16【答案】(1)2m(2)6N()能落到斜面上,第一次落在斜面上的位置距离B点113m解析】(1)设小球离开B点后做平抛运动的时间为,落地点到C点距离为x(2)小球到达B点时受重力mg和竖直向上的弹力N作用,由牛顿第二定律知向=Nmg三m,解得N=6N由牛顿第三定律球到达B点时对圆形轨道的压力大小为6N,方向竖直向(3)如图所示,斜面BEC的倾角=45,CE长为d=h=5m,因为d>x,所以小球离开B点后能落在斜假设小球第一次落在斜面上F点,BF长为L,小球从B点到F点的时Line= gt联立解得t=0.4s,L≈113m

书面表达One posssible versionDear Mr. Smith,I'm Li Hua, a senior 3 student. Learning that your City Culture department is looking for an English editorthis summer, I'm eager to apply for the position.As far as I'm concerned, city culture refers to a citys history, geography, customs, future, and so on. Asa native, I am quite familiar with our citys traditional culture. Besides, I have a good command of ErI' m especially good at English writing. Additionally, Im cooperative and creative, which will give me anadvantage for the position.I'll be greatly honored if I' m chosen for the job. If I'm admitted, please email me at lihua @16com.Yours sincerely,Li Hu

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