【文档说明】江西省九校联盟2023-2024学年高三上学期8月联合考试 物理答案.pdf,共(3)页,274.692 KB,由管理员店铺上传
转载请保留链接:https://www.doc5u.com/view-89a0ff1650c18103a6cf605288f5a9c5.html
以下为本文档部分文字说明:
学科网(北京)股份有限公司物理答案一、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)(1)(2分)(2)C(2分)(3)-2(2分)13、(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)四、计算题(共计37分)14、(10分)(1)最初弹簧压缩量:cmkFx20························(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:FmgSpSp10解得:papp501101························(1分)当重物与弹簧分离时有:
xLSV2························(1分)对活塞和重物受力分析:mgSpSp20解得:pap42108························(1分)由气体状态方程得:222111TVPTVP·····················
··(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:18J20xmgxSPW························
(3分)(其它解法酌情给分)15、(12分)(1)微粒到达A点之前做匀速直线运动,对微粒受力分析得:45sinmgEq························(2分)解得:qmgE22························(1分)
(2)由平衡条件得:45cosmgqvB························(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:Lr2························(1分)又rmv
qvB2························(1分)联立求得:LgqmB2gLv························(2分)gF0ILUD42学科网(北京)股份有限公司(3)微粒匀速直线运动时间:gLvLt221················
········(1分)微粒做匀速圆周运动时间:gLvrt243432························(1分)微粒在复合场中运动得总时间为:gLttt243121··················
······(2分)(其它解法酌情给分)16、(15分)(1)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:20212121mvmgLmv························(1
分)解得:m/s61v························(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LvmFmg21························(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)(2)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02Mvmv························(1分)上升过程中,系统机械能守恒,则:20222212121mvmgLMvmv························(2分)联立解得:m/s22v·········
···············(1分)则:JmvEk42122························(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
xMxLm························(2分)由几何关系可知:222Lyxx)(························(2分)求得小球的轨迹方程为:41412322yx)(························(1分)(其它解法酌情给分)
获得更多资源请扫码加入享学资源网微信公众号www.xiangxue100.com