【AP物理C】【真题】解答题答案C1989.pdf

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【AP物理C】【真题】2004年电磁学解答题答案

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2017ap物理c力学真题答案

2017ap物理c力学真题答案一、单项选择题【答案】a考点：电磁感应现象与电流的磁效应现象分辨a.我国水资源是有限的，所以水能是不可再生能源b.发光二极管主要就是由超导材料制成的c.试电笔在使用时，试电笔上的任何金属都不要接触，否则会有触电危险d.工业上采用超音波探伤仪能够检查出来金属零件内部的裂纹【答案】 d【解析】我国水资源是有限的，但水能是可再生能源，a选项不正确;发光二极管是由半导体材料制成的，b选项不正确;使用试电笔时不能接触笔尖金属体，必须接触笔尾金属体，这样才能形成通路，c选项不正确;故d选项正确，选填d考点：再生能源辨识;发光二极管材料;试电笔的采用;超声波应用领域。

a.使用红外线传输数字信号b.使用超声波传输数字信号c.采用光纤传输数字信号d.采用电磁波传输数字信号【答案】d一、单项选择题a.电路中的电阻变大，灯泡变亮b.电路中的电阻变大，灯泡变暗c.电路中的电阻变大，灯泡变暗d.电路中的电阻变大，灯泡变暗【答案】c【考点定位】滑动变阻器【答案】a【解析】电流表的正确使用方法是：与被测用电器串联;电流必须从电流表的正极流入，负极流出;所测量的电流不能超过电流表的量程;绝对不允许不经过用电器把电流表直接接在电源两极上;电压表的正确使用方法是：把它并联在用电器两端;必须让电流从正接线柱流入，从负接线柱流出(正接线柱与正极相连，负接线柱与负极相连);被测电压不能超过电压表所选的量程，否则会烧坏电压表. a、此图中是电流表，即串联在电路中，且正负接线柱是正确的，故a正确;b、此图中是电流表，相当于导线，所以此时将电流表并联，会发生短路，故b错误;c、此图中是电压表，应该并联在用电器两端，故c错误;d、此图中的电压表虽然并联在用电器两端，但正负接线柱接反了，故d错误;故选a.【考点定位】电流表的采用;电压表的采用a.电流表接电源b.电压表接电源c.导线接电源d.家用电器接电源【答案】b【考点定位】电路的连接a.l1、l2两端实际电压相同b.l1、l2电阻之比是25:38c.l1、l2的电流之比为25 :38d.通过调节电源电压，能使l1、l2同时正常工作【答案】b【解析】【考点定位】并联电路的特点a.电流表与的示数的比值将减小b.电压表的示数与电流表的示数的比值将增大c.电压表的示数与电流表示数的乘积将增大d.电流表与的示数的比值将减小【答案】a【解析】由电路图知道r1与r2并联，a测量干路部分的电流，当滑动变阻器的滑片p向右移动时，r1接入电路的电阻增加，电路总电阻增加，所以a与a2的示数减小，v与a1的示数不变。

大学物理物理c答案

大学物理答案及评分标准（C 卷）一、填空题：1、2m/s -6m/s2、是：保守力做功跟路径无关。

3、ωJ 和221ωJ 4、导体内场强处处为零 5、取向极化和位移极化 6、304r r l Id B d ⨯⋅=μπ 7、M RT 2和M RT 38、R 25和R 23 9、开尔文表述是：不可能从单一的热源吸收热量使之完全变成有用功而不引起其他的变化。

10、频率相同、振动方向相同、位相差恒定。

二、选择题：1、（B ）2、（D ）3、（B ）4、（A ）5（A ）三、判断题：１．（×）２.（×） 3. （×） 4. （×） 5. （×） 6. （√） 7. （×） 8. （×） 9. （×）10. （√）四、解答题：1. 解：（1）根据题意：Kv a -=，所以Kv dt dv -=，分离变量后，Kdt vdv -=，.................................(1分) 积分得，⎰⎰-=t v v Kdt v dv 00，所以有Kt e v t v -=0)(；....................... (3分) 同理，可以求得)1(00Kt e K v x x ---=。

......................................... (1分) （2）根据题意，Kx a =所以， dx Kx dx dtdv ⋅=⋅，积分得⎰⎰=x x v v Kxdx vdv 00；............. (1分) 所以有：)(202202x x K v v -+=.............................................(4分)2. 解：设导体平板的面积为S ，各面的电荷面密度分别为1σ、 2σ、3σ、4σ，根据电荷守恒的条件：A Q S S =+21σσ (1)B Q S S =+43σσ（2）---------------------------- （2分）在金属板内取如图所示的高斯面，根据高斯定理有：032=+σσ （3）---------------------------- (3分)根据场强叠加原理，金属板内某点P 的场强为零：40302012222εσεσεσεσ-+=p E -----------------（3分）联立求解得：SQ Q B A 241+==σσ SQ Q B A 232-=-=σσ----------------------（2分） 3. 解：由于同轴电缆导体内的电流均匀分布，其磁场轴对称分布。

1989年上海高考物理试题及答案

1989年全国普通高等学校招生统一考试上海物理试题考生注意：1．全卷共七大题，在120分钟内完成。

2．第五、六、七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。

有数字计算的问题，答案中必须明确写出数值和单位。

(一)、(32分)每小题4分。

每小题只有一个正确答案，把正确答案前面的字母填写在题后的方括号内。

选对的得4分：选错的或不答的；得0分；选了两个或两个以上的，得4分．填写在方括号外的字母，不作为选出的答案。

（1）下列各组电磁波，按波长由长到短正确排列的是：（A）γ射线、红外线、紫外线、可见光。

（B）红外线、可见光、紫外线、γ射线。

（C）可见光、红外线、紫外线、γ射线。

（D）紫外线、可见光、红外线、γ射线。

（2）下列各组物质，全部都是晶体的是：（A）石英、雪花、沥青。

（B）食盐、橡胶、沥青。

（C）食盐、雪花、石英。

（D）雪花、橡胶、石英。

（3）弹簧振子沿直线作简谐振动。

当振子连续两次经过平衡位置时，振子的（A）加速度相同，动能相同。

（B）动能相同，动量相同。

（C）加速度相同，速度相同。

（D）动量相同，速度相同。

（4）升降机以加速度a竖直向上作匀加速运动。

升降机内的天花板上有一只螺帽相对于地的加速度是（g为重力加速度）：（A）g-a．（B）g+a。

（C）a。

（D）g。

（５）图１为一与电源相接的理想降压变压器。

原线圈中电流为I1，副线圈中电流为I2。

当副线圈中的负载电阻R变小时（A）I2变小，I1变小。

（B）I2变小，I1增大。

（C）I2增大，I1增大。

（D）I2增大，I1变小。

（6）一定质量的理想气体，经历了A—B—C—D的状态变化过程，如图2中的p—V图所示，其中BC段是以p轴和V轴为渐近线的的双曲线。

在p（压强）—T（温度）图上，上述过程可以对应为图3 ：I2R I1图1图2（7）图4所示一通有电流I 的直导线和一矩形导线框平行放置在同一平面上，当线框向哪个方向运动时，才会受到向右的合力。

ap物理c答案

ap物理c答案【篇一：ap物理c】ass=txt>牛顿力学、占整个physics c力学考试的100%其中：a.kinematics ：运动学占18%包括：矢量（ vectors）的概念：既有大小，又有方向；矢量代数（ vector algebra）：矢量和的三角形法则是必须熟练掌握的，最简单的记忆方法就是花萌萌面对两段直的折线路径（对应两段位移矢量之和），她会选择直接连接出发点和终点的直线捷径（等效的对应两个位移矢量和），这样构成了一个矢量和三角形。

矢量的加减，点乘和叉乘，是矢量分析的基础，是我们学习ap物理c的基本数学框架一定要熟练掌握。

矢量在直角坐标系中的分量（components of vectors,coordinate systems）,特别强调的是物理上只会用“右手系”，也就是从x轴到y轴的右手螺旋拇指指向z轴，这个和叉乘的定义是一样的，好记！有了ta，大家在学电磁学的时候就不用左右手的拧麻花了。

ap 物理c还需要掌握柱坐标和球坐标，这在需要柱对称和球对称的积分问题时，就很有用了。

运动学中要用到的三大矢量：位移、速度和加速度（displacement, velocity and acceleration)，特别要注意别把距离（或者叫路程distance），速率（speed）和前面的概念搞混了，后两个概念是标量，只有在一些特殊情况下才和对应矢量的模（大小）相等。

一维运动（motion in one dimension）：一维运动的矢量性就记住有正负的方向就行，对于一维匀加速直线运动，务必掌握其最重要的三个方程：第一求速度的公式，角标i（initial）和f（final）总是代表初和末，这个公式只要从匀加速度等于平均加速度的定义就可以得到：第二个求位移的公式：，这个公式可以理解为保持初速度的匀速运动位移和初速为零的匀加速运动位移之和（第二项在v-t图中是一个三角形面积，底为⊿t，高为a*⊿t）第三个公式是把前面两个公式消去变量⊿t，得：更方便的记忆方法是公式左边用牛顿第二定律f=ma，变成外力做功的形式：f⊿x, 左边多出来的2/m转到右边，右边就正好得到物体动能的变化。

