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从N点运动到C点的时间为1=3X600m60r360°5√3πB1L联立解得t=3E1由几何关系可知LAN=c0s60°在第一象限中运动的时间t1和第二象限中运动的时间t2相等，且AN 2L B LE所以带电小球从A,点出发至回到A,点的过程中所经历的总时间为t总=t十t1十t2联立解得t总=53πB1L,2B,L3E1E√3mg2E1【答案】(1)负电3E1B(2)竖直向下√3E153πB1L2BL3E1E【解析】(1)小球在第一象限中的受力分析如图甲所示E1←quB'E-tXXXXXXXXXX甲带电小球的电性为负电mg=gE,tan 603mgq-3E101qE1=qvB1cos60°2E解得v=B(2)小球若在x轴下方的磁场中做匀速圆周运动，则必须使得电场力与重力二力平衡，即应施加一方向竖直向下的匀强电场，且电场强度大小满足qE=mg解得E=√3E(3)要想让小球恰好与弹性板发生两次碰撞，并且碰撞后返回x轴上方空间匀速运动到A点，则其轨迹如图乙所示E←-由几何关系可知3PD=20NONONOA L=tan 60联立郎得PD=DN=号L则弹性薄板长度至少为PD=名5L3设在x轴下方的磁场磁感应强度大小为B,则满足quB=m RT=2rmqB由几何关系可知R=PD从N点运动到C点的时间为
really oyed ourselves today.第二节书面表达One possible version:Hi,Tom!I'm very interested in a gallery talk held by theBritish Museum and I'd like to invite you to go withme.The gallery talk called Silk Road Traders:Par-thians and Sogdians will be given by an independentspeaker Diana Driscoll this Friday afternoon,March18,2022.It will begin at 13:15 in Room 52 and lastabout 45 minutes.It is suitable for all levels ofknowledge in English.What's more,it is free ofcharge and we can just drop in then.After the talk,we can visit some exhibitions about the ancient SilkRoad.I hope you can go with me.I'm looking forward to your quick response!Yours,Li Hua
