Divers change their body position in midair while rotating about their center of mass. in one dive, the diver leaves the board with her body nearly straight, then tucks into a somersault position.
if the moment of inertia of the diver in a straight position is 14 kg⋅m2 and in a tucked position is 4.0 kg⋅m2 , by what factor is her angular velocity when tucked greater than when straight?
express your answer using two significant figures.

An alpha particle with kinetic energy 12.5 mev 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 = p0 b, where p0 is the magnitude of the initial momentum of the alpha particle and b=1.40×10−12 m. (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 2.) what is the distance of closest approach?