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
Answer:
A (2066,6 N)
Explanation:
Use the Work formula
62.000J = F . 30
62.000/30 = 2066,6 N
The amout of time it took to move the rock doesn´t matter at all.
It is called a distraction variable, We don´t need it to solve the problem it is there just to confuse.
Answer:
1.35 kJ
Explanation:
KE = ½mv² = ½ × 0.030 kg × (300 m·s⁻¹)² = 1350 J = 1.35 kJ