#$#%届高三三月联合测评物理试卷参考答案与解析一!选择题本题共!小题#每小题#分#共#分$题号!#&%'()*!$答案+,-.-,-+.,.,-!/#答案$+!%#解析$根据核反应前后%质量数和电荷数守恒%可以得到0为(1&选项,错误'三种射线穿透能力依!%次是!#&选项+正确'射线来自原子核内部%选项.错误''.的半衰期为(&$年%!分钟几!%乎不影响&服药!分钟后再检测%是为了使药里面的'.充分消化分解%释放到呼吸系统%提高检测准确性&选项-错误&$/#答案$,!#解析$单摆周期公式!2#$&通过单位制检测%只有选项,的单位符合周期单位&选项,槡#正确&%/#答案$-%&#解析$因为卫星,和卫星+的质量未知%所以无法比较两者的动能大小&选项,错误'根据$2'#%$#!&'%可知卫星,的周期小于卫星+的周期&选项+错误'根据扇形面积公式(2)*'和!###!%&)!!(+'!$#2&%可得(2*槡$%'%所以2&选项.错误%选项-正确&''#(,槡'##/#答案$.αβ#解析$根据对称性%光线要在圆内全反射%并回到3点的时间最短%应构成一个边数最少的正多边形&!!!全反射临界角456-22!%如图%当是正三角形时%456!2%!#.%无法发生全反射'当是正方形.槡&#!!槡#!/%槡''时%4562%.%可以发生全反射%符合时间最短条件%%槡#'2)*%.2%所以*2%选项.#)/正确&&/#答案$-####!0!#解析$冰壶从,到+%有7##!#0!2),7)+'冰壶从+到.%有7####0#2$7),%又21%22%##0#)+!所以2槡128!%选项-正确&),物理试卷参考答案与解析!第!页!共页{#{QQABQQIAggAgAIAAABhCUQUQCkIQkBCACKoOxFAAIAAAiQFABAA=}#}书'/#答案$,#.!#解析$变压器的原线圈可视为等效电阻%该电阻与串联&单刀双掷开关接%的滑片!'#'!3'#4.##.!上移%增大%增大%故变压器原线圈电压增大%则电压表示数增大%选项正确'将等效'#!'#,'!.##.!为电源内阻%可知%当时%消耗的功率最大%单刀双掷开关接%的滑片上移时%'!2!'#'#3'#4.##.!与的大小关系不明确%消耗的功率如何变化不确定%选项错误'的滑片不动%单'!!'#'#+'#4.#.!#!'#.#.#7'#刀双掷开关由3接到5时%电压表示数6272%由于.!和.#的数值关.!.!#.!.#!'#8'!'#8'!.#.#.!6系不明%故电压表示数6和电流表示数82的大小变化不确定%选项.(-错误&'#(/#答案$-NBffmgAF45°N#解析$对,(+受力分析如图%门闩刚好启动时%对,在水平方向上有92:456%98#::;4%9%对+!&在竖直方向上有:2:456%9%解得92&#%选项-正确&#()/#答案$+.2#解析$根据点电荷电势关系2;%可知,(+(.(-四个点电荷分别在<和=点的电势标量叠加后$<2=<%选项+正确'根据点电荷电场强度关系72;%可知,(+(.(-四个点电荷分别在<和=$$<#点的电场强度矢量叠加后方向互相垂直%选项.正确&*/#答案$,.2000m!###解析$如图%从桅杆顶部发射的无线电波%直接到达悬崖顶处的接收站波程(!2槡#$$$8!!$7#>2!槡%$!'#>'从桅杆顶部发射的无线电波%经海平面发射后到达悬崖顶处的接收站波程(#2!##!槡#$$$8!!$8#>2槡%$&$'#>'波程差(#7(!2&/(>'两列波在接收站处发生干涉后的减弱%则表现为失去无线电联系%又由于在海平面发射时相位发生!)$$的变化%则到达接收站的波程差%应满足(#7(!2#.%选项,.满足条件%选项,.正确&#物理试卷参考答案与解析!第!#页!共页{#{QQABQQIAggAgAIAAABhCUQUQCkIQkBCACKoOxFAAIAAAiQFABAA=}#}!/#答案$,-)$7#解析$*2$时%两导体棒切割磁感线回路产生总感应电动势72,#)$8#,%电流82%对导#'##&,)$体棒%得>2&)$%选项-正确&##%#%!!/#答案$!!'/($!!#分!#左!!#分!!!!&&!2&!!!&分&*!&*&&*##解析$!!'>>8!%@$/$>>2'/($>>!#因&*3&*5%表明滑块,从右到左做加速运动%即右端偏高%故应该适当抬高气垫导轨的左端%直至&*32&*5&???!!!&若滑块,和滑块+碰撞中动量守恒%应满足&!2&!%得&!2&*!&*&&*#&*!&*&!&!&&*#!$/#答案$!!$/'&!!$/'!%$/''也给分!!#分!#@!!!!分!!$/$!!$也给分!!#分#$?!'$8';!&!!&分%7!%大于!!#分#解析$!!?2$/>>8!/&@$/$!>>2$/'&>>!#改用)@!*挡%'02!