To show that L A L B = α A α B .

Question

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Answer 1

Question

Chapter 20, Problem 52P

(a)

To determine

To show that LALB=αAαB .

(a)

Expert Solution

Explanation of Solution

Introduction:

The difference between the two lengths are constant if the length LA and LB is constant. The coefficient of linear expansions are αA and αB .

Write expression for change in length.

ΔL=LB−LA

Write expression for linear expansion.

ΔL=(LB+αBLBΔT)−(LA+αALAΔT)

Solve above expression.

ΔL=(LB−LA)+(αBLB−αALA)ΔT

Substitute ΔL for LB−LA in above expression and solve.

ΔL=ΔL+(αBLB−αALA)ΔT0=(αBLB−αALA)ΔT0=(αBLB−αALA)LALB=αBαA

Conclusion:

Thus, the equation is proved.

(b)

To determine

The value of LB .

(b)

Expert Solution

Explanation of Solution

Given:

The material A is brass.

The material B is steel.

The value of LA is 250 cm .

Formula used:

Write expression for change in length.

ΔL=LB−LA

Write expression for linear expansion.

ΔL=(LB+αBLBΔT)−(LA+αALAΔT)

Solve above expression.

ΔL=(LB−LA)+(αBLB−αALA)ΔT

Substitute ΔL for LB−LA in above expression and solve.

ΔL=ΔL+(αBLB−αALA)ΔT0=(αBLB−αALA)ΔT0=(αBLB−αALA)

Solve above expression.

LALB=αBαA ..... (1)

Calculation:

Substitute 250 cm for LA , 19×10−6K-1 for αA and 11×10−6 K-1 for αB in equation (1).

250 cmLB=11× 10 −6 K -119× 10 −6K -1LB=430 cm

Conclusion:

Thus, the value of LB is 430 cm .

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Answer 2

Question

Chapter 20, Problem 52P

(a)

To determine

To show that LALB=αAαB .

(a)

Expert Solution

Explanation of Solution

Introduction:

The difference between the two lengths are constant if the length LA and LB is constant. The coefficient of linear expansions are αA and αB .

Write expression for change in length.

ΔL=LB−LA

Write expression for linear expansion.

ΔL=(LB+αBLBΔT)−(LA+αALAΔT)

Solve above expression.

ΔL=(LB−LA)+(αBLB−αALA)ΔT

Substitute ΔL for LB−LA in above expression and solve.

ΔL=ΔL+(αBLB−αALA)ΔT0=(αBLB−αALA)ΔT0=(αBLB−αALA)LALB=αBαA

Conclusion:

Thus, the equation is proved.

(b)

To determine

The value of LB .

(b)

Expert Solution

Explanation of Solution

Given:

The material A is brass.

The material B is steel.

The value of LA is 250 cm .

Formula used:

Write expression for change in length.

ΔL=LB−LA

Write expression for linear expansion.

ΔL=(LB+αBLBΔT)−(LA+αALAΔT)

Solve above expression.

ΔL=(LB−LA)+(αBLB−αALA)ΔT

Substitute ΔL for LB−LA in above expression and solve.

ΔL=ΔL+(αBLB−αALA)ΔT0=(αBLB−αALA)ΔT0=(αBLB−αALA)

Solve above expression.

LALB=αBαA ..... (1)

Calculation:

Substitute 250 cm for LA , 19×10−6K-1 for αA and 11×10−6 K-1 for αB in equation (1).

250 cmLB=11× 10 −6 K -119× 10 −6K -1LB=430 cm

Conclusion:

Thus, the value of LB is 430 cm .

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Answer 3

Question

Chapter 20, Problem 52P

(a)

To determine

To show that LALB=αAαB .

(a)

Expert Solution

Explanation of Solution

Introduction:

The difference between the two lengths are constant if the length LA and LB is constant. The coefficient of linear expansions are αA and αB .

Write expression for change in length.

ΔL=LB−LA

Write expression for linear expansion.

ΔL=(LB+αBLBΔT)−(LA+αALAΔT)

Solve above expression.

ΔL=(LB−LA)+(αBLB−αALA)ΔT

Substitute ΔL for LB−LA in above expression and solve.

ΔL=ΔL+(αBLB−αALA)ΔT0=(αBLB−αALA)ΔT0=(αBLB−αALA)LALB=αBαA

Conclusion:

Thus, the equation is proved.

(b)

To determine

The value of LB .

