2026版四川高三物理高考真题仿真卷（含答案解析与学生作答区）

高三物理高考真题QS01第页2026版四川高三物理高考真题QS01仿真卷Org091（含答案解析与学生作答区）考试时间：90分钟满分：100分适用对象：四川高三物理高考真题型训练答题说明：先检查试卷，按题号规范作答；选择题填入答题栏，计算题写出必要过程和单位。

2026版四川高三物理高考真题QS01仿真卷Org091（含答案解析与学生作答区）姓名：________________班级：________________考号：________________座位号：________________考试时间：90分钟满分：100分答题说明：本试卷共三大题，24小题。请先检查试卷，按题号作答；选择题在答题栏填写选项；填空题在横线上作答；解答题须写出必要公式、代入过程、单位和结论，书写规范。选择题答题栏（每题只有一个正确选项）：123456789101112一、选择题（本大题共12小题，每小题3分，共36分）1.某质点沿直线运动，位移s与时间t的关系为s=2t+3t²（s的单位为m，t的单位为s）。关于该质点在t=1s时的运动情况，下列说法正确的是（3分）A.速度为5m/s，加速度为3m/s²B.速度为8m/s，加速度为6m/s²C.速度为8m/s，加速度为3m/s²D.速度为5m/s，加速度为6m/s²2.质量为60kg的同学站在电梯中。电梯竖直向上运动且正在减速，减速度大小为2m/s²，取g=10m/s²，则地板对该同学的支持力大小为（3分）A.720NB.600NC.480ND.120N3.从离地20m的高处以15m/s的水平速度抛出一小球，不计空气阻力，取g=10m/s²。小球落地瞬间的速度大小为（3分）A.15m/sB.20m/sC.25m/sD.35m/s4.质量为1.0kg的小物块从光滑曲面上距水平面0.80m处由静止滑下，随后压缩水平面上的轻弹簧。弹簧劲度系数为400N/m，取g=10m/s²，最大压缩量为（3分）A.0.10mB.0.20mC.0.40mD.0.80m5.质量为1000kg的汽车通过半径为40m的凸形桥顶，桥顶处车速为10m/s，取g=10m/s²。汽车在桥顶对桥面的压力大小为（3分）A.2500NB.7500NC.10000ND.12500N

6.将电荷量q=+2.0μC的试探电荷从A点移到B点，电场力做功6.0×10^-6J。下列判断正确的是（3分）A.A点电势比B点低3VB.A点电势比B点低12VC.A点电势比B点高12VD.A点电势比B点高3V7.如图示意电路可等效为：电源电压12V，内阻不计；2Ω电阻与由6Ω、3Ω并联组成的支路串联。通过3Ω电阻的电流为（3分）A.1AB.1.5AC.2AD.3A8.矩形导线框向右匀速进入垂直纸面向里的匀强磁场，进入过程中磁通量逐渐增大。线框中的感应电流方向应为（3分）A.逆时针方向B.顺时针方向C.先顺时针后逆时针D.始终为零9.焦距为10cm的薄凸透镜前30cm处放一小物体。成像性质与像距判断正确的是（3分）A.正立放大虚像，像距15cmB.倒立放大实像，像距30cmC.倒立缩小实像，像距15cmD.正立缩小虚像，像距5cm10.一定质量理想气体体积不变，温度由300K升高到450K。若初态压强为p，则末态压强和气体对外做功分别为（3分）A.1.5p，正功B.1.5p，零C.0.67p，零D.0.67p，负功11.某变力F沿位移x方向作用，其F-x关系为：0到4m内F由0线性增至8N，4到6m内F由8N线性减至0。该变力在0到6m内做功为（3分）A.16JB.20JC.24JD.32J12.用多用电表欧姆挡粗测某电阻。关于规范操作，下列说法正确的是（3分）A.每次读数前都必须更换倍率B.换用不同倍率后应重新欧姆调零C.测量时可用手同时接触两表笔金属部分D.指针偏转很小时应换用更小倍率以提高精度二、填空题（本大题共6小题，每小题3分，共18分）13.一质点做直线运动，其速度满足v=4+2t（v的单位为m/s，t的单位为s）。0到3s内位移为__________m，加速度为__________m/s²。（3分）______________________________________________________________________________14.质量2.0kg的物块在水平面上从静止开始运动，受到水平恒力10N作用，通过5.0m后速度达到4.0m/s。该过程中合外力做功为__________J，滑动摩擦力大小为__________N。（3分）______________________________________________________________________________15.电动势为3.0V、内阻为1.0Ω的电源接5.0Ω外电阻。路端电压为__________V，外电阻消耗的功率为__________W。（3分）______________________________________________________________________________16.某金属逸出功为2.0eV，用能量为3.4eV的光子照射时，逸出光电子的最大初动能为__________eV，对应遏止电压为__________V。（3分）______________________________________________________________________________17.双缝干涉实验中，双缝间距d=0.20mm，屏到双缝距离L=1.0m，相邻亮纹间距3.0mm，则光的波长为__________m；若只减小双缝间距，条纹间距将__________。（3分）______________________________________________________________________________18.1.0mol理想气体在等压条件下由300K升至360K，取R=8.3J/(mol·K)。气体对外做功约为__________J；若该气体定容摩尔热容为3R/2，则内能增加约为__________J。（3分）______________________________________________________________________________三、解答题（本大题共6小题，共46分）19.质量m=2.0kg的小滑块以v0=8.0m/s的初速度沿粗糙斜面向上运动。斜面倾角θ=37°，滑块与斜面间动摩擦因数μ=0.25，取g=10m/s²，sin37°=0.60，cos37°=0.80。求：（1）滑块上滑过程的加速度大小；（2）滑块沿斜面上滑的最大距离；（3）滑块返回斜面底端时的速度大小。（7分）学生作答区：____________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________20.某实验小组用电压表和电流表测量一节干电池的电动势E和内阻r，得到如下数据。请根据表中数据求：（1）U与I的函数关系；（2）电源电动势E和内阻r；（3）说明由U-I图像求内阻时应取图线斜率的哪个物理量。（7分）I/A0.100.200.300.40U/V1.451.401.351.30学生作答区：________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________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________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________23.一透明圆筒底部有小灯S，水面到小灯的距离h=0.80m，水的折射率n=4/3。圆筒上方封闭1.0mol理想气体，气体在缓慢加热过程中保持压强不变，温度由300K升至360K。忽略水面位置变化对光路的影响，取R=8.3J/(mol·K)。求：（1）竖直近轴观察时小灯的视深；（2）水中光射向空气的临界角C的正弦值；（3）气体体积变为原来的几倍；（4）气体对外做功。（8分）学生作答区：____________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________24.某电梯从静止开始竖直向上运行，其加速度a随时间t变化如下：0～2s内a=1.5m/s²，2～6s内a=0，6～8s内a=-1.5m/s²。质量为60kg的乘客站在电梯中，取g=10m/s²。求：（1）8s末电梯速度；（2）8s内电梯上升的位移；（3）全过程电梯底板对乘客做的功；（4）全过程底板对乘客做功的平均功率。（8分）学生作答区：_______________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________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2026版四川高三物理高考真题QS01仿真卷Org091（含答案解析与学生作答区）参考答案与解析1.答案：B解析：s=2t+3t²，则v=ds/dt=2+6t，t=1s时v=8m/s；加速度a=dv/dt=6m/s²。2.答案：C解析：电梯向上运动但减速，加速度方向向下。由N-mg=ma，取向上为正，a=-2m/s²，得N=m(g-2)=480N。3.答案：C解析：竖直方向h=1/2gt²，得t=2s；落地时竖直速度vy=gt=20m/s，合速度v=sqrt(15²+20²)=25m/s。4.答案：B解析：机械能守恒：mgh=1/2kx²，x=sqrt(2mgh/k)=sqrt(16/400)=0.20m。5.答案：B解析：桥顶向心加速度向下，mg-N=mv²/R，N=mg-mv²/R=10000-2500=7500N。压力与支持力等大。6.答案：D解析：电场力做功W=q(U_A-U_B)，U_A-U_B=6.0×10^-6/(2.0×10^-6)=3V，故A点电势高3V。7.答案：C解析：6Ω与3Ω并联等效为2Ω，与外部2Ω串联总电阻4Ω，总电流3A，并联支路电压6V，通过3Ω电阻的电流为2A。8.答案：A解析：进入磁场时向里的磁通量增大，感应电流产生向外的磁场阻碍增大。由右手螺旋定则，线框电流为逆时针方向。9.答案：C解析：由1/f=1/u+1/v，1/10=1/30+1