2020—2022英语周报第八期答案

书面表达为现实”,无被动语态,所以将be删除。Dear KatI'm glad to learn that you show great interest in thecoming-of-age ceremony held by our school recently.Now letme tell you something about it in great detailIt was held in the Great Hall of our school between 900 a. m. and 11: 30 a. m. last Friday. At the beginningthe ceremony, all the students present sang the nationalg together. Then followed the wishes from the eldersMeanwhile,the parents also expressed their high expectations of their children. Afterwards, the student representtive delivered a speech on the stage. Finally, the sclheadmaster made a conclusion.78The coming-of-age ceremony is significant and meaningThrough this activityvelop a stronger sense ofhe students, being adultresponsibility and gratitude. TheyS, dehave realized they should Iearn to servethecontnoutons to society in the near futurpeople and makYoursLi Hua

vs,解得如=8m/s,碰擅前系统总动能E=2m×2×10=10.,撞后系统总动能2m m=号×2x(-2y x382J=100J,碰撞过程机械能守恒,碰撞是弹性碰攘,故A项正确;A、B两球碰擅后,B、C两球组成的系统水平方向动量守恒,弹簧恢复原长时B球的速度最小,C球的速度最大,以向左为正方向,从碰撞后到弹簧恢复原长过程,在水平方向,由动量守恒定律得maB=mnva mvc,由机械能守恒定律得立mni=2mn82 2mt,解得m=4m/s,t12m/s〔弹簧恢复原长时C球的速度最大,wm/st=0,不符合实际舍去),由此可知,弹簧恢复原长时C球的速度为12m/s,B球的最小度4m/s,故B项错误,D项正确;BC两球速度弹簧伸长量最大,弹簧弹性势能最大,B、C两球组成的系统在水平方向动量守恒,以向左为正方向,在水平方向由动量守恒定律得msUB=(ms me),由机械能守恒定律得2m=立(mn mk)t E解得弹簧的最大弹性势能E=24J,故C项错误12.AD【解析】A、B两球碰撞过程系统动量守恒,以向左为正方向,由动量守恒定律得mAv=mA0A

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