Answer to Question #105083 in Mechanics | Relativity for Roopali jha

Question #105083
A suture is stretched to 10%. When the stress is released it recovered 35 % of its strain after 5 hr. What is the retardation time in hours. What is the amount of strain recovered (in %) after 6 hrs of its original value.
1
Expert's answer
2020-03-10T11:36:58-0400
"\\frac{l}{l_0}(t)=1-e^{-\\frac{t}{\\tau}}"

1)


"\\frac{l}{l_0}(5)=1-e^{-\\frac{5}{\\tau}}=0.35"

"\\tau=11.6\\ h"

2)


"\\frac{l}{l_0}(6)=1-e^{-\\frac{6}{11.6}}=0.40"

The amount of strain recovered after 6 hrs is 40% of its original value.


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