Answer:
It will become a temporary magnet because the domains will easily realign.
Explanation:
Answer:
The answer is I=70,513kgm^2
Explanation:
Here we will use the rotational mechanics equation T=Ia, where T is the Torque, I is the Moment of Inertia and a is the angular acceleration.
When we speak about Torque it´s basically a Tangencial Force applied over a cylindrical or circular edge. It causes a rotation. In this case, we will have that T=Ft*r, where Ft is the Tangencial Forge and r is the radius
Now we will find the Moment of Inertia this way:
-> 
Replacing we get that I is:
Then 
In case you need to find extra information, keep in mind the Moment of Inertia for a solid cylindrical wheel is:
Answer:
K_a = 8,111 J
Explanation:
This is a collision exercise, let's define the system as formed by the two particles A and B, in this way the forces during the collision are internal and the moment is conserved
initial instant. Just before dropping the particles
p₀ = 0
final moment
p_f = m_a v_a + m_b v_b
p₀ = p_f
0 = m_a v_a + m_b v_b
tells us that
m_a = 8 m_b
0 = 8 m_b v_a + m_b v_b
v_b = - 8 v_a (1)
indicate that the transfer is complete, therefore the kinematic energy is conserved
starting point
Em₀ = K₀ = 73 J
final point. After separating the body
Em_f = K_f = ½ m_a v_a² + ½ m_b v_b²
K₀ = K_f
73 = ½ m_a (v_a² + v_b² / 8)
we substitute equation 1
73 = ½ m_a (v_a² + 8² v_a² / 8)
73 = ½ m_a (9 v_a²)
73/9 = ½ m_a (v_a²) = K_a
K_a = 8,111 J
