According to young modulus equation
E=FL/AΔE=FL/A\DeltaE=FL/AΔL
Where F is applied Force, L is length of the iron bar, A is cross sectional area of the iron bar and
ΔL\Delta LΔL is shorten length
therefore,
ΔL=FL/AE\Delta L=FL/AEΔL=FL/AE
ΔL=2500∗20.006∗1.83∗1011\Delta L=\frac{2500*2}{0.006*1.83 * 10^{11}} \newlineΔL=0.006∗1.83∗10112500∗2
ΔL=4.5537μm\Delta L=4.5537\mu mΔL=4.5537μm
Need a fast expert's response?
and get a quick answer at the best price
for any assignment or question with DETAILED EXPLANATIONS!
Comments