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=2.0×1011 pascals. how far ( δl) would such a string stretch under a tension of 1500 newtons?

15mm

Step-by-step explanation:

we are looking for extension

To get Extension you need the original length and the strain both of which you are given

initial length L = 1.00m

the area A = 0.5mm² = 0.5 mm² = 0.5 x 10⁻⁶ m² ( we are changing to metres squared)

E = 2.0 x 10¹¹ n/m², Young's modulus

P = 1500N, the applied tension

Now to Calculate the stress.

σ = P/A (force/area) = (1500 N)/(0.5 x 10⁻⁶ m²) = 3 x 10⁹ N/m²

Also, Let β =  the stretch of the string.

Then the strain is

ε = β/L (extension/ original length)

By definition, the strain is ε = σ/E = (3 x 10⁹ N/m²)/(2 x 10¹¹ N/m²) = 0.015

Therefore β/(1 m) = 0.015β = 0.015 m = 15 mm

 15 mm

step-by-step explanation:

mean or average is 40.6