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泸州市高2019级高二学年末统一考试理综物理本试卷分第Ⅰ卷（选择题）和第Ⅱ卷（非选择题）两部分，共38题，共300分，共12页，考试时间150分钟。考试结束后，将答题卡交回，试卷自留。注意事项：1.答题前，考生先将自己的姓名、准

考证号码填写清楚，将条形码准确粘贴在条形码区域内。2.选择题必须使用2B铅笔填涂；非选择题必须使用0.5毫米黑色字迹的签字笔书写，字体工整、笔迹清楚。3.请按照题号顺序在各题目的答题区域内作答，超出答题区域书写的答案无

效；在草稿纸、试题卷上答题无效。4.作图可先使用铅笔画出，确定后必须使用黑色字迹的签字笔描黑。5.保持卡面清洁，不要折叠、不要弄破、弄皱，不准使用涂改液、修正带、刮纸刀。可能用到的相对原子质量：H1C12O16S32Ti48Zn65二、选择题（本题

包括8小题，每小题6分，共48分。14~18题中每个小题只有一个选项符合题目要求；19~21题中有多个选项符合题目要求的，全部选对的得6分，选不全的得3分，有选错或不答的得0分）。14．如图所示为氢原子能级示意图。一群处于n=3激发态的氢原子，根据玻尔理论，下列说法中正确的是A．氢

原子不停地向外辐射能量，且仍一直处于第3能级B．围绕氢原子核运动的电子，其轨道半径是一系列连续的值C．从3能级跃迁到2能级辐射出的光子波长大于从2能级跃迁到1能级辐射出的光子波长D．这些氢原子最多可辐射出2种不同频率的光

子15．如图所示，纸面内有菱形ABCD，几何中心为O点。在A、C两点各固定一个电量为Q的正点电荷，选无穷远为零电势点。下列说法中正确的是A．O点的电势为零B．B点和D点的电场强度大小相等，方向也相同C．电子从B点运动到D点的过程中，电势能一直减小D．只在电场力作用下，电子在该电场中可能做匀速圆周运

动16．如图所示，理想变压器原、副线圈的匝数比为11:1，在原线圈输入u=110sin100πt（V）的电压，副线圈接入的两只不同规格灯泡均恰能正常发光。则A．灯泡L1的额定电压是102VB．灯泡L1中电流的方向每秒钟改变50次C．若将滑片向下移动，灯泡L1亮度不变D．若将

滑片向下移动，灯泡L2亮度变亮17．如图所示，虚线a、b、c、d、e是电场中的一组平行等差等势面，实线是电子仅在电场力作用下的运动轨迹，M、N、P、Q分别为运动轨迹与各等势面的交点。已知电子从M点运动到Q点的过程中电场力做功6eV，设b等势面的电势为零。则下列判断正确的是A．c等势面=－

2VB．电子在P点的加速度比在N点的加速度大C．电子经过P点的动能比经过N点的动能小D．电子在P点的电势能比在N点的电势能大18．在竖直向下、磁感强度为B的匀强磁场中，有一质量为m、长度为L的金属棒MN，两端由等长的轻质细线竖直悬挂。若棒中通以恒定电流，

金属棒从静止开始向上摆动，摆动过程中始终保持水平，最高摆到45=的位置，如图所示。不计空气阻力，则金属棒从静止开始摆到最高位置的过程，下列说法中正确的是A．金属棒MN中电流的方向为从M流向NB．细线的拉力先增大后减小C．金属棒MN中的恒定电流mg

IBL=D．安培力对金属棒MN所做的功为22mgL19．如图所示，平行板电容器AB与直流电源连接，B极板接地。一带电油滴在电容器中的P点刚好处于静止状态。现将A极板竖直向下缓慢地移动一小段距离，则下列判断正确的是A．带电油滴将向A板运动B．电容器的电

