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erica [24]
3 years ago
12

calculate the diameter of a silver wire of length 75cm , which is extended by 1.85mm when a 10kg mass is suspended from it's end

. Young modulus for silver is 7.9×10^10N/m²​
Physics
1 answer:
sdas [7]3 years ago
4 0

Answer:0.8\ mm

Explanation:

Given

length of wire l=75\ cm

change in length \Delta l=1.85\ mm

mass of wire m=10\ kg

Young's modulus for silver E=7.9\times 10^{10}\ N/m^2

load on wire F=mg

F=10\times 9.8=98\ kg

change in length is given by

\Delta l=\dfrac{Pl}{AE}

Where A=area of cross-section

A=\dfrac{Pl}{\Delta lE}

A=\dfrac{98\times 0.75}{1.85\times 10^{-3}\times 7.9\times 10^{10}}

A=\dfrac{73.5}{14.615\times 10^{7}}

A=5.029\times 10^{-7}\ m^2

also wire is the shape of cylinder so cross-section is given by

A=\dfrac{\pi d^2}{4}=5.029\times 10^{-7}\ m^2

\Rightarrow d^2=\dfrac{5.029\times 10^{-7}\times 4}{\pi }

\Rightarrow d^2=64.02\times 10^{-8}

\Rightarrow d=8\times 10^{-4}\ m

\Rightarrow d=0.8\ mm

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The potential across the capacitor at t = 1.0 seconds, 5.0 seconds, 20.0 seconds respectively is mathematically given as

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<h3>What is the potential across the capacitor?</h3>

Question Parameters:

A 1. 0 μf capacitor is being charged by a 9. 0 v battery through a 10 mω resistor.

at

  • t = 1.0 seconds
  • 5.0 seconds
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Generally, the equation for the Voltage is mathematically given as

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For t=20s

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V2=4.323v

Therefore, the values of voltages at the various times are

  • t=0.476v
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  • V2=4.323v

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