A nylon string on a tennis racket is under a tension of 285 N . If its diameter is 1.10 mm , by how much is it lengthened from its untensioned length of 29.0 cm ? Use ENylon=5.00×109N/m2.

1.74×10⁻³ m

Explanation:

Applying,

ε = Stress/strain Equation 1

Where ε = Young's modulus

But,

Stress = F/A Equation 2

Where F = Force, A = Area

Strain = e/L Equation 3

e = extension, L = Length.

Substitute equation 2 and 3 into equation 1

ε = (F/A)/(e/L) = FL/eA Equation 4

From the question,

Given: F = 285 N, L = 29 cm = 0.29 m, ε = 5.00×10⁹ N/m²,

A = πd²/4 = 3.14(0.0011²)/4 = 9.4985×10⁻⁶ m²

Substitute these values into equation 4

5.00×10⁹ = (285×0.29)/(9.4985×10⁻⁶×e)

Solve for e

e = (285×0.29)/(5.00×10⁹×9.4985×10⁻⁶)

e = 82.65/4.74925×10⁴

e = 1.74×10⁻³ m

6 cos 34, 6 sin 34 trig