What is the elastic modulus of the wire?

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SUMMARY

The elastic modulus of a wire with a diameter of 0.2 mm that stretches by 0.20% under a 6.28-N force is calculated using the formula Elastic Modulus = F/A/(ΔL/L). The correct calculation reveals that the elastic modulus is 1.0 x 1011 N/m2. Common errors included miscalculating the area and misunderstanding the ratio of change in length to original length.

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  • Understanding of basic physics concepts, specifically stress and strain.
  • Familiarity with the formula for elastic modulus: Elastic Modulus = F/A/(ΔL/L).
  • Knowledge of geometric calculations for the area of a circle: A = πr².
  • Ability to convert units, particularly from millimeters to meters.
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  • Study the derivation and applications of the elastic modulus in materials science.
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  • Explore the differences between elastic and plastic deformation in materials.
  • Investigate the impact of temperature and material composition on elastic modulus.
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joseg707
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Homework Statement


A wire of diameter 0.2mm stretches by 0.20% when a 6.28-N force is applied. What is the elastic modulus of the wire?


Homework Equations


Elastic Modulus=F/A/L/\DeltaL


The Attempt at a Solution


r=.1mm=1x10-4m

6.28N/(2\pi1x10-8/.02=5x109N/m2

The right answer is 1.0x1011.

What am I doing wrong?
 
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joseg707 said:

Homework Equations


Elastic Modulus=F/A/L/\DeltaL
That should be (F/A)/(ΔL/L).
 
Oh, yeah. Well how do I calculate that? I know that the change is .2% if I use .002 or .2 as a value for \DeltaL/L I don't get the right answer. What is wrong with my calculation?
 
joseg707 said:
6.28N/(2\pi1x10-8/.02=5x109N/m2
Area = pi*r². Looks like you have an extra factor of two in there. And ΔL/L = 0.002.
 
Ah! Thank you so much! I don't know why I was thinking are was 2*pi*r^2. Thanks a lot!
 

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