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1、智能提升,1,2,3,4,1,2,3,4,返回,Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0. Copyright 2004-2011 Aspose Pty Ltd.,本题50分 注意：主持人读完题后选手应在150秒内作答！,Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0. Copyright 2004-2011 Aspose Pty Ltd.,如图，一只质量为M的塑料球
2、形容器在A处与水平面接触，它的内部有一根直立的轻弹簧，弹簧下端固定于容器内壁底部，上端系一个带正电，质量为m的小球，加向上的匀强电场后，在弹簧正好恢复原长时，小球恰有最大速度，则当球形容器在A处对桌面压力为零时，小球的加速度a_,Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0. Copyright 2004-2011 Aspose Pty Ltd.,正确答案：,Mg/m,E,由条件可知mg=qE；压力为零时小球在最高点，此时竖直方向的合力为弹簧的弹力即Mg,Evaluation on
3、ly. Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0. Copyright 2004-2011 Aspose Pty Ltd.,本题50分 注意：主持人读完题后选手应在150秒内作答！,Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0. Copyright 2004-2011 Aspose Pty Ltd.,两根光滑绝缘棒在同一竖直面内，两棒与水平面成 角，如下图所示，棒上各穿有一个质量为m、带电量为Q的相
4、同小球，它们在同一高度由静止下滑当两球相距L为_时，小球速度达到最大值,Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0. Copyright 2004-2011 Aspose Pty Ltd.,正确答案：,速度最大时加速度为0即合外力为0。所以有mg=kQ2/L2,Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0. Copyright 2004-2011 Aspose Pty
5、 Ltd.,本题60分 注意：主持人读完题后选手应在150秒内作答！,Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0. Copyright 2004-2011 Aspose Pty Ltd.,如图所示，固定于光滑绝缘水平面上的小球A带正电，质量为2m，另一个质量为m，带负电的小球B以速度 远离A运动时，同时释放小球A，则小球A和B组成的系统在此后的运动过程中，其系统的电势能的最大增量为_,Evaluation only. Created with Aspose.Slides for
6、.NET 3.5 Client Profile 5.2.0.0. Copyright 2004-2011 Aspose Pty Ltd.,正确答案：,释放小球A后由A、B组成的系统动量守恒，mv0=3mv。其中v为系统稳定后速度，此时电势能的增量等于动能的减少量即,Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0. Copyright 2004-2011 Aspose Pty Ltd.,本题60分 注意：主持人读完题后选手应在150秒内作答！,Evaluation only. Crea
7、ted with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0. Copyright 2004-2011 Aspose Pty Ltd.,如下图所示，在竖直放置的半径为R的光滑半圆弧形绝缘细管的圆心处放一点电荷，将质量为m、电荷量为q的小球从圆弧管的水平直径端点A由静止释放，小球沿细管滑到最低点时，对管壁恰好无压力，则放于圆心处O的电荷的电荷量为_,Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0. Copyright 2004-2
8、011 Aspose Pty Ltd.,正确答案：,设放于圆心处O的电荷的电荷量为Q，小球在最低点时应满足 ；再由动能定理可得小球到最低点时满足,Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0. Copyright 2004-2011 Aspose Pty Ltd.,本题30分 注意：主持人读完题后选手应在150秒内作答！,Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0.
9、Copyright 2004-2011 Aspose Pty Ltd.,如图所示为高速摄影机拍摄到的子弹穿过苹果瞬间的照片。该照片经过放大后分析出，在曝光时间内，子弹影响前后错开的距离约为子弹长度的1%2%。已知子弹飞行速度约为500m/s，因此可估算出这幅照片的曝光时间最接近（）,A s B s C s D s,Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0. Copyright 2004-2011 Aspose Pty Ltd.,正确答案：,B s,Evaluation only
10、. Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0. Copyright 2004-2011 Aspose Pty Ltd.,本题30分 注意：主持人读完题后选手应在150秒内作答！,Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0. Copyright 2004-2011 Aspose Pty Ltd.,如图所示，物体A靠在竖直墙面上，在力F作用下，A、B保持静止物体B的受_个力,Evaluation on
11、ly. Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0. Copyright 2004-2011 Aspose Pty Ltd.,正确答案：4个,F,Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0. Copyright 2004-2011 Aspose Pty Ltd.,本题40分 注意：主持人读完题后选手应在150秒内作答！,Evaluation only. Created with Aspose.Sli
12、des for .NET 3.5 Client Profile 5.2.0.0. Copyright 2004-2011 Aspose Pty Ltd.,红宝石激光器的工作物质红宝石含有铬离子的三氧化二铝晶体，利用其中的铬离子产生激光铬离子的能级图如图8所示，E1是基态，E2是亚稳态，E3是激发态，若以脉冲氙灯发出的波长为1的绿光照射晶体，处于基态的铬离子受到激发而跃迁到E3，然后自发地跃迁到E2，释放波长为2的光子，处于亚稳态E2的离子跃迁到基态时辐射出的光就是激光，这种激光的波长为_,Evaluation only. Created with Aspose.Slides for .NET
13、3.5 Client Profile 5.2.0.0. Copyright 2004-2011 Aspose Pty Ltd.,正确答案：,Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0. Copyright 2004-2011 Aspose Pty Ltd.,本题40分 注意：主持人读完题后选手应在150秒内作答！,Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0. Cop
14、yright 2004-2011 Aspose Pty Ltd.,如图所示，光滑水平面上有两个大小相同的钢球A、B，A球的质量大于B球的质量开始时A球以一定的速度向右运动，B球处于静止状态两球碰撞后均向右运动设碰前A球的德布罗意波长为1，碰后A、B两球的德布罗意波长分别为2和3，则1、 2和3关系是_,Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0. Copyright 2004-2011 Aspose Pty Ltd.,正确答案：,解析：由动量守恒定律PA=PA+PB，再由德布罗意波长公式=h/p，得到 或,Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0. Copyright 2004-2011 Aspose Pty Ltd.,
