2022英语周报外延第36期。答案

25.(18分)解:(1)在恒力作用下,设金属棒滑动过程某时刻t速度为n,则感应电动势为E=BL(1分)平行板电容器两极板之间的电势差为U=E(1分)设此时电容器极板上积累的电荷量为Q,按定义有C=2(1分联立可得,Q=CBn(1分)t+△t时刻速度为r同理,此时电容器电荷量Q′=CBn′(1分)在△时间内通过金属棒的电流/=-Q=CBL2-E=CBL(2分)其中a是金属棒的加速度根据牛顿第二定律F一BH=ma(2分所以a=BLC+m(1分)代入数据得a=2m/s2(2分)所以金属棒做初速度为零的匀加速运动根据运动学公式,有v=at= 10 m/s(1分)(2)根据法拉第电磁感应定律E=Bx=20V(2分)根据闭合电路欧姆定律==2A(1分)根据能量守恒,导体棒的动能转化为定值电阻R产生的热量所以Q=m2=50J(2分)

第一节参考范文:Last Friday evening witnessed a breathtaking English Singing Contest in our schoolstadium, where 15 singers competed for"The Best English Singer of the YearThe singers, who had stood out from the first round of selection, gave energeticperformances on stage. Throughout the contest, the atmosphere in the stadium was extremelyexciting, with many students singing along to nearly every song. At last, the award went toZhang Xin from Class 1 Grade 1Everyone agreed that the contest enriched our campus life as well as kindling ourpassion for English learning

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