「2023答案全国100所数学」数学百所名校高三答案

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【答案】AD【解析】甲，乙的角速度相等，根据0=@r得，甲、乙的线连度大小之比为”=2=Rsin30=尽，v:r?Rsin60=3,即甲的线速度3E=m=2大小始终为乙的线连度大小的气，A正确：根据F=mm,得，甲、乙所受向心力大小之比为一m,=32W3即甲所受向心力大小始终为乙所受向心力大小的3倍，B错误；假设当转台角速度大小为！时，陶罐对甲的摩擦力恰好为零，由牛顿第二定律得m1wir1=m1 g tan a,解得w1=Rcos a，假设当转台角速度为2时，陶罐对乙的摩擦力恰好为零，同理可得w?一√Rc0sg,根据4gR<3R<:知，当转台角速度为，晨时，甲、乙所受支持力和重力的合力均大于其所需要的向心力，所以甲、乙均有向内侧运动的趋势，当转台角速度为4g时，乙所受支持力和重力的合力大于其所需要的向心力，乙有向内侧运动的趋势，C错误，D正确。

第二节书面表达With the college entrance examination approaching,inorder to guarantee more study time for senior threestudents,our school plans to cut off one PE class per weekto make time for math tests.However,I don't think it isbeneficial for us.In my opinion,taking physical exercise is of equalimportance for senior three students.First of all,doingsports helps build up our bodies,thus increasing ourresistance.Moreover,we urgently need physical exerciseto help enhance communication among students,avoidingsome psychological problems.In addition,by taking partin physical exercise,our mind gets refreshed.In conclusion,I strongly recommend not reducing ourPE class.