Answer:
502.2 kPa
Explanation:
Stress at any instance, t can be given by
[tex]\sigma (t)=\epsilon_o Ee^{-(t/\tau)}[/tex]
Strain, [tex]\epsilon=\frac {\triangle l}{l}=\frac {5-2}{2}=1.5[/tex]
[tex]E=\frac {Stress}{Strain}=\frac {2}{1.5}=1.333333\approx 1.33[/tex]
Therefore, after 2 hours where t=2 then
[tex]\sigma (2)=1.5\times 1.33 Mpa e^{-(2/1.45)}\approx 502.2 KPa[/tex]
