Physics for Scientists and Engineers, Technology Update (No access codes included)
Physics for Scientists and Engineers, Technology Update (No access codes included)
9th Edition
ISBN: 9781305116399
Author: Raymond A. Serway, John W. Jewett
Publisher: Cengage Learning
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Chapter 17, Problem 17.59AP

(a)

To determine

The speed of the one dimensional compression wave.

(a)

Expert Solution
Check Mark

Answer to Problem 17.59AP

The speed of the one dimensional compression wave is 5.04×103m/s .

Explanation of Solution

Given info: The young’s modulus of steel is 20×1010N/m2 , the density of the steel is 7.86×103kg/m3 , the yield strength of the steel is 400MPa , the length of the rod is 80.0cm and the speed of the rod is 12.0m/s .

Write the expression to calculate the speed of the one dimensional compression wave.

v=Yρ

Here,

Y is the young’s modulus of steel.

ρ is the density of the steel.

Substitute 20×1010N/m2 for Y and 7.86×103kg/m3 for ρ in above expression.

v=20×1010N/m27.86×103kg/m3=5.04×103m/s

Conclusion:

Therefore the speed of the one dimensional compression wave is 5.04×103m/s .

(b)

To determine

The time interval of the wave.

(b)

Expert Solution
Check Mark

Answer to Problem 17.59AP

The time interval of the wave is 1.59×104s .

Explanation of Solution

Given info: The young’s modulus of steel is 20×1010N/m2 , the density of the steel is 7.86×103kg/m3 , the yield strength of the steel is 400MPa , the length of the rod is 80.0cm and the speed of the rod is 12.0m/s .

Write the expression to calculate time interval of the wave.

t=Lv

Here,

L is the length of the rod.

Substitute 5.04×103m/s for v and 80.0cm for L in above expression.

t=80.0cm×1m100cm5.04×103m/s=1.59×104s

Conclusion:

Therefore the time interval of the wave is 1.59×104s .

(c)

To determine

The distance of the travel by the back end of the rod.

(c)

Expert Solution
Check Mark

Answer to Problem 17.59AP

The distance of the travel by the back end of the rod is 1.90×103m .

Explanation of Solution

Given info: The young’s modulus of steel is 20×1010N/m2 , the density of the steel is 7.86×103kg/m3 , the yield strength of the steel is 400MPa , the length of the rod is 80.0cm and the speed of the rod is 12.0m/s .

The expression for the distance of the travel by the back end of the rod.

d=vrt

Here,

vr is the speed of the rod.

Substitute 12.0m/s for vr and 1.59×104s for t in above expression.

d=12.0m/s×1.59×104s=1.90×103m

Conclusion:

Therefore the distance of the travel by the back end of the rod is 1.90×103m .

(d)

To determine

The strain in the rod.

(d)

Expert Solution
Check Mark

Answer to Problem 17.59AP

The strain in the rod is 2.38×103 .

Explanation of Solution

Given info: The young’s modulus of steel is 20×1010N/m2 , the density of the steel is 7.86×103kg/m3 , the yield strength of the steel is 400MPa , the length of the rod is 80.0cm and the speed of the rod is 12.0m/s .

The expression for the strain in the rod.

ε=ΔLL

Here,

ΔL is the change in the length.

Substitute 1.90×103m for ΔL and 80.0cm for L in above expression.

ε=1.90×103m80.0cm×1m100cm=2.38×103

Conclusion:

Therefore the strain in the rod is 2.38×103 .

(e)

To determine

The stress in the rod.

(e)

Expert Solution
Check Mark

Answer to Problem 17.59AP

The stress in the rod is 4.76×108N/m2 .

Explanation of Solution

Given info: The young’s modulus of steel is 20×1010N/m2 , the density of the steel is 7.86×103kg/m3 , the yield strength of the steel is 400MPa , the length of the rod is 80.0cm and the speed of the rod is 12.0m/s .

The expression for the stress in the rod is

σ=Yε

Substitute 20×1010N/m2 for Y and 2.38×103 for ε in above expression.

σ=20×1010N/m2×2.38×103=4.76×108N/m2

Conclusion:

Therefore the stress in the rod is 4.76×108N/m2 .

(f)

To determine

The maximum impact speed of the rod in terms of σy , Y and ρ .

(f)

Expert Solution
Check Mark

Answer to Problem 17.59AP

The maximum impact speed of the rod in terms of σy , Y and ρ is σyρY .

Explanation of Solution

Given info: The young’s modulus of steel is 20×1010N/m2 , the density of the steel is 7.86×103kg/m3 , the yield strength of the steel is 400MPa , the length of the rod is 80.0cm and the speed of the rod is 12.0m/s .

The expression for time is,

t=Lv

Substitute Yρ for v in above expression.

t=LYρ

Thus the time is LYρ .

The expression for change in length is,

ΔL=σyLY

The expression for the maximum impact speed of the rod is,

vmaxi=ΔLt

Substitute σyLY for ΔL and LYρ for t in above expression.

vmaxi=σyLYLYρ=σyρY

Conclusion:

Therefore the maximum impact speed of the rod in terms of σy , Y and ρ is σyρY .

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Chapter 17 Solutions

Physics for Scientists and Engineers, Technology Update (No access codes included)

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