Answer:
0.29 m
Explanation:
9 mm = 0.009 m in diameter
Cross-sectional area [tex]A = \pi d^2/4 = \pi * 0.009^2/4 = 6.36\times 10^{-5} m^2[/tex]
Let the tensile modulus of Nickel [tex]E = 170 \times 10^9Pa[/tex].
The elongation of the rod can be calculated using the following formula:
[tex]\Delta L = \frac{F L}{A E} = \frac{6283*50}{6.36\times 10^{-5} * 170 \times 10^9} = \frac{314150}{1081200} = 0.29 m[/tex]