大学物理c的试题及答案

大学物理c的试题及答案一、选择题（每题2分，共20分）1. 下列哪个选项是牛顿第一定律的描述？A. 物体在没有外力作用下，总保持静止或匀速直线运动状态B. 物体的加速度与作用力成正比，与质量成反比C. 物体的加速度与作用力成正比，与质量成正比D. 物体在任何情况下都保持静止或匀速直线运动状态答案：A2. 光在真空中的传播速度是多少？A. 299,792,458 m/sB. 299,792,458 km/sC. 299,792,458 km/hD. 299,792,458 m/h答案：A3. 以下哪个是电场强度的定义？A. 电场力与电荷的比值B. 电荷与电场力的比值C. 电场力与电场强度的比值D. 电场强度与电荷的比值答案：A4. 根据热力学第一定律，系统内能的增加等于系统吸收的热量与对外做的功之和。

A. 正确B. 错误答案：A5. 电磁波的频率与波长的关系是？A. 频率与波长成正比B. 频率与波长成反比C. 频率与波长无关D. 频率与波长成正比，但只在特定条件下成立答案：B6. 根据麦克斯韦方程组，变化的磁场会产生什么？A. 变化的电场B. 恒定的电场C. 恒定的磁场D. 没有影响答案：A7. 欧姆定律描述的是电流、电压和电阻之间的关系，其表达式为？A. I = V/RB. I = R/VC. V = I * RD. R = V/I答案：A8. 以下哪个选项是描述波的干涉现象？A. 两个波相遇时，振幅相加B. 两个波相遇时，振幅相减C. 两个波相遇时，振幅不变D. 两个波相遇时，振幅消失答案：A9. 根据量子力学，电子在原子中的运动状态是由什么决定的？A. 电子的电荷B. 电子的质量C. 电子的能级D. 电子的动量答案：C10. 根据相对论，当物体的速度接近光速时，其质量会如何变化？A. 质量不变B. 质量增加C. 质量减少D. 质量消失答案：B二、填空题（每题2分，共20分）1. 根据牛顿第二定律，物体的加速度与作用力成_______，与物体的质量成_______。

AP physics C 2004 真题多项选择试题题目

AP physics C 2004 真题多项选择试题题目Questions 40-4136. Three 1/2 μF capacitors are connected in series as shown in the diagram above. Thecapacitance of the combination is (A) 0.1 μF (B) 1 μF (C) 2/3 μF(D) ? μF (E) 1/6 μF37. A hair dryer is rated as 1200 W, 120 V. Its effective internal resistance is(A) 0.1 Ω (B) A particle of charge +e and mass m moves with 10 Ω (C) 12Ω speed v perpendicular to a uniform magnetic field (D) 120 Ω (E) B directed into the page. The path of the particle is 1440 Ω a circl eof radius r, as shown above. 40. Which of the following correctly gives thedirection of motion and the equationrelating v and r ?Direction Equation(A) Clockwise eBr = mv 2 (B) Clockwise eBr = mv 38. A point charge+Q is inside an uncharged (C) Counterclockwise eBr = mv conducting spherical shell that in turn is near 2(D) Counterclockwise eBr = mv several isolated point charges, as shown above. 22(E) Counterclockwise eBr = mv The electric field at point P inside the shell depends on the magnitude of 41. The period of revolution of the particle (A) Q only is (B) the charge distribution on the sphere only (A) mr/eB (B) meB/(C) Q and the charge distribution on the sphere (D) all of the point charges (C) 2πm/eB (D) 2/,meB(E) all of the point charges a nd the charge (E)2/,mreBdistribution on the sphere 42. A 20 μF parallel-plate capacitor is fully charged to 30 39. In a certain region, the electric field along V. The energy stored in the capacitor is most nearly the x-axis is given by 3-3-4(A) 9 x 10 J (B) 9 x 10 J (C) 6 x 10 J 2E = ax + b, where a = 40 V/m -4-7(D) 2 x 10 J (E) 2 x 10 J and b = 4 V/m. The potentialdifference between the origin 43. A potential difference V is maintained between and x = 0.5 m is two large, parallel conducting plates. An (A) -36 V (B) -7 V (C) -3 V (D) 10 V electron starts from rest on the surface of one (E) 16 V plate and accelerates toward the other. Its speed as it reaches the second plate isproportional to(A) 1/V(B)1VV(C)(D) V 2(E) V44. A wire of radius R has a current I uniformly 48. Two conducting cylindrical wires are made outdistributed across its cross-sectional area. of the same material. Wire X has twice theAmpere's law is used with a concentric length and twice the diameter of wire Y. Whatcircular path of radius r, with r < R, to /R of their resistances?is the ratio Rxycalculate the magnitude of the magnetic (A) 1/4 (B) ? (C) 1 (D) 2 (E) 4field B at a distance r from the center of the wire. Which of the following equationsresults from a correct application ofAmpere's law to this situation? 22(A) B(2πr) = μI (B) B(2πr) =μI(r/R) 00(C) B(2πr) = 0 (D) B(2πR) = μI (E) B(2πR) 022= μI(r/R) 0Questions 45-4649. A solid metallic sphere of radius R has chargeQ uniformly distributed on its outer surface. Agraph of electric potential V as a function ofposition r is shown above. Which of thefollowing graphs best represents the magnitudeof the electric field E as a function of position r Particles of charge Q and -4Q are located on the for this sphere? x-axis as shown in the figure above. Assumethe particles are isolated from all other charges.45. Which of the following describes the directionof the electric field at point P ?(A) +x (B) +y (C)-y(D) Components in both the -x- and+y-directions(E) Components in both the+x- and -y-directions46. At which of the labeled points on the x-axis is the electric field zero?(A) A (B) B (C) C (D) D (E) E47. When the switch S is open in the circuitshown above, the reading on the ammeter Ais 2.0 A. When the switch is closed, thereading on the ammeter is(A) doubled(B) increased slightly but not doubled(C) the same(D) decreased slightly but not halved (E) halved50. Two parallel wires, each carrying a current I, repel each other with a force F. If bothcurrents are doubled, the force of repulsion is (A) 2F (B) F (C) 4F22(D) F (E) 8F 4251. A circular current-carrying loop lies so that the plane of the loop is perpendicular to a constant 54. A conducting loop of wire that is initially magnetic field of strength B. Suppose that the around a magnet is pulled away from the radius R of the loop could be made to increase magnet to the right, as indicated in the figure with time t so that R = at, where a is a constant. above, inducing a current in the loop. What is What is the magnitude of the emf that would be the direction of the force on the magnet and generated around the loop as a function of t ? 2the direction of the magnetic field at the (A) 2πBat(B) 2πBat (C) 2πBt 223center of the loop due to the (D) πBat (E) (π/3)Bat induced current? Direction ofMagnetic Field atDirection of Center of Loop dueForce on the Magnet to Induced Current(A) To the right To the right(B) To the right To the left(C) To the left To the right 52. The figures above show parts of two circuits, (D) To the left To the left(E) No direction; To the left each containing a battery of emf ε and internal the force is zero. resistance r. The current in each battery is 1 A, but the direction of the current in one battery is opposite to that in the other. If the potential differences across the batteries' terminals are 10 V and 20 V as shown, what are the valuesof ε and r ? 55. A square loop of wire carrying a current I is (A) ε = 5 V, r = 15 Ω initially in the plane of the page and is located in a uniform magnetic field B that points (B) ε =IOV, r=100 Ω toward the bottom of the page, as shown above. (C) ε = 15 V, r = 5 Ω Which of the following shows the correct initialrotati on of the loop due to the force exerted on (D) ε = 20 V, r = 10 Ωit by the magnetic field? (E) The values cannot be computed unless the complete circuits are shown.53. A charged particle can move withconstant velocity through a regioncontaining both an electric field and amagnetic field only if the(A) electric field is parallel to the magnetic field(B) electric field is perpendicular to the magnetic field(C) electric field is parallel to the velocity vector(D) magnetic field is parallel to the velocity vector(E) magnetic field is perpendicular to the velocity vectorQuestions 59-6156. In the circuit shown above, the equivalent The diagram above shows equipotential lines resistance of the three resistors is produced by an unknown charge distribution. A, (A) 10.5 Ω (B) 15Ω (C) 20 ΩB, C, D, and E are points in the plane. (D) 50 Ω (E) 115 Ω59. Which vector below best describes the direction Questions 57-58of the electric field at point A ?(A) (B) (C) (D)(E) None of these; the field is zero.60. At which point does the electric field havethe greatest magnitude? (A) A As shown in the figure above, six particles, each (B) B with charge +Q, are held fixed and ate equally(C) C spaced around the circumference of a circle of(D) D radius R.(E) E57. What is the magnitude of the resultantelectric field at the center of the circle? 61. How much net work must be done by anexternal force to move a -1 μC point charge 6Q(A) 0 (B) (C) from rest at point C to rest at point E ? 24,,R0(A) -20 μJ(B) -10 μJ 23Q32Q (D) 22(C) 10 μJ ,,,,4R4R00(D) 20 μJ 3Q(E) 30μJ (E) 2,,2R 058. With the six particles held fixed, how muchwork would be required to bring a seventhparticle of charge + Q from very far awayand place it at the center of the circle?23Q6Q(A) 0 (B) (C) 2,,,,24RR00229Q3Q(D) (E) ,,,,2R00R62. One of Maxwell's equations can be written 65. A physics problem starts: "A solid sphere hascharge distributed uniformly throughout. . . " d,as. This equation expresses ,,,Eds,It may be correctly concluded that the dt(A) electric field is zero everywhere inside the the fact that sphere (A) a changingmagnetic field produces (B) electric field inside the sphere is the same an electric field as the electric field outside (B) a changing electric field produces a (C) electric potential on the surface of the magnetic field sphere is not constant (C) the net magnetic flux through a (D) electric potential in the center of the sphere closed surface depends on the is zero current inside (E) sphere is not made of metal (D) the net electric flux through a closed surface depends on the charge inside Questions 66-67 relate to the circuit represented (E) electric charge is conserved below. The switch S, after being open for a long time, is then closed., ,12 V4 HS63. The plates of a parallel-plate capacitor of cross 66. What isthe current in the circuit after the sectional area A are separated by a distance d, switch has been closed a long time? as shown above. Between the plates is a (A) 0 A dielectric material of constant K. The plates are (B) 1.2 A connected in series with a variable resistance R (C) 2 A and a power supply of potential difference V. (D) 3 A The capacitance C of this capacitor will (E) 12 A increase if which of the following isdecreased? (A) A (B) R (C) K 67. What is the potential difference across the (D) d (E) V resistor immediately after the switch is closed? (A) 0 V(B) 2 V(C) 7.2 V(D) 8 V(E) 12 V68. A uniform spherical charge distribution hasradius R.. Which of the following is true of the electric field strength due to this charge 64. The currents in three parallel wires, X, Y, and Z, distribution at a distance r from the center of each have magnitude l and are in the directions the charge? shown above. Wire Y is closer to wire X than to (A) It is greatest when r = 0. wire Z. The magnetic force on wire Y is (B) It is greatest when r = R/2. (A) zero (B) into the page (C) out of the page (C) It is directly proportional to r when r > (D) toward the bottom of the page R. (E) toward the left (D) It is directly proportional to r when r < R. 2(E) It is directlyproportional to r.69. When a negatively charged rod is brought near, but does not touch, the initially uncharged electroscope shown above, the leaves spring apart (I). When the electroscope is then touched with a finger, the leaves collapse (II). When next the finger and finally the rod are removed, the leaves spring apart a second time (III). The charge on the leaves is(A) positive in both I and III(B) negative in both I and III(C) positive in I, negative in III(D) negative in I, positive in III(E) impossible to determine in either I or III70. A sheet of copper in the plane of the page is connected to a battery as shown above, causing electrons to drift through the copper toward the bottom of the page. The copper sheet is in a magnetic field B directed intothe page. P and P are points at the edges of 12the strip. Which of the following statements is true?(A) P is at a higher potential than P. 12(B) P is at a higher potential than P. 21(C) P and P are at equal positive potential. 12(D) P and P are at equal negative potential. 12(E) Current will cease to flow in the copper sheet.。

