山东省重点中学2022-2023学年高二上学期期末考试物理答案

DOC
  • 阅读 10 次
  • 下载 0 次
  • 页数 3 页
  • 大小 75.500 KB
  • 2024-10-09 上传
  • 收藏
  • 违规举报
  • © 版权认领
下载文档3.00 元 加入VIP免费下载
此文档由【管理员店铺】提供上传,收益归文档提供者,本网站只提供存储服务。若此文档侵犯了您的版权,欢迎进行违规举报版权认领
山东省重点中学2022-2023学年高二上学期期末考试物理答案
可在后台配置第一页与第二页中间广告代码
山东省重点中学2022-2023学年高二上学期期末考试物理答案
可在后台配置第二页与第三页中间广告代码
山东省重点中学2022-2023学年高二上学期期末考试物理答案
可在后台配置第三页与第四页中间广告代码
在线阅读已结束,您可下载此文档进行离线阅读 已有0人下载 下载文档3.00 元
/ 3
  • 收藏
  • 违规举报
  • © 版权认领
下载文档3.00 元 加入VIP免费下载
文本内容

【文档说明】山东省重点中学2022-2023学年高二上学期期末考试物理答案.doc,共(3)页,75.500 KB,由管理员店铺上传

转载请保留链接:https://www.doc5u.com/view-56db2e665b28a857848c3638e38bfbd9.html

以下为本文档部分文字说明:

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△AO

E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:2225sin5OEiOEA

E==+·······················(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/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:20mvBqvr=······················································(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。

管理员店铺
管理员店铺
管理员店铺
  • 文档 467379
  • 被下载 24
  • 被收藏 0
若发现您的权益受到侵害,请立即联系客服,我们会尽快为您处理。侵权客服QQ:12345678 电话:400-000-0000 (支持时间:9:00-17:00) 公众号
Powered by 太赞文库
×
确认删除?