A nylon string on a tennis racket is under tension of 295 N. If its diameter is 1.05 mm, by how much is it lengthened from its untensioned length of 31.0 cm? Use E nylon = 5.00 x 10^9 N/m^2.
Added by Danielle M.
Step 1
Given diameter = 1.05 mm = 1.05 x 10^-3 m Radius = diameter / 2 = 1.05 x 10^-3 m / 2 = 5.25 x 10^-4 m Area = πr^2 = π(5.25 x 10^-4)^2 = 2.17 x 10^-7 m^2 Show more…
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