1989年第3届初中物理竞赛初赛试卷(解析版)

1989年上海市第三届初中物理竞赛初赛试题说明：下列各题中，有且只有一个选项符合题意。

请把符合题意的选项前的编号填写在各题后的括号内。

每小题2.5分，满分为100分。

答题时间为90分钟。

1.用杆秤称物体质量，如果秤砣发生缺损，若不考虑其它因素，则结果是(A)称出的质量仍等于物体真实质量；(B)称出的质量一定大于物体真实质量；(C)称出的质量一定小于物体真实质量；(D)以上三种情况都有可能。

【B】秤砣发生缺损，质量变小，平衡同样重量的物体时，力臂变大，从而使读数比原正确值大，即称出的质量一定大于物体真实质量2.就图1中的物体乙的受力情况，正确的说法是(A)乙只受重力和桌面支持力的作用；(B)乙只受甲的重力和桌面支持力的作用；(C)乙受到甲的压力、重力和桌面的支持力的作用；(D)乙受甲、乙两个物体的重力和桌面支持力的作用。

【C】隔离物体分析受力：乙受到三个力3.船在静水中的速度为V，河水流速为v，船顺流由甲地到乙地，再逆流返回甲地，所用时间是t1，该船在静水中往返同样距离所用时间为t2。

则(A)t1=t2；(B)t1>t2；(C)t1<t2；(D)没有数据，无法判断，【B】设甲地到乙地距离为XX/(V+U)+X/(V-U)=t1X=t1/(V+U)+t1/(V-U)平均速度=2X/t1=（V^2-U^2）/Vt2=2X/V=(V^2-U^2)*t1/V^2=(1-U^2/V^2)t1明显t1>t24.用铁和铜各做一个实心球，则不可能发生的情况是：(A)铁球的体积和质量都比铜球大；(B)铁球的体积和质量都比铜球小；(C)铁球的体积比铜球大，铁球质量比铜球小；(D)铁球的体积比铜球小，铁球质量比铜球大。

