Answer:
The Value is 
Explanation:
The explanation is shown on the first uploaded image
The law of conservation of momentum<span> states that for two objects colliding in an isolated system, the total </span>momentum<span> before and after the collision is equal. Momentum should be conserved. Hope this answers the question. Have a nice day.</span>
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.
A truck is moving with less velocity in the direction in which the truck is moving earlier because the truck has more momentum.
<h3 /><h3>In which direction the truck moves?</h3>
A truck is moving with the velocity of 10 m/s in the same direction in which the truck is moving earlier because the truck has more mass so it has more momentum. Due to collision, the velocity of the truck is slow down but can't be stopped because of high momentum in the truck.
So we can conclude that a truck is moving with less velocity in the direction in which the truck is moving earlier because the truck has more momentum.
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(a) The work done by the force applied by the tractor is 79,968.47 J.
(b) The work done by the frictional force on the tractor is 55,977.93 J.
(c) The total work done by all the forces is 23,990.54 J.
<h3>
Work done by the applied force</h3>
The work done by the force applied by the tractor is calculated as follows;
W = Fd cosθ
W = (5000 x 20) x cos(36.9)
W = 79,968.47 J
<h3>Work done by frictional force</h3>
W = Ffd cosθ
W = (3500 x 20) x cos(36.9)
W = 55,977.93 J
<h3>Net work done by all the forces on the tractor</h3>
W(net) = work done by applied force - work done by friction force
W(net) = 79,968.47 J - 55,977.93 J
W(net) = 23,990.54 J
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