Physical Chemistry Plus Mastering Chemistry With Etext -- Access Card Package (3rd Edition) (engel Physical Chemistry Series)
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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, Ao is area of cross section, ΔL is change in length , Lo is length and g is acceleration due to gravity.

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Physical Chemistry Plus Mastering Chemistry With Etext -- Access Card Package (3rd Edition) (engel Physical Chemistry Series)

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