英语周报2021-2022高二外研版第8期答案

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5.解析:(1)粒子在电场中做类平抛运动,由动能定理得Ed=2my2-2m………(2分=2×103C/kg…………………………………(2分(2)设粒子从O点进入磁场时的速度方向与x轴的夹角为0,则(1分)解得6=45°,即粒子沿着OB方向进入磁场粒子进入磁场后做匀速圆周运动,磁感应强度为B1,半径为n由几何关系粒子轨迹对应的圆心角为120Rtanbo=(1分解得n=2m(1分)在磁场中由牛顿第二定律qvB1mU(2分)B1=T………………(2分)(3)粒子进入磁场后做匀速圆周运动,磁感应强度为B2,半径为r2,周期为T,依题意可知,与刚性圆环碰撞2次后回到O点时运动时间最短,依据轨迹的对称性和几何关系。tan30°=R(1分)tan 30=Rr2=5(1分由牛顿第二定律UBzn(2分)解得B2=12T(1分)粒子在磁场中圆周运动的周期T=22(1分)粒子在磁场中运动时间t=2T=√6x×10-3s(2分)

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英语周报2021-2022高二外研版第8期答案

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