【D】因为ρ铁小于ρ铜, 所以当铁的体积小于铜的时候质量必小于铜5.一根两端开口的玻璃管竖直插入水银槽内。

再注入高度为h1的某种液体，结果使管内水银面下降下h2。

如果水银密度为ρ0，则该液体密度ρ应等于(A)ρ0(h1+h2)；(B)ρ0h2/h1；(C)ρ0h1/h2；(D)ρ0(h1-h2)。

高考40 年 1989年高考物理试题答案及解析

1989年全国普通高等学校招生统一考试物理试题答案及解析江苏省特级教师戴儒京解析第I卷（选择题）一、本题中每小题给出的几个说法中,只有一个是正确的．(1)设地球表面的重力加速度为g0,物体在距地心4R(R是地球半径)处,由于地球的作用而产生的加速度为g,则g/g0为:A．1．B．1/9．C．1/4．D．1/16．(2)若单摆的摆长不变,摆球的质量增加为原来的4倍,摆球经过平衡位置时的速度减小为原来的1/2,则单摆振动的A．频率不变,振幅不变．B．频率不变,振幅改变．C．频率改变,振幅改变．D．频率改变,振幅不变．(4)两辆汽车在同一平直路面上行驶,它们的质量之比m1∶m2=1∶2,速度之比v1∶v2=2∶1．当两车急刹车后,甲车滑行的最大距离为s1,乙车滑行的最大距离为s2．设两车与路面间的滑动摩擦系数相等,不计空气阻力,则A．s1∶s2=1∶2．B．s1∶s2=1∶1．C．s1∶s2=2∶1．D．s1∶s2=4∶1．(5)一轻弹簧上端固定,下端挂一重物,平衡时弹簧伸长了4厘米．再将重物向下拉1厘米,然后放手,则在刚释放的瞬间重物的加速度是(g取10米/秒2)A．2．5米/秒2．B．7．5米/秒2．C．10米/秒2．D．12．5米/秒2．(6)一质量为m的小球,用长为l的轻绳悬挂于O点．小球在水平拉力F作用下,从平衡位置P点很缓慢地移动到Q点(如图所示),则力F所做的功为A．mglcosθ．B．mgl(1-cosθ)．C．Flsinθ．D．Flθ．(7)在图示的电路中,已知电容C=2微法,电源电动势ε=12伏特,内电阻不计,R1∶R2∶R3∶R4=1∶2∶6∶3．则电容器极板a所带的电量为A．-8×10-6库．B．4×10-6库．C．-4×10-6库．D．8×10-6库．(8)在下页图所示的电路中,当滑线变阻器的滑动触点向b端移动时,A．伏特表V的读数增大,安培表A的读数减小．B．伏特表V和安培表A的读数都增大．C．伏特表V和安培表A的读数都减小．D．伏特表V的读数减小,安培表A的读数增大．(9)一矩形通电线框abcd,可绕其中心轴OO′转动,它处在与OO′垂直的匀强磁场中(如图)．在磁场作用下线框开始转动,最后静止在平衡位置．则平衡后A．线框四边都不受磁场的作用力．B．线框四边受到指向线框外部的磁场作用力,但合力为零．C．线框四边受到指向线框内部的磁场作用力,但合力为零．D．线框的一对边受到指向线框外部的磁场作用力,另一对边受到指向线框内部的磁场作用力,但合力为零．(10)一架飞机水平地匀速飞行．从飞机上每隔1秒钟释放一个铁球,先后共释放4个．若不计空气阻力,则四个球A．在空中任何时刻总是排成抛物线;它们的落地点是等间距的．B．在空中任何时刻总是排成抛物线;它们的落地点是不等间距的．C．在空中任何时刻总在飞机正下方排成坚直的直线;它们的落地点是等间距的．D．在空中任何时刻总在飞机正下方排成竖直的直线;它们的落地点是不等间距的．(11)p-T图上的图线abc表示一定质量的理想气体的状态变化过程,此过程在p-V图上的图线应为(12)在下列四个日光灯的接线图中(S为起辉器,L为镇流器),正确的是二、(10分)每小题2分．本题中每小题给出的几个说法中,有一个或几个是正确的．把正确的说法全选出来．(13)在光滑水平面上有三个完全相同的小球排成一条直线．2、3小球静止,并靠在一起,1球以速度v0射向它们(如图)．设碰撞中不损失机械能,则碰后三个小球的速度可能值是(14)红色、绿色和黄色的三束平行光分别沿主轴射向同一个玻璃凸透镜,通过透镜后会聚到主轴上,会聚点到光心的距离分别是f红、f绿、f黄,则A．f红=f绿=f黄．C．f绿<f黄<f红．B．f红<f黄<f绿．D．f红>f绿>f黄．(15)在光滑水平地面上有一木板,一木棒可沿水平轴O转动,其下端B搁在木板上,而整个系统处于静止状态(如图)．现在用水平力F向左推木板,但木板仍未动．由此可以得出结论:施力F后,木板和木棒之间的正压力A．变大．B．不变．C．变小．D．条件不足,不能判断如何改变．(16)玻尔在他提出的原子模型中所做的假设有:A．原子处于称为定态的能量状态时,虽然电子做加速运动,但并不向外辐射能量．B．原子的不同能量状态与电子沿不同的圆轨道绕核运动相对应,而电子的可能轨道的分布是不连续的．C．电子从一个轨道跃迁到另一轨道时,辐射(或吸收)一定频率的光子．D．电子跃迁时辐射的光子的频率等于电子绕核做圆周运动的频率．(17)一个点电荷,从静电场中的a点移到b点,其电势能的变化为零,则A．a、b两点的场强一定相等．B．该点电荷一定沿等势面移动．C．作用于该点电荷的电场力与其移动方向总是垂直的．D．a、b两点的电势一定相等．(18)一平行板电容器C,极板是水平放置的,它和三个可变电阻及电源联接成如图所示的电路．今有一质量为m的带电油滴悬浮在两极板之间静止不动．要使油滴上升,可采用的办法是A．增大R1．B．增大R2．C．增大R3．D．减小R2．(19)一平行板电容器充电后与电源断开,负极板接地．在两极板间有一正电荷(电量很小)固定在P点,如下图所示．以E表示两极板间的场强,U表示电容器的电压,W表示正电荷在P点的电势能．若保持负极板不动,将正极板移到图中虚线所示的位置,则A．U变小,E不变．B．E变大,W变大．C．U变小,W不变．D．U不变,W不变．(20)对于一定质量的理想气体,在下列各种过程中,可能发生的过程是A．气体膨胀对外做功,温度升高．B．气体吸热,温度降低．C．气体放热,压强增大．D．气体放热,温度不变．第Ⅱ卷(非选择题共60分)三、(24分)每小题3分,把答案填写在题中横线上空白处,不要求写出演算过程．(21)中子的质量为1．0087u,质子的质量为1．0073u,氘核的质量为2．0136u．中子和质子结合成氘核时释放的能量为焦耳．计算结果取两位有效数字．1u=1．7×10-27千克．(22)图中A、B是体积相同的气缸,B内有一导热的、可在气缸内无摩擦滑动的、体积不计的活塞C,D为不导热的阀门．起初,阀门关闭,A内装有压强p1=2．0×105帕,温度T1=300K的氮气．B内装有压强p2=1．0×105帕,温度T2=600K的氧气．阀门打开后,活塞C向右移动,最后达到平衡．以V1和V2分别表示平衡后氮气和氧气的体积,则V1∶V2= ．(假定氧气和氮气均为理想气体,并与外界无热交换,连接气缸的管道体积可忽略)(23)下左图中实线是一列简谐波在某一时刻的波形图线．虚线是0．2秒后它的波形图线．这列波可能的传播速度是．(24)某一用直流电动机提升重物的装置,如图所示．重物的质量m=50千克,电源的电动势ε=110伏特,不计电源内阻及各处的摩擦．当电动机以v=0．90米/秒的恒定速度向上提升重物时,电路中的电流强度I=5安培,由此可知电动机线圈的电阻R= 欧姆．(25)在测定匀变速直线运动的加速度的实验中,用打点计时器记录纸带运动的时间．计时器所用电源的频率为50赫．图为做匀变速直线运动的小车带动的纸带上记录的一些点,在每相邻的两点中间都有四个点未画出．按时间顺序取0、1、2、3、4、5六个点,用米尺量出1、2、3、4、5点到0点的距离分别是(单位:厘米),由此可得小车的加速度的大小为米/秒2,方向．8．78 16．08 21．87 26．16 28．94(26)电阻为R的矩形导线框abcd,边长ab=L,ad=h,质量为m,自某一高度自由落下,通过一匀强磁场,磁场方向垂直纸面向里,磁场区域的宽度为h(如下页图所示)．若线框恰好以恒定速度通过磁场,线框中产生的焦耳热是．(不考虑空气阻力)(27)质量为m的运动员站在质量为m/2的均匀长板AB的中点,板位于水平地面上,可绕通过B点的水平轴转动,板的A端系有轻绳,轻绳的另一端绕过两个定滑轮后,握在运动员手中．当运动员用力拉绳时,滑轮两侧的绳都保持在竖直方向,如图所示．要使板的A端离开地面,运动员作用于绳的最小拉力是．(28)一个房间的地面面积是15米2,高3分．试估算该房间内空气的质量．已知空气的平均摩尔质量是2．9×10-2千克/摩尔．答: 千克．四、(10分)在测定金属的电阻率的实验中,金属导线长约0．8米,直径小于1毫米,电阻在5欧左右．实验步骤如下:(29)用米尺测量金属导线的长度,测三次,求出平均值L．在金属导线三个不同的位置上用测量直径,求出平均值d．(30)用伏安法测量金属导线的电阻R．试把下图中所给的器材连接成测量R的合适的线路．图中安培表的量程为0．6安,内阻接近1欧;伏特表的量程为3伏,内阻为几千欧;电源的电动势为6伏;变阻器的阻值为0～20欧．在闭合电键前,变阻器的滑动触点应处于正确位置．(31)用上面测得的金属导线长度L、直径d和电阻R,可根据电阻率的表达式ρ=算出所测金属的电阻率．五、(26分)本题共有三个计算题,要求写出主要的文字说明、方程式和演算步骤．只写出最后答案,而未写出主要演算过程的,不能得分．有数值计算的题,答案中必须明确写出数值和单位．(32)(9分)如图所示,AB、CD是两根足够长的固定平行金属导轨,两导轨间的距离为L,导轨平面与水平面的夹角是θ．在整个导轨平面内都有垂直于导轨平面斜向上方的匀强磁场,磁感应强度为B．在导轨的AC端连接一个阻值为R 的电阻．一根垂直于导轨放置的金属棒ab,质量为m,从静止开始沿导轨下滑,求ab棒的最大速度．要求画出ab棒的受力图．已知ab与导轨间的滑动摩擦系数μ,导轨和金属棒的电阻都不计．(33)(8分)把一个点光源放在焦距为f的凸透镜的焦点上,在透镜的另一侧2倍焦距处放一个垂直于主轴的光屏,在光屏上看到一个半径为R的光亮的圆．现保持透镜和光屏不动,而在主轴上移动点光源,若要使光屏上亮圆的半径缩为R/2,则这个点光源应移到什么位置上?(34)(9分)一个质量为m、带有电荷-q的小物体,可在水平轨道Ox上运动,O端有一与轨道垂直的固定墙．轨道处于匀强电场中,场强大小为E,方向沿Ox轴正向,如图所示．小物体以初速v0从x0点沿Ox轨道运动,运动时受到大小不变的摩擦力f作用,且f<qE;设小物体与墙碰撞时不损失机械能,且电量保持不变,求它在停止运动前所通过的总路程s．1989年全国普通高等学校招生统一考试物理参考答案及解析一、全题30分,每小题2分．答错的或不答的,都给0分． (1)D ．【解析】02mg R Mm G == ，mg R Mm G 2)4(=，1610=g g 选D . (2)B ．【解析】glT π2=,l 不变，T 不变，T f 1= ,f 不变。