容减小C．电容器极板上的电荷量增加D．电容器中的P点电势不变20．如图甲所示，N匝矩形闭合导线框在匀强磁场中绕与磁感线方向垂直的轴线匀速转动，当轴线在ad和bc边的中线处，线框中产生的交变电流的图像如图乙所示。下列说法中正

确的是A．若线框绕cd边以相同角速度匀速转动，线框的感应电动势将变大B．若线框绕ab边以相同角速度匀速转动，线框的感应电动势将不变C．0.4st=时，矩形线框刚好垂直于中性面D．在中性面只将匝数减半，穿过矩形线框的磁通量不

变21．如图所示，在纸面内半径为R的圆形区域中充满了垂直纸面向外的匀强磁场，AO与水平方向的夹角为30°。现有氢的同位素粒子从A点沿水平方向以大小为0v的速度垂直射入磁场，其离开磁场时，速度方向刚好改

变了180°；氢的另一同位素粒子以大小为0v的速度从C点沿CO方向垂直射入磁场。已知的电荷量为e，质量为m，不计粒子的重力和两粒子间的相互作用。下列说法中正确的是A．粒子竖直向下射出磁场B．两粒子在磁场中运动的时间相同C．该匀强磁场的磁感应强度0mvBeR=D．两粒

子从圆形边界的射出点和圆形区域圆心O构成的三角形面积234SR=11H21H11H21H第II卷（非选择题共174分）三、非选择题：包括必考题和选考题两部分。第22题～第32题为必考题，每个试题考生都必须作答。第33题～第38题为选考题，考生根

据要求作答。（一）必考题（共129分）22.（6分）某同学用图甲所示装置测量磁场的磁感应强度。含有大量一价的正、负离子导电液体流过水平管道，管道长为l、宽为d、高为h，置于竖直向上的匀强磁场中。管道上下两面是绝缘板，两侧面M、N是电阻可忽略的导体板

，两导体板与开关S、电阻箱R、灵敏电流表G（内阻为Rg）连接。管道内始终充满导电液体，忽略导电液体的电阻，液体以恒定速度v通过。闭合开关S，调节电阻箱的取值，记下相应的电流表读数。（1）若侧面M导体板带正电，则液体通过管道的方向为_________（选填“自左向右”或“自右向左”）

。（2）如图乙所示为灵敏电流表G在某次指针所指的位置，其读数为_________μA。（3）若某次电阻箱接入电路的阻值R与相应的电流表读数I，则磁感应强度B=（用题中物理量字母I、R、Rg、d、v表示，且均为国际单位）。23.（9分）如图所示，为某物理兴趣小组改装电流表的电路。已知毫安表表头

内阻为60Ω，满偏电流为2mA，R1和R2为定值电阻。若毫安表可改装成量程为4mA和15mA的电流表，则（1）图中表笔A应为色（选填“红”或“黑”）。（2）选择开关S与（选填“a”或“b”）端相接时，电流表的量程为15mA。（3）由题给出的条件和数据，可以求出R1=Ω

，R2=Ω。（4）若表头标记的内阻不准，造成改装后的电流表示数比标准电流表示数偏小，则表头内阻的真实值应（选填“大于”或“小于”）60Ω。24.（12分）如图所示，足够长光滑导轨MN、PQ与水平面间的夹角θ=37°，导轨间距L=0.5m，两轨道间连接定值电

阻R=8Ω，磁感应强度B=1T的匀强磁场竖直向下穿过斜面，质量m=0.1kg、长度L=0.5m的导体杆垂直导轨放置，在外力F作用下导体杆由静止开始沿斜面向上，以a=2m/s2的加速度做匀加速直线运动。导体杆与导轨电阻均不计且始终接触良好，已知sin

370.6=，重力加速度g=10m/s2。求：（1）导体杆从静止开始沿斜面向上运动4s的过程中，通过电阻R的电荷量；（2）在4s末，外力F对导体杆做功的功率。25.（20分）如图所示，质量m1=1kg的木板静止在倾角θ=37°足够

