【AP物理C】【真题】解答题答案C1977

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【AP物理C】【真题】2003年力学解答题答案

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 membership association whose mission is to prepare, inspire, and connect students to college and opportunity. Founded in 1900, the association is composed of more than 4,300 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 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 equity and excellence, and that commitment is embodied in all of its programs, services, activities, and concerns. For further information, visit Copyright © 2003 College Entrance Examination Board. All rights reserved. College Board, Advanced Placement Program, AP, AP Vertical Teams, APCD, Pacesetter, Pre-AP, SAT, Student Search Service, and the acorn logo are registered trademarks of the College Entrance Examination Board. AP Central is a trademark owned by the College Entrance Examination Board. PSAT/NMSQT is a registered 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. Other products and services may be trademarks of their respective owners. For the College Board’s online home for AP professionals, visit AP Central at .

2018ap物理c力学大题

2018ap物理c力学大题
2018年AP物理C力学大题主要涉及以下几个方面：
1. 力学基本概念，题目可能涉及牛顿运动定律、摩擦力、弹簧力、重力等基本力学概念。

学生需要理解这些概念并能够运用它们
解决问题。

2. 动量和能量，题目可能涉及动量守恒定律、动能、势能、机
械能守恒等内容。

学生需要能够分析物体的动能和势能变化，并理
解动量守恒的应用。

3. 旋转运动，题目可能涉及刚体的平衡、转动惯量、角动量等
内容。

学生需要理解刚体平衡的条件，能够计算刚体的转动惯量并
应用相关知识解决问题。

4. 振动和波动，题目可能涉及弹簧振子、简谐振动、波速、波
长等内容。

学生需要理解振动的特点、波动的传播规律，并能够运
用相关公式解决问题。

总体来说，2018年AP物理C力学大题涵盖了力学的基本概念、
动量和能量、旋转运动、振动和波动等多个方面的内容。

学生在备考时需要扎实掌握这些知识点，并能够灵活运用到解决实际问题的能力。

希望这些信息能够对你有所帮助。

【AP物理C】【真题】解答题 C1973

AP® Physics C错误！未找到引用源。

1973 Free Response QuestionsThese 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 membership association dedicated to preparing, inspiring, and connecting students to college and opportunity. Founded in 1900, the association is composed 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®, thePSAT/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.APIEL is a trademark owned by the College Entrance Examination Board. PSAT/NMSQT is a registered 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.1973M1. A horizontal force F is applied to a small block of mass m1 to make it slide along the top of a larger block of mass m2 and length l. The coefficient of friction between the blocks is μ. The larger block slides without friction along a horizontal surface. The blocks start from rest with the small block at one end of the larger block, as shown.a. On the diagrams below draw all of the forces acting on each block. Identify each force.b. Find the acceleration of each block. a1 and a2, relative to the horizontal surface.c. In terms of 1, a1, and a2, find the time t needed for the small block to slide off the endof the larger block.d. Find an expression for the energy dissipated as heat because of the friction betweenthe two blocks.1973M2. A 30-gram bullet is fired with a speed of 500 meters per second into a wall.a. If the deceleration of the bullet is constant and it penetrates 12 centimeters into the wall, calculate theforce on the bullet while it is stopping.b. If the deceleration of the bullet is constant and it penetrates 12 centimeters into the wall, how muchtime is required for the bullet to stop?c. Suppose, instead, that the stopping force increases from zero as the bullet penetrates. Discuss themotion in comparison to the case for a constant deceleration.1973M3. A ball of mass m is attached by two strings to a vertical rod. as shown above. The entire system rotates at constant angular velocity ω about the axis of the rod.a. Assuming ω is large enough to keep both strings taut, find the force each string exerts onthe ball in terms of ω, m, g, R, and θ.b. Find the minimum angular velocity, ωmin for which the lower string barely remains taut.1973E1. The plates of an isolated parallel plate capacitor are pulled apart very slowly by a forceF. Each plate has charge q and area A. Assume that edge effects are negligible, i.e., thespacing x is much smaller than the plate dimensions.a. Determine the change in capacitance as x is increased by dxb. Determine the change of stored energy in the capacitor as x is increased by dxc. How is the force F related to the change in stored energy ? Determine F in terms of x, q,and A.1973E2. A surveyor attempts to use a compass below a power line carrying a steady currentof 103 amperes. The compass Is 6.0 meters directly below the wire.a. If the horizontal component of the Earth’s field is 0.10 gauss, could the power linedisturb the compass reading? Give a quantitative argument.b. Suppose, instead, the current were 103 amperes of 60 cycles per second alternatingcurrent. Would the compass reading be disturbed? Explain your answerqualitatively in terms of the properties of the compass.1973E3. In a uniform magnetic field B directed vertically downward. a metal bar of mass m is released from rest and slides without friction down a track inclined at an angle , as shown above.The electrical resistance of the bar between its two points of contact with the track is R: the trackhas negligible resistance. The width of the track is l.a. Show on the diagram the direction of the current in the sliding bar.b. Denoting by v the instantaneous speed with which the bar is sliding down the incline,determine an expression for the magnitude of the current in the bar.c. Determine an expression for the force exerted on the bar by the magnetic field.学习使人进步d. Determine an expression for the terminal velocity of the sliding bar.。

