2022英语周报七年级新目标ylg答案

参考范文:One possible versionDear Sir or MadamImLi Hua, a Senior Two student. I have read your advertisement for volunteers of the Sino-American YouthAmbassadorsonline Program. I'm writing to apply for participation in this activity.I consider myself a qualified candidate for the several reasons. First of all, a good command of English will make itpossible for me to communicate with fore gA members fluently. Besides, QIm)an outgoing person and have rich experiencein voluntary jobs, which can help get along well with American friends,in the'programI would like to present the long history and diverse culture of China to our friends and also learn more about Americaand its peopleI would appreciate it if you could consider my application and accept me as a member of the program. Lookingforward to your early reply.Yours sincerely,Li Hua

25.考查目标本题考查了带电粒子在匀强电场、匀强磁场中的运动,考查考生的分析综合能力和应用数学处理物理问题的能力。思路点拨(1)通过几何知识得出粒子a在场中的偏转半径,由向心力公式求碰场I的磁感应强度大小B,通过题中已知的粒子a在电场中的水平位移和竖直位移,由类平抛运动公式的水平和竖直分运动求得电场强度大小E;(2)粒子α在赡场Ⅰ中运动的时间可由圆周运动的周期公式的偏转角求得,在电场中的运动时间可由曲线运动中的等时性求得,在磁场Ⅱ中秃由题意得到粒子α在磁场Ⅱ中的偏转角和偏转率径,从而求得磁场Ⅱ的磁感应强度大小,再由圆周运动周期公式和偏转角求得时最后相加得到总时间;3)通过几何关系求得粒子b经过y轴正方向时的位置,同(2)中类似,求得粒子b在磁场Ⅱ中的运动半径,进而求得粒子b在磁场Ⅱ中的水平位移,由粒子b进出磁场Ⅱ的位置坐标差求得其离开磁场Ⅲ时的参考答案(1)粒子a从P点进入磁场后,经过y轴正方向上与圆心等高的M点,则粒子在磁场中做圆周运动的半径为r=L(1分)又华B解得B===粒子a进入电场后做类平抛运动,水平方向上有2L=nt(1分)竖直方向上有L=解得E=2)设粒子a离开电场时的速度大小为v,与x轴的夹角为a,则粒子a在电场运动过程中由动能定理有设粒子a在磁场区域Ⅱ中做圆周运动的半径为R,则2Rsin a =2/(1分)又gB=m解得a=-,B'=(1分)粒子a在圆形磁场I中运动的时间t1=(1分)在电场中运动的时间t2=(1分)3 2m 37在磁场区域Ⅱ中运动的时间3=4B'20(1分)则粒子a由P点运动到第一次离开磁场区域Ⅱ的时间1=4 2 1=(2m 2)L(1分)(3)设粒子b在磁场中做圆周运动的圆心为O,离开磁场时的位置为M′,因粒子b在圆内的运动半径为L,易知POM′O为棱形,由于PO竖直,因而M'O′也为竖直方向,故粒子b离开磁场时速度一定沿x轴正方向,运动轨迹如图所示。由题图可知粒子b到达y轴的坐标y=L Lim(120°-90°)=3L(1分)设粒子b离开电场时的速度大小为与x轴正"方向的夹角为a',则由动能定理有gEl=my- mo(1分)设粒子在磁场区域Ⅱ中做圆周运动的半径为R',则有geB=mr(1分)粒子b第一次离开磁场Ⅱ区域的位置坐标x=n0t'-2R'sina'(1分)解得x=0(1分)

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