Question

. If the stress produced in stretching a wire is 5.00 x 10 6 N/m2 by an applied force of

10.0 N, what is the cross-sectional area of the wire?  

Answer 1

$$\sigma=\frac{F}{A}$$

$$A=\frac{F}{\sigma}=\frac{10.0\;\rm N}{5.00*10^6\:\rm N/m^2}\\ =2*10^{-6}\:\rm m^2=2\: mm^2$$