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=p0b, 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? repeat for b=1.50×10−13 m .

Let v1, v2, w be three linearly independent vectors in r 3 . that is, they do not all lie on the same plane. for each of the following (infinite) set of vectors, carefully sketch it in r 3 , and determine whether or not it is a vector space (i.e., a subspace of r 3 ). explain your reasoning

You are given a vector a = 125i and an unknown vector b that is perpendicular to a. the cross-product of these two vectors is a × b = 98k. what is the y-component of vector b?

1.)the isotope cobalt-60 has a nuclear mass of 59.933820 u calculate the mass defect of cobalt-60 using the following information. mass of proton: 1.007825 u mass of neutron: 1.008665 u 1 u = 931.5 mev 2.)the isotope cobalt-60 has a nuclear mass of 59.933820 u calculate the binding energy of cobalt-60 using the following information. mass of proton: 1.007825 u mass of neutron: 1.008665 u 1 u = 931.5 mev 3.)the isotope cobalt-60 has a nuclear mass of 59.933820 u calculate the binding energy per nucleon of cobalt-60 using the following information. mass of proton: 1.007825 u mass of neutron: 1.008665 u 1 u = 931.5 mev