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2022－2023学年度第一学期期末学业水平检测高二物理答案及评分标准一、单项选择题：本大题共8小题，每小题3分，共24分。1．A2．D3．B4．A5．B6．D7．C8．C二、多项选择题：本大题共4小题，每小题4分，共16

分，选不全得2分，有选错得0分。9．AC10．AB11．ACD12．BC三、非选择题13．（6分）（1）C（1分）；摆角过大，已不能将小球的运动看作简谐运动。（1分）（2）偏小（2分）（3）4π2(l1－l2)T12－T22或4π2(

l2－l1)T22－T12（2分）14．（9分）（1）×1（2分）；30（2分）（2）丙（2分）；C（1分）（3）偏小（1分）；电压表的分流（1分）15．（8分）（1）由图像可知：T＝6s，波长λ＝12m········

············································（2分）则波的传播速度：v＝2T=m/s·········································

···············（2分）（2）质点位置与时间的关系式为：y＝Asin2πTt···········································（1分）当t＝9.5s，y＝－1，因此坐标为（3m，－1cm）····

·····························（1分）（3）当t＝9.5内质点走过的路程s＝6A+1＝13cm········································（2分）评分标准：第1问，4分；第2问，2分；第3问

，2分。共8分。16．(10分)（1）在平衡位置：有F－kx0＝0，x0＝Fk··················································（2分）规定向左为正，当BC向左偏离

平衡位置为x时回复力（即合外力）为F回＝F－k（x+x0）综上解得F回＝－kx，因此得证。·····················································（2分）（2）撤去F后，B、C回到弹簧

原长处：由简谐振动的知识可知：弹簧压缩到最短时，BC前进的位移s＝2x0········（1分）Fs＝21122mv····················································

····················（1分）解得：v1＝Fsm···········································································（1分）到弹簧最长时AB共速：mv1＝3mv2··

·················································（1分）根据能量守恒：EP＝12mv12－32mv22··································

···············（1分）解得：EP＝2F23k··········································································

··（1分）评分标准：第1问，4分；第2问，6分。共10分。17．(12分)光路图如图所示（1）由几何知识可知△AOE与△BEF相似AOBFOEEF=且OE＋EF＝14cm·（1分）解得OE＝6cm··············

······（1分）设玻璃砖厚为d△AOE与△DMH相似5AOdOEHM−=，解得：d＝4cm··（2分）（2）入射角的正弦：2225sin5OEiOEAE==+·······················（1分）折射角的正弦：2255ENsinrENNH==+·················

·····（1分）折射率：sin2sininr==················································（2分）（3）光在玻璃砖内地传播速度8=1.510cv

n=m/s····················（1分）经过的路程：22=25cms2ENd=+··························（1分）所用的时间：-108510s3stv==·····

·····························（2分）评分标准：第1问，4分；第2问，4分；第3问，4分。共12分。18．(15分)（1）扇形电场的电势差为：(22CCCURRR=−−−=）······

·········（1分）粒子在加速电场中根据动能定理有：212qUmv=···········（1分）y/cmx/cmABCDO34141810－5－3rENHMGFi解得速率：CqvRm=·····························

··················（1分）粒子在磁场中的运动轨迹如右图所示，可得粒子在磁场中做匀速圆周运动的半径：03tan30RrR==····················································（1分）由：2

0mvBqvr=······················································（1分）解得：0233CqRmBqR=···············································（1分）（2）

打到荧光屏上的竖直高度为：02tan6023yRR==········（2分）（3）当竖直穿出磁场时，粒子在磁场中运动的半径为R由于：rvU∝∝可知加速电压：U′＝13U·······································（2分）可得：－Cx－(－CR)＝

13.C2R····································（1分）解得：x＝65R························································（1分）（4）

扫过的区域形状右图所示222minπ-1+122SRRR=−=（）（）··························（3分）评分标准：第1问，6分；第2问，2分；第3问，4分；第4问，3分。共15分。