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第1页/共2页学科网（北京）股份有限公司大庆市2025届高三年级第一次教学质量检测物理试题答案及评分标准1.A2.B3.D4.C5.A6.C7.D8.AD9.AB10.BCD11.（6分）（每空2分）（1）B（

2）a（3）小于12.（8分）（（1）（2）每空1分，其余每空2分）（1）4.90（2）不需要（5）�u����������（6）s∙�gh�h（或s∙g�h）（7）m1v1=m1v3+m2v213.（10分）解：（1）对活塞受力分析得：P1S=mg+P0S········

··········································2分解得P1=1.2×105Pa根据玻意耳定律得：P0SL0=P1SL····················································2分解得L=5cm······

·································································1分（2）图甲和图乙中气缸内空气的温度为T0=t+273K=300K·············

······················1分根据盖-吕萨克定律得：���������······················································2分解得：T=360K·

···································································2分14.（13分）解：根据牛顿第二定律，对背包列式：��h������

���ܿh�����·························································1分得：����h�（或�����）·····································

····················1分根据牛顿第二定律，对滑雪者列式：��h���������ܿh�����·························································1分得：���

�h��（或�����）··························································1分（2）设滑道AB段的长度为x，滑雪者从顶端到B点所用时间为t，背包的加速度为1a，滑雪者的加速度为2a，初速度为0v，则有对背包：�������u

��···························································1分对滑雪者：�����u������··································

·······················1分{#{QQABLYqAggggQIAAABhCUwW4CgKQkAAAAYgOgBAEMAIAAAFABAA=}#}第2页/共2页学科网（北京）股份有限公司解得：t=1s··

··········································································1分x=4m············································

································1分（3）到达B点时背包和滑雪者的速度大小分别为������u����݉h······························································1分�����u����6�݉h··

····························································1分滑雪者拎起背包过程中背包和滑雪者动量守恒，两者的共同速度为v，则有12()mvMvmMv················

·············································2分解得���hm�݉h···································································1分15.（17分）

解：（1）粒子通过速度选择器时，根据平衡条件，有qvBqE··································2分可得离子速度为����····························································

·1分（2）离子经过静电分析器，根据电场力提供向心力��ʹ������································2分解得�ʹ�������································

·····································1分在磁分析器中，洛伦兹力提供向心力有���ʹ������······································2分解得�ʹ������·································

····································1分（3）离子在圆形磁场中做匀速圆周运动的周期为T��π���·································1分离子在圆形磁场中运动时间������π����············

···································1分离子离开圆形磁场到达到硅片阶段做匀速直线运动����h������������·····················1分加上电场后，离子在复合场中做螺旋线运动，平行平面方向不发生变化

，垂直平面方向即沿电场方向做匀加速运动:�E���··································································1分在复合场中沿电场方向位移���������得���π������m���······

································1分离开复合场时沿电场方向速度������得���������m�·····································1分离开复合场时沿电场方向做匀速运动，位移

为�������得����������m��·······················1分MN间距为����u���π������m���+�������m�����u������mm��········

·····························1分{#{QQABLYqAggggQIAAABhCUwW4CgKQkAAAAYgOgBAEMAIAAAFABAA=}#}