Answer:
1.+160.0567km
2.-0.102km
Step-by-step explanation:
Given initial length L1=160Km
L2=?
Initial temperature=10.2°C
Final temperature=32.1°c
Coefficient of linear expansivity a= 1.62*10^-5per Celsius
1: a=L2-L1/L1*∆T
(L2-160/160*21.9)=1.62*10^-5
L2-160=0.0567
L2= 0.0567+160
L2=160.0567Km
2.∆L=? at -28.7°C
Also a=∆L/L1∆T
We're ∆L= aL1∆T
1.65*10^-5*160*(-28.7-10.2)
=-0.102km