-λt
Substitute the given values
N(t) = 20e^-(0.179)(300)
N(t) = 20e^(-53.7)
N(t) = 20(4.7885)
N(t) =143.055
To know how much of the original material that is active, we will find N(t)/N0 = 143.055/20 = 7.152
About 7 times the original material is still radioactive