【AP物理C】【真题】2003年力学解答题答案

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大学物理C课后答案

习题5题5-2图题5-2图5-2 两小球的质量都是m ，都用长为l 的细绳挂在同一点，它们带有相同电量，静止时两线夹角为2θ，如题5--2图所示.设小球的半径和线的质量都可以忽略不计，求每个小球所带的电量.解: 如题5-2图示⎪⎩⎪⎨⎧===220)sin 2(π41sin cos θεθθl q F T mg T e解得 θπεθtan 4sin 20mg l q =5-4 长l ＝15.0 cm 的直导线AB 上均匀地分布着线密度95.010C m λ-=⨯的正电荷.试求：(1)在导线的延长线上与导线B 端相距1 5.0a cm =处P 点的场强；(2)在导线的垂直平分线上与导线中点相距2 5.0d cm =处Q 点的场强. 解：如题5-4图所示题5-4图(1)在带电直线上取线元x d ，其上电量q d 在P 点产生场强为20)(d π41d x a xE P -=λε222)(d π4d x a xE E l l P P -==⎰⎰-ελ]2121[π40l a l a +--=ελ)4(π220l a l-=ελ用15=l cm ，9100.5-⨯=λ1m C -⋅, 5.12=a cm 代入得21074.6⨯=P E 1C N -⋅ 方向水平向右(2)同理 2220d d π41d +=x xE Q λε 方向如题5-4图所示由于对称性⎰=lQx E 0d ，即Q E只有y 分量，∵ 22222220d d d d π41d ++=x x xE Qyλε22π4d d ελ⎰==lQyQy E E ⎰-+2223222)d (d l l x x2220d4π2+=l lελ以9100.5-⨯=λ1cm C -⋅, 15=l cm ,5d 2=cm 代入得21096.14⨯==Qy Q E E 1C N -⋅，方向沿y 轴正向5-7 半径为1R 和2R (21R R >)的两无限长同轴圆柱面，单位长度上分别带有电量λ和－λ，试求：(1) 1r R <；(2) 12R r R <<；(3) 2r R >处各点的场强.解: 高斯定理0d ε∑⎰=⋅qS E s取同轴圆柱形高斯面，侧面积rl S π2=则 rl E S E Sπ2d =⋅⎰对(1) 1R r <0,0==∑E q(2) 21R r R << λl q =∑∴ rE 0π2ελ=沿径向向外(3) 2R r >=∑q∴ 0=E5-9 如题5-9图所示，在A ，B 两点处放有电量分别为＋q ，－q 的点电荷，AB 间距离为2R ，现将另一正试验点电荷0q 从O 点经过半圆弧移到C 点，求移动过程中电场力做的功. 解: 如题5-9图示0π41ε=O U 0)(=-RqR q 0π41ε=O U )3(R qR q -Rq 0π6ε-= ∴ Rqq U U q A o C O 00π6)(ε=-=题5-9图题5-10图5-10 如题5-10图所示的绝缘细线上均匀分布着线密度为λ的正电荷，两段直导线的长度和半圆环的半径都等于R .试求环中心O 点处的场强和电势.解: (1)由于电荷均匀分布与对称性，AB 和CD 段电荷在O 点产生的场强互相抵消，取θd d R l =则θλd d R q =产生O 点Ed 如图，由于对称性，O 点场强沿y 轴负方向题5-10图θεθλππcos π4d d 2220⎰⎰-==R R E E yR 0π4ελ=［)2sin(π-2sin π-］R0π2ελ-=(2) AB 电荷在O 点产生电势，以0=∞U⎰⎰===AB200012ln π4π4d π4d R R x x x x U ελελελ 同理CD 产生 2ln π402ελ=U 半圆环产生 0034π4πελελ==R R U∴ 0032142ln π2ελελ+=++=U U U U O 86)35251(5021=+=+=U U U AB V 习题66-5 在真空中，有两根互相平行的无限长直导线L 1和L 2，相距0.10 m ，通有方向相反的电流，120A I =，210A I =，如题6-5图所示.A ，B 两点与导线在同一平面内.这两点与导线L 2的距离均为5.0 cm.试求A ，B 两点处的磁感应强度，以及磁感应强度为零的点的位置.题6-5图解：如题6-5图所示,A B方向垂直纸面向里42010102.105.02)05.01.0(2-⨯=⨯+-=πμπμI I B A T(2)设0=B在2L 外侧距离2L 为r 处则02)1.0(220=-+rI r Iπμπμ 解得 1.0=r m6-7 设题6-7图中两导线中的电流均为8 A ，对图示的三条闭合曲线a ，b ，c ，分别写出安培环路定理等式右边电流的代数和.并讨论：(1)在各条闭合曲线上，各点的磁感应强度B 的大小是否相等？ (2)在闭合曲线c 上各点的B 是否为零？为什么？题6-7图解： ⎰μ=⋅al B 08d⎰μ=⋅bal B 08d⎰=⋅cl B 0d(1)在各条闭合曲线上，各点B的大小不相等．(2)在闭合曲线C 上各点B 不为零．只是B 的环路积分为零而非每点0=B．题6-10图6-10 如题6-10图所示，在长直导线AB 内通以电流120A I =，在矩形线圈CDEF 中通有电流210A I =，AB 与线圈共面，且CD ，EF 都与AB 平行.已知a ＝9.0 cm ，b ＝20.0 cm ，d ＝1.0 cm ，求：(1)导线AB 的磁场对矩形线圈每边所作用的力； (2)矩形线圈所受合力和合力矩.解：(1)CD F方向垂直CD 向左，大小4102100.82-⨯==dI bI F CD πμ N 同理FE F方向垂直FE 向右，大小5102100.8)(2-⨯=+=a d I bI F FE πμ NCF F方向垂直CF 向上，大小为⎰+-⨯=+πμ=πμ=ad dCF dad I I r r I I F 5210210102.9ln 2d 2 N ED F方向垂直ED 向下，大小为5102.9-⨯==CF ED F F N(2)合力ED CF FE CD F F F F F+++=方向向左，大小为4102.7-⨯=F N合力矩B P M m⨯=∵ 线圈与导线共面∴ B P m//0=M．题6-12图6-12 一长直导线通有电流120A I =，旁边放一导线ab ，其中通有电流210A I =，且两者共面，如题6-12图所示.求导线ab 所受作用力对O 点的力矩. 解：在ab 上取r d ，它受力ab F ⊥d 向上，大小为rI rI F πμ2d d 102= F d 对O 点力矩F r M ⨯=d Md 方向垂直纸面向外，大小为r I I F r M d 2d d 210πμ== ⎰⎰-⨯===ba bar I I M M 6210106.3d 2d πμ m N ⋅题6-13图6-13 电子在47010T B -=⨯的匀强磁场中作圆周运动，圆周半径r ＝3.0 cm.已知B 垂直于纸面向外，某时刻电子在A 点，速度v 向上，如题6-13图所示.(1)试画出这电子运动的轨道；(2)求这电子速度v 的大小； (3)求这电子的动能k E . 解：(1)轨迹如图题6-13图(2)∵ rv m evB 2=∴ 7107.3⨯==m eBrv 1s m -⋅ (3) 162K 102.621-⨯==mv E J05.1===H H B o r μμμ T习题77-1 一半径r ＝10 cm 的圆形回路放在B ＝0.8 T 的均匀磁场中，回路平面与B 垂直.当回路半径以恒定速率=80drdtcm/s 收缩时，求回路中感应电动势的大小. 解: 回路磁通 2πr B BS m ==Φ感应电动势大小40.0d d π2)π(d d d d 2====trr B r B t t m Φε V题7-37-3 如题7-3图所示，在两平行载流的无限长直导线的平面内有一矩形线圈.两导线中的电流方向相反、大小相等，且电流以d Id t 的变化率增大，求：(1)任一时刻线圈内所通过的磁通量； (2)线圈中的感应电动势. 