Explanation:
It is given that,
Mass of the rim of wheel, m₁ = 7 kg
Mass of one spoke, m₂ = 1.2 kg
Diameter of the wagon, d = 0.5 m
Radius of the wagon, r = 0.25 m
Let I is the the moment of inertia of the wagon wheel for rotation about its axis.
We know that the moment of inertia of the ring is given by :


The moment of inertia of the rod about one end is given by :

l = r


For 6 spokes, 
So, the net moment of inertia of the wagon is :


So, the moment of inertia of the wagon wheel for rotation about its axis is
. Hence, this is the required solution.
It stops accelerating when the air resistance is equal to its weight.
That's (m•g)
= (2 kg) • (9.8 m/s^2)
= 19.6 newtons
The boy's momentum is 160 kg*m/s north.
The formula of momentum is p = mv, where p is momentum.
p = 40 kg * 4m/s north
p =160 kg*m/s north<span>Thank you for posting your question. I hope you found what you were after. Please feel free to ask me more.</span>
The strength of the electric and magnetic fields there is no physical "distance" of oscillation here. nothing is actually moving up and down if you draw light as a sinusoidal wave, the up and down motion is the strength of the EM fields cheers