Answer:
Young-modulus=3.7*10^8N/m^2
Explanation:
We are not given the read position on the scale as this value represent the new length, which is needed to compute the change in length "e"
However, let us use a value of 4m to represent the value on the scale
Step one:
given data
original length=3.5m
diameter d= 0.7mm in meters= 7*10^-4m
Area A= πd^2/4
A= 3.142* (7*10^-4)^2/4
A=3.142*4.9*10^-7/4
A= 3.85×10-7m^2
Force F= 20N.
original length =3.5m
change in length = 4-3.5= 0.5m
Step two:
[tex]young-modulus= \frac{stress}{strain}[/tex]
[tex]stress= \frac{F}{A}[/tex]
[tex]strain= \frac{l}{L}[/tex]
l= change in length = 4-3.5= 0.5m
L= original length =3.5m
solving for the stress
[tex]stress= \frac{20}{3.85*10^-^7}\\\\stress= 5.19*10^7[/tex]
stress= 5.19*10^7N/m^2
solving for strain
[tex]strain= \frac{0.5}{3.5}\\strain= 0.14[/tex]
[tex]Young-modulus= \frac{5.19*10^7}{0.14}\\\\ Young-modulus=3.7*10^8[/tex]
Young-modulus=3.7*10^8N/m^2
