12-46 할리데이 11판 솔루션 일반물리학

$$ \begin{cases} m&=95\ut{kg}\\ L&=15\ut{m}\\ 2R&=9.6\ut{mm}\\ \Delta L &= 1.3\ut{cm}\\ \end{cases} $$ $${F\over A}=E{\Delta L\over L},$$ $$ \begin{aligned} E&={F\over A}\cdot{L\over\Delta L}\\ &={mgL\over \pi R^2\Delta L}\\ &=\frac{185546875000 g}{39 \pi }\\ &\approx 1.4851141642012506\times10^{10}\ut{N/m^2}\\ &\approx 1.5\times10^{10}\ut{N/m^2}\\ &\approx 15\ut{GN/m^2}\\ \end{aligned} $$