A thin uniform rod (length = 1.2 m, mass = 2.0 kg) is pivoted about a horizontal, frictionless pin through one end of the rod. (The moment of inertia of the rod about this axis is ML2/3.) The rod is released when it makes an angle of 37° with the horizontal. What is the angular acceleration of the rod at the instant it is released (in rad/s^2)?

A: 9.8 rad/s2

B: 7.4 rad/s2

C: 8.4 rad/s2

D: 5.9 rad/s2

E: 6.5 rad/s2

I think the answer is A 9.8rad/s2

electrical potential energy =0.072 j

explanation:

electrical potential energy is given by

u= q δv

q= charge=6 c

δv= potential difference and is given by

δv= e d

e= electric field

d= distance=5 mm=0.005 m

δv=2.4 (0.005)=0.012 v

now electric potential energy= u= q δv

u=6 (0.012)

u=0.072 j

blue light is scattered in different directions by the tiny molecules of air. blue is scattered more than other colours because it has a shorter distance to travel which results in smaller waves. this is why we see a blue sky most of the time.