试卷库首页 高二试卷 高二生物上

辽宁省大连市2023-2024学年高二上学期1月期末考试+物理+PDF版含答案【KS5U+高考】.pdf

期末试卷 辽宁省 2023 2024 格式PDF   11页   下载2433   2024-01-24   收藏1732   点赞3761   免费试卷
温馨提示:当前文档最多只能预览 5 页,若文档总页数超出了 3 页,请下载原文档以浏览全部内容。
辽宁省大连市2023-2024学年高二上学期1月期末考试+物理+PDF版含答案【KS5U+高考】.pdf 第1页
辽宁省大连市2023-2024学年高二上学期1月期末考试+物理+PDF版含答案【KS5U+高考】.pdf 第2页
辽宁省大连市2023-2024学年高二上学期1月期末考试+物理+PDF版含答案【KS5U+高考】.pdf 第3页
辽宁省大连市2023-2024学年高二上学期1月期末考试+物理+PDF版含答案【KS5U+高考】.pdf 第4页
辽宁省大连市2023-2024学年高二上学期1月期末考试+物理+PDF版含答案【KS5U+高考】.pdf 第5页
剩余6页未读, 下载浏览全部
高二物理 答案 第 1页 共 4页 大连市 2023 ~ 2024 学年度第一学期期末考试 高二物理 参考答案 一、选择题 : 本题共 10 小题 ,共 46 分 。在每小题给出的四个选项中,第 1~ 7题只有 一项符合题目要求,每个小题 4分;第 8~ 10 题有多项符合题目要求, 每小题 6分, 全 部选对的得 6分,选对但不全的得 3分,有错选或不答的得 0分 。 1 2 3 4 5 6 7 8 9 10 B C D A D A C AD BC BCD 二、实验题(本题共 2小题,共 14分) 11.( 1) 1.875 或 15 8 (2 分 )( 2) AC (2 分 )( 2) 左 (1 分 )、是 (1 分 ) 12.( 1) 如下图 (共 3分,画出 V2给 1分,画出 R1给 1分,电路 1分 ) ( 3) 0.34 (1 分 ) ( 4) 1.41 — 1.43(1 分 )、 0.837 — 0.915 (1 分 ) ( 4) 偏小 (1 分 )、 偏小 (1 分 ) 三、计算题(本题共 3小题,共 40 分 。其中 13 题 10分、 14 题 14分、 15 题 16分。 解 答应写出必要的文字说明、方程式和重要的演算步骤,只写出最后答案的不能得分,有 数值计算的题答案中必须明确写出数值和单位 。) 13. (10 分 ) ( 1)从左向右或从 p向 q ········································································· (1 分 ) ( 2)线圈产生电动势的最大值 Em=nBS ω ······················································ (1 分 ) e=Emcos ωt ····························································································· (1 分 ) e=nBS ωcos ωt ························································································· (1 分 ) ( 3)线圈产生电动势的有效值 2 Em E ······················································· (1 分 ) 电路中电流的有效值   E I R r ···································································· (1 分 ) 电路中产生的焦耳热 2( )   Q I R rT ····························································· (1 分 ) 2  T ································································································ (1 分 ) 由题意,线圈匀速转动,则外力做的功等于电路中产生的焦耳热 W=Q ················ (1 分 ) (拟合图像参考图) 2{#{QQABYYaEogAIAAJAABhCUQUoCACQkACCACoOxAAMMAAAgBNABCA=}#} 高二物理 答案 第 2页 共 4页 外力驱动线圈转动一 周 所做的功 2 2 2     nB S W R r ············································· (1 分 ) 14. (14 分 ) ( 1) 在匀强电场中 , 粒子做类平抛运动 , 轨迹如图中 AO 段所示 x方向由运动学公式 0 2Lvt ····································································· (1 分 ) y方向由运动学公式 2 1 2 Lat ··································································· (1 分 ) 由牛顿第二定律得 F a m  电 qE F  电 ···················································· (共 1分 ) 联立解得 qL mv E 2 20  ·················································································· (1 分 ) ( 2) 设 粒子在 O点的速度 v与 x轴负方向夹角为 α 粒子在 O点沿 y轴负方向分速度的大小 yv at  ·············································· (1 分 ) 2 2 0 y v v v   0 tan yv v   ····························································· (共 1分 ) 解得 0 2v v 45    粒子在第三象限做匀速圆周运动的 轨迹为 OP段, 半径为 R 由牛顿第二定律得 2v qBv m R  ·································································· (1 分 ) 解得 0 2mv R qB  由几何关系得 2 sin45 OP R   ·································································· (1 分 ) 0 2mv OP qB  故 P点坐标为( 0、 0 2mv qB  ) ····································································· (1 分 ) ( 3) 解法一:粒子在第四象限做匀速圆周运动的 轨迹为 PQ 段, 半径为 R1,转过的圆 心角为 β,通过的弧长为 l 由几何关系得 R1=2 R ················································································ (1 分 ) 3 4    ······························································································· (1 分 ) 1 2 2      lR ························································································ (1 分 ) 粒子在第四象限运动的时间  l t v ······························································ (1 分 ) 联立解得 3 2 m t qB   ··················································································· (1 分 ){#{QQABYYaEogAIAAJAABhCUQUoCACQkACCACoOxAAMMAAAgBNABCA=}#} 高
辽宁省大连市2023-2024学年高二上学期1月期末考试+物理+PDF版含答案【KS5U+高考】.pdf