A 1.53-kg piece of iron is hung by a vertical ideal spring. When perturbed slightly, the system is moves up and down in simple harmonic oscillations with a frequency of 1.95 Hz and an amplitude of 7.50 cm. If we choose the total potential energy (elastic and gravitational) to be zero at the equilibrium position of the hanging iron, what is the total mechanical energy of the system

Neutrons are placed in a magnetic field with magnitude 2.30 t. part a part complete what is the energy difference between the states with the nuclear spin angular momentum components parallel and antiparallel to the field? δe δ e = 2.77×10−7 ev previous answers correct part b part complete which state is lower in energy: the one with its spin component parallel to the field or the one with its spin component antiparallel to the field? which state is lower in energy: the one with its spin component parallel to the field or the one with its spin component antiparallel to the field? parallel antiparallel previous answers correct part c part complete how do your results compare with the energy states for a proton in the same field (δe=4.05×10−7ev)? how do your results compare with the energy states for a proton in the same field this result is smaller than but comparable to that found in the example for protons. this result is greater than but comparable to that found in the example for protons. previous answers correct part d the neutrons can make transitions from one of these states to the other by emitting or absorbing a photon with energy equal to the energy difference of the two states. find the frequency of such a photon. f f = mhz previous answersrequest answer incorrect; try again; 5 attempts remaining