物理答案-2026太原高考一模物理答案_2024-2026高三（6-6月题库）_2026年03月高三试卷_260328山西省太原市2026届高三年级模拟考试（一）（全科）_2026太原高考一模物理

太原市 2026 年高三年级模拟考试（一） 物理参考答案与评分建议 一、选择题：本题共7小题，每小题4分，共28分。在每小题给出的四个选项中，只有一 项是符合题目要求的。 题目 1 2 3 4 5 6 7 答案 D D C A D C B 二、多项选择题：本题包含3小题，每小题6分，共18分。在每小题给出的四个选项中， 至少有两个选项正确，全部选对的得6分，选对但不全的得3分，有选错的得0分。 题目 8 9 10 答案 AC AD BD 三、实验题：本题包含2小题，共16分。 11.（8分） （1）0.650（3分） d （2） （2分） t 1 (M 5m)d2 （3） （3分） 2mkL 12.（8分） L （1）  （2分） S 1 （2） （2分） I b （2分） aS b R （2分） S 0四、计算题：本题包含3小题，共38分。解答应写出必要的文字说明、方程式和重要演算 步骤，只写出最后答案的不能得分。有数值计算的题，答案中必须明确写出数值和单位。 13.（9分） （1）瓶内气体发生等容变化 p p 1  2 ···············································································（3分） T T 1 2 T t 273K 400K 1 1 T t 273K 300K 2 2 p 1.5atm··········································································（1分） 2 （2）瓶内原有气体，据理想气体状态方程 pV pV 1  3 ··········································································（2分） T T 1 2 m V-V 漏  ·········································································（2分） m V 原 m 1 漏  ···············································································（1分） m 3 原 14.（12分） （1）S拨向1，电容器充电 q C  ·················································································（2分） E q CE ················································································（1分） （2）S拨向2，电容器放电，线框加速运动并切割磁感线产生感应电动势， 直到电容器两极间电压与线框的感应电动势相等时，电容器放电结束，线 框获得的速度最大，此时电容器的电压为U,电容器的电荷量为q 1 U  BLv··············································································（1分） q C  1 U 对于线框，由动量定理2iLBt mv0································································（1分） q qq 2it·····························································（1分） 1 CEBL v ······································································（1分） CB2L2 m 1 （3）根据图像可得电容器储能公式为 CU2 ·······························（1分） 2 根据能量守恒定律 1 1 1 CE2  CU2  mv2 Q ················································· （2分） 总 2 2 2 PQ边在加速过程中发出的热量 1 Q  Q ·············································································（1分） 总 2 mCE2 Q  ·································································（1分） 4(mCB2L2) 15.（17分） （1）离子带负电····································································（3分） （2）离子从ab边界进入磁场到N点做匀速圆周运动，根据几何条件， 离子在磁场中的轨迹半径为r 1 r  R··················································································（1分） 1 v2 qv B m 1 ··········································································（1分） 1 r 1 离子从N点到P点做类平抛运动 qE ma ·············································································（1分） 1 1R vt ················································································（1分） 11 1 R  at2 ·············································································（1分） 2 11 1 v  BkR,t  1 1 Bk E  E 2kRB2 ·····································································（1分） 1 （3）离子从距M点0.8R处射入磁场，做匀速圆周运动，半径为r ， 2 Rsinr 0.8R·································································（1分） 2 Rcosr  R·····································································（1分） 2 sin2cos21 r 0.2R 2 离子做匀速圆周运动周期为T ，运动时间为t 磁1 磁1 v2 qv B m 2 2 r 2 v 0.2BkR 2 2πr T  2 磁1 v 2π t  ············································································（1分） 磁1 2Bk 离子从C点进入电场到Q点做类平抛运动，运动时间为t ，OC与OM的 2 夹角θ=37° RsinR v t ···································································（1分） 2 2 1 0.8R  kE t2 ······································································（1分） 2 2 2 从Q点离开电场的速度大小为v ，方向与OQ夹角为α 3 v v  2 3 cos 8 t  ，45，v 0.2 2BkR 2 Bk 3 离子从Q点离开电场进入QON直线上方磁场做匀速圆周运动，运动半径为r ， 3 周期为T ，运动时间为t ，根据几何条件 磁2 磁2 r  2R··············································································（1分） 3 2πr T  3 磁2 v 3 3 t  T 磁2 4 磁2 15π t  ············································································（1分） 磁2 2Bk 离子从N点返回电场到D点做类斜抛运动，运动时间为t ，从D点垂直电场 3 离开，据运动的对称性 8 t t  3 2 Bk 离子第二次离开电场，从D点到ab边界做匀速圆周运动，用时t ，据运动 磁3 的对称性 π t t  磁3 磁1 2Bk t t t t t t 总 磁1 2 磁2 3 磁3 17π32 t  ·······································································（1分） 总 2Bk