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.
1 kg=100000 cg
2 kg=200000 cq
If mass is the quantity then kg is the S.I
2 kg=2kg
By Newton's second law, the net force on the object is
∑ <em>F</em> = <em>m</em> <em>a</em>
∑ <em>F</em> = (2.00 kg) (8 <em>i</em> + 6 <em>j</em> ) m/s^2 = (16.0 <em>i</em> + 12.0 <em>j</em> ) N
Let <em>f</em> be the unknown force. Then
∑ <em>F</em> = (30.0 <em>i</em> + 16 <em>j</em> ) N + (-12.0 <em>i</em> + 8.0 <em>j</em> ) N + <em>f</em>
=> <em>f</em> = (-2.0 <em>i</em> - 12.0 <em>j</em> ) N
Answer:
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Explanation:
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Answer:
It covers distance of 9.15 football fields in the said time.
Explanation:
We know that

Thus distance covered in blinking of eye =

Thus no of football fields=