Answer:
I = I₀ + M(L/2)²
Explanation:
Given that the moment of inertia of a thin uniform rod of mass M and length L about an Axis perpendicular to the rod through its Centre is I₀.
The parallel axis theorem for moment of inertia states that the moment of inertia of a body about an axis passing through the centre of mass is equal to the sum of the moment of inertia of the body about an axis passing through the centre of mass and the product of mass and the square of the distance between the two axes.
The moment of inertia of the body about an axis passing through the centre of mass is given to be I₀
The distance between the two axes is L/2 (total length of the rod divided by 2
From the parallel axis theorem we have
I = I₀ + M(L/2)²
Answer:
Explanation:
When a standing wave is formed with six loops means the normal mode of the wave is n=6, the frequency of the normal mode is given by the expression:
Where 
is the length of the string and 
the velocity of propagation. Use this expression to find the value of .
The velocity of propagation is given by the expression:
Where 
is the desirable variable of the problem, the linear mass density, and 
is the tension of the cord. The tension is equal to the weight of the mass hanging from the cord:
With the value of the tension and the velocity you can find the mass density:
Answer:
So coefficient of kinetic friction will be equal to 0.4081
Explanation:
We have given mass of the block m = 0.5 kg
The spring is compressed by length x = 0.2 m
Spring constant of the sprig k = 100 N/m
Blocks moves a horizontal distance of s = 1 m
Work done in stretching the spring is equal to 
This energy will be equal to kinetic energy of the block
And this kinetic energy must be equal to work done by the frictional force
So 
So coefficient of kinetic friction will be equal to 0.4081
Answer:
Explanation:
separation between two gaps, d = 5 cm
angle between central and second order maxima, θ = 0.52°
use
d Sinθ = n λ
n = 2
0.05 x Sin 0.52° = 2 x λ
λ = 2.27 x 10^-4 m
λ = 226.9 micro metre