$/$'07'0%7!金属丝长度对应的电阻%则'可得''故有&0'02'6262#0;2?#'$8'$?!'$8'$!##%7$?!'$8';'%可得#'2$?!'$8'%77'0!%若考虑电压表内阻%则在相同0时实际总电流偏大%实际电压相对62偏小%电压表的测量'$8'值6偏大%670图像中直线的斜率;偏大%故'的测量值偏大&!%@#解析$!!由查理定律1!1#2++++++++++++++++++++++++++++++++++++!&分!!!#物理试卷参考答案与解析!第!&页!共页{#{QQABQQIAggAgAIAAABhCUQUQCkIQkBCACKoOxFAAIAAAiQFABAA=}#}解得1#2!!%$>>AB+++++++++++++++++++++++++++++!#分!#由玻意耳定律1#A!2!18&1A!81A++++++++++++++++++++++++++++!&分解得12#$$>>AB++++++++++++++++++++++++++++++!#分!#@#解析$!!C通过,点%由牛顿第二定律#)+917&++++++++++++++++++++++++++++++++!!分'C由,点运动到.点%由机械能守恒定律!#!#&)+2&)-8&##'+++++++++++++++++++++++++++++!!分##C从.点平抛%有02)-*++++++++++++++++++++++++++++++++++++!!分!#'2#*#++++++++++++++++++++++++++++++++++!!分#解得02$@%>+++++++++++++++++++++++++++++++++!!分!#C由-点运动到,点%由动能定理!#!#&)C2&)+8�C+++++++++++++++++++++++++++++!!分###D由-点运动到7点%由动能定理!#%)52%#0D++++++++++++++++++++++++++++++++!!分##且0C80D20+++++++++++++++++++++++++++++++++!!分解除锁定时%由动量守恒定律$2&)C7%)D+++++++++++++++++++++++++++++++++!#分解得)C2&>,4!)D2!>,4对D%由动量定理#%#*2%)5++++++++++++++++++++++++++++++++++!!分解得*2$@%4+++++++++++++++++++++++++++++++++!!分!&由能量守恒定律%弹簧释放的弹性势能!#!#742&)C8%)D++++++++++++++++++++++++++++++!#分##解得742$@&E++++++++++++++++++++++++++++++++!!分!&@#解析$!!设粒子在匀强磁场中做圆周运动半径为<%由)#,2)2&+++++++++++++++++++++++++++++++++++!#分<解得<2#?如图!C所示%设粒子在磁场中运动轨迹与吸收板左侧相切时%该粒子打到吸收板上的时间最长%有?!456!22++++++++++++++++++++++++++++++++++!#分<#(可得!2'物理试卷参考答案与解析!第!%页!共页{#{QQABQQIAggAgAIAAABhCUQUQCkIQkBCACKoOxFAAIAAAiQFABAA=}#}粒子在磁场中圆周运动的周期#$2!+++++++++++++++++++++++++++++++!#分#$$&解得*>2+++++++++++++++++++++++++++++++++!!分&,200vrrαβPv图C?!!#设粒子运动轨迹与吸收板右侧相切时%则有-45622<#(故2++++++++++++++++++++++++++++++++++++!#分'发射速度方向在(8!8范围内的粒子能打在吸收板上表面被板吸收%板吸收的粒子数与发射粒子总数之比为(8!82++++++++++++++++++++++++++++++++++!#分)#(#F2++++++++++++++++++++++++++++++++++++!!分)&!&吸收板吸收的粒子数与发射数之比)不变%则!与角不变%由!!有<2#?不变%++++++++++++++++++++++++++++++!!分vαr01r0βdv图D如图!D%当粒子发射后%运动半个圆周打在吸收板的最左端%有#!#<#7?#2+++++++++++++++++++++++++++++++!#分!#)#又,2)2&<可得-$#槡!?+++++++++++++++++++++++++++++++!#分物理试卷参考答案与解析!第!页!共页{#{QQABQQIAggAgAIAAABhCUQUQCkIQkBCACKoOxFAAIAAAiQFABAA=}#}