Answer:
string stretched is 1.02 cm
Explanation:
given data
length = 80-cm
diameter = 1.0-mm
tension = 2000 N
solution
we get here string stretched that will be as and here
we know that young modulus for steel = 200 × [tex]10^{9}[/tex]
so here stress will be
stress = y × strain .............1
that is express as
[tex]\frac{force}{area} = \frac{Y \Delta L}{L}[/tex]
ΔL = [tex]\frac{0.80*2000}{\pi * 0.0005^2*200*10^9}[/tex]
ΔL = 0.0102 m
ΔL = 1.02 cm
so string stretched is 1.02 cm
The distance through which the string was stretched is equal to 1.1 cm.
Given the following data:
Young modulus for steel = [tex]200 \times 10^9[/tex]
Radius = [tex]\frac{Diameter}{2} = \frac{0.1}{2} = 0.05 \;cm[/tex]
To determine the distance through which the string was stretched:
Mathematically, stress is given by the formula:
[tex]Stress = \frac{Force }{Area}[/tex] ....equation 1.
We would relate the tensional stress experienced by the steel guitar string to its strain as follows:
[tex]Stress = Y \times strain\\\\\frac{Force }{Area}=Y \times \frac{\Delta L}{L} \\\\\frac{Force }{Area}=\frac{Y\Delta L}{L}\\\\\Delta L = \frac{Force \; \times\; L}{Area\; \times \;Y }[/tex]....equation 2.
Also, the area of a guitar string is given by the formula:
[tex]Area = \pi r^2 \\\\Area = 3.142 \times 0.05^2\\\\Area = 3.142 \times 0.0025[/tex]
Area = 0.007 [tex]cm^2[/tex]
Substituting the parameters into eqn. 2, we have:
[tex]\Delta L = \frac{2000 \times 80}{0.07 \times 200 \times 10^9} \\\\\Delta L = \frac{160000}{14 \times 10^9}\\\\\Delta L = 1.1 \; cm[/tex]
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