(b)

Expert Solution

Explanation of Solution

Given:

The material A is brass.

The material B is steel.

The value of LA is 250 cm .

Formula used:

Write expression for change in length.

ΔL=LB−LA

Write expression for linear expansion.

ΔL=(LB+αBLBΔT)−(LA+αALAΔT)

Solve above expression.

ΔL=(LB−LA)+(αBLB−αALA)ΔT

Substitute ΔL for LB−LA in above expression and solve.

ΔL=ΔL+(αBLB−αALA)ΔT0=(αBLB−αALA)ΔT0=(αBLB−αALA)

Solve above expression.

LALB=αBαA ..... (1)

Calculation:

Substitute 250 cm for LA , 19×10−6K-1 for αA and 11×10−6 K-1 for αB in equation (1).

250 cmLB=11× 10 −6 K -119× 10 −6K -1LB=430 cm

Conclusion:

Thus, the value of LB is 430 cm .

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Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Answer 4

Question

Chapter 20, Problem 52P

(a)

To determine

To show that LALB=αAαB .

(a)

Expert Solution

Explanation of Solution

Introduction:

The difference between the two lengths are constant if the length LA and LB is constant. The coefficient of linear expansions are αA and αB .

Write expression for change in length.

ΔL=LB−LA

Write expression for linear expansion.

ΔL=(LB+αBLBΔT)−(LA+αALAΔT)

Solve above expression.

ΔL=(LB−LA)+(αBLB−αALA)ΔT

Substitute ΔL for LB−LA in above expression and solve.

ΔL=ΔL+(αBLB−αALA)ΔT0=(αBLB−αALA)ΔT0=(αBLB−αALA)LALB=αBαA

Conclusion:

Thus, the equation is proved.

(b)

To determine

The value of LB .

(b)

Expert Solution

Explanation of Solution

Given:

The material A is brass.

The material B is steel.

The value of LA is 250 cm .

Formula used:

Write expression for change in length.

ΔL=LB−LA

Write expression for linear expansion.

ΔL=(LB+αBLBΔT)−(LA+αALAΔT)

Solve above expression.

ΔL=(LB−LA)+(αBLB−αALA)ΔT

Substitute ΔL for LB−LA in above expression and solve.

ΔL=ΔL+(αBLB−αALA)ΔT0=(αBLB−αALA)ΔT0=(αBLB−αALA)

Solve above expression.

LALB=αBαA ..... (1)

Calculation:

Substitute 250 cm for LA , 19×10−6K-1 for αA and 11×10−6 K-1 for αB in equation (1).

250 cmLB=11× 10 −6 K -119× 10 −6K -1LB=430 cm

Conclusion:

Thus, the value of LB is 430 cm .

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Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Answer 5

Chapter 20, Problem 52P

(a)

To determine

To show that LALB=αAαB .

(a)

Expert Solution

Explanation of Solution

Introduction:

The difference between the two lengths are constant if the length LA and LB is constant. The coefficient of linear expansions are αA and αB .

Write expression for change in length.

ΔL=LB−LA

Write expression for linear expansion.

ΔL=(LB+αBLBΔT)−(LA+αALAΔT)

Solve above expression.

ΔL=(LB−LA)+(αBLB−αALA)ΔT

Substitute ΔL for LB−LA in above expression and solve.

ΔL=ΔL+(αBLB−αALA)ΔT0=(αBLB−αALA)ΔT0=(αBLB−αALA)LALB=αBαA

Conclusion:

Thus, the equation is proved.

(b)

To determine

The value of LB .

(b)

Expert Solution

Explanation of Solution

Given:

The material A is brass.

The material B is steel.

The value of LA is 250 cm .

Formula used:

Write expression for change in length.

ΔL=LB−LA

Write expression for linear expansion.

ΔL=(LB+αBLBΔT)−(LA+αALAΔT)

Solve above expression.

ΔL=(LB−LA)+(αBLB−αALA)ΔT

Substitute ΔL for LB−LA in above expression and solve.

ΔL=ΔL+(αBLB−αALA)ΔT0=(αBLB−αALA)ΔT0=(αBLB−αALA)

Solve above expression.

LALB=αBαA ..... (1)

Calculation:

Substitute 250 cm for LA , 19×10−6K-1 for αA and 11×10−6 K-1 for αB in equation (1).

250 cmLB=11× 10 −6 K -119× 10 −6K -1LB=430 cm

Conclusion:

Thus, the value of LB is 430 cm .

