Let's look at the details of eddy-current braking. a square loop, length l on each side, is shot with velocity v0 into a uniform magnetic field b. the field is perpendicular to the plane of the loop. the loop has mass m and resistance r, and it enters the field at t = 0s. assume that the loop is moving to the right along the x-axis and that the field begins at x = 0m.

Astone dropped into a well is heard to strike the water after 8seconds. find the depth of the well if the velocity of sound is 350m/s

Distinguish between a transverse wave and a longitudinal wave.

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).