【AP物理C】【真题】解答题C2005

2005年高考理综北京卷物理部分第Ⅰ卷单项选择题14.下列关于热现象的说法，正确的是DA.外界对物体做功，物体的内能一定增加B.气体的温度升高，气体的压强一定增大C.任何条件下，热量都不会由低温物体传递到高温物体D.任何热机都不可能使燃料释放的热量完全转化为机械能15.在下列各组的两个现象中都表现出光具有波动性的是CA.光的折射现象、色散现象B.光的反射现象、干涉现象C.光的衍射现象、偏振现象D.光的直线传播现象、光电效应现象16.为纪念爱因斯坦对物理学的巨大贡献，联合国将2005年定为“国际物理年”。

对于爱因斯坦提出的质能方程E=mc2，下列说法中不正确的是DA.E=mc2表明物体具有的能量与其质量成正比B.根据ΔE＝Δmc2可计算核反应的能量C.一个质子和一个中子结合成一个氘核时释放能量，表明此过程出现了质量亏损D.E=mc2中的E是发生核反应中释放的核能17.一列简谐机械横波某时刻的波形图如图所示，波源的平衡位置坐标为x=0。

当波源质点处于其平衡位置上方且向下运动时，介质中平衡位置坐标x=2m的质点所处位置及运动情况是AA.在其平衡位置下方且向上运动B.在其平衡位置下方且向下运动C.在其平衡位置上方且向上运动D.在其平衡位置上方且向下运动18.正弦交流电源与电阻R、交流电压表按图1所示的方式连接，R=100Ω，交流表的示数是10V。

图2是交变电源输出电压u随时间t变化的图象，则AA.通过R 的电流i R 随时间t 变化的规律是i R = 2 cos100πt (A)B.通过R 的电流i R 随时间t 变化的规律是i R = 2 cos50πt (A)C.R 两端的电压u R 随时间t 变化的规律是u R =5 2 cos100πt (V)D.R 两端的电压u R 随时间t 变化的规律是u R =5 2 cos50πt (V)19.一人看到闪电12.3s 后又听到雷声。

已知空气中的声速约为330m/s~340m/s ，光速为3×108m/s ，于是他用12.3除以3很快估算出闪电发生位置到他的距离为4.1km 。

2005年全国高考物理试题全集12套（一）目录2005年高考物理试题上海卷 (2)2005年上海物理参考答案 (8)2005年高考广东物理试题 (10)2005年高考广东物理试题参考答案及评分标准 (14)2005年江苏省高考综合考试理科综合试卷 (20)2005年江苏省高考综合考试理综试卷参考答案 (21)2005普通高等学校春季招生考试理科综合能力测试（北京卷） (24)2005普通高等学校春季招生考试理科综合能力测试（北京卷）参考答案 272005年普通高等学校招生全国统一考试（辽宁卷）（物理部分） (29)2005年普通高等学校招生全国统一考试（辽宁卷）（物理部分）参考答案 (30)2005年普通高等学校招生全国统一考试（广东卷）（物理部分） (31)2005年高考物理试题上海卷一．(20分)填空题．本大题共5小题，每小题4分．答案写在题中横线上的空白处或指定位置，不要求写出演算过程．本大题中第l、2、3小题为分叉题。

分A、B两类，考生可任选一类答题．若两类试题均做。

一律按A类题计分．A类题(适合于使用一期课改教材的考生)1A．通电直导线A与圆形通电导线环B固定放置在同一水平面上，通有如图所示的电流时，通电直导线A受到水平向＿＿＿的安培力作用．当A、B中电流大小保持不变，但同时改变方向时，通电直导线A所受到的安培力方向水平向＿＿＿＿．、S2发出的波的波峰位置，2A．如图所示，实线表示两个相干波源S则图中的＿＿＿＿＿点为振动加强的位置，图中的＿＿＿＿＿点为振动减弱的位置．3A．对“落体运动快慢”、“力与物体运动关系”等问题，亚里士多德和伽利略存在着不同的观点．请完成下表：B类题(适合于使用二期课改教材的考生)2B．正弦交流电是由闭合线圈在匀强磁场中匀速转动产生的．线圈中感应电动势随时间变化的规律如图所示，则此感应电动势的有效值为＿＿＿＿V，频率为＿＿＿＿Hz．3B．阴极射线是从阴极射线管的阴极发出的高速运动的粒子流，这些微观粒子是＿＿＿＿＿．若在如图所示的阴极射线管中部加上垂直于纸面向里的磁场，阴极射线将＿＿＿＿＿(填“向上”“向下”“向里”“向外”)偏转．公共题(全体考生必做) B类题(适合于使用二期课改教材的考生)4．如图，带电量为+q的点电荷与均匀带电薄板相距为2d，点电荷到带电薄板的垂线通过板的几何中心．若图中a点处的电场强度为零，根据对称性，带电薄板在图中b点处产生的电场强度大小为＿＿＿＿＿＿，方向＿＿＿＿＿＿．(静电力恒量为k)5．右图中图线①表示某电池组的输出电压一电流关系，图线②表示其输出功率一电流关系．该电池组的内阻为＿＿＿＿＿Ω．当电池组的输出功率为120W 时，电池组的输出电压是＿＿＿＿＿V ．二．(40分)选择题．本大题共8小题，每小题5分．每小题给出的四个答案中，至少有一个是正确的．把正确答案全选出来，并将正确答案前面的字母填写在题后的方括号内．每一小题全选对的得5分；选对但不全，得部分分；有选错或不答的，得O 分．填写在方括号外的字母，不作为选出的答案． 6．2005年被联合国定为“世界物理年”，以表彰爱因斯坦对科学的贡献．爱因斯坦对物理学的贡献有(A)创立“相对论”． (B)发现“X 射线”．(C)提出“光子说”．(D)建立“原子核式模型”．7．卢瑟福通过实验首次实现了原子核的人工转变，核反应方程为4141712781He N O H +→+，下列说法中正确的是(A)通过此实验发现了质子． (B)实验中利用了放射源放出的γ射线．(C)实验中利用了放射源放出的α射线． (D)原子核在人工转变过程中，电荷数可能不守恒． 8．对如图所示的皮带传动装置，下列说法中正确的是(A)A 轮带动B 轮沿逆时针方向旋转． (B)B 轮带动A 轮沿逆时针方向旋转． (C)C 轮带动D 轮沿顺时针方向旋转． (D)D 轮带动C 轮沿顺时针方向旋转．9．如图所示，A 、B 分别为单摆做简谐振动时摆球的不同位置．其中，位置A 为摆球摆动的最高位置，虚线为过悬点的竖直线．以摆球最低位置为重力势能零点，则摆球在摆动过程中 (A)位于B 处时动能最大．(B)位于A 处时势能最大．(C)在位置A 的势能大于在位置B 的动能． (D)在位置B 的机械能大于在位置A 的机械能．10．如图所示的塔吊臂上有一可以沿水平方向运动的小车A ，小车下装有吊着物体B 的吊钩．在小车A与物体B 以相同的水平速度沿吊臂方向匀速运动的同时，吊钩将物体B 向上吊起，A 、B 之间的距离以22d H r =- (SI)(SI 表示国际单位制，式中H 为吊臂离地面的高度)规律变化，则物体做(A)速度大小不变的曲线运动． (B)速度大小增加的曲线运动． (C)加速度大小方向均不变的曲线运动． (D)加速度大小方向均变化的曲线运动．11．如图所示，A 是长直密绕通电螺线管．小线圈B 与电流表连接，并沿A 的轴线OX 从D 点自左向右匀速穿过螺线管A ．能正确反映通过电流表中电流，随工变化规律的是12．在场强大小为E 的匀强电场中，一质量为m 、带电量为q 的物体以某一初速沿电场反方向做匀减速直线运动，其加速度大小为0.8qE/m ，物体运动S 距离时速度变为零．则 (A)物体克服电场力做功qES (B)物体的电势能减少了0.8qES (C)物体的电势能增加了qES (D)物体的动能减少了0.8qES13．A 、B 两列波在某时刻的波形如图所示，经过t ＝T A 时间(T A 为波A 的周期)，两波再次出现如图波形，则两波的波速之比VA ：VB 可能是 (A)1：3 (B)1：2 (C)2：1(D)3：1三．(32分)实验题．14．(6分)部分电磁波的大致波长范围如图所示．若要利用缝宽与手指宽度相当的缝获得明显的衍射现象，可选用___________波段的电磁波，其原因是＿＿＿＿＿＿＿＿＿＿＿＿＿＿＿＿＿＿＿＿＿＿＿＿＿＿＿＿＿＿＿＿＿＿＿＿＿＿＿＿＿＿＿＿＿＿＿＿＿＿＿＿＿＿＿。

