福建省11月名校联盟优质校2025届高三半期考物理试卷

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Writing 3Dear JamesI am glad to have received the book Biogra phy o/Lincoln you bought for me. I like it very much and amwriting specially to express my thanks for your kindness.You know, Lincoln is my hero. He came from a poorfamily and had only several monthseducationHowever, he became a present of the United States. Whata great man! When I received the book, I felt more thanexcited. That was not only because you sent me a birthdaresent but also because it was the exact book I wantedYou are so thoughtful!Thanks again and best wishes to youYLi h

3.【解析】(1)在研究向心力的大小F与质量m、角速度和半径r之间的关系时我们主要用到了物理学中的控制变量法,故选C(2)图中所示,若两个钢球质量和运动半径相等,则是在研究向心力的大小F与角速度o的关系,故选C(3)根据F=mo2r可知若两个钢球质量m和运动半径r相等,图中标尺上黑白相间的等分格显示出钢球1和钢球2所受向心力的比值为1:9,可知两轮的角速度之比为1:3,根据U=oR可知,因为变速轮塔和变速轮塔2是皮带传动,边缘线速度相等,则与皮带连接的变速轮塔1和变速轮塔2的半径之比为3:故选B

1.BC【解析】释放物块C前,细线对物块A、B的作用力为零物块A、B整体受力平衡则有kx1=2mosin30°,解得x;=Ing弹簧的弹性势能为Ekm"g2·释放物块C的瞬间,对物块C,根据牛顿第二定律有-Fn=2a1,对物块A、B有Fn+k2mosin=2ma1,解得a1-=,F1=2mg5·A错误B正确;物块A、B分离时物块A、B之间的相互作用力为0.且物块A、B的加速度相同.分别对物块A、B、C受力分析,对物块A有kx:-ngsin30°=ma2,对物块B有F1-msin30°=ma:对物块C有知g一F=2a,解得a:=0,Fn=2x:=2k,弹簧的弹性势能Emg8-,从释放C到A、B分离过程中,A、B、C移动的距离为△r=x1-x2次…系统的机械能守恒则有方×2×v2+2m·Arsin6-2gΔr=E一E,解得物块A、B分离时物块A的速度大小为v=ngC正确,D错误。故选BC10k

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