Answer 6

Chapter 20, Problem 52P

(a)

To determine

To show that LALB=αAαB .

(a)

Expert Solution

Explanation of Solution

Introduction:

The difference between the two lengths are constant if the length LA and LB is constant. The coefficient of linear expansions are αA and αB .

Write expression for change in length.

ΔL=LB−LA

Write expression for linear expansion.

ΔL=(LB+αBLBΔT)−(LA+αALAΔT)

Solve above expression.

ΔL=(LB−LA)+(αBLB−αALA)ΔT

Substitute ΔL for LB−LA in above expression and solve.

ΔL=ΔL+(αBLB−αALA)ΔT0=(αBLB−αALA)ΔT0=(αBLB−αALA)LALB=αBαA

Conclusion:

Thus, the equation is proved.

Answer 7

Chapter 20, Problem 52P

(a)

To determine

To show that LALB=αAαB .

Answer 8

(a)

Expert Solution

Explanation of Solution

Introduction:

The difference between the two lengths are constant if the length LA and LB is constant. The coefficient of linear expansions are αA and αB .

Write expression for change in length.

ΔL=LB−LA

Write expression for linear expansion.

ΔL=(LB+αBLBΔT)−(LA+αALAΔT)

Solve above expression.

ΔL=(LB−LA)+(αBLB−αALA)ΔT

Substitute ΔL for LB−LA in above expression and solve.

ΔL=ΔL+(αBLB−αALA)ΔT0=(αBLB−αALA)ΔT0=(αBLB−αALA)LALB=αBαA

Conclusion:

Thus, the equation is proved.

Answer 9

(a)

Expert Solution

Answer 10

(a)

Expert Solution

Answer 11

Expert Solution

Answer 12

(b)

To determine

The value of LB .

(b)

Expert Solution

Explanation of Solution

Given:

The material A is brass.

The material B is steel.

The value of LA is 250 cm .

Formula used:

Write expression for change in length.

ΔL=LB−LA

Write expression for linear expansion.

ΔL=(LB+αBLBΔT)−(LA+αALAΔT)

Solve above expression.

ΔL=(LB−LA)+(αBLB−αALA)ΔT

Substitute ΔL for LB−LA in above expression and solve.

ΔL=ΔL+(αBLB−αALA)ΔT0=(αBLB−αALA)ΔT0=(αBLB−αALA)

Solve above expression.

LALB=αBαA ..... (1)

Calculation:

Substitute 250 cm for LA , 19×10−6K-1 for αA and 11×10−6 K-1 for αB in equation (1).

250 cmLB=11× 10 −6 K -119× 10 −6K -1LB=430 cm

Conclusion:

Thus, the value of LB is 430 cm .

Answer 13

(b)

To determine

The value of LB .

Answer 14

(b)

Expert Solution

Explanation of Solution

Given:

The material A is brass.

The material B is steel.

The value of LA is 250 cm .

Formula used:

Write expression for change in length.

ΔL=LB−LA

Write expression for linear expansion.

ΔL=(LB+αBLBΔT)−(LA+αALAΔT)

Solve above expression.

ΔL=(LB−LA)+(αBLB−αALA)ΔT

Substitute ΔL for LB−LA in above expression and solve.

ΔL=ΔL+(αBLB−αALA)ΔT0=(αBLB−αALA)ΔT0=(αBLB−αALA)

Solve above expression.

LALB=αBαA ..... (1)

Calculation:

Substitute 250 cm for LA , 19×10−6K-1 for αA and 11×10−6 K-1 for αB in equation (1).

250 cmLB=11× 10 −6 K -119× 10 −6K -1LB=430 cm

Conclusion:

Thus, the value of LB is 430 cm .

Answer 15

(b)

Expert Solution

Answer 16

(b)

Expert Solution

Answer 17

Expert Solution

Answer 18

Ch. 20 - Prob. 1P Ch. 20 - Prob. 2P Ch. 20 - Prob. 3P Ch. 20 - Prob. 4P Ch. 20 - Prob. 5P Ch. 20 - Prob. 6P Ch. 20 - Prob. 7P Ch. 20 - Prob. 8P Ch. 20 - Prob. 9P Ch. 20 - Prob. 10P

To show that L A L B = α A α B .

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To show that LALB=αAαB .

Explanation of Solution

The value of LB .

Explanation of Solution

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Chapter 20 Solutions