【文档说明】云南省昭通市云天化中学教研联盟2023-2024学年高二下学期7月期末考试 物理 PDF版含答案.pdf,共(9)页,1.776 MB,由小赞的店铺上传
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{#{QQABYYKEoggAAIBAAAhCQQX4CgEQkBEACSgOQEAEoAAAABFABAA=}#}{#{QQABYYKEoggAAIBAAAhCQQX4CgEQkBEACSgOQEAEoAAAABFABAA=}#}{#{QQABYYKEoggAAIBAAAhC
QQX4CgEQkBEACSgOQEAEoAAAABFABAA=}#}{#{QQABYYKEoggAAIBAAAhCQQX4CgEQkBEACSgOQEAEoAAAABFABAA=}#}{#{QQABYYKEoggAAIBAAAhCQQX4CgEQk
BEACSgOQEAEoAAAABFABAA=}#}{#{QQABYYKEoggAAIBAAAhCQQX4CgEQkBEACSgOQEAEoAAAABFABAA=}#}答案第1页,共3页高二物理期末考试参考答案:12345678910DBDBBBCBCADBC11.(1)22271
68~73aaa102VVNSV(2)BC(每空2分)12.(1)AE(2)0.085~0.0870.25(3)cb(空每空2分最后一问3分)13.(1)37.5cmHg;(2)21.5cm【详解】(1)以右管封闭的气体为研究对象,若右管管口封闭,右管中封闭气体的压强为p0=75cmHg..
............................................1分气柱的长度为L1=7.5cm;当左右管中水银面相平,右管中气体压强为p1,气柱的长度为2115cm2hLL.............................
.................1分0112pLSpLS..............................................2分解得p1=37.5cmHg....................
..........................1分(2)以左管封闭的气体为研究对象,开始时左管气体的压强为p2=p0+h=90cmHg.......................................
.......1分气柱的长度为L3=10cm当左右管中水银面相平时,气体的压强为p1,设气柱长为L4,则2314pLSpLS...........................................
...2分解得L4=24cm..............................................2分则活塞上移的距离为14321.5cm2hhLL..............................................2分14.(1)5010m/
sv;(2)110mr;(3)63110s6t【详解】(1)根据动能定理可得20012Uqmv............................................2分{#{QQABYYKEoggAAIBAAAhCQQX4Cg
EQkBEACSgOQEAEoAAAABFABAA=}#}答案第2页,共3页解得175002522501.610m/s10m/s1.610Uqvm............................................1分(2
)在偏转电场中做类平抛运动,可得01Lvt............................................2分根据Eqam...........................
.................1分可得510310m/s3yEqLvatmv............................................1分故出偏转电场的速度为22502310m/s3yvvv..........
..................................1分在磁场中根据2vBqvmr...........................................2分可得在磁场中运动的轨道半径为2551317231.610103m10m203101.6103mvrBq
............................................1分(3)进入磁场时速度方向与竖直方向的夹角为03sin2vv......................
......................1分可得60根据几何关系可知在磁场中偏转角度为60,故在磁场中运动时间为256231711212121.6103s10s66666203101.6103rmtTvBq
...............................1分故从A点进入电场到最终离开磁场的运动时间为666123310s10s110s66ttt...
.........................................2分15.(1)22sin)(LBrRmgvM(2)442232sin)()(sinLBrRRgmrRmgRxQR
.(3)sin)()(2222rRmgxLBLBrRmt【详解】(1)金属棒产生的感应电动势为EBLv..............................................1分
由闭合电路的欧姆定律可得EIRr..............................................1分金属棒受到的安培力为F=BIL...........................................
...1分联立得22BLvFRr当金属棒所受合外力为零时,速度最大,有mgsinθ=F..............................................1分{#{QQABYYKEoggAAIBAAAhCQQX
4CgEQkBEACSgOQEAEoAAAABFABAA=}#}答案第3页,共3页解得v=mg(R+r)sinθB2L2..............................................1分(2)根据能量守恒
定律,可得21sin2mmgxmvQ..............................................2分根据串联规律有RRQQRr..............................................1分联立方程,解
得442232sin)()(sinLBrRRgmrRmgRxQR..............................................1分(3)通过电阻的电荷量为qIt.............
.................................1分电路中的平均感应电流为EIRr..............................................1分平均感应电动势为BlxEt.....................
.........................1分由动量定理可得mgsinθt-LIBtmv.............................................2分解得sin)()(2222rRmgxLBL
BrRmt.............................................1分{#{QQABYYKEoggAAIBAAAhCQQX4CgEQkBEACSgOQEAEoAAA
ABFABAA=}#}