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AlekseyPX
3 years ago
10

Consider a metal bar of initial length L and cross-sectional area A. The Young's modulus of the material of the bar is Y. Find t

he "spring constant" k of such a bar for low values of tensile strain.
Physics
1 answer:
RUDIKE [14]3 years ago
6 0

Answer:

Spring constant = YA / L

Explanation:

Let F be the force being applied on cross sectional area A of metal bar due to which an extension of ΔL is obtained in the wire of length L then

stress = F / A

strain = ΔL /L

Young's modulus = ( F / A ) / (ΔL /L)

Y = ( F L / A ΔL )

( F / ΔL ) = ( YA / L )

Spring constant = ( F / ΔL )

Spring constant = YA / L

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The answer is B increase the resistance of the circuit
5 0
3 years ago
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
sineoko [7]

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,

                 P=\frac{F}{A}

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,

             P=1atm=1.013*10^5Pascals

  • From this, the value of weight will be,

            F=mg=P*A=1.013*10^5*4.5=4.56*10^5N

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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4 0
2 years ago
Two large parallel conducting plates carrying opposite charges of equal magnitude are separated by 2.20 cm. If the surface charg
bekas [8.4K]

Answer:

5.3\times 10^3 N/C

Explanation:

We are given that

Distance between plates=d=2.2 cm=2.2\times 10^{-2} m

1 cm=10^{-2} m

\sigma=47nC/m^2=47\times 10^{-9}C/m^2

Using 1 nC=10^{-9} C

We have to find the magnitude of E in the region between the plates.

We know that the electric field for parallel plates

E=\frac{\sigma}{2\epsilon_0}

E_1=\frac{\sigma}{2\epsilon_0}

E_2=\frac{\sigma}{2\epsilon_0}

E=E_1+E_2

E=\frac{\sigma}{2\epsilon_0}+\frac{\sigma}{2\epsilon_0}=\frac{\sigma}{\epsilon_0}

Where \epsilon_0=8.85\times 10^{-12}C^2/Nm^2

Substitute the values

E=\frac{47\times 10^{-9}}{8.85\times 10^{-12}}

E=5.3\times 10^3 N/C

Hence, the magnitude of E in the region between the plates=5.3\times 10^3 N/C

5 0
3 years ago
What scientist first discovered that white light is a mixture of a rainbow spectrum of light rays
dsp73

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.

7 0
4 years ago
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A sample contains 20 kg of radioactive material. The decay constant of the material is 0.179 per second. If the amount of time t
Nataly [62]

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

4 0
3 years ago
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