The mass required to extend the length of the fiber by 10% needs to be determined. If the Young’s modulus of muscle fiber is 2.80 × 10 7 , the length and diameter of the mass M suspended is 3.25 cm and 0,125 cm respectively. Concept Introduction: The mass of the suspended can be calculated using the young’s modulus is as follows: m = E A o Δ L L o g Here, E is Young’s modulus, A o is area of cross section, Δ L is change in length , L o is length and g is acceleration due to gravity.

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Answer 1

Question

Chapter 2, Problem 2.41NP

Interpretation Introduction

Interpretation: The mass required to extend the length of the fiber by 10% needs to be determined. If the Young’s modulus of muscle fiber is 2.80×107 , the length and diameter of the mass M suspended is 3.25 cm and 0,125 cm respectively.

Concept Introduction: The mass of the suspended can be calculated using the young’s modulus is as follows:

m=EAoΔLLog

Here, E is Young’s modulus, A_o is area of cross section, ΔL is change in length , Lo is length and g is acceleration due to gravity.

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Answer 2

Question

Chapter 2, Problem 2.41NP

Interpretation Introduction

Interpretation: The mass required to extend the length of the fiber by 10% needs to be determined. If the Young’s modulus of muscle fiber is 2.80×107 , the length and diameter of the mass M suspended is 3.25 cm and 0,125 cm respectively.

Concept Introduction: The mass of the suspended can be calculated using the young’s modulus is as follows:

m=EAoΔLLog

Here, E is Young’s modulus, A_o is area of cross section, ΔL is change in length , Lo is length and g is acceleration due to gravity.

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Answer 3

Question

Chapter 2, Problem 2.41NP

Interpretation Introduction

Interpretation: The mass required to extend the length of the fiber by 10% needs to be determined. If the Young’s modulus of muscle fiber is 2.80×107 , the length and diameter of the mass M suspended is 3.25 cm and 0,125 cm respectively.

Concept Introduction: The mass of the suspended can be calculated using the young’s modulus is as follows:

m=EAoΔLLog

Here, E is Young’s modulus, A_o is area of cross section, ΔL is change in length , Lo is length and g is acceleration due to gravity.

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Answer 4

Question

Chapter 2, Problem 2.41NP

Interpretation Introduction

Interpretation: The mass required to extend the length of the fiber by 10% needs to be determined. If the Young’s modulus of muscle fiber is 2.80×107 , the length and diameter of the mass M suspended is 3.25 cm and 0,125 cm respectively.

Concept Introduction: The mass of the suspended can be calculated using the young’s modulus is as follows:

m=EAoΔLLog

Here, E is Young’s modulus, A_o is area of cross section, ΔL is change in length , Lo is length and g is acceleration due to gravity.

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Answer 5

Chapter 2, Problem 2.41NP

Interpretation Introduction

Interpretation: The mass required to extend the length of the fiber by 10% needs to be determined. If the Young’s modulus of muscle fiber is 2.80×107 , the length and diameter of the mass M suspended is 3.25 cm and 0,125 cm respectively.

Concept Introduction: The mass of the suspended can be calculated using the young’s modulus is as follows:

m=EAoΔLLog

Here, E is Young’s modulus, A_o is area of cross section, ΔL is change in length , Lo is length and g is acceleration due to gravity.

Answer 6

Chapter 2, Problem 2.41NP

Interpretation Introduction

Interpretation: The mass required to extend the length of the fiber by 10% needs to be determined. If the Young’s modulus of muscle fiber is 2.80×107 , the length and diameter of the mass M suspended is 3.25 cm and 0,125 cm respectively.

Concept Introduction: The mass of the suspended can be calculated using the young’s modulus is as follows:

m=EAoΔLLog

Here, E is Young’s modulus, A_o is area of cross section, ΔL is change in length , Lo is length and g is acceleration due to gravity.

Answer 7

Chapter 2, Problem 2.41NP

Interpretation Introduction

Interpretation: The mass required to extend the length of the fiber by 10% needs to be determined. If the Young’s modulus of muscle fiber is 2.80×107 , the length and diameter of the mass M suspended is 3.25 cm and 0,125 cm respectively.

Concept Introduction: The mass of the suspended can be calculated using the young’s modulus is as follows:

m=EAoΔLLog

Here, E is Young’s modulus, A_o is area of cross section, ΔL is change in length , Lo is length and g is acceleration due to gravity.

Answer 8

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Answer 9

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Answer 10

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Answer 11

Ch. 2 - Prob. 2.1CP Ch. 2 - Prob. 2.2CP Ch. 2 - Prob. 2.3CP Ch. 2 - Prob. 2.4CP Ch. 2 - Prob. 2.5CP Ch. 2 - Prob. 2.6CP Ch. 2 - Prob. 2.7CP Ch. 2 - Prob. 2.8CP Ch. 2 - Prob. 2.9CP Ch. 2 - Prob. 2.10CP

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