解: 以向外磁通为正则(1) ]ln [lnπ2d π2d π2000dad b a b Ilr l r Ir l r Iab b ad d m +-+=-=⎰⎰++μμμΦ (2) tIb a b d a d l t d d ]ln [ln π2d d 0+-+=-=μΦε习题88-1 质量为10×10－3 kg 的小球与轻弹簧组成的系统，按20.1cos(8)3x t ππ=+(SI)的规律做谐振动，求：(1)振动的周期、振幅、初位相及速度与加速度的最大值；(2)最大的回复力、振动能量、平均动能和平均势能，在哪些位置上动能与势能相等？ (3)t 2＝5 s 与t 1＝1 s 两个时刻的位相差. 解：(1)设谐振动的标准方程为)cos(0φω+=t A x ，则知：3/2,s 412,8,m 1.00πφωππω===∴==T A 又 πω8.0==A v m 1s m -⋅ 51.2=1s m -⋅2.632==A a m ω2s m -⋅(2) N 63.0==m m a FJ 1016.32122-⨯==m mv E J 1058.1212-⨯===E E E k p当p k E E =时，有p E E 2=，即)21(212122kA kx ⋅= ∴ m 20222±=±=A x (3) ππωφ32)15(8)(12=-=-=∆t t8-2 一个沿x 轴做简谐振动的弹簧振子，振幅为A ，周期为T ，其振动方程用余弦函数表出.如果t ＝0时质点的状态分别是：(1)x 0＝－A ；(2)过平衡位置向正向运动；(3)过2Ax =处向负向运动； (4)过x =处向正向运动.试求出相应的初位相，并写出振动方程. 解：因为 ⎩⎨⎧-==0000sin cos φωφA v A x将以上初值条件代入上式，使两式同时成立之值即为该条件下的初位相．故有)2cos(1πππφ+==t T A x)232cos(232πππφ+==t T A x)32cos(33πππφ+==t T A x)452cos(454πππφ+==t T A x8-3 一质量为10×10－3 kg 的物体做谐振动，振幅为24 cm ，周期为4.0 s ，当t ＝0时位移为＋24 cm.求：(1)t ＝0.5 s 时，物体所在的位置及此时所受力的大小和方向； (2)由起始位置运动到x ＝12 cm 处所需的最短时间； (3)在x ＝12 cm 处物体的总能量. 解：由题已知 s 0.4,m 10242=⨯=-T A∴ 1s rad 5.02-⋅==ππωT又，0=t 时，0,00=∴+=φA x 故振动方程为m )5.0cos(10242t x π-⨯=(1)将s 5.0=t 代入得0.17m m )5.0cos(102425.0=⨯=-t x πN102.417.0)2(10103232--⨯-=⨯⨯⨯-=-=-=πωxm ma F方向指向坐标原点，即沿x 轴负向． (2)由题知，0=t 时，00=φ，t t =时 3,0,20πφ=<+=t v A x 故且 ∴ s 322/3==∆=ππωφt (3)由于谐振动中能量守恒，故在任一位置处或任一时刻的系统的总能量均为J10 1.7) 24 .0()2(10102121214223222--⨯=⨯⨯⨯===πωAmkAE8-5 题8-5图为两个谐振动的x－t曲线，试分别写出其谐振动方程.题8-5图解：由题8-5图(a)，∵0=t时，s2,cm10,,23,0,0===∴>=TAvx又πφ即1srad2-⋅==ππωT故m)23cos(1.0ππ+=txa由题8-5图(b)∵0=t时，35,0,20πφ=∴>=vAx1=t时，22,0,0111ππφ+=∴<=vx又ππωφ253511=+⨯=∴πω65=故mtxb)3565cos(1.0ππ+=习题9 机械波9-4 已知波源在原点的一列平面简谐波，波动方程为y＝A cos (Bt－Cx)，其中A，B，C为正值恒量.求：(1)波的振幅、波速、频率、周期与波长；(2)写出传播方向上距离波源为l处一点的振动方程；(3)任一时刻，在波的传播方向上相距为d的两点的位相差.解: (1)已知平面简谐波的波动方程)cos(CxBtAy-=(0≥x)将上式与波动方程的标准形式)22cos(λππυxt A y -=比较，可知：波振幅为A ，频率πυ2B =，波长C πλ2=，波速CB u ==λυ，波动周期BT πυ21==．(2)将l x =代入波动方程即可得到该点的振动方程)cos(Cl Bt A y -=(3)因任一时刻t 同一波线上两点之间的位相差为 )(212x x -=∆λπφ将d x x =-12，及Cπλ2=代入上式，即得 Cd =∆φ．9-5 沿绳子传播的平面简谐波的波动方程为y ＝0.05cos(10πt －4πx )，式中x ，y 以m 计，t 以s 计.求：(1)波的波速、频率和波长；(2)绳子上各质点振动时的最大速度和最大加速度；(3)求x ＝0.2 m 处质点在t ＝1 s 时的位相，它是原点在哪一时刻的位相？这一位相所代表的运动状态在t ＝1.25 s 时刻到达哪一点？解: (1)将题给方程与标准式)22cos(x t A y λππυ-=相比，得振幅05.0=A m ，频率5=υ1-s ，波长5.0=λm ，波速5.2==λυu 1s m -⋅． (2)绳上各点的最大振速，最大加速度分别为ππω5.005.010max =⨯==A v 1s m -⋅ 222max 505.0)10(ππω=⨯==A a 2s m -⋅(3)2.0=x m 处的振动比原点落后的时间为08.05.22.0==u x s 故2.0=x m ,1=t s 时的位相就是原点(0=x )，在92.008.010=-=t s 时的位相，即 2.9=φπ．设这一位相所代表的运动状态在25.1=t s 时刻到达x 点，则825.0)0.125.1(5.22.0)(11=-+=-+=t t u x x m9-7 如题9-7图所示，S 1和S 2为两相干波源，振幅均为A 1，相距λ4，S 1较S 2位相超前π2，求：题9-7图(1)S 1外侧各点的合振幅和强度；(2)S 2外侧各点的合振幅和强度.解：（1）在1S 外侧，距离1S 为1r 的点，1S 2S 传到该P 点引起的位相差为πλλππφ=⎥⎦⎤⎢⎣⎡+--=∆)4(2211r r 0,0211===-=A I A A A（2）在2S 外侧.距离2S 为1r 的点，1S 2S 传到该点引起的位相差.0)4(2222=-+-=∆r r λλππφ2121114,2A A I A A A A ===+=9-9 一驻波方程为y ＝0.02cos 20x cos 750t (SI)，求：(1)形成此驻波的两列行波的振幅和波速； (2)相邻两波节间距离. 解: (1)取驻波方程为t uxA y πυπυ2cos 2cos 2= 故知 01.0202.0==A m 7502=πυ，则πυ2750=， 202=uπυ∴ 5.37202/7502202=⨯==πππυu 1s m -⋅(2)∵314.01.020/2====πυπυυλu m 所以相邻两波节间距离157.02==∆λx m习题10 波动光学10-4 在杨氏双缝实验中，双缝间距d ＝0.20 mm ，缝屏间距D ＝1.0 m .试求：(1)若第2级明条纹离屏中心的距离为6.0 mm ，计算此单色光的波长； (2)求相邻两明条纹间的距离.解: (1)由λk dDx =明知，λ22.01010.63⨯⨯=， ∴ 3106.0-⨯=λmm oA 6000=(2) 3106.02.010133=⨯⨯⨯==∆-λd D x mm10-5 在双缝装置中，用一很薄的云母片(n ＝1.58)覆盖其中的一条缝，结果使屏幕上的第7级明条纹恰好移到屏幕中央原零级明纹的位置.若入射光的波长为550 nm ，求此云母片的厚度.解: 设云母片厚度为e ，则由云母片引起的光程差为e n e ne )1(-=-=δ按题意λδ7=∴ 610106.6158.1105500717--⨯=-⨯⨯=-=n e λm 6.6=m μ10-7 在折射率n 1＝1.52的镜头表面涂有一层折射率n 2＝1.38的MgF 2增透膜，如果此膜适用于波长λ＝550 nm 的光，问膜的厚度应取何值？解: 设光垂直入射增透膜，欲透射增强，则膜上、下两表面反射光应满足干涉相消条件，即λ)21(22+=k e n ),2,1,0(⋅⋅⋅=k∴ 222422)21(n n k n k e λλλ+=+=)9961993(38.14550038.125500+=⨯+⨯=k k o A 令0=k ，得膜的最薄厚度为996oA ．当k 为其他整数倍时，也都满足要求．。

