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Ksju [112]
3 years ago
5

A spring with spring constant of 26 N/m is stretched 0.22 m from its equilibrium position. How much work must be done to stretch

it an additional 0.12 m
Physics
1 answer:
Evgen [1.6K]3 years ago
8 0

Answer:

1.503 J

Explanation:

Work done in stretching a spring = 1/2ke²

W = 1/2ke²........................... Equation 1

Where W = work done, k = spring constant, e = extension.

Given: k = 26 N/m, e = (0.22+0.12), = 0.34 m.

Substitute into equation 1

W = 1/2(26)(0.34²)

W = 13(0.1156)

W = 1.503 J.

Hence the work done to stretch it an additional 0.12 m = 1.503 J

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The mean free path of a helium atom in helium gas at standard temperature and pressure is 0.2 um.What is the radius of the heliu
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Where:

\lambda=0.2\mu m=0.2(10)^{-6}m

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P=1atm=101.3kPa=101.3(10)^{3}Pa=101.3(10)^{3}J/m^{3} absolute standard pressure

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d=\sqrt{\frac{RT}{\sqrt{2}\pi\lambda N_{A}P}}    (2)

d=\sqrt{\frac{(8.3145J/mol.K)(273.115K)}{\sqrt{2}\pi(0.2(10)^{-6}m)(6.0221(10)^{23}/mol)(101.3(10)^{3}J/m^{3})}}    (3)

d=2.0467(10)^{-10}m    (4)

If the radius is half the diameter:

r=\frac{d}{2}  (5)

Then:

r=\frac{2.0467(10)^{-10}m}{2}  (6)

r=1.0233(10)^{-10}m  (7)

However, we were asked to find this radius in nanometers. Knowing 1nm=(10)^{-9}m:

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Finally:

r=0.10233nm This is the radius of the helium atom in nanometers.

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