The length of the spring 2 due to arrangement of the equal masses is determined as 0.238 m.
<h3>Extension of the spring 2</h3>
The extension of the spring 2 is determined from the net force acting on spring 2.
F(net) = Kx
<em>Net force on spring 2 = sum of forces acting downwards - sum of forces acting upwards.</em>
Net force on spring 2 = (force on 2 + force on 3) - (force on 1)
F(net) = (m₂ + m₃)g - (m₁)g
F(net) = (6.67 + 6.67)9.8 - (6.67)9.8
F(net) = 130.732 - 65.366
F(net) = 65.366 N
x = F/k
x = (65.366)/(8130)
x = 0.00804 m
The length of spring 2 = x + 0.23 m
= 0.00804 m + 0.23 m = 0.238 m
Thus, the length of the spring 2 due to arrangement of the equal masses is determined as 0.238 m.
Learn more about extension of elastic material here: brainly.com/question/26838571
Answer:
Molybdenum and seaborgium
Given:
u = 6.5 m/s, initial velocity
a = 1.5 m/s², acceleration
s = 100.0 m, displacement
Let v = the velocity attained after the 100 m displacement.
Use the formula
v² = u² + 2as
v² = (6.5 m/s)² + 2*(1.5 m/s²)*(100 m) = 342.25 (m/s)²
v = 18.5 m/s
Answer: 18.5 m/s
The smallest level of ordination in living things. Every living thing is made up of atoms. Atoms are made up of protons, neutrons, and electrons.
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
