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
<span>(M G H)=(0.5 x 9.8 x 10) = 49 joules.</span>
Answer:
there are many types of family but main is 2
Explanation:
1). joint family 2). nucleic family joint family has mother father grandfather grandmother uncle aunty children. and nucliec family has mother father and childrens
Answer: 1175 J
Explanation:
Hooke's Law states that "the strain in a solid is proportional to the applied stress within the elastic limit of that solid."
Given
Spring constant, k = 102 N/m
Extension of the hose, x = 4.8 m
from the question, x(f) = 0 and x(i) = maximum elongation = 4.8 m
Work done =
W = 1/2 k [x(i)² - x(f)²]
Since x(f) = 0, then
W = 1/2 k x(i)²
W = 1/2 * 102 * 4.8²
W = 1/2 * 102 * 23.04
W = 1/2 * 2350.08
W = 1175.04
W = 1175 J
Therefore, the hose does a work of exactly 1175 J on the balloon
