2022 英语周报 七年级 牛津SYL 2答案

第二节书面表达【写作思路】根据试题要求,考生需要写一篇记叙文,记述身边一位值得尊敬和爱戴的人。首先考生需要介绍自己最尊敬和爱戴的人是谁,简单地介绍一下他(她)。然后,陈述为什么他(她)值得尊敬和爱戴。最后,可以对前文进行一个总结,使文章结构严谨。本文的写作重点在"尊敬和爱戴的原因",考生可围绕这个要点适当拓展,从性格特点、处事风格等方面展开写作。【范文赏读】The most beloved and respected person around me is my teacher, Ms LiThough she has been teaching English for twenty years, she is still passionateabout teaching. She is kind and considerate towards us just like our dear motherWe all respect her because she always tries new ways to make her classes livelyand interesting. Hardworking and knowledgeable, she is one of the best teachersin our school. When we have a problem, we will turn to her for help. She alwaystalks to us patiently and helps us to find a solution. In our eyes, she is not onlyour teacher, but also our best friend. Those are why she deserves our respect【亮点词句】亮点词汇: beloved, respected, be passionate about, be consideratetowards, turn to sb. for help, in one's eyes亮点句式:Why引导的表语从句( why she deserves our respect)

15.(1)0.48m(2)2s【解析】(1)乙向下做匀加速直线运动,加速度大小为2gcos30° m:sin30°=12.5m/s2(1分)甲向下做匀减速直线运动,加速度大小为a:=u: m,gcos 30" AI(n m:)g cos 30-mI gsin 30设甲和乙刚好达到共同速度时速度为v1,所用时间为t1,根据运动学公式可得(1分)甲的位移为x甲=D1:-2ai=0.84m(1分)乙的位移为xz=a1!i=0.36乙相对甲滑行的最大距离△(2)当甲和乙刚好达到共同速度的瞬间启动传送带时,对甲、乙系统,有p1(m1 m2)gcos30°=(m1 m2)gsin30°(1分)所以甲、乙与传送带共速前,甲、乙一起做匀速直线运动,运动时间为(]分)甲、乙一起运动的位移x1=v;t2=3m(1分)甲、乙与传送带共速后,甲、乙与传送带一起做匀加速直线运动,根据运动学公式有L=vts at(1分)解得t3=1s(1分)从启动传送带到甲的下端运动到传送带底端N点所用的时间t=t2 t3=2s(1分)

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