太原市2025年高三模拟考试（一）物理答案

太原市2025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。题号1234567答案DBDABDA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。题号8910答案BCACABD三、实验题：本题包含2小题，共15分。请将正确答案填在题中横线上或按要求作答。11.(6分）kbr（1）A（2分）（2）（2分）（2分）rkg12.（9分）（1）分压式（1分）（2）截止（或极限）（1分）（3）Oa（1分）（4）Ob（1分）3.91015（3.6~4.31015）（2分）ke（1分）（1分）（5）右（1分）四、计算题：本题包含3小题，共39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。（）到达时速度为13.1Mbbv1v2Fmgm1···················································································1分NRFNF压2mg分v1gR····························································································1方法一：M从aa到bb，依能量守恒1mgRmv2Q·················································································1分21Qr1N·······················································································1分Q2r21QmgR························································································1分N4方法二：M从aa到bb，依动能定理12mgRW安mv0······································································1分21W安Q分QQMQN2QN·············································································11QmgR························································································1分N4（2）M到达bb时，N解除锁定，M减速，N加速，最后分别做匀速运动N切割磁感线的有效长度为L,vM、vN水平向右分B2LvMBLvN····················································································1M减速过程，依动量定理，以右为正分I2LBtmvMmv1······································································1N加速过程，依动量定理，以右为正分ILBtmvN0··············································································11vgR························································································1分M52vgR························································································1分N514.（1）电子在间距为rn的轨道上运动时e2v2分k2m························································································3rnrn2nhrmvn2πke2v2mrnn2h2r······················································································1分n2π2ke2m（2）方法一：电子偶素中电子的总动能、势能分别为1E2mv2·····················································································1分k21e2分Ekk·························································································12rne2分Epk··························································································1rnEnEkEpπ2k2e4mE···················································································1分nn2h2方法二：电子偶素中电子或正电子的动能、势能分别为1Emv2··························································································1分k21e2分Ekk·························································································14rne2分Epk··························································································1rnEn2EkEpπ2k2e4mE···················································································1分nn2h2（3）电子偶素由第一激发态跃迁到基态时，放出光子的频率为分hE2E1························································································1π2k2e4mE···················································································1分24h2π2k2e4mE···················································································1分1h23π2k2e4m······················································································1分4h315.（1）方法一：火箭在点火预热阶段喷出气体，火箭对气体的平均作用力为F，依动量定理，以下为正分Ft0mv00······················································································3火箭获得的平均推力FFF································································································1分mvF0·····························································································1分t0方法二：火箭在点火预热阶段喷出气体，火箭对气体的平均作用力为F，分Fm0a·····························································································2va0································································································1分t0火箭获得的平均推力FFF································································································1分mvF0·····························································································1分t0（2）运载物、燃料在喷出气体的过程中动量守恒，以上为正分Mv23mv10················································································33mvv1···························································································1分2M（3）方法一：以地面为参考系，以上为正，第一级燃料燃烧后，运载物获得的速度大小为，v运1依动量守恒定律（分(M2m)v运1m-vv运1)0····························································1第二级燃料燃烧后，运载物获得的速度大小为v运2（分(Mm)v运2m-vv运2)(M2m)v运1·············································1第三级燃料燃烧后，运载物获得的速度大小为v3分Mv3m(vv3)(Mm)v运2······························································1mv解得：v运1M3mmvmvv运2M3mM2mmvmvmvv···················································2分3M3mM2mMm方法二：以火箭为参考系，以上为正，第一级燃料燃烧后，运载物获得的速度增量为，第二级燃v1料燃烧后，运载物获得的速度增量为，第三级燃料燃烧后，运载物获得的速度增量为，v2v3依动量守恒定律分(M2m)v1m（-vv1)0·································