Answer:
1411.8 N/m
Explanation:
From Hooke's law;
F= Ke
Where
F= force on the spring
K= force constant
e = extension
But e= 8.50 × 10^-2m
F= weight = 12.0 kg × 10 = 120 N
K = F/e = 120/8.50 × 10^-2
K= 1411.8 N/m
This is in the thermosphere which is at an altitude of 85-520km
Answer:
(a) Elongation of the rod==5.61×10⁻⁹m
(b) Change in diameter=1.640×10⁻⁸m
Explanation:
Given data
Diameter d=78 in=1.9812 m
Cross Area is:

Applied Load P=17 KN=17×10³N
E=29 × 106 psi=1.99947961×10¹¹Pa
Stress and Strain in x direction
Stress in x direction
σ=P/A

σ=5517.25 Pa
Strain in x direction
ε=σ/E

ε=2.76×10⁻⁸
Part (a)
Elongation of the rod=Lε
=(0.2032)(2.76×10⁻⁸)
Elongation of the rod==5.61×10⁻⁹m
Part(b) Change in diameter
Strain in y direction
ε₁= -vε
ε₁= -(0.30)(2.76×10⁻⁸)
ε₁=-8.28×10⁻⁹
Change in diameter=d×ε₁
Change in diameter=(1.9812m)×(-8.28×10⁻⁹)
Change in diameter=1.640×10⁻⁸m
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