The moment of inertia of a point mass about an arbitrary point is given by:
I = mr²
I is the moment of inertia
m is the mass
r is the distance between the arbitrary point and the point mass
The center of mass of the system is located halfway between the 2 inner masses, therefore two masses lie ℓ/2 away from the center and the outer two masses lie 3ℓ/2 away from the center.
The total moment of inertia of the system is the sum of the moments of each mass, i.e.
I = ∑mr²
The moment of inertia of each of the two inner masses is
I = m(ℓ/2)² = mℓ²/4
The moment of inertia of each of the two outer masses is
I = m(3ℓ/2)² = 9mℓ²/4
The total moment of inertia of the system is
I = 2[mℓ²/4]+2[9mℓ²/4]
I = mℓ²/2+9mℓ²/2
I = 10mℓ²/2
I = 5mℓ²
Answer:
Explanation:
Magnitude of force per unit length of wire on each of wires
= μ₀ x 2 i₁ x i₂ / 4π r where i₁ and i₂ are current in the two wires , r is distance between the two and μ₀ is permeability .
Putting the values ,
force per unit length = 10⁻⁷ x 2 x i x 2i / ( 6 x 10⁻³ )
= .67 i² x 10⁻⁴
force on 3 m length
= 3 x .67 x 10⁻⁴ i²
Given ,
8 x 10⁻⁶ = 3 x .67 x 10⁻⁴ i²
i² = 3.98 x 10⁻²
i = 1.995 x 10⁻¹
= .1995
= 0.2 A approx .
2 i = .4 A Ans .
Answer:
c. Case iii
Explanation:
the ball will experience the largest change in case iii
where 
is the gravitational constant and is about 
:D
A) 8.11 m/s
For a satellite orbiting around an asteroid, the centripetal force is provided by the gravitational attraction between the satellite and the asteroid:
where
m is the satellite's mass
v is the speed
R is the radius of the asteroide
h is the altitude of the satellite
G is the gravitational constant
M is the mass of the asteroid
Solving the equation for v, we find
where:
Substituting into the formula,
B) 11.47 m/s
The escape speed of an object from the surface of a planet/asteroid is given by
where:
Substituting into the formula, we find:
