Answer:
The extent to which it would stretch is [tex]\Delta L = 0.015 \ m[/tex]
Explanation:
From the question we are told that
The initial length is [tex]L = 1.00m[/tex]
The area is [tex]A = 0.500 mm^2 = \frac{0.500}{1 *10^6} = 0.500*10^6 \ m^2[/tex]
The Young modulus of the steel is [tex]Y = 2.0*10^{11} Pa[/tex]
The tension is [tex]T =1500 N[/tex]
The Young modulus is mathematically represented as
[tex]Y = \frac{\sigma}{e}[/tex]
Where [tex]\sigma[/tex] is the stress which is mathematically represented as
[tex]\sigma = \frac{F}{A}[/tex]
Substituting values
[tex]\sigma = \frac{1500}{0.500*10^{-6}}[/tex]
[tex]\sigma = 3.0*10^9 N/m^2[/tex]
And e is the strain which is mathematically represented as
[tex]e = \frac{\Delta L}{L }[/tex]
Where [tex]\Delta L[/tex] The extension of the steel string
Substituting these into the equation above
[tex]Y = \frac{3.0*10^9}{\frac{\Delta L}{L} }[/tex]
Substituting values
[tex]2.0 *10^{11} = \frac{3.0*10^9}{\frac{\Delta L}{L} }[/tex]
[tex]\Delta L = \frac{3.0*10^9 * 1}{2.0 *10^{11}}[/tex]
[tex]\Delta L = 0.015 \ m[/tex]