【AP物理C】【真题】2004年电磁学解答题答案

Байду номын сангаас
The 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 4,500 schools, colleges, universities, and other educational organizations. Each year, the College Board serves over three 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. For further information, visit Copyright © 2004 College Entrance Examination Board. All rights reserved. College Board, Advanced Placement Program, AP, AP Central, AP Vertical Teams, APCD, Pacesetter, Pre-AP, SAT, Student Search Service, and the acorn logo are registered trademarks of the College Entrance Examination Board. PSAT/NMSQT is a registered trademark of the College Entrance Examination Board and National Merit Scholarship Corporation. Educational Testing Service and ETS are registered trademarks of Educational Testing Service. Other products and services may be trademarks of their respective owners. For the College Board’s online home for AP professionals, visit AP Central at .

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。

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。

【AP物理C】【真题】2004年力学解答题答案

The 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 4,500 schools, colleges, universities, and other educational organizations. Each year, the College Board serves over three 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. For further information, visit Copyright © 2004 College Entrance Examination Board. All rights reserved. College Board, Advanced Placement Program, AP, AP Central, AP Vertical Teams, APCD, Pacesetter, Pre-AP, SAT, Student Search Service, and the acorn logo are registered trademarks of the College Entrance Examination Board. PSAT/NMSQT is a registered trademark of the College Entrance Examination Board and National Merit Scholarship Corporation. Educational Testing Service and ETS are registered trademarks of Educational Testing Service. Other products and services may be trademarks of their respective owners. For the College Board’s online home for AP professionals, visit AP Central at .

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物理c】真题】解答题c2000

【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.。

【AP物理C】【真题】2003年力学解答题答案

【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。

【AP物理C】【真题】解答题 C1975

AP® Physics C错误!未找到引用源。

1975 Free Response QuestionsThe materials 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 materials and any copies made of them 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 membership association dedicated to preparing, inspiring, and connecting students to college and opportunity. Founded in 1900, the association is composed 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®, thePSAT/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.APIEL is a trademark owned by the College Entrance Examination Board. PSAT/NMSQT is a registered 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.1975M1. A sphere of mass m is released from rest. As it falls, the air exerts a retarding force on the sphere that is proportional to the sphere's velocity ( F R = -kv). Neglect the buoyancy force of the air.a. On the circles below draw vectors representing the forces acting on the spherei. just after it is released andii. after it has been falling- for a long time and reached terminal velocity. Give each vector a descriptive label(i) (ii)b. Determine the terminal velocity of the spherec. Draw the following three graphs for the sphere's motion clearly showing significant features of themotion just after the sphere is released as well as after a long time.i. Acceleration as a function of time ii. Velocity as a function of time iii. Position as a function of time.AccelerationVelocityPositionView From Above1975M2. A bicycle wheel of mass M (assumed to be concentrated at its rim) and radius R is mounted horizontally so it may turn without friction on a vertical axle. A dart of mass m o is thrown with velocity v o as shown above and sticks in the tire.a. If the wheel is initially at rest, find its angular velocity ω after the dart strikes.b. In terms of the given quantities, determine the ratio:final kinetic energy of the systeminitial kinetic energy of the system1975M3. A uniform chain of mass M and length l hangs from a hook in the ceiling. The bottom link is now raised vertically and hung on the hook as shown above on the right.a. Determine the increase in gravitational potential energy of the chain by considering the change inposition of the center of mass of the chain.b. Write an equation for the upward external force F(y) required to lift the chain slowly as afunction of the vertical distance y.c. Find the work done on the chain by direct integration of ⎰Fdx.1975E1. Two stationary point charges +q are located on the y-axis as shown above. A third charge +q is brought in from infinity along the x-axis.a. Express the potential energy of the movable charge as a function of its position on the x-axis.b. Determine the magnitude and direction of the force acting on the movable charge when it islocated at the position x = lc. Determine the work done by the electric field as the charge moves from infinity to the origin.1975E2. In the diagram above, V = 100 volts; C1 = 12 microfarads; C2 = 24 microfarads; R =10 ohms. Initially, C l and C2 are uncharged, and all switches are open.a. First, switch S1 is closed. Determine the charge on C l when equilibrium is reached.b. Next S1 is opened and afterward S2 is closed. Determine the charge on C1 when equilibrium is againreached.c. For the equilibrium condition of part (b), determine the voltage across C1.d. S2 remains closed, and now S1 is also closed. How much additional charge flows from the battery?1975E3. A long straight conductor lies in the plane of a rectangular loop of wire as shown above. The total resistance of the loop is R. The current in the long straight conductor increases at a constant rate dI/dt.a. Indicate on the diagram the direction of the induced current in the loop and explain yourreasoning.b. Determine the magnitude of the current assuming the self-inductance of the loop may beneglected.。

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。

【AP物理C】【真题】2006年力学解答题答案

AP® Physics C: Mechanics2006 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 5,000 schools, colleges, universities, and other educational organizations. Each year, the College Board serves seven 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.© 2006 The 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 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 Board. PSAT/NMSQT is a registered trademark of the College Board and National Merit Scholarship Corporation. All 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: .AP® PHYSICS C MECHANICS2006 SCORING GUIDELINESGeneral Notes About 2006 AP Physics Scoring Guidelines1.The solutions contain the most common method of solving the free-response questions and theallocation of points for this solution. Some also contain a common alternate solution. 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 1 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. However, when students are asked to derive an expression, it is normally expected that they will begin by writing one or more fundamental equations, such as those given on the AP Physics exam equation sheet. See pages 21–22 of the AP Physics Course Description for a description of the use of such terms as “derive” and “calculate” on the exams, and what is expected for each.4.The scoring guidelines typically show numerical results using the value 2g=,but use of9.8m s210m s is of course also acceptable. Solutions usually show numerical answers using both values when they are significantly different.5.Strict rules regarding significant digits are usually not applied to numerical answers. However, in somecases answers containing too many digits may be penalized. In general, two to four significant digits are acceptable. Numerical answers that differ from the published answer due to differences in rounding throughout the question typically receive full credit. Exceptions to these guidelines usually occur 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 and 20.278). Rounding to three digits will lose the accuracy required to determine the difference in the numbers, and some credit may be lost.AP ® PHYSICS C MECHANICS 2006 SCORING GUIDELINESQuestion 115 points total Distributionof points(a) 4 pointsFor the block force diagram:For correctly labeled horizontal force (friction to the left, no other forces or vectors) 1 point For correctly labeled vertical forces (normal up and weight down; gravity alone notaccepted)1 point For the slab force diagram:For correctly labeled horizontal force (friction to the right, no other forces or vectors) 1 point For correctly labeled vertical forces (normal up and combined weight down; combinedweight can be shown with two arrows or identifying weight as orS B W W ++S B M g M g ; gravity alone not accepted)1 point(b) and (c)These two parts were scored together because of the different approaches that could be used to answer them.Momentum approach to part (b); Newton’s second law and kinematics approach to part (c)(b) 3 pointsFor any statement of conservation of momentum 1 point No net external forces act on the two-block system, so linear momentum is conserved. For a correct momentum equation 1 point()0u u =+B B S M M M f00.50 kg4.0 m s 0.50 kg +3.0 kg u u ==+B f B S M M MFor the correct answer 1 point 0.57 m u =fDistribution of points Momentum approach (continued)(c) 6 pointsFor a correct expression for the friction force (awarded if found in the solution to any of parts (a) through (d)) 1 pointm =f mg or m =f NFor correct substitution of =B m M for the friction force on the block1 point m =B f M gFor recognizing that the friction force on the slab is equal in magnitude to the friction force on the block and for an equation relating this force to the acceleration of the slab1 point=S S f M aFor a correct expression for the acceleration of the slab or its numerical value1 point 20.33 m s B S SM ga M m == For a correct kinematic equation for the slab1 point 2202u u =+f S a x , where00u = 2222u u m ==f f S S B M x a M gFor correct substitutions consistent with earlier values 1 point ()()()220.57 m s 3.0 kg20.200.50 kg 9.8 m s =xx = 0.49 m or 0.50 m, depending on use of 29.8 or 10 m s g = and where substitution androunding took placeDistribution of points Newton’s second law and kinematics approach to part (b); kinematics approach to part (c)(b) 7 pointsFor a correct expression for the friction force (awarded if found in the solution to any of parts (a) through (d)) 1 pointm =f mg or m =f NFor correct substitution of =B m M1 point m =B f M gFor recognizing that the friction force on the slab is equal in magnitude to the friction force on the block and for an equation relating this force to the acceleration of the slab1 point=S S f M aFor a correct expression for the acceleration of the slab or its numerical value1 point 20.33 m s m ==B S SM g a MFor a correct expression for the acceleration of the block or its numerical value1 point 22.0m s m m ===B B BM g a g MFor a solution of the following simultaneous kinematic equations for the block andthe slab, such as by setting the times equal and solving for u f 1 point 0u u =-f a t B for the block u =f S a t for the slab()()20220.33 m s 4.0 m s 0.33 m s + 2.0 m s S f S B a a a u u ==+For the correct answer 1 point 0.57 m f u =(c) 2 pointsFor a correct kinematic equation for the slab1 point 2202u u =+f S a x , where00u = 2222u u m ==f fS S B M x a M gFor correct substitutions consistent with earlier values 1 point ()()()220.57 m s 3.0 kg20.200.50 kg 9.8 m s =xx = 0.49 m or 0.50 m, depending on use of 29.8 or 10 m s g = and where substitution androunding took placeDistribution of pointsx(d) 2 pointsFor a correct expression for the work done1 pointm ==B W Fd M g OR 212u D ==S f W K M For consistent substitution from parts (b) and (c)1 point ()()(20.200.5 kg 9.8 m s 0.50 m =W ) OR ()()213.0 kg 0.57 m s 2=W W = 0.49 J (or W = 0.50 J using 210 m s g =)AP ® PHYSICS C MECHANICS 2006 SCORING GUIDELINESQuestion 215 points total Distributionof points(a) 1 pointFor indicating that F vs. 2xor vs. x should be graphed, or other equivalent correctresponse (Must clearly specify two variables in order to earn this point.) 1 point(b) 2 pointsFor a correct column label, including units1 point For calculated values that match what is indicated in (a)1 point Note: If answer to (a) was incorrect or incomplete, (b) received no credit.Example using F vs.2x x (m) F (N) ()22m x0.05 4 0.0025 0.10 17 0.010 0.15 38 0.023 0.20 68 0.040 0.25 1060.063(c) 3 pointsFor appropriate linear axes scales 1 point For correct axes labels 1 point For plotting the points1 point Note: Axes and scales must match answer in (a). However, if (a) was incorrect orincomplete, points were awarded in (c) if graph was executed correctly. If (a) was blank or didn’t include any variables, no credit was awarded for (b) or (c).Example using data aboveAP ® PHYSICS C MECHANICS 2006 SCORING GUIDELINESQuestion 2 (continued)Distributionof points(d) 2 pointsFor indication of a correct relationship between the coefficient A and the slope for the values graphed in (c)1 pointFor correct units and no more than four significant figures on value of A 1 point Example using data in the table and two points on the line in the graph2=F Ax , so A is equal to the slope of the F vs. 2x line.32222100 N 50 N slope 1.710N m 0.060 m 0.030 mD D -====¥-F A xNotes:This part stated to “calculate,” so an answer with correct units and significant figures butwith no work shown earned 1 point.Since all the data points are on the best-fine line, additional credit was not awarded for a correctly drawn best-fine line or for use of points on the line instead of data points.(e) 4 pointsUsing the definition of work=ÚW F dxFor correct substitution of F (x ) into the integral for work 1 point For correct limits on the integral 1 point For correct evaluation of the integral1 point ()()(0.10 m32320110.10 m 1.710 N/m 1.010 m 33W Ax dx A -===¥¥Ú)33For the correct answer with correct units1 point 0.57 J =WNote: This part stated to “calculate,” so a correct answer with correct units, but with nowork shown, earned 1 point.(f) 3 pointsFor an appropriate expression of conservation of energy or the work-energy theorem1 point For a correct expression for K and substitution of W from part (e), expressed algebraically ornumerically1 point 21u D ==W K mu == For a value of u consistent with the value of W in (e), with correct units 1 point 1.5 m s u =Note: This part stated to “calculate,” so an answer consistent with (e) and with correctunits, but with no work shown, earned 1 point.AP ® PHYSICS C MECHANICS 2006 SCORING GUIDELINESQuestion 315 points total Distribution of points(a) and (b)These two parts were scored together because of the different approaches that could be used to answer them. The parts could be answered in either order.Approach using translational and rotational dynamics(a) 5 pointsFor use of Newton’s 2ndlaw in both translational and rotational forms 1 point=Âcm F ma andt a =Âcm I For a correct equation applying Newton’s second law in translational form1 point sin q -=cm Mg f MaFor a correct equation applying Newton’s second law in rotational form 1 point a =cm f R IFor a correct relationship between linear and angular acceleration for rolling withoutslipping 1 point a =cm cm aSubstituting for I and into the rotational equation abovea cm 2=cma f R MR R=cm f MaSubstituting this expression for f into the equation for translational motion above sin q -=cm cm Mg Ma Ma For the correct answer1 point sin 2q =cm ga(b) 3 pointsFor a correct kinematic equation containing a and u 1 point2202u u D =+a x ,0u = For correct substitution of the expression for acceleration from part (a)1 pointFor correct substitution of the distance traveled1 point 22sin 2u q Êˆ=Ë¯gLu =AP ® PHYSICS C MECHANICS 2006 SCORING GUIDELINESQuestion 3 (continued)Distribution of points Approach using torque about point of contact between hoop and ramp and parallel axis theorem(a) 5 pointsFor use of Newton’s 2ndlaw in rotational form and the parallel axis theorem1 point ta =Âcm I and2=+cm I I Mh For a correct rotational inertia about the point of contact using the parallel axis theorem 1 point 222=+=I MR MR MR 2For a correct torque about the point of contact 1 point sin t q =ÂRMgFor a correct relationship between linear and angular acceleration for rolling withoutslipping 1 point a =cm cm a RSubstituting for , I , and a into the rotational equation abovet Âcm 2sin 2cma RMg MR Rq = For the correct answer1 point sin 2q =cm ga(b) 3 pointsFor a solution to part (b) as in the previous approach with points allotted similarly3 pointsDistributionof pointsApproach using conservation of energy and kinematics, working part (b) first(b) 5 pointsFor a statement of conservation of energy containing potential and kinetic energy terms1 point D =+rot trans U K KFor a correct expression for the potential energy change 1 point For correct translational and rotational kinetic energies1 point 2211sin 22q u =+w MgL M IFor a correct relationship between linear and angular velocity for rolling without slipping1 point Ru w =Substituting expressions for I and into the energy equation above w ()()22211sin u q u =+MgL M MR2211sin 22q u u u =+=gL 2For the correct answer 1 pointu =(a) 3 pointsFor a correct kinematic relationship1 point 2202u u D =+a x ,00u = For correct substitution of the expression for velocity 1 point For correct substitution of the distance traveled 1 point sin 2q =cm gL a Lsin 2q =cm gaDistributionof points (c) 4pointsApplying the kinematic equation for distance as a function of time to the vertical motion22H gt=1 pointt=For use of zero acceleration in calculation of the horizontal distance traveled 1 point u=xx tFor correct substitution of u from part (b)x1 pointFor correct substitution of t from previous calculation1 pointd==d(d) 3pointsFor checking the space next to “Greater than” 1 point For a sufficiently detailed justification containing no incorrect statements. Such an answer logically concludes, at a minimum, that the linear speed or velocity at the bottom of theramp is greater for the disk because the rotational inertia of the disk is less. It is notnecessary to state that the time of fall is the same.2 pointsOne point was awarded for a minimal or partially correct answer.No justification points were awarded if the space next to “Greater than” was not checked.Examples of 2-point answers:A disk will have smaller rotational inertia and will therefore have a greater rotationalvelocity. This will lead to a greater translational velocity, and a greater distance x.The rotational inertia is less than the hoop, causing greater acceleration and more finalspeed at the end of the table.The acceleration when 2=I MR is ()23sing q, so the disk will be moving faster atthe bottom of the ramp and will travel farther.Examples of 1-point answers:A disk has a larger rotational inertia, so it will have a greater kinetic energy and willtherefore land farther from the ramp.The moment of inertia for the disk is smaller, thus its rotational velocity is bigger,causing it to go further.Less energy will be used to spin the disk than the hoop, and I of the disk is less than I of the hoop.。

