Answer with Explanation:
We are given that
Diameter=d=22.6 cm
Mass,m=426 g=
1 kg=1000 g
Radius,r=
1m=100 cm
Height,h=5m

a.By law of conservation of energy






Where 
b.Rotational kinetic energy=
Rotational kinetic energy=8.35 J
Answer:
The end of the neutral rod which is the closest part to the charged rod would acquire a negative charge.
Explanation:
One of the rods is positively charged and one of them is neutral.
And the important part is that <u>they do not touch one another</u>, but get close to each other.
In this case, the end of the neutral rod which is the closest part to the charged rod would acquire a negative charge. This is because of the Coulomb's Law. The opposite charges exert an attractive force to each other. The positive charges attract the negative charges on the neutral rod.
Answer:

Explanation:
As we know that the resultant of two vectors is given as

here we know that



now we have



Answer:
Propels in the opposite direction
Explanation: