高中物理演示课件生活中的圆周运动 50页

第五章新知预习·巧设计名师课堂·一点通要点一要点二第7节创新演练·大冲关随堂检测归纳小结课下作业综合提升要点三Evaluationonly.CreatedwithAspose.Slidesfor.NET3.5ClientProfile5.2.0.0.Copyright2004-2011AsposePtyLtd.Evaluationonly.CreatedwithAspose.Slidesfor.NET3.5ClientProfile5.2.0.0.Copyright2004-2011AsposePtyLtd.\nEvaluationonly.CreatedwithAspose.Slidesfor.NET3.5ClientProfile5.2.0.0.Copyright2004-2011AsposePtyLtd.Evaluationonly.CreatedwithAspose.Slidesfor.NET3.5ClientProfile5.2.0.0.Copyright2004-2011AsposePtyLtd.\nEvaluationonly.CreatedwithAspose.Slidesfor.NET3.5ClientProfile5.2.0.0.Copyright2004-2011AsposePtyLtd.Evaluationonly.CreatedwithAspose.Slidesfor.NET3.5ClientProfile5.2.0.0.Copyright2004-2011AsposePtyLtd.\n1.了解圆周运动在日常生活中的实际应用。2．会在具体问题中分析向心力的来源。3．掌握处理圆周运动综合题目的基本方法。4．知道什么是离心现象，知道离心运动的应用和防止。Evaluationonly.CreatedwithAspose.Slidesfor.NET3.5ClientProfile5.2.0.0.Copyright2004-2011AsposePtyLtd.Evaluationonly.CreatedwithAspose.Slidesfor.NET3.5ClientProfile5.2.0.0.Copyright2004-2011AsposePtyLtd.\nEvaluationonly.CreatedwithAspose.Slidesfor.NET3.5ClientProfile5.2.0.0.Copyright2004-2011AsposePtyLtd.Evaluationonly.CreatedwithAspose.Slidesfor.NET3.5ClientProfile5.2.0.0.Copyright2004-2011AsposePtyLtd.\n[读教材·填要点]1．铁路的弯道(1)火车车轮的结构特点：火车的车轮有凸出的轮缘，且火车在轨道上运行时，有凸出轮缘的一边在两轨道内侧，这种结构特点有助于固定火车运动的轨迹，如图5－7－1所示。图5－7－1Evaluationonly.CreatedwithAspose.Slidesfor.NET3.5ClientProfile5.2.0.0.Copyright2004-2011AsposePtyLtd.Evaluationonly.CreatedwithAspose.Slidesfor.NET3.5ClientProfile5.2.0.0.Copyright2004-2011AsposePtyLtd.\n(2)火车转弯时存在的问题：如果铁路弯道的内外轨一样高，外侧车轮的轮缘挤压外轨，使外轨发生弹性形变，外轨对轮缘的弹力就是火车转弯的向心力，如图5－7－2所示。但火车太大，靠这种办法得到向心力，轮缘与外轨间的太大，铁轨和车轮极易受损。图5－7－2质量相互作用力Evaluationonly.CreatedwithAspose.Slidesfor.NET3.5ClientProfile5.2.0.0.Copyright2004-2011AsposePtyLtd.Evaluationonly.CreatedwithAspose.Slidesfor.NET3.5ClientProfile5.2.0.0.Copyright2004-2011AsposePtyLtd.\n(3)铁路弯道的特点：①转弯处略高于。②铁轨对火车的支持力FN不是竖直向上的，而是斜向弯道。③铁轨对火车的支持力与火车所受重力的合力指向轨道的，它提供了火车做圆周运动的。外轨内轨内侧圆心向心力Evaluationonly.CreatedwithAspose.Slidesfor.NET3.5ClientProfile5.2.0.0.Copyright2004-2011AsposePtyLtd.Evaluationonly.CreatedwithAspose.Slidesfor.NET3.5ClientProfile5.2.0.0.Copyright2004-2011AsposePtyLtd.\n2．拱形桥(1)关于汽车过拱形桥问题，用图表概括如下：mg－FNFN－mgEvaluationonly.CreatedwithAspose.Slidesfor.NET3.5ClientProfile5.2.0.0.Copyright2004-2011AsposePtyLtd.Evaluationonly.CreatedwithAspose.Slidesfor.NET3.5ClientProfile5.2.0.0.Copyright2004-2011AsposePtyLtd.\n减小增大(2)汽车在凸形桥的最高点处于失重状态，在凹形桥的最低点处于超重状态。Evaluationonly.CreatedwithAspose.Slidesfor.NET3.5ClientProfile5.2.0.0.Copyright2004-2011AsposePtyLtd.Evaluationonly.CreatedwithAspose.Slidesfor.NET3.5ClientProfile5.2.0.0.Copyright2004-2011AsposePtyLtd.\nmg－FN失重Evaluationonly.CreatedwithAspose.Slidesfor.NET3.5ClientProfile5.2.0.0.Copyright2004-2011AsposePtyLtd.Evaluationonly.CreatedwithAspose.Slidesfor.NET3.5ClientProfile5.2.0.0.Copyright2004-2011AsposePtyLtd.\n(2)对失重现象的认识：航天器内的任何物体都处于状态，但并不是物体不受重力。正因为受到重力作用才使航天器连同其中的乘员环绕地球转动。4．离心运动(1)定义：物体沿切线飞出或做的逐渐的运动。(2)原因：向心力突然消失或合外力不足以提供所需。(3)应用：洗衣机的脱水筒、离心制管技术。(4)危害：汽车转弯车速过大会造成事故；砂轮、飞轮转速过高时会破裂酿成事故完全失重远离圆心向心力Evaluationonly.CreatedwithAspose.Slidesfor.NET3.5ClientProfile5.2.0.0.Copyright2004-2011AsposePtyLtd.Evaluationonly.CreatedwithAspose.Slidesfor.NET3.5ClientProfile5.2.0.0.Copyright2004-2011AsposePtyLtd.\n[试身手·夯基础]1．在水平铁路转弯处，往往使外轨略高于内轨，这是为了(　　)A．减小火车轮子对外轨的挤压B．减小火车轮子对内轨的挤压C．使火车车身倾斜，利用重力和支持力的合力提供转弯所需的向心力D．限制火车向外脱轨Evaluationonly.CreatedwithAspose.Slidesfor.NET3.5ClientProfile5.2.0.0.Copyright2004-2011AsposePtyLtd.Evaluationonly.CreatedwithAspose.Slidesfor.NET3.5ClientProfile5.2.0.0.Copyright2004-2011AsposePtyLtd.\n解析：火车轨道建成外高内低，火车转弯时，轨道的支持力与火车的重力两者的合力指向弧形轨道的圆心。若合力大于火车转弯所需的向心力(火车速度较小时)，则火车轮缘挤压内侧铁轨；若合力等于所需向心力(火车速度刚好等于规定速度时)，则火车不挤压铁轨；若合力小于所需向心力(火车速度较大时)，则火车挤压外侧铁轨。所以这种设计主要是为了减少对铁轨的挤压破坏，故A、B、C正确。答案：ABCEvaluationonly.CreatedwithAspose.Slidesfor.NET3.5ClientProfile5.2.0.0.Copyright2004-2011AsposePtyLtd.Evaluationonly.CreatedwithAspose.Slidesfor.NET3.5ClientProfile5.2.0.0.Copyright2004-2011AsposePtyLtd.\n2．下列哪些现象是为了防止物体产生离心运动(　　)A．汽车转弯时要限制速度B．转速很高的砂轮半径不能做得太大C．在修筑铁路时，转弯处内轨要低于外轨D．离心水泵工作时答案：ABCEvaluationonly.CreatedwithAspose.Slidesfor.NET3.5ClientProfile5.2.0.0.Copyright2004-2011AsposePtyLtd.Evaluationonly.CreatedwithAspose.Slidesfor.NET3.5ClientProfile5.2.0.0.Copyright2004-2011AsposePtyLtd.\n答案：BEvaluationonly.CreatedwithAspose.Slidesfor.NET3.5ClientProfile5.2.0.0.Copyright2004-2011AsposePtyLtd.Evaluationonly.CreatedwithAspose.Slidesfor.NET3.5ClientProfile5.2.0.0.Copyright2004-2011AsposePtyLtd.\n4．如图5－7－3所示，质量m＝2.0×104kg的汽车以不变的速率先后驶过凹形桥面和凸形桥面，两桥面的圆弧半径均为20m。如果桥面承受的压力不得超过3.0×105N，则：(1)汽车允许的最大速率是多少？(2)若以所求速度行驶，汽车对桥面的最小压力是多少？(g取10m/s2)图5－7－3Evaluationonly.CreatedwithAspose.Slidesfor.NET3.5ClientProfile5.2.0.0.Copyright2004-2011AsposePtyLtd.Evaluationonly.CreatedwithAspose.Slidesfor.NET3.5ClientProfile5.2.0.0.Copyright2004-2011AsposePtyLtd.\n答案：(1)10m/s　(2)105NEvaluationonly.CreatedwithAspose.Slidesfor.NET3.5ClientProfile5.2.0.0.Copyright2004-2011AsposePtyLtd.Evaluationonly.CreatedwithAspose.Slidesfor.NET3.5ClientProfile5.2.0.0.Copyright2004-2011AsposePtyLtd.\nEvaluationonly.CreatedwithAspose.Slidesfor.NET3.5ClientProfile5.2.0.0.Copyright2004-2011AsposePtyLtd.Evaluationonly.CreatedwithAspose.Slidesfor.NET3.5ClientProfile5.2.0.0.Copyright2004-2011AsposePtyLtd.\n图5－7－4Evaluationonly.CreatedwithAspose.Slidesfor.NET3.5ClientProfile5.2.0.0.Copyright2004-2011AsposePtyLtd.Evaluationonly.CreatedwithAspose.Slidesfor.NET3.5ClientProfile5.2.0.0.Copyright2004-2011AsposePtyLtd.\n2．明确圆周平面虽然外轨高于内轨，但整个外轨是等高的，整个内轨是等高的。因而火车在行驶的过程中，重心的高度不变，即火车重心的轨迹在同一水平面内。故火车的圆周平面是水平面，而不是斜面。火车的向心加速度和向心力均是沿水平面而指向圆心。Evaluationonly.CreatedwithAspose.Slidesfor.NET3.5ClientProfile5.2.0.0.Copyright2004-2011AsposePtyLtd.Evaluationonly.CreatedwithAspose.Slidesfor.NET3.5ClientProfile5.2.0.0.Copyright2004-2011AsposePtyLtd.\n3．速度与轨道压力的关系(1)当火车行驶速度v等于规定速度v0时，所需向心力仅由重力和支持力的合力提供，此时内外轨道对火车无挤压作用。(2)当火车行驶速度v与规定速度v0不相等时，火车所需向心力不再仅由重力和支持力的合力提供，此时内外轨道对火车轮缘有挤压作用，具体情况如下：①当火车行驶速度v＞v0时，外轨道对轮缘有侧压力。②当火车行驶速度v＜v0时，内轨道对轮缘有侧压力。Evaluationonly.CreatedwithAspose.Slidesfor.NET3.5ClientProfile5.2.0.0.Copyright2004-2011AsposePtyLtd.Evaluationonly.CreatedwithAspose.Slidesfor.NET3.5ClientProfile5.2.0.0.Copyright2004-2011AsposePtyLtd.\n[名师点睛]汽车、摩托车赛道拐弯处，高速公路转弯处设计成外高内低，也是尽量使车受到的重力和支持力的合力提供向心力，以减小车轮受到地面施加的侧向挤压。Evaluationonly.CreatedwithAspose.Slidesfor.NET3.5ClientProfile5.2.0.0.Copyright2004-2011AsposePtyLtd.Evaluationonly.CreatedwithAspose.Slidesfor.NET3.5ClientProfile5.2.0.0.Copyright2004-2011AsposePtyLtd.\n1．有一列重为100t的火车，以72km/h的速率匀速通过一个内外轨一样高的弯道，轨道半径为400m。(g取10m/s2)(1)试计算铁轨受到的侧压力；(2)若要使火车以此速率通过弯道，且使铁轨受到的侧压力为零，我们可以适当倾斜路基，试计算路基倾斜角度θ的正切值。Evaluationonly.CreatedwithAspose.Slidesfor.NET3.5ClientProfile5.2.0.0.Copyright2004-2011AsposePtyLtd.Evaluationonly.CreatedwithAspose.Slidesfor.NET3.5ClientProfile5.2.0.0.Copyright2004-2011AsposePtyLtd.\n[思路点拨]解答本题时应注意以下两个方面：(1)内外轨一样高时，外轨对轮缘的侧压力提供火车转弯的向心力。(2)火车通过弯道所受侧向压力为零时，重力和铁轨对火车的支持力的合力提供火车转弯的向心力。Evaluationonly.CreatedwithAspose.Slidesfor.NET3.5ClientProfile5.2.0.0.Copyright2004-2011AsposePtyLtd.Evaluationonly.CreatedwithAspose.Slidesfor.NET3.5ClientProfile5.2.0.0.Copyright2004-2011AsposePtyLtd.\n[答案](1)105N　(2)0.1Evaluationonly.CreatedwithAspose.Slidesfor.NET3.5ClientProfile5.2.0.0.Copyright2004-2011AsposePtyLtd.Evaluationonly.CreatedwithAspose.Slidesfor.NET3.5ClientProfile5.2.0.0.Copyright2004-2011AsposePtyLtd.\n解决此类问题时，首先要明确火车转弯做的是圆周运动，其次要找准圆周运动的平面及圆心位置，理解向心力的来源是物体所受的合外力。Evaluationonly.CreatedwithAspose.Slidesfor.NET3.5ClientProfile5.2.0.0.Copyright2004-2011AsposePtyLtd.Evaluationonly.CreatedwithAspose.Slidesfor.NET3.5ClientProfile5.2.0.0.Copyright2004-2011AsposePtyLtd.\n图5－7－5Evaluationonly.CreatedwithAspose.Slidesfor.NET3.5ClientProfile5.2.0.0.Copyright2004-2011AsposePtyLtd.Evaluationonly.CreatedwithAspose.Slidesfor.NET3.5ClientProfile5.2.0.0.Copyright2004-2011AsposePtyLtd.\n[名师点睛](1)在离心现象中并不存在离心力，是外力不足以提供其做圆周运动所需向心力而引起的，是惯性的一种表现形式。(2)做离心运动的物体，并不是沿半径方向向外远离圆心。Evaluationonly.CreatedwithAspose.Slidesfor.NET3.5ClientProfile5.2.0.0.Copyright2004-2011AsposePtyLtd.Evaluationonly.CreatedwithAspose.Slidesfor.NET3.5ClientProfile5.2.0.0.Copyright2004-2011AsposePtyLtd.\n2．下列关于离心现象的说法中正确的是(　　)A．当物体所受的离心力大于向心力时产生离心现象B．做匀速圆周运动的物体，当它所受的一切力都消失时，它将做背离圆心的圆周运动C．做匀速圆周运动的物体，当它所受的一切力都突然消失时，它将沿切线做直线运动D．做匀速圆周运动的物体，当它所受的一切力都突然消失时，它将做曲线运动Evaluationonly.CreatedwithAspose.Slidesfor.NET3.5ClientProfile5.2.0.0.Copyright2004-2011AsposePtyLtd.Evaluationonly.CreatedwithAspose.Slidesfor.NET3.5ClientProfile5.2.0.0.Copyright2004-2011AsposePtyLtd.\n[思路点拨]解答本题时应注意以下两个方面：(1)物体做离心运动的条件；(2)做匀速圆周运动的物体，向心力突然消失时的运动方向。[解析]向心力是根据效果命名的，做匀速圆周运动的物体所需要的向心力，是它所受的某个力或几个力的合力提供的，因此，它并不受向心力和离心力的作用。它之所以产生离心现象是由于F合＝F向＜mω2r，故A错。物体做匀速圆周运动时，若它所受到的力都突然消失，根据牛顿第一定律，它从这时起做匀速直线运动，故C正确，B、D错。[答案]CEvaluationonly.CreatedwithAspose.Slidesfor.NET3.5ClientProfile5.2.0.0.Copyright2004-2011AsposePtyLtd.Evaluationonly.CreatedwithAspose.Slidesfor.NET3.5ClientProfile5.2.0.0.Copyright2004-2011AsposePtyLtd.\n(1)物体提供的力不足以提供向心力时做离心运动；(2)离心后物体可以做直线运动，也可以做曲线运动。Evaluationonly.CreatedwithAspose.Slidesfor.NET3.5ClientProfile5.2.0.0.Copyright2004-2011AsposePtyLtd.Evaluationonly.CreatedwithAspose.Slidesfor.NET3.5ClientProfile5.2.0.0.Copyright2004-2011AsposePtyLtd.\n物体在竖直平面内做圆周运动时，通常受弹力和重力两个力的作用，物体做变速圆周运动，我们只研究物体在最高点和最低点时的两种情形，具体情况又可分为以下两种：Evaluationonly.CreatedwithAspose.Slidesfor.NET3.5ClientProfile5.2.0.0.Copyright2004-2011AsposePtyLtd.Evaluationonly.CreatedwithAspose.Slidesfor.NET3.5ClientProfile5.2.0.0.Copyright2004-2011AsposePtyLtd.