Two particles are fixed to an x axis: particle 1 of charge −1.50 ✕ 10−7 c at x = 6.00 cm, and particle 2 of charge +1.50 ✕ 10−7 c at x = 27.0 cm. midway between the particles, what is their net electric field? (express your answer in vector form.) e = −244897.96i correct: your answer is correct. n/c

You have been called to testify as an expert witness in a trial involving a head-on collision. car a weighs 1515 lb and was traveling eastward. car b weighs 1125 lb and was traveling westward at 44.0 mph. the cars locked bumpers and slid eastward with their wheels locked for 22.5 ft before stopping. you have measured the coefficient of kinetic friction between the tires and the pavement to be 0.750 . how fast (in miles per hour) was car a traveling just before the collision? (this problem uses english units because they would be used in a u. s. legal proceeding.)

Suppose that an electromagnetic wave which is linearly polarized along the x−axis is propagating in vacuum along the z−axis. the wave is incident on a conductor which is placed at z > 0 region of the space. the conductor has conductivity σ, magnetic permeability µ and electric permittivity ε. (a) find the characteristic time for the free charge density which dissipates at the conductor. (b) write the maxwell equations and derive the wave equation for a plane wave propagating in a conductor. (c) find the attenuation distance at which the incident amplitude reduces to e ^−1 of its initial value. (d) find the electric and magnetic fields inside the conductor. 8 (e) find the power loss per area of the incident electromagnetic wave at the surface of conductor.

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).