1989年第4届全国中学生物理竞赛决赛试题及参考答案

第4届全国中学生物理竞赛决赛试题一、如图所示，da 、cb 为相距l 米的平行导轨（电阻很小，可以忽略）。

a 、b 间连接一个固定电阻，阻值为R 。

长直细杆MN 可以按任意角θ架在平行导轨上，并以匀速v 滑动（平移），v 的方向和da 平行。

杆MN 有电阻，每米长的电阻值为R 。

整个空间充满匀强磁场，磁感应强度的大小为B ，方向垂直纸面（dabc 平面）向里。

（1）求固定电阻R 上消耗的电功率为最大时角θ的值。

（2）求杆MN 上消耗的电功率为最大时角θ的值。

二、（1）质量为M 的静止原子，由于内部能级变化（21E E →）而发射出一个光子。

已知221()Mc E E >>-，求原子的反冲（后退）动能T 与光子能量E 的比值。

（注意，能量为E 的光子，其动量为/E c ，c 为光速。

）（2）在绝对温度T 时，电磁场的光子平均能量约为03/RT N ，R 为气体常量，0N 为阿伏加德罗常数。

假设在宇宙发展的初期，光子间的碰撞能够产生质子、反质子（即一对光子经过碰撞而转变成一个质子和一个反质子）。

试估算当时温度的数量级（二位有效数字）。

（3）气体分子的直径约为10210m -⨯，试估算标准状况下近邻气体分子间的平均距离l 与分子直径d 的比值（二位有效数字）。

附：83.010/c m s =⨯，质子质量271.710p m kg -=⨯，8.3/()R J mol K =⋅，2310 6.010N mol -=⨯。

三、对于太阳系中的行星运动，天文观测发现了如下的事实〈称为开普勒三定律）： 1. 各个行星分别在大小不同的椭圆轨道上围绕太阳运动，太阳的位置是椭圆的一个焦点。

（第一定律）2. 对于每个行星来说，太阳至行星的联线（图中FP 线）在每一个单位时间内扫过的面积（称为面积速度）相等。

（第二定律）3. 行星椭圆轨道的半长轴的三次方与公转周期的平方的比值，对于各个行星来说是相同的。

（第三定律）行星轨道如图所示，P 为行星，F 为焦点（太阳），a 、b 、2c 分别为半长轴、半短轴和焦距，O 为椭圆中心。

【AP物理C】【真题】2001年电磁学解答题答案

AP Physics C: Electricity and Magnetism2001 Scoring Guidelinesemployment 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 is composed of more than 3,900 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™, the Advanced Placement Program® (AP®), and Pacesetter®. The College Board is committed to the principles of equity and excellence, and that commitment is embodied in all of itsprograms, services, activities, and concerns.Copyright © 2001 by College Entrance Examination Board. All rights reserved. College Board, Advanced Placement Program, AP, and the acorn logo are registeredtrademarks of the College Entrance Examination Board.General Notes about 2001 AP Physics Solutions1. 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.3. An 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.4. Implicit statements of concepts normally receive credit. For example, if use of the equationexpressing a particular concept is worth one point, and a solution contains the application of the equation to the problem but does not separately list the basic equation, the point is still awarded.Copyright © 2001 by College Entrance Examination Board. All rights reserved.Copyright © 2001 by College Entrance Examination Board. All rights reserved.Question 115 points total1. (a) 4 points Distribution of PointsFor a correct formula for determining the electric field1 point 22ˆˆ or ==åkQ kQr rE r E r Summing the contributions to the field from the four charges, letting fields directed upward be positive and fields directed downward be negative (full credit also given for using opposite convention as long as answers were consistent):()()()()()()()()()()()()99992222333391030910309103091030 310210210310××××=−++−××××EFor correct substitutions shown in the above equation 1 point 30,000 N C 67,500 N C 67,500 N C 30,000 N C =−++−E 75,000 N C =E , directed upwardFor correct magnitude (()660810 or 8104π−−×=×E k ò also accepted) 1 point For correct direction, either stated or shown by an upward directed arrow 1 pointNotes: If wrong signs were used in the substitution, the point for correct magnitude wasnot awarded. If calculation was done using only the two real charges, a maximum of 3 points was awarded as follows: 1 point for the formula, 1 point for the calculation, 1 point for direction.1. (b) i. 1 pointFor correctly indicating direction, such as by an upward directed arrow at point 2P , 1 point1. (b) ii. 2 pointsFor correctly checking the space in front of “Less”1 point For correct justification, such as “2P is farther from all the charges than 1P , so the net fieldis less.” If student chose to work out the actual magnitude of the field at 2P (which isabout 45,000 N/C), the justification point was awarded for any calculated numerical value less than 75,000 N/C.1 pointNote: No points were awarded for (b) ii. if the wrong space was checked.Copyright © 2001 by College Entrance Examination Board. All rights reserved.Question 1 (cont.)1. (c) i. 2 points Distribution of PointsThe potential at 1P is 0, which can be determined without calculation from symmetryconsiderations.For stating 0=V or for just 0 2 pointsAlternate SolutionAlternate points For a correct formula for determining the potential1 point =åkQ V r30303030 03000200020003000æö=−+−=ç÷èøV kFor the correct answer 1 point1. (c) ii. 2 pointsThe potential at 2P is 0, which can also be determined without calculation from symmetryconsiderations.For stating 0=V or for just 0 2 pointsAlternate SolutionAlternate points For a correct formula for determining the potential1 point =åkQ V r0=−+−=V k For the correct answer1 point1. (d) 2 pointsFor a correct formula for determining the potential1 point =åkQ V rFor correct substitution of values (no credit lost for failing to convert kilometers to meters)1 point ()()()()3333303030303110211021103110æö=−+−ç÷ç÷−×−×+×+×èøP V k881.1210 V 10 V =−×≈−P VCopyright © 2001 by College Entrance Examination Board. All rights reserved.Question 1 (cont.)1. (e) 2 points Distribution of PointsFor a correct formula for the potential energy (must include a summation sign)1 point ,=åi j i jijq q U k r or =åU qVFor correct substitution without regard to sign error (no credit lost for failing to convert kilometers to meters) 1 point101.610 J =−×U()()()()()()()9333333303030303030303030303030910 J 1051061041051010U −−−−−−éù=×+++++êú××××ëûCopyright © 2001 by College Entrance Examination Board. All rights reserved.Question 215 points total2. (a) 4 points Distribution of PointsThere were three methods generally used to solve this problem.Method 1.For a correct method based on determining the time constant using values from the graph 60 min 3600 s τ=≈=RC2 points For correct substitution of values with proper units1 point -63600 s 8.010 Fτ==×R CFor answer consistent with values used 1 point 84.510=×ΩRMethod 2.0−=t RC V V eFor using the above equation with given value of C and values for V , 0V , and t from thegraph with t correlating with V2 points Example: 010 V, and 2 V at 100 min ===V V t For correct substitution of values1 point ()()66000 s 810 F210−−×=R e()()()6ln 2106000810−=×RFor answer consistent with values used 1 point 84.710=×ΩRMethod 3.Find a correct relationship that depends on the slope of the graph:Example: ==V V R i dQ dtBut =dQ CdVSo ()=VR C dV dtFor estimating dV dt by computing ∆∆V t for a particular value of V 2 points For substituting values in equation above 1 point For answer consistent with values used 1 pointNote: The value used for V must be that at the point where the slope is taken and clearlyindicated.Copyright © 2001 by College Entrance Examination Board. All rights reserved.Question 2 (cont.)2. (b) 3 points Distribution of PointsFor a correct equation for the capacitance1 point 0κ=A C dòFor correct algebraic solution for area A and substitution of variables1 point ()()()64-12208.010 F 1.010 m 5.68.8510C N m κ−−•××==×Cd A òFor the correct answer 1 point 216 m =A2. (c) 3 pointsFor a correct equation for the resistance1 point ρ=L R AFor correct algebraic solution for resistivity H and substitution of variables1 point ()()82-44.510 16m 1.010 mρ×Ω==×RA L For answer consistent with values used 1 point 137.210 m ρ•=×ΩCopyright © 2001 by College Entrance Examination Board. All rights reserved.Question 2 (cont.)2. (d) 4 points Distribution of PointsThere were four general methods for solving this problem. Solutions were scored as follows: For using a correct method 2 points For substituting appropriate values 1 point For answer consistent with values obtained in earlier parts 1 point Method 1:∆=∆Q C V , and substitute values obtained from graphExample: ()()68.010 F 10 V −=×i Q and ()()68.010 F 2 V −=×f Q So 56.410 C or 64 C Q −∆=×mMethod 2:6000 6000 6000 36008010V 4.510−−===×Ωòòòt ttt RC t VQ I dt edt e dt R()()6000 s 802.22103600−−=×−t Q e56.510 C or 65 C Q −=×mMethod 3:()01−=−t Q CV e()()()()866000 s 4.510 8.010 F 68.010 F 10 V 1−−×Ω×−æö=×−ç÷èøQ e 56.510 C or 65 C Q −=×mMethod 4:Determine the area under the curve of I vs.t , which is the V vs.t graph shown with 0=I V R .Each block has area equal to 61.310−=×Vt R C. Estimating 47 blocks under the curve gives56.310 C or 63 C Q −=×mUnit point: For correct units given on answers for three of the four parts 1 pointCopyright © 2001 by College Entrance Examination Board. All rights reserved.Question 315 points total3. (a) 2 points Distribution of PointsUsing Ohm’s law: =V IRFor correct equation for I1 point ε=I RFor correctly indicating the current direction on the diagram or in the answer space, suchas by stating that it is clockwise, or to the left, or by showing an arrow pointing left 1 point3. (b) 4 pointsFor indicating on the diagram or in the answer space a direction opposite to the answerin part (a). If part (a) does not contain a direction, then for an indication that the direction is to the right or by showing an arrow pointing right. 1 point For a complete justification3 points Full credit awarded for an answer that indicated the right-hand rule to obtain themagnetic field directed out of the page at the rod, and then used the cross product to obtain that the force on the rod is up2 points partial credit awarded for an answer that just stated the rule that antiparallelcurrents repel or that just stated I ℓ x B and the right-hand rule1 point partial credit awarded for an answer that just stated the right-hand rule orI ℓ x B or some fragment with some correct element3. (c) 4 pointsFor indicating that the gravitational force will be equal to I ℓ x B 1 point F = I ℓ x B = mgFor giving the correct equation for the magnetic field1 point 02µπ=iB rFor correctly substituting in the first equation above the values for B and for I from part (a)1 point 02µπε=l c mg rRI For the correct answer1 point 02πµε=l c mg rRICopyright © 2001 by College Entrance Examination Board. All rights reserved.Question 3 (cont.)3. (d) 5 points Distribution of PointsFor the correct expression for φ1 point φ•=òd B A , where 02µπ=cI B x, and x is the vertical distance from the cable Letting =l dA dx and substituting the values for B above and for c I from part (c):02 2µπφπµε+=òl l r d r mgrR dx x For correct limits of integration1 point For correct substitution of the values consistent with previous answers 1 point For correct integration1 point ln φε+=r dr mgrRxFor the correct answer1 point ln φε+æö=ç÷èømgrRr d r。

【C语言】中的版本规范（C89C99等）

【C语⾔】中的版本规范（C89C99等）C语⾔中的版本ISO:国际标准化组织（International Organization for Standardization，ISO）简称ISO，是⼀个全球性的，是国际标准化领域中⼀个⼗分重要的组织。

K&R C：C语⾔由Dennis M. Ritchie在1973年设计和实现。

从那以后使⽤者逐渐增加。

到1978年Ritchie和Bell实验室的另⼀位程序专家Kernighan合写了著名的《The C Programming Language》，将C语⾔推向全世界，许多国家都出了译本，国内有⼀些C语⾔书就是这本书的翻译或者编译。

由这本书定义的C语⾔后来被⼈们称作 K&R C。

C89 (ANSI C）：随着C语⾔使⽤得越来越⼴泛，出现了许多新问题，⼈们⽇益强烈地要求对C语⾔进⾏标准化。

这个标准化的⼯作在美国国家标准局（ANSI）的框架中进⾏（1983-1988），最终结果是1988年10⽉颁布的ANSI标准X3.159-1989，也就是后来⼈们所说的ANSI C标准。

由这个标准定义的C语⾔被称作ANSI C。

1989 年，草案被 ANSI 正式通过成为美国国家标准，被称为 C89 标准C90 (ISO C)：随后，《The C Programming Language》第⼆版开始出版发⾏，书中内容根据 ANSI C(C89)进⾏了更新。

