Answer
a. $1.4 \times 10^6 N/m^2$
b. $6.9 \times 10^{-6}$
c. $6.6 \times 10^{-5}m$
Work Step by Step
a. The stress is $\frac{F}{A}=\frac{(1700 kg)(9.80 m/s^2)}{0.012m^2}\approx1.4 \times 10^6 N/m^2$
b. The strain is $\frac{stress}{E}=\frac{1388333N/m^2}{200\times 10^9 N/m^2}=6.9 \times 10^{-6}$
c. Use the definition of strain to find the amount by which it is lengthened.
$$\Delta \mathcal{l}=\mathcal{l}_o(strain)$$
$$=(9.50m)(6.9 \times 10^{-6})=6.6 \times 10^{-5}m$$