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
辽宁大连市2023-2024学年高二上学期1月期末考试+物理+(含参考答案)试卷文档在线免费下载.pdf
