An alpha particle with kinetic energy 10.0MeV makes a collision with lead nucleus, but it is not aimed" at the center of the lead nucleus, and has an initial nonzero angular momentum (with respect to the stationary lead nucleus) of magnitude L pob where po is the magnitude of the initial momentum of the alpha particle and b 1.30x10 12m (Assume that the lead nucleus remains stationary and that it may be treated as a point charge. The atomic number of lead is 82. The alpha particle is a helium nucleus, with atomic number What is the distance of closest approach?

Ithink it is letter b, i may be wrong but i think that is the answer

W= fd
where f is an applied force and d is the distance over which it is applied. when a parachuting skydiver is at constant velocity, he is experiencing "terminal velocity", the point at which his acceleration down is canceled out by air resistance. that means that the resisting force of the air is equal to his weight, but in the opposite direction. with this, we can solve for the work done by the force.
w = fd
w = (92.0)(9.81 m/s^2) (325)
w= 2.93 x 10^5 j
the value for work in this case is positive because it is pushing up on the skydiver, who has a downward force of gravity.
therefore the answer is 2.93 x 10^5 j