Answer:
The period of the resulting oscillatory motion is 0.20 s.
Explanation:
Given that,
Spring constant 
We need to calculate the time period
The object is at rest and has no elastic potential but it does has gravitational potential.
If the object falls then the the gravitational potential change in to the elastic potential.
So,
Where,h = distance
k = spring constant
Put the value into the formula
Using formula of time period
Put the value into the formula
Hence, The period of the resulting oscillatory motion is 0.20 s.
Answer:
0.0327 m
Explanation:
m = 2 kg
ω = 24 rad/s
A = 0.040 m
Let at position y, the potential energy is twice the kinetic energy.
The potential energy is given by
U = 1/2 m x ω² x y²
The kinetic energy is given by
K = 1/2 m x ω² x (A² - y²)
Equate both the energies as according to the question
1/2 m x ω² x y² = 2 x 1/2 m x ω² x (A² - y²)
y² = 2 A² - 2 y²
3y² = 2A²
y² = 2/3 A²
y = 0.82 A = 0.82 x 0.040 = 0.0327 m
Answer:
Explanation:
Total momentum of the system before the collision
.5 x 3 - 1.5 x 1.5 = -0.75 kg m/s towards the left
If v be the velocity of the stuck pucks
momentum after the collision = 2 v
Applying conservation of momentum
2 v = - .75
v = - .375 m /s
Let after the collision v be the velocity of .5 kg puck
total momentum after the collision
.5 v + 1.5 x .231 = .5v +.3465
Applying conservation of momentum law
.5 v +.3465 = - .75
v = - 2.193 m/s
2 ) To verify whether the collision is elastic or not , we verify whether the kinetic energy is conserved or not.
Kinetic energy before the collision
= 2.25 + 1.6875
=3.9375 J
kinetic energy after the collision
= .04 + 1.2 =1.24 J
So kinetic energy is not conserved . Hence collision is not elastic.
3 ) Change in the momentum of .5 kg
1.5 - (-1.0965 )
= 2.5965
Average force applied = change in momentum / time
= 2.5965 / 25 x 10⁻³
= 103.86 N
Answer:
Newton's F=ma, which means the force (F) acting on an object is equal to the mass (m) of an object times its acceleration (a)
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