2005年高考物理试题上海卷考生注意：1．答卷前，考生务必将姓名、准考证号、校验码等填写清楚．2．本试卷共10页，满分150分．考试时间120分钟．考生应用蓝色或黑色的钢笔或圆珠笔将答案直接写在试卷上．3．本试卷一、四大题中，小题序号后标有字母A 的试题，适合于使用一期课改教材的考生；标有字母B 的试题适合于使用二期课改教材的考生；其它未标字母A 或B 的试题为全体考生必做的试题．不同大题可以选择不同的A 类或B 类试题，但同一大题的选择必须相同．若在同一大题内同时选做A 类、B 类两类试题，阅卷时只以A 类试题计分．4．第19、20、2l 、22、23题要求写出必要的文字说明、方程式和重要的演算步骤．只写出最后答案，而未写出主要演算过程的，不能得分．有关物理量的数值计算问题，答案中 一．(20分)填空题．本大题共5小题，每小题4分．答案写在题中横线上的空白处或指定位置，不要求写出演算过程．本大题中第l 、2、3小题为分叉题。

分A 、B 两类，考生可任选一类答题．若两类试题均做。

一律按A 类题计分．A 类题(适合于使用一期课改教材的考生)1A ．通电直导线A 与圆形通电导线环B 固定放置在同一水平面上，通有如图所示的电流时，通电直导线A 受到水平向＿＿＿的安培力作用．当A 、B 中电流大小保持不变，但同时改变方向时，通电直导线A 所受到的安培力方向水平向＿＿＿＿．2A ．如图所示，实线表示两个相干波源S 1、S 2发出的波的波峰位置，则图中的＿＿＿＿＿点为振动加强的位置，图中的＿＿＿＿＿点为振动减弱的位置．3A ．对“落体运动快慢”、“力与物体运动关系”等问题，亚里士多德和伽利略存在着不同的观点．请完成下表：B 类题(适合于使用二期课改教材的考生)1B ．右面是逻辑电路图及其真值表，此逻辑电路为＿＿门电路，在真值表中X 处的逻辑值为＿＿＿＿＿＿．2B ．正弦交流电是由闭合线圈在匀强磁场中匀速转动产生的．线圈中感应电动势随时间变化的规律如图所示，则此感应电动势的有效值为＿＿＿＿V ，频率为＿＿＿＿Hz ．3B ．阴极射线是从阴极射线管的阴极发出的高速运动的粒子流，这些微观粒子是＿＿＿＿＿．若在如图所示的阴极射线管中部加上垂直于纸面向里的磁场，阴极射线将＿＿＿＿＿(填“向上”“向下”“向里”“向外”)偏转．公共题(全体考生必做) B 类题(适合于使用二期课改教材的考生)4．如图，带电量为+q 的点电荷与均匀带电薄板相距为2d ，点电荷到带电薄板的垂线通过板的几何中心．若图中a 点处的电场强度为零，根据对称性，带电薄板在图中b 点处产生的电场强度大小为＿＿＿＿＿＿，方向＿＿＿＿＿＿．(静电力恒量为k)5．右图中图线①表示某电池组的输出电压一电流关系，图线②表示其输出功率一电流关系．该电池组的内阻为＿＿＿＿＿Ω．当电池组的输出功率为120W 时，电池组的输出电压是＿＿＿＿＿V ． 二．(40分)选择题．本大题共8小题，每小题5分．每小题给出的四个答案中，至少有一个是正确的．把正确答案全选出来，并将正确答案前面的字母填写在题后的方括号内．每一小题全选对的得5分；选对但不全，得部分分；有选错或不答的，得O 分．填写在方括号外的字母，不作为选出的答案． 6．2005年被联合国定为“世界物理年”，以表彰爱因斯坦对科学的贡献．爱因斯坦对物理学的贡献有、(A)创立“相对论”． (B)发现“X 射线”． (C)提出“光子说”．(D)建立“原子核式模型”．7．卢瑟福通过实验首次实现了原子核的人工转变，核反应方程为4141712781He N O H +→+，下列说法中正确的是 (A)通过此实验发现了质子．(B)实验中利用了放射源放出的γ射线． (C)实验中利用了放射源放出的α射线．(D)原子核在人工转变过程中，电荷数可能不守恒． 8．对如图所示的皮带传动装置，下列说法中正确的是 (A)A 轮带动B 轮沿逆时针方向旋转． (B)B 轮带动A 轮沿逆时针方向旋转． (C)C 轮带动D 轮沿顺时针方向旋转． (D)D 轮带动C 轮沿顺时针方向旋转．9．如图所示，A 、B 分别为单摆做简谐振动时摆球的不同位置．其中，位置A 为摆球摆动的最高位置，虚线为过悬点的竖直线．以摆球最低位置为重力势能零点，则摆球在摆动过程中(A)位于B 处时动能最大．(B)位于A 处时势能最大．(C)在位置A 的势能大于在位置B 的动能． (D)在位置B 的机械能大于在位置A 的机械能．10．如图所示的塔吊臂上有一可以沿水平方向运动的小车A ，小车下装有吊着物体B 的吊钩．在小车A 与物体B 以相同的水平速度沿吊臂方向匀速运动的同时，吊钩将物体B 向上吊起，A 、B 之间的距离以22d H r =- (SI)(SI 表示国际单位制，式中H 为吊臂离地面的高度)规律变化，则物体做(A)速度大小不变的曲线运动． (B)速度大小增加的曲线运动． (C)加速度大小方向均不变的曲线运动． (D)加速度大小方向均变化的曲线运动．11．如图所示，A 是长直密绕通电螺线管．小线圈B 与电流表连接，并沿A 的轴线OX 从D 点自左向右匀速穿过螺线管A ．能正确反映通过电流表中电流，随工变化规律的是12．在场强大小为E 的匀强电场中，一质量为m 、带电量为q 的物体以某一初速沿电场反方向做匀减速直线运动，其加速度大小为0.8qE/m ，物体运动S 距离时速度变为零．则 (A)物体克服电场力做功qES (B)物体的电势能减少了0.8qES (C)物体的电势能增加了qES (D)物体的动能减少了0.8qES13．A 、B 两列波在某时刻的波形如图所示，经过t ＝T A 时间(T A 为波A 的周期)，两波再次出现如图波形，则两波的波速之比V A ：V B 可能是 (A)1：3(B)1：2 (C)2：1 (D)3：114．(6分)部分电磁波的大致波长范围如图所示．若要利用缝宽与手指宽度相当的缝获得明显的衍射现象，可选用___________波段的电磁波，其原因是＿＿＿＿＿＿＿＿＿＿＿＿＿＿＿＿＿＿＿＿＿＿＿＿＿＿＿＿＿＿＿＿＿＿＿＿＿＿＿＿。

2005年普通高等学校招生全国统一考试理科综合物理部分试题（北京卷）14.下列关于热现象的说法，正确的是 A .外界对物体做功，物体的内能一定增加 B .气体的温度升高，气体的压强一定增大C .任何条件下，热量都不会由低温物体传递到高温物体D .任何热机都不可能使燃料释放的热量完全转化为机械能 15.在下列各组的两个现象中都表现出光具有波动性的是A .光的折射现象、色散现象B .光的反射现象、干涉现象C .光的衍射现象、偏振现象D .光的直线传播现象、光电效应现象16.为纪念爱因斯坦对物理学的巨大贡献，联合国将2005年定为“国际物理年”。

对于爱因斯坦提出的质能方程E=mc 2，下列说法中不正确的是 A .E ＝mc 2表明物体具有的能量与其质量成正比 B .根据ΔE ＝Δmc 2可计算核反应的能量C .一个质子和一个中子结合成一个氘核时释放能量，表明此过程出现了质量亏损D .E ＝mc 2中的E 是发生核反应中释放的核能17.一列简谐机械横波某时刻的波形图如图所示，波源的平衡位置坐标为x ＝0。

当波源质点处于其平衡位置上方且向下运动时，介质中平衡位置坐标x ＝2m 的质点所处位置及运动情况是 A .在其平衡位置下方且向上运动 B ..在其平衡位置下方且向下运动 C .在其平衡位置上方且向上运动 D .在其平衡位置上方且向下运动18.正弦交流电源与电阻R 、交流电压表按图1所示的方式连接，R ＝100Ω，交流表的示数是10V 。

图2是交变电源输出电压u 随时间t 变化的图象，则1A .通过R 的电流i R 随时间t 变化的规律是i R ＝ 2 cos100πt (A)B .通过R 的电流i R 随时间t 变化的规律是i R ＝ 2 cos50πt (A)C .R 两端的电压u R 随时间t 变化的规律是u R ＝5 2 cos100πt (V)D .R 两端的电压u R 随时间t 变化的规律是u R ＝5 2 cos50πt (V)19.一人看到闪电12.3s 后又听到雷声。

2005高考试题及答案一、选择题（每题3分，共30分）1. 下列哪项不是细胞的基本结构？A. 细胞膜B. 细胞质C. 细胞核D. 细胞壁答案：D2. 光合作用中，光能转化为化学能的场所是：A. 叶绿体B. 线粒体C. 核糖体D. 内质网答案：A3. 以下哪个选项是正确的化学方程式？A. 2H2 + O2 → 2H2OB. 2H2 + O2 → 2HOC. 2H2 + O2 → 2H2O2D. 2H2 + O2 → H2O + H2O2答案：A4. 根据牛顿第三定律，以下说法正确的是：A. 作用力和反作用力大小相等，方向相反B. 作用力和反作用力大小不相等，方向相反C. 作用力和反作用力大小相等，方向相同D. 作用力和反作用力大小不相等，方向相同答案：A5. 在下列选项中，哪一项是正确的物理量单位？A. 速度的单位是米每秒（m/s）B. 质量的单位是千克（kg）C. 力的单位是牛顿（N）D. 所有选项都是正确的答案：D6. 下列关于DNA复制的描述，错误的是：A. DNA复制是半保留性的B. DNA复制需要DNA聚合酶C. DNA复制发生在细胞分裂间期D. DNA复制是全保留性的答案：D7. 以下哪个选项是正确的化学反应速率？A. 反应速率与反应物浓度无关B. 反应速率与温度无关C. 反应速率与催化剂无关D. 所有选项都是错误的答案：D8. 以下哪个选项是正确的电磁波谱？A. 无线电波、微波、红外线、可见光、紫外线、X射线、伽马射线B. 无线电波、红外线、可见光、紫外线、X射线、伽马射线、微波C. 微波、红外线、可见光、紫外线、X射线、伽马射线、无线电波D. 微波、无线电波、红外线、可见光、紫外线、X射线、伽马射线答案：A9. 以下哪个选项是正确的元素周期表的排列规律？A. 按照原子序数递增排列B. 按照原子量递增排列C. 按照电子数递增排列D. 按照质子数递增排列答案：A10. 下列哪项不是生态系统中的生产者？A. 植物B. 动物C. 藻类D. 细菌答案：B二、填空题（每题5分，共30分）1. 细胞膜的主要组成成分是________和________。

【AP物理C】【真题】解答题C2000AP? Physics C2000 Free response QuestionsThe mat erials included in these files are intended for use by AP teachers for course and exam preparation in the classroom; permission for any other use must be sought from the Advanced Placement Program?. Teachers may reproduce them, in whole or in part, in limited quantities, for face-to-face teaching purposes but may not mass distribute the materials, electronically or ot herwise. These mat erials and any copies made of t hem may not be resold, and the copyright notices must be retained as they appear here. This permission does not apply to any third-party copyrights contained herein.These materials were produced by Educational Testing Service? (ETS?), which develops and administers the examinations of the Advanced Placement Program for the College Board. The College Board and Educational Testing Service (ETS) are dedicated to the principle of equal opportunity, and their programs, services, and employment policies are guided by that principle.The College Board is a national nonprofit membershipassociation dedicated to preparing, inspiring, and connecting students to college and opportunity. Founded in 1900, the association iscomposed of more than 4,200 schools, colleges, universities, and other educational organizations. Each year, the College Board serves over three million students and their parents, 22,000 high schools, and 3, 500 colleges, through major programs and services in college admission, guidance, assessment, financial aid, enrollment, and teaching and learning・ Among its best-known programs are the SAT?, the PSAT/NMSQT?, and the Advanced Placement Program? (AP?). The College Board is committed to the principles of equity andexcellence, and that comm it men t is embodied in allof its programs, services, activities, and concerns.API EL is a trademark owned by the College Ent rance Examination Board・ PSAT/NMSQT is a registeted trademark jointly owned by the College Entrance Examination Board and the National Merit Scholarship Corporation. Educational Testing Service and ETS are registered trademarks of Educational Testing Service.Copyright ? 2000 by College Entrance Examination Board.All rights reserved.College Board, Advanced Placement Program, AP, SAT, and the acorn logo are registered trademarks of the College Entrance Examination Board.(2000 Ml) You are conducting an experiment to measure the acceleration due to gravity gu at an unknown location. In the measurement apparatus, a simple pendulum swings past a photogate located at the pendulum^ s lowest point, which records the time tlO for the pendulum to undergo 10 full oscillations. The pendulum consists of a sphere of mass m at the end of a st ring and has a leng th 1. There are four versions of this apparatus, each with a different length. All four are at the unknown location, and the data shown below are sent to you during the experiment.tio T T2 ? (s) (s) (s2) (cm) 12 18 21 32 a. For each pendulum, calculate the period T and the square of the period. Use a reasonable number of signifies nt figures. Enter these results in the table above.b.On the axes below, plot the square of the period versus the length of the pendulum. Draw a best-fit straight linefor this data. c. Assuming that each pendulum undergoessmall amplitude oscillations, from your fit determine the experimental value gexp of the acceleration due to gravity at this unknown location. Justify your answer.d.If the measurement apparatus allows a determination of gexp that is accurate to within 4%, is your experimentalvalue in agreement with the value m/s2 ? Justify your answer.e.Someone informs you that the experimental appara tus is in fac t neat Ear th's surface, but that the experiment hasbeen conducted inside an elevator with a constant acceleration a. Assuming that your experimental value g is exact, determine the magnitude and direction of the elevatoH s acceleration.Copyright ? 2000 by College Entrance Examination Board. All rights reserved.College Board, Advanced Placement Program, AP, SAT, and the acorn logo are registeted trademarks of the College Entrance Examination Board.2000M2. A rubber ball of mass m is dropped from a cliff. As the ball falls, it is subject to air drag (a resistiveforce caused by the air). The drag force on the ball has magnitude bv2, where b is a constant drag coefficient and v is the instantaneous speed of the ball. The drag coeff icie nt b is direc tly proportional to the cross-sectional area of the ball and the density of the air and does not depend on the mass of the ball. As the ball falls, its speed approaches a constant value called the terminal speed.a.On the figure below, draw and label all the forces on the ball at some instant before it reaches terminal speed.b.State whether the magnitude of the acceleration of the ball of mass m increases, decreases, or remains the sameas the ball approaches terminal speed. Explain.c.Write, but do NOT solve, a differential equation for the instantaneous speed v of the ball in terms of time t, thegiven quantities, and fundamentai constants.d.Det ermine the terminal speed vt in t erms of the given quantities and fundamental constants.e.Determine the energy dissipated by the drag force during the fall if the ball is released at height h and reachesits terminal speed before hitting the ground, in terms of the given quantities and fundamental constants.2000M3・ A pulley of radius R1 and rotational inertiaII is mounted on an axle with negligible friction. A light cord passing over the pulley has two blocks of mass m attached to either end, as shown above. Assume that the cord does not slip on the pulley. Det ermine the answers to parts (a) and (b) in t erms of m, Rl, II, and fundamental constants, a. Determine the tension T in the cord.b.One block is now removed from the right and hung on the left・ When the system is released from rest, the threeblocks on the left accelerate downward with an accelera tion g/3 • Det ermine the following, i. The tension T3 in the section of cord supporting the three blocks on the left ii. The tension T1 in the section of cord supporting the single block on the right iii. The rotational inertia II of the pulleyCopyright ? 2000 by College Entrance Examination Board. All rights reserved.College Board, Advanced Placement Program, AP, SAT, and the acorn logo are registered trademarks of the CollegeEntrance Examination Board.c.The blocks are now removed and the cord is tied into a loop, which is passed around the original pulley and asecond pulley of radius 2R1 and rotational inertia 1611. The axis of the original pulley is attached to a motor that rotates it at angular speed ?1, which in turn causes the larger pulley to rotate. The loop does not slip on the pulleys. Det ermine the following in terms of II, RI, and ?1. i. The angular speed ?2 of the larger pulleyii. The angular momentum L2 of the larger pulley iii. The totai kinetic energy of the system2000E1. Lightbulbs A, B, and C are connected in the circuit shown above.a.List the bulbs in order of t heir brigh tn ess, from brightest to least bright・ If any bulbs have the same brightness,state which ones. Justify your answer.Now a switch S and a mH indue tor are added to the circuit; as shown above. The switch is closed at time t二0・b.Det ermine the curre nts IA, IB, and IC for the following times. i. Immediately after the switch is closedii. A long time after the switch is closedCopyright ? 2000 by College Entrance Examination Board. All rights reserved.College Board, Advanced Placement Program, AP, SAT, and the acorn logo are registered trademarks of the College Entrance Examination Board.c.On the axes below, sketch the magnitude of the potential difference VL across the inductor as a function of time,from immediately after the switch is closed untila long time after the switch is closed.d.Now consider a similar circuit with an uncharged ?F capacitor instead of the induetor, as shown above. Theswitch is again closed at time t 二0・ On the axes below, sketch the magnitude of the potential difference Vcap across the capacitor as a function of time, from immediately after the switch is closed until a long time after the switch is closed.Copyright ? 2000 by College Entrance Examination Board. All rights reserved.College Board, Advanced Placement Program, AP, SAT, and the acorn logo are registered trademarks of the College Entrance Examination Board.2000E2・Three particles, A, B, and C, have equal positive charges Q and are held in place at the vertices of anequilater&l triangle with sides of length 1, as shown in the figures below. The dot ted lines represe nt the bisec tors for each side. The base of the t riangle lies on the x-axis, and the altitude of the triangle lies on the y-axis.a.i.Point Pl, the intersection of the three bisec to rs, loca tes the geometric center of the triangle and is one pointwhere the electrie field is zero. On Figure 1 above, draw the elec trie field vec tors EA, EB, and EC at P, due to each of the three charges. Be sure your arrows are drawn to reflect the relative magnitude of the fields・ii.Another poi nt where the elec trie field is zero ispoint P2 at (0, y2). On Figure 2 above, draw electrie field vectors EA, EB, and EC at P2 due to each of the three point charges. Indicate below whether the magnitude of each of these vectors is greater than, less than, or the same as for point Pl.EA EB EC Greater than at Pl Less than at Pl The same as at Pl b. Explain why the x-component of the total electric field is zero at any point on the y-axis.c.Write a general expression for the electrie potential V at any point on the y-axis inside the trianglein terms of Q,1, and y.d.Describe how the answer to part (c) could be used to determine the y-coordinates of points Pl and P2 at whichthe electric field is zero. (You do not need toactually determine these coordinates.)Copyright ? 2000 by College Entrance Examination Board. All rights reserved.College Board, Advanced Placement Program, AP, SAT, and the acorn logo are registered trademarks of the College Entrance Examination Board.2000E3. A capacitor consists of two conducting, coaxial, cylindrical shells of radius a and b, respec ti vely, and leng th L » b. The space bet ween the cylinders is filled with oil that has a dielectric constant K .Initially bothcylinders are uncharged, but then a battery is used to charge the capacitor, leaving a charge +Q on the inner cylinder and -Q on the outer cylinder, as shown above. Let r be the radial distance from the axis of the capacitor.ing Gauss，s 1aw, determine the electrie field midway along the length of the cylinder for the following valuesof r, in t erms of the given qua ntities and fundamental constants. Assume end effects are negligible. i.a b. Determine the following in terms of the givenquantities and fundamentai constants.i. The potential difference across the capacitor ii. The capacitance of this capacitorc.Now the capacitot is discharged and the oil is drained from it. As shown above, a battery of emf ? is connectedto opposite ends of the inner cylinder and a battery of emf 3? is connected to opposite ends of the outer cylinder. Each cylinder has resistance R. Assume that end effects and the contributions to the magnetic field from the wires are negligible. Using Ampere" s law, determine the magnitude Bof the magnetic field midway along the length of the cylinders due to the current in the cylinders for the following values of r. i. a Copyrigh t ? 2000 by College Ent rance Examina tion Board. All rights reserved.College Board, Advanced Placement Program, AP, SAT, and the acorn logo are registered trademarks of the College Entrance Examination Board.AP? Physics C2000 Free response QuestionsThe mat erials included in these files are intended for use by AP teachers for course and exam preparation in the classroom; permission for any other use must be sought from the Advanced Placement Program?. Teachers may reproduce them, in whole or in part, in limited quantities, for face-to-face teaching purposes but may not mass distribute the materials,electronically or otherwise. These mat erials and any copies made of t hem may not be resold, and the copyright notices must be retained as they appear here. This permission does not apply to any third-party copyrights contained herein.These materials were produced by Educational Testing Service? (ETS?), which develops and administers the examinations of the Advanced Placement Program for the College Board. The College Board and Education&l Testing Service (ETS) are dedicated to the principle of equal opportunity, and theirprograms, services, and employment policies are guided by that principle.The College Board is a national nonprofit membership association dedicated to preparing, inspiring, and connecting students to college and opportunity. Founded in 1900, the association iscomposed of more than 4,200 schools, colleges, universities, and other educational organizations. Each year, the College Board serves over three million students and their parents，22,000 high schools, and 3, 500 colleges, through major programs and services in college admission, guidance, assessment, financial aid, enrollment, and teaching andlearning・ Among its best-known programs are the SAT?, the PSAT/NMSQT?, and the Advanced Placement Program? (AP?). The College Board is committed to the principles of equity andexcellence, and that commitment is embodied in all of its programs, services, activities, and concerns.API EL is a trademark owned by the College Ent rance Examination Board・PSAT/NMSQT is a registeted trademark jointly owned by the College Entrance Examination Board and the National Merit Scholarship Corporation. Educational Testing Service and ETS are registered trademarks of Educational Testing Service.Copyright ? 2000 by College Entrance Examination Board. All rights reserved.College Board, Advanced Placement Program, AP, SAT, and the acorn logo are registered trademarks of the College Entrance Examination Board.(2000 Ml) You are conducting an experiment to measure the acceleration due to gravity gu at an unknown location. In the measurement apparatus, a simple pendulum swings past a photogate located at the pendulum^ s lowest point, which records the time tlO for the pendulum to undergo 10 fulloscillations. The pendulum consists of a sphere of mass m at the end of a st ring and has a length 1. There are four versions of this apparatus, each with a different length. All four are at the unknown location, and the data shown below are sent to you during the experiment.tio T T2 ? (s) (s) (s2) (cm) 12 18 21 32 a. For each pendulum, calculate the period T and the square of the period. Use a reasonable number of significant figures. Enter these results in the table above.b.On the axes below, plot the square of the period versus the length of the pendulum. Draw a best-fit straight linefor this data. c. Assuming that each pendulum undergoes small amplitude oscillations, from your fit determine the experimentai value gexp of the acceleration due to gravity at this unknown location. Justify your answer.d.If the measurement apparatus allows a determination of gexp that is accurate to within 4%, is your experimentalvalue in agreement with the value m/s2 ? Justify your answer.e.S omeone informs you that the experimental apparatus isin fact near Earth^ s surface, but that the experiment has been conducted inside an elevator with a constant acceleration a. Assuming that your experimentai value g is exact, determine the magnitude and direction of the elevatoH s acceleration.Copyright ? 2000 by College Entrance Examination Board. All rights reserved.College Board, Advanced Placement Program, AP, SAT, and the acorn logo are registered trademarks of the College Entrance Examination Board.2000M2. A rubber ball of mass m is dropped from a cliff. As the ball falls, it is subject to air drag (a resistive force caused by the air). The drag force on the ball has magnitude bv2, where b is a const&nt drag coefficient and v is the instantaneous speed of the ball. The drag coefficient b is directly proportional to the cross-sectional area of the ball and the density of the air and does not depend on the mass of the ball. As the ball falls, its speed approaches a constant value called the terminal speed.a.On the figure below, draw and label all the forces on the ball at some inst&nt before it reaches terminal speed.b.State whether the magnitude of theacceleration of the ball of mass m increases, decreases, or remains the sameas the ball approaches terminal speed. Explain.c.Write, but do NOT solve, a differential equation for the instantaneous speed v of the ball in terms of time t, thegiven quantities, and fundamentai constants.d・Determine the terminal speed vt in terms ofthe given quantities and fundamental constants.e. Determine the energy dissipated by the drag force during the fall if the ball is released at height h and reachesits terminal speed before hitting the ground, in terms of the given quantities and fundamental constants.2000M3・ A pulley of radius R1 and rotational inertiaII is mounted on an axle with negligible friction. A light cord passing over the pulley has two blocks of mass m attached to either end, as shown above. Assume that the cord does not slip on the pulley.Determine the answers to parts (a) and (b) in terms of m, Rl, II, and fundamental constants. a. Determine the tension T in the cord.b.One block is now removed from the right and hung on the left・ When the system is released from rest, the threeblocks on the left accelerate downward with an accelera tion g/3 • Det ermine the followin g, i. The tension T3 in the section of cord supporting the three blocks on the left ii. The tension T1 in the section of cord supporting the single block on the right iii. The rotation&l inertia II of the pulleyCopyright ? 2000 by College Entrance Examination Board. All rights reserved.College Board, Advanced Placement Program, AP, SAT, and the acorn logo are registered trademarks of the College Entrance Examination Board.c.The blocks are now removed and the cord is tied into a loop, which is passed around the original pulley and asecond pulley of radius 2R1 and rotational inertia 1611. The axis of the original pulley is attached to a motor that rotates it at angular speed ?1, which in turn causes the larger pulley to rotate. The loop does not slip on the pulleys.Det ermine the following in terms of II, RI, and ?1. i. The angular speed ?2 of the larger pulleyii. The angular momentum L2 of the larger pulley iii. The total kinetic energy of the system2000E1. Lightbulbs A, B, and C are connected in the circuit shown above.a.List the bulbs in order of t heir brigh tn ess, from brightest to least bright. If any bulbs have the same brightness,state which ones. Justify your answer.Now a switch S and a mH indue tor are added to the circuit; as shown above. The switch is closed at time t 二0.b.Det ermine the curre nts IA, IB, and IC for the following times. i. Immediately after the switch is closed ii. A long time after the switch is closedCopyright ? 2000 by College Entrance Examination Board. All rights reserved.College Board, Advanced Placement Program, AP, SAT, and the acorn logo are registered trademarks of the College Entrance Examination Board.c.On the axes below, sketch the magnitude of thepotenti&l difference VL across the inductor as a function of time,from immediately after the switch is closed unt订a long time after the switch is closed.d.Now consider a similar circuit with an uncharged ?F capacitor instead of the inductor, as shown above. Theswitch is again closed at time t = 0. On the axes below, sketch the magnitude of the potential difference Vcap across the capacitor as a function of time, from immediately after the switch is closed until a long time after the switch is closed.Copyright ? 2000 by College Entrance Examination Board. All rights reserved.College Board, Advanced Placement Program, AP, SAT,and the acorn logo are registered trademarks of theCollege Entrance Examination Board.。

2005年普通高等学校招生全国统一考试(全国Ⅱ卷)14.B【解析】对物体进行受力分析，如图所示，由牛顿第二定律得mgsinθ−Fcosθ=ma有a=gsinθ−Fcosθm，所以当F逐渐减小时，物体的加速度一定变大，故B正确.15.AD【解析】由折射定律n=sinisinr可知，两单光的入射角相同，但a折射角小，所以n a>n b，又由光在介质中的传播速度公式v=cn可知,v a<v b，故AD正确.16.AC【解析】等压压缩时，外界对气体做功，由热力学第二定律ΔU=W+Q知，如果其他对外界放热，且放出的热量大于外界对气体做的功，则ΔU<0,气体温度降低，故A正确；同理可判断C正确，BD错误.17.C【解析】从能级图可以知道一群氢原子从处于第四能级向低能级跃迁时最多可以发出6种不同频率的光子来，由能级中给出的能量可以知道从第四能级跃迁到第三能级和从第三能级跃迁到第二能级时发出光子的能量比2.22eV小，只有这两种情况不能打出光子来，所以共有四种情况可以从金属表面打出光电子来，故C正确.18.BD【解析】由牛顿第二定律知，月地间相互作用的万有引力是提供月球绕地球做圆周运动的向心力，由F=G m1m2R2=m24π2T2R可以求出地球的质量，但不能求出月球的质量，故A错误B正确；R是月地间的距离，地球的半径也不可能求出，故C错误；由v=2πRT可知能求出月球绕地球的运行的速度大小，故D正确.19.AD【解析】从波的图形中可以直接读出波长为4cm，但不能读出周期,故A正确，B错误；从传播方向可以判断出各点在这一时刻的各质点速度方向，从而知道经过四分之一周期后的质元位置和速度，故C错误，D正确.20.C【解析】由右手定则或楞次定律可以确定开始时刻感应电流的方向从c→d→a→b→c，与规定的正方向相同，而且该时刻感应电流为最大，故C正确.21.ACD【解析】由点电荷产生电场的叠加及正负电荷产生电场方向，以及利用用平行四边形定则来确定合场强的方向，四种情况如图所示，分析可知ACD正确.22.(1)42.12;(2)①连线如图所示;②43.【解析】(1)主尺读数为4.2cm=42mm,游标尺上第6个刻度和主尺上某一刻度对齐,游标尺读数为6×0.02mm=0.12mm,故读数为42mm+6×0.02mm=42.12mm.(2)①连线如图所示;②由闭合电路欧姆定律得，当开关K1闭合、K2断开时,E=I1(R1+R A+R2),当K1、K2均闭合时,E=I2(R1+R A)将数据代入得R A=43Ω.23.Fℎ−12(m A+m B)v2−m B gℎ.【解析】由题意知，在此运动过程中,B的重力势能的增量为m B gℎ，A、B动能的增量为12(m A+m B)v2，恒力F 所做的功为Fℎ，用W表示A克服摩擦力所做的功，由功能关系得Fℎ−W=12(m A+m B)v2+m B gℎ,解得W=Fℎ−12(m A+m B)v2−m B gℎ.24.见解析.【解析】已知带电质点受到的电场力为qE,方向沿z轴正方向；质点受到的重力为mg，沿z轴的负方向.假设质点在x轴上做匀速运动，则它受的洛伦兹力必沿z轴正方向(当v沿x轴正方向)或沿z轴负方向(当v沿x轴负方向)，要质点做匀速运动必分别有qvB+qE=mg①，或qE=qvB+mg②；假设质点在y轴上做匀速运动，则无论沿y轴正方向还是负方向，洛伦兹力都为0，要质点做匀速运动必有qE=mg③;假设质点在z轴上做匀速运动，则它受的洛伦兹力必平行于z轴，而电场力和重力都平行于z轴，三力的合力不可能为0，与假设矛盾，故质点不可能在z轴上做匀速t运动.25.12μg(MLm√g2ℎ−v0)2.【解析】设t为A从离开桌面至落地经历的时间,V表示刚碰后A的速度，A碰后做平抛运动，由平抛运动的规律得ℎ=12gt2①，L=Vt②，设v为刚碰后B的速度的大小，由动量守恒得mv0=MV−mv③，设B后退的距离为l，由功能关系得μmgl=12mv2④，由以上各式联立解得l=12μg (MLm√g2ℎ−v0)2⑤.。

AP® Physics C: Electricity and Magnetism2005 Scoring GuidelinesThe College Board: Connecting Students to College SuccessThe College Board is a not-for-profit membership association whose mission is to connect students to college success and opportunity. Founded in 1900, the association is composed of more than 4,700 schools, colleges, universities, and other educational organizations. Each year, the College Board serves over three and a half million students and their parents, 23,000 high schools, and 3,500 colleges through major programs and services in college admissions, guidance, assessment, financial aid, enrollment, and teaching and learning. Among its best-known programs are the SAT®, the PSAT/NMSQT®, and the Advanced Placement Program® (AP®). The College Board is committed to the principles of excellence and equity, and that commitment is embodied in all of its programs, services, activities, and concerns.Copyright © 2005 by College Board. All rights reserved. College Board, AP Central, APCD, Advanced Placement Program, AP, AP Vertical Teams, Pre-AP, SAT, and the acorn logo are registered trademarks of the College Entrance Examination Board. Admitted Class Evaluation Service, CollegeEd, Connect to college success, MyRoad, SAT Professional Development, SAT Readiness Program, and Setting the Cornerstones are trademarks owned by the College Entrance Examination Board.PSAT/NMSQT is a registered trademark of the College Entrance Examination Board and National Merit Scholarship Corporation. Other products and services may be trademarks of their respective owners. Permission to use copyrighted College Board materials may be requested online at: /inquiry/cbpermit.html.Visit the College Board on the Web: .AP Central is the official online home for the AP Program and Pre-AP: .AP® PHYSICS C ELECTRICITY & MAGNETISM2005 SCORING GUIDELINESGeneral Notes About 2005 AP Physics Scoring Guidelines1. The solutions contain the most common method(s) of solving the free-response questions and theallocation of points for these solutions. Other methods of solution also receive appropriate credit for correct work.2. Generally, double penalty for errors is avoided. For example, if an incorrect answer to part (a) iscorrectly substituted into an otherwise correct solution to part (b), full credit will usually be awarded.One exception to this may be cases when the numerical answer to a later part should be easilyrecognized as wrong, e.g., a speed faster than the speed of light in vacuum.3. Implicit statements of concepts normally receive credit. For example, if use of the equation expressing aparticular concept is worth one point, and a student’s solution contains the application of that equation to the problem but the student does not write the basic equation, the point is still awarded.2g,but use of9.8m s=4. The scoring guidelines typically show numerical results using the value210m s is of course also acceptable.5. Numerical answers that differ from the published answer due to differences in rounding throughout thequestion typically receive full credit. The exception is usually when rounding makes a difference in obtaining a reasonable answer. For example, suppose a solution requires subtracting two numbers that should have five significant figures and that differ starting with the fourth digit (e.g., 20.295 and20.278). Rounding to three digits will lose the accuracy required to determine the difference in thenumbers, and some credit may be lost.Copyright © 2005 by College Entrance Examination Board. All rights reserved.Visit (for AP professionals) and /apstudents (for AP students and parents).2AP ® PHYSICS C ELECTRICITY & MAGNETISM2005 SCORING GUIDELINESQuestion 115 points total Distribution of points Copyright © 2005 by College Entrance Examination Board. All rights reserved.Visit (for AP professionals) and /apstudents (for AP students and parents).(a)(i) 2 pointsFor indicating that the electric field magnitude is greatest at point C 1 point For a correct justification 1 point For example: Field lines are drawn closer together where the field is greater. Note: No credit was awarded for the justification if an incorrect point was chosen.(ii) 2 pointsFor indicating that the electric potential is greatest at point A 1 point For a correct justification 1 pointFor example: The field along is toward the right. The field points in thedirection of decreasing potential, so A must be at the highest potential.0.6 m y = Note: No credit was awarded for the justification if an incorrect point was chosen. (b)(i) 4 pointsFor indicating that the electron moves to the left, stated explicitly or implied 1 point For indicating that the speed increases 1 point For indicating that the acceleration is directed to the left, stated explicitly or implied 1 point For indicating that the magnitude of the acceleration decreases 1 point Example of a good answer: The force on an electron is opposite to the field, so it will move left. The field is weaker to the left so the acceleration will decrease. As long as there is a force on the electron, its speed will continue to increase to the left.(ii) 3 pointsFor using conservation of energy with U q V = 1 point 21 2m q u D =Vu = For correct substitution of values into either equation above 1 pointu =For the correct answer1 point 61.910 m s u =¥Note: Substitution point was awarded if correct answer was indicated .3AP ® PHYSICS C ELECTRICITY & MAGNETISM2005 SCORING GUIDELINESQuestion 1 (continued)Distributionof points(c) 2 pointsV E r D =- 20 V 0.01 mE =For the correct answer with correct units 1 point 2000 V m E = or 2000 N CFor the correct assumption that the field is close enough to uniform in this region todo a calculation as if it were1 point(d) 2 pointsFor drawing a curved line concave up or concave right that passes through point D andat least three electric field lines1 point For drawing the curved line perpendicular to at least three field lines1 pointCopyright © 2005 by College Entrance Examination Board. All rights reserved.Visit (for AP professionals) and /apstudents (for AP students and parents).4AP ® PHYSICS C ELECTRICITY & MAGNETISM2005 SCORING GUIDELINESQuestion 215 points total Distribution of points Copyright © 2005 by College Entrance Examination Board. All rights reserved.Visit (for AP professionals) and /apstudents (for AP students and parents).(a) 3 pointsThe current through the inductor is zero immediately after the switch is closed ().0L I = Using Ohm’s law V IR =For the correct substitution of the emf e for the voltage V1 point For the correct substitution of the total resistance tot R1 point 12tot R R R =+()12init I R R e =+For the correct answer 1 point ()12init I R R e =+(b) 3 pointsFor the correct application of the loop theorem to an appropriate loop of the circuit 1 point For example, using the right-hand loop containing L and 2R 20R L V V +=22R init V R I =L dI V L dt =- 2init dIR I L dt=This equation could also be obtained directly by recognizing that L and 2R are in paralleland have the same voltage across them.For the correct substitution of the current obtained in part (a) 1 point 212dI L R e = For the correct answer1 point ()212R dI dt R R L e =+5AP ® PHYSICS C ELECTRICITY & MAGNETISM2005 SCORING GUIDELINESQuestion 2 (continued)Distributionof points(c) 2 pointsAfter a long time the current is constant, so .0L V = 20L R V V ==, so a constant current goes through resistor 1 and the inductor.1batt R V V =For the correct substitution of both voltage and resistance, using Ohm’s law for 1R V1 point batt 1I R e =For the correct answer 1 point batt 1I R e =(d) 4 pointsFor a graph that rises asymptotically1 point This point must be earned in order to obtain any of the following points.For starting the line above zero1 point For starting the line at the lower limit determined in (a) 1 point For approaching the upper limit determined in part (c)1 point(e) 3 pointsThe current calculated in part (c) that was going through the inductor now goesthrough only resistor 2.For correct application of the loop theorem1 point 2R L I =I , where L I equals determined in (c)batt I For correct substitution of both currents, using Ohm’s law for 2R I with a resistance 2R1 point 221R V R R e=For a correct final answer 1 point 221R V R R e =Copyright © 2005 by College Entrance Examination Board. All rights reserved.Visit (for AP professionals) and /apstudents (for AP students and parents).6AP ® PHYSICS C ELECTRICITY & MAGNETISM2005 SCORING GUIDELINESQuestion 315 points total Distributionof pointsCopyright © 2005 by College Entrance Examination Board. All rights reserved.Visit (for AP professionals) and /apstudents (for AP students and parents).(a) 5 pointsTrial Position of End Q (cm) Measured Magnetic Field (T)(directed from P to Q )n (turns/m) 1 40 49.7010-¥250 2 50 47.7010-¥200 3 60 46.8010-¥ 167 4 80 44.9010-¥ 1255 10044.0010-¥100Dividing 100 turns by the length of the spring will yield the number of turns per meter.For each correct value of n1 point each Two points were deducted for using more than three significant figures.Some students used 0B nI m = and the theoretical value ()7410T m m p -0=¥i to solvefor n . Since the question did not have any indication of using the data to obtain an experimental value for 0m until part (c), full credit for the question could be earned for this approach.(b) 2 pointsFor correctly plotting the data from the chart1 point For a best-fit straight line through the plotted data points, with points both above andbelow the line1 point7AP ® PHYSICS C ELECTRICITY & MAGNETISM2005 SCORING GUIDELINESQuestion 3 (continued)Distributionof points(c) 6 points0S B nI m =For correctly relating the slope of the graph to 0I m1 point 0 slope of line SB I nm D D ==For correctly finding the slope1 point For using at least one point from the graph in the calculation (i.e., not using two pointsfrom the chart that are not on the best-fit line)1 point From the graph shown here()()4449.510 4.510 T 5.010 T130 turns m 240110 turns mS B n D D ---¥-¥¥==- For correctly substituting the obtained slope into the equation for 0m 1 point For correctly substituting the given value of I1 point 401 5.010 T3.0 A 130 turns mex m -¥= ()60 1.310 T m A ex m -=¥i For the correct units1 point(d) 2 pointsFor a correct percent error formula1 point percent error = ()000100exm m m -For using the value of 0ex m from part (c)1 pointpercent error =()()()()767410 T m A 1.310 T m A100410T m Ap p ---¥-¥¥i i ipercent error = ()0.035100- percent error = 3. 5%Copyright © 2005 by College Entrance Examination Board. All rights reserved.Visit (for AP professionals) and /apstudents (for AP students and parents).8。

