六五文档>基础教育>试卷>23济宁期末-物理答案
23济宁期末-物理答案
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2022~2023学年度第一学期高三质量检测高三物理参考答案123456789101112CBACDDDCACBDCDBC13.(6分)(1)2(2)大于(每空2分)2cbgt02gt02c14.(8分)(1)CE(2)A(3)0.56(0.53∽0.58均可)(每空2分)15.(7分)解析:(1)对小球由牛顿第二定律得0.2mgmgma...............................................(1分)t小球上升过程中,在竖直方向上vsina....................................................(1分)02解得t1s....................................................................................................................(1分)()小球水平方向做匀速运动,则(分)2dv0cost...............................................1解得d8m.................................................................................................................(1分)小球上升过程中,在竖直方向上2(分)v0sin2ah.................................................1解得h1.5m..............................................................................................................(1分)16.(9分)解析:()线框的边进入磁场时产生的电动势为(分)1abEBLv0........................................1r电路的总电阻Rr..............................................................................................(1分)2E通过ab边的电流为I........................................................................................(1分)R根据平衡条件BILmg...........................................................................................(1分)解得0.45................................................................................................................(1分)(2)由于ab、cd、ef三个边的电阻一样,磁场宽度与ad和de相等,所以穿过磁场过程中,始终只有一个边在切割磁感线,且电路中的总电阻保持不变。B2L2v线框穿过磁场过程中,由动量定理得tmvmv............................(1分)R0B2L23L即.-mvmv..........................................................................................(1分)R011由能量守恒定律Qmv2mv2........................................................................(1分)202解得Q=1.25J..........................................................................................................(1分)17.(14分)解析:(1)设滑块A的质量为m,在斜面上运动的加速度大小为aA。对A由牛顿第二定律mgsinθ=maA...............................................................................(1分)2设下滑L1碰撞前的速度为,则(分)Av0v02aAL1.......................................................1对滑块A、B碰撞过程由动量守恒和能量守恒得(分)mv0mvA2mvB.........................................................................................................1121212mvmv2mv.....................................................................................(.1分)202A2B解得vA=-2m/s,vB=4m/s..............................................................................................(1分)即A的速度大小为2m/s,方向沿斜面向上;B的速度大小为4m/s,方向沿斜面向下。(2)滑块B下滑过程,由牛顿第二定律有2mgsinθ-µ·2mgcosθ=ma,解得a=0,即第一次碰撞后B向下做匀速运动。.........................................................................................(1分)设滑块经与挡板相撞,则L2(分)Bt1t1........................................................................1vB滑块B沿斜面上升过程,对滑块B匀减速过程由牛顿第二定律2mgsinθ+µ·2mgcosθ=2maB.........................................................................................(1分)设沿斜面向上匀减速为零的时间为,则vB(分)Bt2t2.............................................1aB7ttts.............................................................................................................(1分)126vB(3)t2时间内B沿斜面向上滑动的距离为xt............................................(1分)B221t时间内A的位移为xvtat2.....................................................................(1分)AA2A则(分)xAxBL2............................................................................................................12所以第二次碰撞在Q点,即dxm................................................................(1分)B3118.解析:(1)粒子在加速电场中,由动能定理qUmv20.............................(1分)202qU解得v......................................................................................................(1分)0mv2在静电分析器中,由牛顿第二定律得qEm0................................................(1分)L2U解得E.............................................................................................................(1分)L(2)设粒子恰好能打在点M时圆周运动的半径为R。22由几何关系可得L32(分)RLR..........................................................122v2由牛顿第二定律得qvBm0.................................................................................(1分)00R12mU解得B...................................................................................................(1分)0Lq()粒子在加速电场中运动的时间2L(分)3t1.............................................................1v0粒子在偏转电场中运动的时间12L(分)t2............................................................14v03L设粒子在磁场中做圆周运动转动角度为θ,则sin2,解得θ=60°............(1分)R粒子在磁场中运动的时间12L(分)t3.....................................................................16v0125Lm粒子从P运动到M的时间tttt...............................(.1分)12362qU22(4)长方体中的磁感应强度为BBxByB0...................................................(1分)所以所有粒子在磁场中的运动轨迹半径均为L。..........................................................(1分)L由几何关系可知,一个周期内粒子在荧光屏上留下的光斑轨迹是以r为半径,以坐2标为(,,)的点为圆心的圆(如图所示)。(分)003..............................................1L2L2轨迹方程为22,(分)xyz=3.............................................................................14L2

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