2022英语周报高三第7答案

第二节One possible version:Dear RussellI,m glad to inform you that our school will hold anEnglish humorous story speech contest in the center hallthis Sunday in order to develop our sense of humor andpractice our spoken English. I' d like to invite you to takert in ItEvery participant is requested to prepare a humorousstory in English by themselves. So choose your story cau-tiously. The funnier your story is, the better it is. You areexpected to deliver your speech within five minutes. Youshould tell your story at a proper pace with a clear voiceLooking forward to your participation. I'm sure youwill be one of the winnersYoursLi Hu

12.AD〖解析】A、B两球碰撞过程系统劲量守恒,以艮向左为正方向,由动量守恒定律得mAvo=mA0A+m2vB,解得vB=8m/s,碰撞前系统总动能Ek=1mn=1×2×102J=100,碰撞后系统总动能Ek=2mAn+mv=t×2×(-2)2J+3×82J=100J,碰撞过程机械能守恒,碰撞是弹性碰撞,故A项正确;A、B两球碰撞后,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m0a=m2i'+meve,由机械能守恒定律得FRU2mBU8 2+2mvi,解得2=4mkt=12m/s(弹簧恢复原长时C球的速度最大,vB′=8m/s,v=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球的最小速度为4m/s,故B项错误,D项正确;B、C两球速度相等时弹簧伸长量最大,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mv=(m+mc)vBC,由机械能守恒定律得m1nh=(mB+m)h+E,解得弹簧的最大弹性势能E=24J,故C项错误。

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