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钦州市2020年秋季学期教学质量监测参考答案高二物理一、单项选择题（24分）题号12345678答案CDACBDAB二、多项选择题（16分）题号9101112答案BCACBDAD三、实验题（17分）13.（每空2分，共8分）（1）甲（2）ABDFH（3）

1.50.714.（每空3分，共9分）（1）A1如图所示（2）x910URI=−（Ω）四、计算题（43分）15.（8分）解析：设滑块与N点的距离为L，由动能定理可得，qEL－μmgL－mg·2R＝12mv2－0·····························3分小滑

块在C点时，mg＝mv2R································································3分解得v＝2m/s，L＝20m·············

······················································2分16.（10分）解析：（1）根据闭合电路欧姆定律，12302A1410EIrRR===++++·

·············2分220VUIR==·············································1分（2）电源的输出功率：WrIEIP562=−=···············

························2分（3）电场强度：20V200V/m0.1mUEd===········································1分粒子做类似平抛运动，

有：tvL0=················································1分221aty=·············································

·································1分qEam=············································································

··1分联立解得：mm1.28m101.28y-3==·············································1分17.（12分）解析：(1)粒子在电场中加速，由动能定理有：212qUmv=

····················2分解得：2qUvm=······································································2分(2)粒子在磁场中的运动轨迹刚好与PQ相切时的轨道半径，是粒子

从边界MN离开磁场最大轨道半径，如图由几何知识得：+=30sinrrd········································2分粒子在磁场中做圆周运动，洛伦兹力提供向心力，2mvqvBr=···2分得：322mUBd

q=························2分粒子最终从磁场边界MN离开磁场，磁感应强度：322mUBdq···········2分18.（13分）解析：（1）根据法拉第电磁感应定律，得感应电动势为：BLvE=·················1分根据

闭合电路的欧姆定律，得感应电流为：EIRr=+····················1分导体棒受到的安培力为：FBIL=安···········································1分导体棒做匀速运动，处于平衡状态，由平衡条件得：sinF

mgBIL=+···········································2分代入数据解得：smv/4=························································1分（2）

金属棒运动过程，由能量守恒定律得：21sin2FsmgsmvQ=++····3分电阻R产生的热量为：RRQQRr=+···········································2分代入数据解得：1.

28RQ=J·······················································2分