A 12.3 kg box is being pulled up a rough surfaced incline which makes an angle of 38° with the horizontal, with a force of 223 N. The coefficient of kinetic friction is 0.325.

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