The magnetic moment m⃗ of a current loop is defined as the vector whose magnitude equals the area of the loop times the magnitude of the current flowing in it (m=IA), and whose direction is perpendicular to the plane in which the current flows. Find the magnitude, F, of the force on the loop from Part A in terms of the magnitude of its magnetic moment. Express F in terms of m, I2, a, d, and μ0.

A6,600 kg train car moving at +2.0 m/s bumps into and locks together with one of mass 5,400kg moving at -3.0m/s. what is their final velocity? +0.50 m/s -1.0 m/s +2.5 m/s -0.25 m/s

What is relative velocity? in which condition is relative velocity known by adding the velocity of the first body and that of the second body

Part f - example: finding two forces (part i) two dimensional dynamics often involves solving for two unknown quantities in two separate equations describing the total force. the block in (figure 1) has a mass m=10kg and is being pulled by a force f on a table with coefficient of static friction îľs=0.3. four forces act on it: the applied force f (directed î¸=30â above the horizontal). the force of gravity fg=mg (directly down, where g=9.8m/s2). the normal force n (directly up). the force of static friction fs (directly left, opposing any potential motion). if we want to find the size of the force necessary to just barely overcome static friction (in which case fs=îľsn), we use the condition that the sum of the forces in both directions must be 0. using some basic trigonometry, we can write this condition out for the forces in both the horizontal and vertical directions, respectively, as: fcosî¸â’îľsn=0 fsinî¸+nâ’mg=0 in order to find the magnitude of force f, we have to solve a system of two equations with both f and the normal force n unknown. use the methods we have learned to find an expression for f in terms of m, g, î¸, and îľs (no n).