2022——2023 学年第一学期期末教学质量监测 高二物理答案

2022-2023学年第一学期期末教学质量监测试题（卷）
高二物理参考答案
一、选择题：共12小题，每小题4分，共48分。

在每小题给出的四个选项中，第1-8题只有一个选项符合题目要求，第9-12题有多项符合题目要求。

全部选对的得4分，选对但不全的得2分，有选错的得0分。

题号1
23456789101112答案D C A C B D B A BD AC BC ACD
二、实验题：共16分。

13．(1)偏转（2分）(2)不偏转（2分）偏转（2分）
(3)穿过闭合回路的磁通量发生变化（2分）
14．(1)
（2分）
(2)旧干电池内阻较大，容易测量，实验误差小（2分）
(3)1.48(1.46～1.49均可)（2分） 1.36(1.32～1.38均可)（2分）
三、计算题：共36分。

解答时写出必要的文字说明、方程式和重要演算步骤，只写最后答案不得分。

15．解析
(1)电动机额定状态下工作时的功率P ＝UI ＝48×12W ＝576W·························（3分）
(2)电动机的热功率
P 热＝P －P 出＝576W －350W ＝226W·························································（2分）
由电动机的热功率P 热＝I 2r ·······································································（2分）
代入数据得：r =1.57Ω··············································································（1分）
16．解析(1)粒子在加速器中的加速过程，根据功能关系有
qU ＝12
m v 2······························································································（2分）解得U ＝m v 22q
··························································································（1分）(2)粒子在速度选择器中的匀速过程，电场力与洛伦兹力平衡，有qE ＝q v B ·········（1分）
解得B ＝E v
·····························································································（1分）根据左手定则及带正电粒子所受洛伦兹力方向可知，磁感应强度方向垂直纸面向里（1分）
17．解析(1)小球处于静止平衡状态
小球所受电场力方向与电场强度方向相反，则小球带负电·······························（1分）由平衡可得qE ＝mg tan 37°········································································（1分）解得q ＝1.0×10－6C················································································（1分）
(2)剪断轻绳后，小球所受合外力F 合＝mg cos 37°由牛顿第二定律，小球加速度a ＝F 合m
··························································（1分）联立得a ＝12.5m/s 2··················································································（1分）
(3)剪断轻绳2s 后，小球位移x ＝12
at 2··························································（1分）电场力对小球做的功W ＝qEx ·sin 37°···························································（1分）联立得W ＝0.45J····················································································（1分）
18．解析轨迹如图所示
(1)在电场中，沿着电场线方向
a ＝qE m ＝v 202h
·····························································································（2分）根据v 2y ＝2ah ··························································································（1分）解得v y ＝v 0····························································································（1分）粒子在A 点的速度为
v ＝v 2y ＋v 20＝2v 0···················································································（1分）
则进入磁场时速度方向和水平方向的夹角tan θ＝v y v 0
＝1···································（1分）解得θ＝45°····························································································（1分）
(2)粒子在电场中运动有v y ＝at ···································································（1分）垂直电场方向的位移x OA ＝v 0t ····································································（1分）根据几何知识得2r sin 45°＝x OA ··································································（1分）解得r ＝2h ···························································································（1分）
根据q v B ＝m v 2r
·······················································································（2分）解得B ＝m v 0qh ··························································································（1分）。