\nEvaluationonly.CreatedwithAspose.Slidesfor.NET3.5ClientProfile5.2.0.0.Copyright2004-2011AsposePtyLtd.Evaluationonly.CreatedwithAspose.Slidesfor.NET3.5ClientProfile5.2.0.0.Copyright2004-2011AsposePtyLtd.\nEvaluationonly.CreatedwithAspose.Slidesfor.NET3.5ClientProfile5.2.0.0.Copyright2004-2011AsposePtyLtd.Evaluationonly.CreatedwithAspose.Slidesfor.NET3.5ClientProfile5.2.0.0.Copyright2004-2011AsposePtyLtd.\n3．如图5－7－6所示，质量为m的小球置于正方体的光滑盒子中，盒子的边长略大于球的直径。某同学拿着该盒子在竖直平面内做半径为R的匀速圆周运动，已知重力加速度为g，空气阻力不计，问：(1)要使盒子在最高点时盒子与小球之间恰好无作用力，则该盒子做匀速圆周运动的周期为多少？图5－7－6Evaluationonly.CreatedwithAspose.Slidesfor.NET3.5ClientProfile5.2.0.0.Copyright2004-2011AsposePtyLtd.Evaluationonly.CreatedwithAspose.Slidesfor.NET3.5ClientProfile5.2.0.0.Copyright2004-2011AsposePtyLtd.\n[思路点拨]解答本题时应注意以下两点：(1)小球沿半径方向的合力提供向心力。(2)小球沿运动方向的加速度为零。Evaluationonly.CreatedwithAspose.Slidesfor.NET3.5ClientProfile5.2.0.0.Copyright2004-2011AsposePtyLtd.Evaluationonly.CreatedwithAspose.Slidesfor.NET3.5ClientProfile5.2.0.0.Copyright2004-2011AsposePtyLtd.\nEvaluationonly.CreatedwithAspose.Slidesfor.NET3.5ClientProfile5.2.0.0.Copyright2004-2011AsposePtyLtd.Evaluationonly.CreatedwithAspose.Slidesfor.NET3.5ClientProfile5.2.0.0.Copyright2004-2011AsposePtyLtd.\n设小球受盒子右侧面的作用力为F，受上侧面的作用力为FN，根据牛顿运动定律知在水平方向上有F＝ma＝4mg，在竖直方向上有FN＋mg＝0，即FN＝－mg，F为正值，FN为负值，所以小球对盒子的右侧面和下侧面有作用力，大小分别为4mg和mg。Evaluationonly.CreatedwithAspose.Slidesfor.NET3.5ClientProfile5.2.0.0.Copyright2004-2011AsposePtyLtd.Evaluationonly.CreatedwithAspose.Slidesfor.NET3.5ClientProfile5.2.0.0.Copyright2004-2011AsposePtyLtd.\nEvaluationonly.CreatedwithAspose.Slidesfor.NET3.5ClientProfile5.2.0.0.Copyright2004-2011AsposePtyLtd.Evaluationonly.CreatedwithAspose.Slidesfor.NET3.5ClientProfile5.2.0.0.Copyright2004-2011AsposePtyLtd.\n1．(对应要点一)铁路转弯处的弯道半径r是由地形决定的。弯道处要求外轨比内轨高，其内外轨高度差h的设计不仅与r有关，还与火车在弯道上的行驶速率v有关。下列说法正确的是(　　)A．v一定时，r越小，要求h越大B．v一定时，r越大，要求h越大C．r一定时，v越小，要求h越大D．r一定时，v越大，要求h越大Evaluationonly.CreatedwithAspose.Slidesfor.NET3.5ClientProfile5.2.0.0.Copyright2004-2011AsposePtyLtd.Evaluationonly.CreatedwithAspose.Slidesfor.NET3.5ClientProfile5.2.0.0.Copyright2004-2011AsposePtyLtd.\n答案：ADEvaluationonly.CreatedwithAspose.Slidesfor.NET3.5ClientProfile5.2.0.0.Copyright2004-2011AsposePtyLtd.Evaluationonly.CreatedwithAspose.Slidesfor.NET3.5ClientProfile5.2.0.0.Copyright2004-2011AsposePtyLtd.