读后续写Possible version lI stared at the paper and then looked around. I was cold. Although I could not see my sistersnor the new house, which was hidden from my sight by a grey wall, I got the feeling that I was notalone. Walking along the shore was a tall familiar figure who looked lost in his thoughts. I lookedgain at the paper in my hand, closed my eyes and wishedAs /opened my eyes, I found my father was standing right beside me. Happy birthday, sonhe said, patting my shoulder gently. I smiled up at him and took his hand. Looking back, life fromthat day on marked the start of a close bond and a strong friendship with my father. What was mybirthday surprise? Wishes do come truePossible version 2I stared at the paper and then looked around. On the backside, there was a map. I noticed thatit was clearly leading me in a direction, but who could have sent this? I stood there in silence andstudied it. I decided to follow the map and it led me down the beach until I reached a staircase. Iooked back and could no longer see my siblings. I continued to follow the point on the map, untilI unexpectedly lost my footing and fell zksq down the stepsAs I opened my eyes, I found my father was standing right beside me. Happy birthday, sonhe said, patting my shoulder gently. Then he asked how I had come upon this place? I explained tohim that I had found the bottle in the ocean and when I opened it had the words on it. Shocked bythe message, my father realized what I had found I threw this bottle into the ocean many years agowhen I was alone. he said. I used to come to this exact place to get away from everyone else. NowI get to share this experience with you. What a birthday surprise

3B【解析】粒子在加速电场中由动能定理有qU1=2,解得v=/qU,设偏转电场中的极板长度为L,平行板电容器间距为b,粒子在偏转电场中运动时的偏转位移为y.则有、、=1,,L2 mb voU,L-AU,b由此可见粒子射出电场的偏转位移与比荷无关比荷不同的粒子射入磁场的位置M相同,A错误;设粒子在加速电场中运动的时间为t1,加速电场两极板间的2me距离为{,根据一a1i=l,解得t粒子在偏转电场中运动的时网小~上粒子的比荷越大,t1、t2越小.粒子在电场中运动的总吋间也越小D错误;带电粒子在电场中做类平抛运动,可将射出电场的粒子速度v分解成水平方向和竖直方向,设出射速度与水平夹角为0,则有cos粒子在磁场中做匀速圆周运动,设运动轨迹对应的半径为R,由几何关系可得,半径与直线MN夹角正好等于日,根据几何关系R=cos0,解得d=2Ke带电粒子在匀强磁场中运动的半径为R=“,解得d=B正B确:粒子进入磁场由功能关系可得U1+qb·y=m2,解得2qU, QULm2mU,b,则R=1/2mU1, nULB√q2vUb2·可知粒子在磁场中运动的轨迹与U2有关,C错误。故选B。

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