標題:
AL physics mechanics & E-field
此文章來自奇摩知識+如有不便請留言告知
發問:
1. A spaceship burns and moves with constant acceleration in a straight line. How would be the graph best represents the variation of its momentum p with time (p-t graph)2. Two bodies X & Y are moving with constant velocities in the directions indicated by the arros. At time t=0, they are at the positions... 顯示更多 1. A spaceship burns and moves with constant acceleration in a straight line. How would be the graph best represents the variation of its momentum p with time (p-t graph) 2. Two bodies X & Y are moving with constant velocities in the directions indicated by the arros. At time t=0, they are at the positions shown. At time t = 2s, the magnitude of the vleocity of Y relative to X is? http://i213.photobucket.com/albums/cc101/Glamorous_Premonition/88b5c232.jpg 3. A potential difference V is applied between two large parallel plates, distance s apart. An electron of mass m & charge -e starts from rest at the negative plate and travels across the gap to the positive plate. THe time taken is? 4. A builder using a pulley system to lift a bucket of cement of weight 150N exerts a steady force F and pulls 30m of rope through the system in order to raise the bucket 10m. The friction in the system is small but NOT negligible. Why is the value of the force F is most probably between 50 N and 150 N? Please explain these questions Thanks!
最佳解答:
1. Assume the mass of fuel burnt is small compare with the spaceship, momentum p increases with velocity v.Since v = at, where a is the acceleration and t is time takenp = mv = matthe graph of p against t is a straight line starting at the origin with +ve slope ma.2. Velocity of Y relative to that of X, Vr = Vy - Vx where Vy and Vx are the velocity vectors of Y ad X respectively Using right angle triangle, Vr = square-root[4^2 + 6^2] m/s = 7.21 m/s 3. Force on electron = e(V/s)acceleration of electron = e(V/s)/m = eV/smUse equation of motion: s = ut + (1/2)at^2hence, s = (1/2).(eV/sm).t^2t = s.square-root[2m/eV] 4. By conservation of energy, F.(30) = 150 x 10 + Wfwhere Wf is the energy expended on friction F = [1500/30 + Wf/30] N If there is no friction, Wf = 0 and F = 1500/30 N = 50 N If F = 150 N, then Wf = 3000 J when compared with the useful energy out of 1500 J, a friction that dissipates energy of 3000 J cannot be regarded as "small"
其他解答:
留言列表