\n2．(对应要点二)下列有关洗衣机中脱水筒的脱水原理的说法正确的是(　　)A．水滴受离心力作用而背离圆心方向甩出B．水滴受到向心力，由于惯性沿切线方向甩出C．水滴受到的离心力大于它受到的向心力，而沿切线方向甩出D．水滴与衣服间的附着力小于它所需要的向心力，于是水滴沿切线方向甩出Evaluationonly.CreatedwithAspose.Slidesfor.NET3.5ClientProfile5.2.0.0.Copyright2004-2011AsposePtyLtd.Evaluationonly.CreatedwithAspose.Slidesfor.NET3.5ClientProfile5.2.0.0.Copyright2004-2011AsposePtyLtd.\n解析：根据离心运动的特点知，水滴的离心现象是由于水滴与衣服间的附着力小于水滴运动所需要的向心力，即提供的向心力不足，所以水滴沿切线方向甩出，正确选项为D。答案：DEvaluationonly.CreatedwithAspose.Slidesfor.NET3.5ClientProfile5.2.0.0.Copyright2004-2011AsposePtyLtd.Evaluationonly.CreatedwithAspose.Slidesfor.NET3.5ClientProfile5.2.0.0.Copyright2004-2011AsposePtyLtd.\n3．(对应要点三)乘坐如图5－7－7所示游乐园的过山车时，质量为m的人随车在竖直平面内沿圆周轨道运动，下列说法正确的是(　　)A．车在最高点时人处于倒坐状态，全靠保险带拉住，若没有保险带，人一定会掉下去B．人在最高点时对座位仍可能产生压力，但压力一定小于mgC．人在最高点和最低点时的向心加速度大小相等D．人在最低点时对座位的压力大于mg图5－7－7Evaluationonly.CreatedwithAspose.Slidesfor.NET3.5ClientProfile5.2.0.0.Copyright2004-2011AsposePtyLtd.Evaluationonly.CreatedwithAspose.Slidesfor.NET3.5ClientProfile5.2.0.0.Copyright2004-2011AsposePtyLtd.\n答案：DEvaluationonly.CreatedwithAspose.Slidesfor.NET3.5ClientProfile5.2.0.0.Copyright2004-2011AsposePtyLtd.Evaluationonly.CreatedwithAspose.Slidesfor.NET3.5ClientProfile5.2.0.0.Copyright2004-2011AsposePtyLtd.\n4．(对应要点三)如图5－7－8所示，细杆的一端与一小球相连，可绕过O点的水平轴自由转动。现给小球一初速度，使它做圆周运动。图中a、b分别表示小球轨道的最低点和最高点，则杆对球的作用力可能是(　　)A．a处为拉力，b处为拉力B．a处为拉力，b处为推力C．a处为推力，b处为拉力D．a处为推力，b处为推力图5－7－8Evaluationonly.CreatedwithAspose.Slidesfor.NET3.5ClientProfile5.2.0.0.Copyright2004-2011AsposePtyLtd.Evaluationonly.CreatedwithAspose.Slidesfor.NET3.5ClientProfile5.2.0.0.Copyright2004-2011AsposePtyLtd.\n答案：ABEvaluationonly.CreatedwithAspose.Slidesfor.NET3.5ClientProfile5.2.0.0.Copyright2004-2011AsposePtyLtd.Evaluationonly.CreatedwithAspose.Slidesfor.NET3.5ClientProfile5.2.0.0.Copyright2004-2011AsposePtyLtd.\nEvaluationonly.CreatedwithAspose.Slidesfor.NET3.5ClientProfile5.2.0.0.Copyright2004-2011AsposePtyLtd.Evaluationonly.CreatedwithAspose.Slidesfor.NET3.5ClientProfile5.2.0.0.Copyright2004-2011AsposePtyLtd.\n点击下图进入“课下作业综合提升”Evaluationonly.CreatedwithAspose.Slidesfor.NET3.5ClientProfile5.2.0.0.Copyright2004-2011AsposePtyLtd.Evaluationonly.CreatedwithAspose.Slidesfor.NET3.5ClientProfile5.2.0.0.Copyright2004-2011AsposePtyLtd.