1990 年，在 ISO/IEC JTC1/SC22/WG14 (ISO/IEC 联合技术第 I 委员会第 22 分委员会第 14 ⼯作组) 的努⼒下，ISO 批准了 ANSI C 成为国际标准。

于是 ISO C(⼜称为 C90) 诞⽣了。

除了标准⽂档在印刷编排上的某些细节不同外，ISO C(C90) 和 ANSI C(C89) 在技术上完全⼀样。

C95：之后，ISO 在 1994、1996 年分别出版了 C90 的技术勘误⽂档，更正了⼀些印刷错误，并在1995 年通过了⼀份 C90 的技术补充，对 C90 进⾏了微⼩的扩充，经过扩充后的 ISO C 被称为 C95。

AP-物理C-力学-2021真题卷1(2)

Universal gravitational constant,
G
6.67 1011 N<m2 kg2
Universal gas constant, R 8.31 J (mol<K)
Acceleration due to gravity at Earth’s surface,
g 9.8 m s2
ADVANCED PLACEMENT PHYSICS C TABLE OF INFORMATION
CONSTANTS AND CONVERSION FACTORS
Proton mass, mp 1.67 1027 kg Electron charge magnitude, e 1.60 1019 C
Neutron mass, mn 1.67 1027 kg
1 electron volt, 1 eV 1.60 1019 J
Electron mass, me 9.11 1031 kg
Speed of light, c 3.00 108 m s
Avogadro’s number, N0 6.02 1023 mol1
dpG dt
Ô G
J
G F dt
D pG
GG p mv
G
G
Ff m FN
Ô DE W
G F
<
d
rG
Байду номын сангаас
K 1 mÃ2 2
P
dE dt
P
G F
<
vG
DUg mgDh
ac
Ã2 r
w2r
tG
rG

G F
aG Ç tG tGnet

2000年ap物理c力学答案

2000年ap物理c力学答案1、C.电源的电动势与外电路无关(正确答案)D.电源电动势等于内电压答案解析：ABC都正确，D选项电源的电动势= 电源两端没有接用电器时，用电压表测得的电压。

此时，E全部加在内阻r上（即：全部电压都分给内阻r）当电源接入电路中时，全部电压（电动势E）分为两部分：①内阻r分得的电压Ur （内电压）②外部电路分得的电压U （外电压）所以：“电动势为什么等于内外电压之和”即：E = Ur + U2、26．下列现象中，属于升华的现象是（）[单选题] *A．夏天，冰棍周围冒“白气”B．冬天，玻璃窗上结冰花C．衣箱中的樟脑丸逐渐变小(正确答案)D．夏天，水缸外壁“出汗”3、当0℃的冰熔化成0℃的水时，温度和内能都不变[判断题] *对错(正确答案)答案解析：温度不变，内能增大4、3．屋檐滴下的水滴下落可视为自由落体运动．[判断题] *对(正确答案)错5、错竹筷漂浮在水面上，是由于筷子受到的浮力大于自身重力[判断题] *对错(正确答案)答案解析：漂浮时浮力等于重力6、88．如图为甲、乙两种物质的m﹣V图像，下列说法中正确的是（）[单选题] *A．体积为15cm3的乙物质的质量为30g(正确答案)B．甲的质量一定比乙的质量大C．甲、乙体积相同时，乙的质量是甲的2倍D．甲、乙质量相同时，甲的体积是乙的2倍7、30．如图，我国首款国际水准的大型客机C919在上海浦东机场首飞成功，标志着我国航空事业有了重大突破。

它的机身和机翼均采用了极轻的碳纤维材料。

这种材料的优点是（）[单选题] *A．密度大B．密度小(正确答案)C．熔点低D．硬度小8、72．学习质量和密度的知识后，小明同学想用天平、量筒和水完成下列实验课题，你认为不能够完成的是（）[单选题] *A．测量牛奶的密度B．鉴别金戒指的真伪C．鉴定铜球是否空心D．测一捆铜导线的长度(正确答案)9、水平桌面上的文具盒在水平方向的拉力作用下，沿拉力的方向移动一段距离，则下列判断正确的是（）[单选题]A．文具盒所受拉力做了功(正确答案)B．文具盒所受支持力做了功C．文具盒所受重力做了功D．没有力对文具盒做功10、用丝绸摩擦过的玻璃棒能吸引纸屑，说明玻璃棒有磁性[判断题] *对错(正确答案)答案解析：玻璃棒带电可以吸引轻小物体11、69．两种不同材料制成的大小相同的实心球甲、乙，在天平右盘中放入4个甲球，在左盘中放入5个乙球，这时天平刚好平衡，且游码没有移动，则可知（）[单选题] *A．甲球和乙球质量之比为5：1B．甲球和乙球质量之比为4：5C．甲球和乙球密度之比为5：4(正确答案)D．甲球和乙球密度之比为4：512、8．将耳朵贴在长铁水（管中有水）管的一端，让另外一个人敲击一下铁水管的另一端。

大学物理八九章部分习题解答

大学物理八九章部分习题解答-CAL-FENGHAI-(2020YEAR-YICAI)_JINGBIAN第8章磁场8-10一均匀密绕直螺线管的半径为，单位长度上有匝线圈，每匝线圈中的电流为，用毕奥—萨伐尔定律求此螺线管轴线上的磁场。

分析：由于线圈密绕，因此可以近似地把螺线管看成一系列圆电流的紧密排列，且每一匝圆电流在轴线上任一点的磁场均沿轴向。

解：取通过螺线管的轴线并与电流形成右旋的方向（即磁场的方向）为x 轴正向，如习题8-10图解（a ）所示。

在螺线管上任取一段微元dx ，则通过它的电流为dI nIdx =，把它看成一个圆线圈，它在轴线上O 点产生的磁感应强度dB 为2022322()R nIdxdB R x μ=+ 由叠加原理可得，整个螺线管在O 点产生的磁感应强度B 的大小为212022322()x Lx R nIdxB dB R x μ==+⎰⎰0212212221221[]2()()nIx x R x R x μ=-++ 由图可知12122212221212cos os ()()x x R x R x ββ==++ c ，代入上式并整理可得 021(cos cos )2nIB μββ=-式中12ββ和分别为x 轴正向与从O 点引向螺线管两端的矢径r 之间的夹角。

讨论：（1）若螺线管的长度远远大于其直径，即螺线管可视为无限长时，20β=，1βπ=，则有nI B 0μ=上式说明，无限长密绕长直螺线管内部轴线上各点磁感应强度为常矢量。

理论和实验均证明：在整个无限长螺线管内部空间里，上述结论也适用。

即无限长螺线管内部空间里的磁场为均匀磁场，其磁感应强度B 的大小为0nI μ，方向与轴线平行；（2）若点O 位于半无限长载流螺线管一端，即12πβ=，20β＝或12πβ=，2βπ＝时，无论哪一种情况均有nI B 021μ=------(8-19) 习题8-10图解可见半无限长螺线管端面中心轴线上磁感应强度的大小为管内的一半；综上所述，密绕长直螺线管轴线上各处磁感应强度分布见习题8-10图解（b ）所示，从图中也可看出，长直螺线管内中部的磁场可以看成是均匀的。

物理c试题及答案

物理c试题及答案一、选择题（每题5分，共20分）1. 光在真空中的传播速度是（）A. 2.99×10^8 m/sB. 3.00×10^8 m/sC. 3.01×10^8 m/sD. 2.98×10^8 m/s2. 根据牛顿第三定律，作用力和反作用力（）A. 总是大小相等，方向相反B. 总是大小相等，方向相同C. 总是大小不等，方向相反D. 总是大小不等，方向相同3. 一个物体从静止开始做匀加速直线运动，第1秒内通过的位移是x1，第2秒内通过的位移是x2，那么x2与x1的关系是（）A. x2 = 2x1B. x2 = 3x1C. x2 = 4x1D. x2 = 5x14. 一个质量为m的物体在水平面上受到一个恒定的力F作用，若物体与水平面之间的摩擦力为f，则物体的加速度a为（）A. a = (F - f) / mB. a = (F + f) / mC. a = F / mD. a = f / m二、填空题（每题5分，共20分）1. 根据万有引力定律，两个物体之间的引力与它们的质量成正比，与它们之间的距离的平方成反比。

若两个物体的质量分别为m1和m2，它们之间的距离为r，则它们之间的引力大小为__________。

2. 电容器的电容C与其极板面积A和极板间的距离d的关系为C = __________，其中ε0是真空的电容率。

3. 在理想气体状态方程中，PV = nRT，其中n表示气体的摩尔数，R 是气体常数，T是气体的绝对温度。

若温度从T1升高到T2，气体的压强P和体积V的变化关系为PV^__________ = nRT1。

4. 根据欧姆定律，电流I与电压V和电阻R之间的关系为I =__________。

三、计算题（共60分）1. 一个质量为2kg的物体从静止开始，受到一个大小为10N的恒定力作用，求物体在力作用下10秒内的位移。

（15分）2. 一个电容器的电容为4μF，当它充电到6V时，储存的电荷量是多少？（15分）3. 一个理想气体在绝热过程中，体积从V1减少到V2，温度从T1降低到T2。

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AP® Physics C
1989 Scoring Guidelines
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