Since Young's modulus is E=FSΔllE =\frac {\frac {F} {S}} {\frac {\Delta l} {l}}E=lΔlSF , where FFF is the force of action on the body, SSS is the cross-sectional area, lll is the body length ,Δl\Delta lΔl is the extension, then
F=E×S×Δll=20×1011×π×(1.6×10−32)2×10−36≈670.21F=\frac{E\times S \times \Delta l}{l}=\frac{20\times10^{11}\times \pi\times (\frac{1.6\times 10^{-3}}{2})^2 \times 10^{-3}}{6}\approx670.21F=lE×S×Δl=620×1011×π×(21.6×10−3)2×10−3≈670.21 N.
Need a fast expert's response?
and get a quick answer at the best price
for any assignment or question with DETAILED EXPLANATIONS!
Comments