物理答案一、单选题(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,只有一项符合题目要求)题号1234567答案BCDDCBA二、多选题(本题共4小题,每小题5分,共20分,在每小题给出的四个选项中,有多项符合题目要求全部选对的得5分,选对但不全的得3分,有选错的得0分)题号891011答案ACDADCDBD三、实验题(共计15分)12、(6分)(2分)(2)C(2分)(3)-2(2分)(9分)(1)B(1分);E(1分)(2)C(2分)(3)5.900(2分)(4)(2分);偏大(1分)计算题(共计37分)(10分)(1)最初弹簧压缩量: ························(1分)封闭气体,初始状态:T1=T0=300KV1=SL ························(1分)对活塞和重物受力分析:解得:· ·······················(1分)当重物与弹簧分离时有: ························(1分)对活塞和重物受力分析:解得: ························(1分)由气体状态方程得: ·······················(1分)联立上式解得:T2=120K ························(1分)(2)从开始到物体刚离开弹簧时,活塞向左移动x=20cm外界对气体做功为:························(3分)(其它解法酌情给分)(12分)微粒到达A点之前做匀速直线运动,对微粒受力分析得:························(2分)解得:························(1分)由平衡条件得:························(1分)电场大小和方向改变后,微粒所受得重力和电场力平衡,微粒在洛伦兹力作用下做匀速圆周运动,由几何知识可得:························(1分)又························(1分)联立求得:························(2分)微粒匀速直线运动时间:························(1分)微粒做匀速圆周运动时间:························(1分)微粒在复合场中运动得总时间为:························(2分)(其它解法酌情给分)(15分)设小球能通过最高点,且此时得速度为v1,上升过程中,小球机械能守恒,则:························(1分)解得:························(1分)设小球达到最高点,轻杆对小球的作用力为F,方向竖直向下,则························(1分)解得:F=4N························(1分)由牛顿第三定律可知,小球对轻杆的作用力大小为4N,方向竖直向上············(1分)解除锁定后,设小球通过最高点时的速度为v2,此时滑块的速度为v,在上升的过程中系统水平方向上动量守恒,以水平向右方向为正方向,则有························(1分)上升过程中,系统机械能守恒,则:························(2分)联立解得:························(1分)则:························(1分)(3)上升过程中,滑块向左运动,小球水平方向向右运动,当小球位置坐标为(x,y)时,此时滑块运动的位移为x,则························(2分)由几何关系可知:························(2分)求得小球的轨迹方程为:························(1分)(其它解法酌情给分)公众号:黑洞视角
江西省九校联盟2023-2024学年高三上学期8月联考物理考试参考答案
2023-11-24
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