安徽省江南十校2024-2025学年高二上学期12月联考物理答案

2024年“江南十校”高二12月份联考物理参考答案1.答案：C【思路点拨】：麦克斯韦预言了电磁波，赫兹通过实验证实了电磁波的存在。故本题选C。2.答案：B【思路点拨】：根据感应电流产生的条件，A选项中线框磁通量BS,始终不变，故错误；B选项中磁场为非匀强磁场，线框中的磁通量发生变化，故正确；C、D选项中线圈中磁通量始终为零，故错误。故本题选B。3.答案：C【思路点拨】：根据场强叠加原理，B、C两电荷在b处的场强方向与A电荷在b处的场强方向相同，均竖直向下，而B、C两电荷在a处的场强方向与A电荷在a处的场强方向相反，故b点场强大于a处场强，故A、B错误。b到o的电场方向均竖直向下，沿着电场方向电势不断降低，则b点电势高于o点电势，故D错误。电势是标量，根据对称性可知，e点电势等于f点电势，故C正确。所以本题选C。4.答案：A【思路点拨】：根据电场线垂直于等势线，做一电场线与运动轨迹相交，根据曲线运动轨迹和受力特点以及带电粒子的电性，可知云层带负电，避雷针带正电，故B错。根据电场线由高电势指向低电势，故N点电势高于M点电势，因带电粒子带负电，所以带电粒子在M点的电势能大于N点的电势能，故A正确。带电粒子在M点受力方向为该点电场线切向的反方向，故C错误。越靠近避雷针场强越大，加速度越大，故D错。所以本题选A。5.答案：D【思路点拨】：AB.由图线得，测试者呼出的气体中酒精浓度越低，气敏电阻的阻值越大，根据欧姆定律，电阻的阻值越大，电压表示数越大、电流表示数越小，电源的效率越大，AB错误；因为电源的内阻未知，无法确定电源的输出功率的变化情况，C错误；因为气敏R1c2E电阻的阻值随酒精浓度c成反比4，根据闭合电路欧姆定律得I1，R2c1R1R0rEI24R2R0r4R2R0r3R0r3R0rI2，解得44，D正R2R0rI1R2R0r4R2R0r4R2R0r确。故本题选D。第1页共6页{#{QQABIQQUggiAQABAARhCQwWwCgAQkhGAASgOQBAMMAIAiBNABAA=}#}6.答案：B【思路点拨】：将电荷P从C移动到B，电场力做功为WCBUCBqCBq，得WW0.6J，所以克服电场力做功0.6J，故A错误。由题可得，UCA8V，所以CBCAq112V，过B点做AC的垂线，交于D点，根据几何关系可得ADABAC，A241则有UU2V，可得D0VB，可知BD为等势线，则电场方向由C指向A，大DA4CAUU8小为ECACAN/C20N/C，故B正确，C错误。因为Eq0.2J，CA2AB20.2PAA故D错误。所以本题选B。7.答案：B【思路点拨】：A选项中平面磁通量BScos，所以错误；B选项中电通量qESk4r24kq，所以正确；C选项中由于穿过球面的电场线条数未发生变Er2化，电通量不变，所以错误；D选项中球面的磁感应线穿进和穿出条数相等，磁通量为零，所以错误。故本题选B。8.答案：A【思路点拨】：由于球壳内部电场强度为零，所以试探电荷在内部移动时电场力不做功，故电势能不变化，球壳外部电场可等效为点电荷产生的电场，试探电荷由球壳表面向外移动时电场力做正功，电势能渐少，由于电场力越来越小，移动相同位移过程中电场力做功越来越少，电势能减少越来越慢。故本题选A。9.答案：BD【思路点拨】：断开开关S，场强不变，故小球恰好能运动至小孔N处，所以A错误。若仅S将B板下移，根据公式Cr，电容减小；由于二极管的作用可以阻止电容器上的电量4kdQ4kQ流出，故电量不变，根据C，U=Ed，得到E，故场强不变，故到达小孔N时，UrS重力做功小于电场力做功，可知未达到小孔N时速度已经减为零返回了，所以B正确；若S仅将B板上移，根据公式Cr，电容增加，电容器要充电；由于电压U一定，根据4kdU=Ed，电场强度增加；故到达小孔N时，重力做功小于电场力做功，可知未达到小孔N时速度已经减为零返回了，所以C错误；若将滑动变阻器的滑片上移，分压增加，故电容器的电压增加，故如果能到达小孔N，重力做功小于电场力做功，可知未达到小孔N时速度已经减为零返回了,故D正确。故本题选BD。第2页共6页{#{QQABIQQUggiAQABAARhCQwWwCgAQkhGAASgOQBAMMAIAiBNABAA=}#}10.答案：BCD4【思路点拨】：由题意，电场力大小FEqmg方向：水平向右,对小球由a到c的过程，31由动能定理得F0.5RRmgRmv2解得v2gR,在C点运用牛顿第二定律可得：2ccmv210v2NFc，解得Nmg.A错误。hcR,HRR2R。B正确。小球离开c点R32g后竖直方向做竖直上抛运动，设小球离开c点到其轨迹最高点所需的时间为t，有0vcgt，小球沿水平方向做初速度为0的匀加速运动，加速度为a，则有：由牛顿第二定律可知F414ag，此过程小球沿电场方向位移为xat2R，电场力做功为m32334WF0.5RRxmgR，因为电场力做正功，所934以电势能减少量为EmgR。C正确。由题意可P9知，重力与电场力合力方向与竖直方向夹角53°，斜向下如图，沿着合力与垂直合力方向建立坐标系，将C点速度沿着两个方向分解，y方向做匀减速直线运动，x方向做匀速直线运动，此速度为离开42gRC后的最小速度vvsin53，D正确。故本题选BCD。xC511.每空2分(1)红10(2)偏大【思路点拨】：（1）由“红入黑出”可知，b端应与红表笔连接；开关S断开与闭合时相比较断开时欧姆表的内阻大，即中值电阻大，所以开关S闭合时欧姆表的倍率是10。（2）由闭合电路欧姆定律知，当内阻变大，但此表仍能调零，说明欧姆表的内阻不变，当电池电动势变小，通过表头的电流变小，指针的由原来的位置向左偏移，测量结果将比真实值偏大；12.每空2分（1）A20~2mA（只写2mA也给分）（2）（3）0.91（0.89~0.93都给分）3.7×102Ω（3.6~3.8×102Ω都给分）【思路点拨】：(1)实验电路中最大电流约1~2mA，则需选择电流表A2，并用定值电阻R0对第3页共6页{#{QQABIQQUggiAQABAARhCQwWwCgAQkhGAASgOQBAMMAIAiBNABAA=}#}200μA900Ω其量程进行扩充；扩充后的量程为I200μA2000μA2mA，故0~2mA。100Ω(2)电路如图所示900100（3）改装后的电流表内阻为Ω90Ω，设电流表的电流为I，干路电流为I总900100I900E11010r90I总I10I，根据闭合电路欧姆定律得I总，解得R100Rr90IEE1016510310r90根据图像得，5103，解得E0.91V，r3.7102ΩE10000E13.（1）rM3，P输出2.52W（2）会被烧坏UrM【思路点拨】：（1）当R56时，由闭合电路欧姆定律得：···（2分）ErrMR带入数据得：rM3······································（1分）当R9时由闭合电路欧姆定律：UMEI（rR）······································（1分）2电动机得输出功率：P输出IUMIrM······································（1分）解得：P输出2.52W······································（2分）（2）若被卡住，由闭合电路欧姆定律：EI（1RrrM）···························（1分）2电动机得热功率：P热I1rM······································（2分）解得：P热2.56W2W，会被烧坏······································（2分）'1kQq14.（1）FFmg（2）v3gRNN2R2C【思路点拨】：（1）如图，有几何关系可得：3RAC2BC·····························(2分)cosBC2Rcos································(1分)3所以：cos······························（1分）2第4页共6页{#{QQABIQQUggiAQABAARhCQwWwCgAQkhGAASgOQBAMMAIAiBNABAA=}#}1kQq由B点受力分析可得：FF库sinmgsinmg··················(2分)N2R2'1kQq所以：FFmg······························（2分）NN2R2'1kQq（第（1）问如果学生用负电荷计算，得到结果FFmg也给分）NN2R2（2）根据点电荷的电场特征可知，BC所在的圆是一个等势面，所以小球从B到C电场力做的总功为零，由几何关系可得BC的竖直高度为：3hR······························（1分）BC2小球从B到C，根据动能定理可得：311mgRmv2mv2······························（2分）22C2B解得：（1分）vC3gR······························L315.（1）v（2）m60,yL（3）30TcosB2【思路点拨】：(1)粒子水平方向做匀速直线运动：Lv①············································································(1分)xTvvx②······················································································(1分)cosLv··············································································(1分)Tcos(2)逆向思维，沿上边界曲线Ⅰ运动的粒子可视为从B垂直于荧光屏向左做类平抛运动：Eq3La③·········································································(1分)mT2在O点处：3LvaT··············································································(1分)yTvy分tanm3············································································(1)vx分m60······················································································(1)第5页共6页{#{QQABIQQUggiAQABAARhCQwWwCgAQkhGAASgOQBAMMAIAiBNABAA=}#}13yaT2L··········································································(1分)B22(3)设曲线Ⅱ的上的一点Q坐标为(x,y),经过Q的粒子打在