长、固定的光滑斜面上，木板下端上表面与半径R=2m的固定光滑圆弧轨道在A点相切，圆弧轨道最高点D与圆心O等高，C为最低点。OC右侧区域有竖直向上的匀强电场，电场强度E=2×103N/C,电荷量q=-0.01C、质量m2=1.95kg的小滑块静止在长木板顶端，现有一颗质量m0=

0.05kg的子弹以v0=240m/s的速度击中滑块并留在其中，运动过程中小滑块恰好不从木板上端滑出且电量不变，已知小滑块和木板之间的动摩擦因数μ=0.75，sin37°=0.6，重力加速度g=10m/s2。求：（1）小滑块下滑到长木板底端时的速度；（2）小滑块第一次

经过C点时对轨道的压力；（3）从开始运动到小滑块第一次返回斜面到达最高点的过程中，小滑块与木板间摩擦产生的热量。（二）选考题（共45分）33.【物理—选修3-3】(15分)（1）（5分）下列说法正确的是_____

___（填正确答案标号。选对1个得2分，选对2个得4分，先对3个得5分；每选错1个扣3分，最低得分为0分）。A．可以从单一热源吸收热量，使其全部转化成内能而不引起其他变化B．所有晶体都有固定熔点C．做布朗运动的花粉颗粒，温度越高，运动越剧烈D．单位时间内气体分子对容器壁单位面积上碰撞次数

减少，气体的压强一定减小E．由于液体表面层的分子间距离大于液体内部分子间的距离，所以液体表面具有收缩的趋势（2）（10分）如图所示，开口向上的绝热汽缸固定在水平地面上，用一横截面积S＝30cm2的轻质绝

热活塞封闭了一定质量的理想气体，开始时，封闭气体的温度t1＝27℃，活塞到缸底的距离L1＝100cm；现将质量为m的物体放在轻质活塞上，重新达到平衡时，轻质活塞下降了20cm，气体的温度变为t2＝127℃。已知外界大气压p0＝1.0×105Pa不变，取

重力加速度g＝10m/s2，不计一切摩擦。（ⅰ）求物体的质量m；（ⅱ）若使气体降温，轻质活塞又下降了40cm，求此时气体的温度t3。34.【物理—选修3-4】（15分）（1）（5分）两个相干波源S1、S2产生的简谐波在同一种均匀介质中相遇，某时刻两波叠加情况如

图所示。图中实线表示波峰，虚线表示波谷，M和N分别为ab和cd的中点。已知波源S1的振幅为4cm，波源S2的振幅为2cm，下列判断正确的是（填正确答案标号。选对1个得2分，选对2个得4分，先对3个得5分；每选错1个扣3分，最低得分为0分）。A．图示位置

M点的振幅为6cmB．a点一直处于波谷C．再经过半个周期，a点将运动到c点处D．M点的振幅是N点的三倍E．N点经过半个周期，路程为4cm（2）（10分）如图所示，一底面半径为R的圆柱形透明体，O为圆心，一细光束在横截面内从AO边上的A

点以θ=60°的入射角入射，经玻璃折射后到B点，B点到AO的距离为。已知光在真空中的传播速度为c，不考虑光线在射出透明体时的反射。求：（ⅰ）圆柱形透明体的折射率n；（ⅱ）光线从进入透明体到离开透明体所经

历的时间t。泸州市高2019级高二学年末统一考试物理试题参考答案第Ⅰ卷（共54分）选择题（本题包括8小题，每小题6分，共48分。每个小题所给出的四个选项中，有一个或多个是符合题目要求的。全部选对的得6分，选不全的得3

分，有选错或不答的得0分）。题号1415161718192021答案CDCABACBDBD第II卷非选择题（共62分）22.（6分）（1）自左向右···························································

····························································2分（2）160······························································

·································································2分1432R（3）(凡是满足这种关系式都可以)·····················

··········································2分23.（9分）（1）红··················································

·················································································1分（2）a····························································

·········································································2分（3）16，44····························

·······································································（每空2分）4分（4）小于································································

·······························································2分24.（12分）解：(1)设导体杆在4s内运动的位移为x，电荷量为q，由运动学公式得·····································

···························································································1分·····································

···········································································1分·········································

······················································································1分······················································

················································································1分·············································

··························································································1分代入数据得··········

·······················································································································1分（

2）设导体杆受到的安培力为F0，功率为P，由法拉第电磁感应定律知························································································

····························1分················································································

·····················································1分······················································································

···········································1分····························································································1分··

·································································································································1分代

入数据得····························································································································1分25.（20分）解：（1）设碰后

小滑块的速度为1v，由小滑块与子弹组成的系统动量守恒可知12000)(vmmvm+=········································································

··························································································2分16m/sv=子弹与小滑块沿木板向下运动的过程中，由牛顿第二定律可得

202020()sin37()cos37()mmgmmgmma+−+=+··············································2分代入0.75=，得dvRR

Ig)(+221atx=Δcos37BLx=Ent=EIR=qIt=0.8Cq=cos37EBLv=REI=0FBIL=0sin37cos37FmgFma−−=FvP=7.68WP=a=0即子弹与小滑块做匀速直线运动到木板底

端的速度为16m/sv=·····························································································

································1分（2）设子弹与小滑块到C点的速度为v2，由动能定理可知0220202202111()(cos37)()()22mmgRRmmvmmv+−=+−+···························

······························2分在C点，由提供物块做圆周运动的向心力公式可知220202()()vNqEmmgmmR−−+=+······················

·······························································································2分再由牛

顿第三定律知'84NNN==·····························································································

························1分（2）设子弹与小滑块返回斜面向上运动的加速度为0a，木板的加速度为1a，经过时间t与木板达到共同速度3v，由牛顿第二定律知0202020()sin37()cos37()

mmgmmgmma+++=+················································································1分0002111()cos37sin37mmgmgma+−=················

······························································1分tatav101=−················································

·········································································1分1s3t=···························································

········································································1分tav13=··············································

·····················································································1分之后，子弹、滑块和木板一起

向上做匀减速运动，直至速度为零，到达最高点tvvx2311+=··················································································

·······································1分tvx232=····················································

·············································································1分21xxx−=········································

···················································································1分子弹与滑块在木板上来回运动，摩擦产生的热为002()cos372Qmmgx=+··················

······································································1分24JQ=·······························

·································································································1分33.（15分）（1）（填正确

答案标号。选对1个得2分，选对2个得4分，先对3个得5分；每选错1个扣3分，最低得分为0分）BCE·································································································

··········································5分（2）解：（i）设初状态气体的温度为T1，末状态气体的温度T2、压强为P1,活塞到缸底的距离L2由平衡条件可知，P1S=P0S+mg··············

··················································································································

·······················································2分根据理想气体状态方程·························································

··················································································································

········2分代入数据得m=20kg···························································································

····································································································2分（

ii）设末状态气体的温度T3，活塞到缸底的距离L3，由等压变化知3322TSLTSL=··································································································

··························2分代入数据得373Ct=−····························································

································································2分34.（15分）（1）（填正确答案标号。选对1个得2分，选对2个得4分，先对3个得5分；每选错1个扣3分，最低得分为0分）ADE·······

··············································································································

···························5分（2）解：（i）由几何关系知RRBOC23sin=······································································

··············································1分060=BOC由于OBA是等边三角形则060=BAC···································

·······················································································1分由折射定律知0030sin60sin=

n····················································································································

·············1分3=n············································································································

······························1分（ii）由透明体产生全反射时的临界角公式知1sinCn=·······································································

······························································1分光线在B点产生全反射由光在介质中传播速度公式知vcn=·····················

·······················································································································2分011212PSLPSLTT

=由几何关系知23Rs=····························································································

··············································1分则光在介质中的传播时间s=vt····························································

·····························1分cRt233=·······························································

··············1分