As per the question the uncharged pith ball is taken towards the electrostatic field of a charged rod.
It is told that the side of the pith ball is closest to the negatively charged end of the rod .
The pith ball was initially neutral.The electrons in the atom were oriented in their respective orbits.
Whenever it is taken towards the negative end of the rod,the electrons present in the ball will not stick to their orbit .Basically the closest electrons will suffer electrostatic repulsive force and will come out their orbits.They will move towards the other end of the pith ball.Hence the closest side of the pith ball will acquire positive charges and other side will be negative.
The bound charges nearest to the charged rod is called induced positive charges and bound negative charges present at the other end are called free bound charges.The induced charges and free charges are equal in amount.Hence the pith ball is neutral.
Hence from above we see that only option B is right.
Formula for velocity:
V(t) = Vi + at
Where V(t) is velocity at time t, Vi is initital velocity, and a is acceleration.
Solving for a:
V(8) = Vi + a(8)
V(8) = 0 (car has stopped)
0 = 24 + 8a
-24 = 8a
a = -3 m/s/s
Answer:
Her angular velocity when tucked is greater than when straight by a factor of 0.23
Explanation:
Moment of inertia (I) = mr^2 = mv^2/w^2
m is mass of the diver
v is diver's linear velocity
w is her angular velocity
When straight, I = 14 kg.m^2
mv^2/w^2 = 14
w^2 = mv^2/14
w = sqrt(mv^2/14) = 0.27sqrt(mv^2)
When tucked, I = 4 kg.m^2
w^2 = mv^2/4
w = sqrt(mv^2/4) = 0.5sqrt(mv^2)
Her angular velocity when tucked is greater than when straight by 0.23 (0.5 - 0.27 = 0.23)
