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
1) 0N... friction opposes the motion of an object, since the block is at rest there is no motion thus no friction
2) F=ma
= (5.5kg)(30m/s)
=165 N
Answer:
Difference in height = 7.5 cm
Explanation:
We are given;.
Height of ethyl alcohol;h2 = 20 cm = 0.2 m
Density of glycerin: ρ1 = 1260 kg/m³
Density of ethyl alcohol; ρ2 = 790 kg/m³
To get the difference in height, the pressure at the top of the open end must be equal to the pressure at the point where the liquids do not mix since both points will be at different levels after the pouring.
Thus;
P1 = P2
Formula for pressure is; P = ρgh
Thus;
ρ1 × g × h1 = ρ2 × g × h2
g will cancel out to give;
ρ1 × h1 = ρ2× h2
Making h1 the subject, we have;
h1 = (ρ2× h2)/ρ1
h1 = (790 × 0.2)/1260
h1 = 0.125 m
Difference in height will be;
Δh = h2 - h1
Δh = 0.2 - 0.125
Δh = 0.075 m = 7.5 cm
Answer: this isnt really helping me
Explanation:
