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mezya [45]
4 years ago
5

A cylindrical Nickel rod (9 mm diameter, 50 m long) is pulled in tension with a load of 6,283 N. What would the elongation of th

e rod be under this load?
Physics
1 answer:
Norma-Jean [14]4 years ago
6 0

Answer:

0.29 m

Explanation:

9 mm = 0.009 m in diameter

Cross-sectional area A = \pi d^2/4 = \pi * 0.009^2/4 = 6.36\times 10^{-5} m^2

Let the tensile modulus of Nickel E = 170 \times 10^9Pa.

The elongation of the rod can be calculated using the following formula:

\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

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