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E=5*10^9


E=Stress/Strain=(F/A)/(delta(L)/initial(L))

therefor, delta(L)/initial(L)=(F/A)/E =(275/(((1*10^-3)/2)*pi)/5*10^9 =0.0700 m


delta(L)/initial(L)=0.0700 m

delta(L)=initial(L)*0.700=0.3*0.0700=0.021 m

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Q: If a nylon string on a tennis racket is under a tension of 275 N If it's diameter is 1.00 mm by how much is it lengthened from its untensioned length of 30.0 cm?
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