Young's modulus - calculating delta L

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SUMMARY

The discussion focuses on calculating the elongation (Delta L) of a steel guitar string under tension using Young's modulus. Given a steel string with an initial length of 1.00 meter, a cross-sectional area of 0.500 square millimeters, and a Young's modulus of 2.0 × 1011 pascals, the correct formula to determine the stretch is Delta L = F/k, where k = YA/L. The correct calculation reveals that the string stretches 15.0 mm under a tension of 1500 Newtons, correcting a previous unit conversion error regarding area.

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1. Consider a steel guitar string of initial length L=1.00 meter and cross-sectional area A=0.500 square millimeters. The Young's modulus of the steel is Y=2.0 \times 10^{11} pascals. How far ( Delta L) would such a string stretch under a tension of 1500 Newtons?
Use two significant figures in your answer. Express your answer in millimeters.




2. Y = kL/A
k = YA/L




3. k = YA/L = (2*10^11 N/m^2)*(.0005m^2) / 1m = 1*10^8 N/m

deltaL = F/k = 1500N/(1*10^8 N/m) = 1.5*10^-5m = .015 mm

I made a mistake with units somewhere but I can't spot it.
 
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Remember that when converting from square millimeters to square meters, you have to square the factor that you are changing by.

Since 1mm = 0.001m, 1mm2 = 0.001m * 0.001m = 0.000001m2

Your problem is that you are dividing by 0.0005m2 instead of 0.0000005m2, so your answer is exactly three orders of magnitude too small.
 
Thanks!
 

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