2005年高考各地试题汇编（一）力学、热学部分一、 力 物体的平衡【B 】（2005辽宁综合卷）36、两光滑平板MO 、NO 构成一具有固定夹角θ0=75°的V 形槽，一球置于槽内，用θ表示NO 板与水平面之间的夹角，如图5所示。

若球对板NO 压力的大小正好等于球所受重力的大小，则下列θ值中哪个是正确的？A 、15°B 、30°C 、45°D 、60°二、 直线运动（2005江苏理综卷）35、 (9分)如图所示为车站使用的水平传送带装置的示意图。

绷紧的传送带始终保持3.Om ／s 的恒定速率运行，传送带的水平部分AB 距水平地面的高度为A=0.45m 。

现有一行李包(可视为质点)由A 端被传送到B 端，且传送到月端时没有被及时取下，行李包从B 端水平抛出，不计空气阻力，g 取lOm ／s 2(1)若行李包从B 端水平抛出的初速v ＝3.Om ／s ，求它在空中运动的时间和飞出的水平距离；(2)若行李包以v 。

＝1.Om ／s 的初速从A 端向右滑行，包与传送带间的动摩擦因数μ＝0.20，要使它从B 端飞出的水平距离等于(1)中所求的水平距离，求传送带的长度L 应满足的条件。

解析：（1）设行李包在空中运动时间为t ，飞出的水平距 离为s ，则 12h g t = s ＝vt代入数据得：t ＝0.3s s ＝0.9m（2）设行李包的质量为m ，与传送带相对运动时的加速度为a ，则 滑动摩擦力F mg ma μ==代入数据得：a ＝2.0m/s 2要使行李包从B 端飞出的水平距离等于（1）中所求水平距离，行李包从B 端飞出的水平抛出的初速度v=3.0m/s设行李被加速到时通过的距离为s 0，则 22002as v v =－ ⑦代入数据得s 0＝2.0m ⑧ 故传送带的长度L 应满足的条件为：L ≥2.0m ⑨三、 牛顿运动定律【BC 】（2005广东物理卷）1、一汽车在路面情况相同的公路上直线行驶，下面关于车速、惯性、质量和滑行路程的讨论，正确的是：Ａ、车速越大，它的惯性越大 Ｂ、质量越大，它的惯性越大Ｃ、车速越大，刹车后滑行的路程越长 Ｄ、车速越大，刹车后滑行的路程越长，所以惯性越大【D 】（2005北京春季理综）20、如图，一个盛水的容器底部有一小孔。