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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=1.510cvn=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/cmABCDO
34141810-5-3rENHMGFi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:20mvBqvr=·····
·················································(1分)解得:0233CqRmBqR=···············································(1分)(2)打到荧光屏上的竖直高度为:0
2tan6023yRR==········(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。