ap物理c知识点梳理

ap物理c知识点梳理
AP物理C是高中物理课程中的最高难度课程之一，它涵盖了力学、电学和热学等领域的知识。

以下是一些AP物理C的重要知识点：力学：
1. 牛顿三定律：力的性质和效应
2. 动量：动量定理和碰撞
3. 能量：功和能量定理
4. 旋转：力矩、角动量和角动量守恒
5. 万有引力：开普勒定律和万有引力定律
电学：
1. 电场：电势和电场力
2. 电路：欧姆定律和基本电路元件（电池、电阻、电容和电感等）
3. 静电场：库仑定律和高斯定理
4. 磁场：磁场力和安培环路定理
5. 时间变化的电磁场：法拉第电磁感应定律和最大韦尔斯定理
热学：
1. 温度和热力学系统：温度、热量和热力学状态方程
2. 热传递：热传导、对流和辐射
3. 热力学第一、二定律：热机和热泵、热力学第一定律和第二定律的应用
4. 统计物理学：热力学中的统计分布和温度
这些知识点在AP物理C考试中都有可能出现，学生们应该对这些知识点进行深入的学习和理解。

【AP物理C】【真题】解答题 C1997

AP® Physics C1997 Free response QuestionsThese 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 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 is composed 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 commitment is embodied in all of its programs, services, activities, and concerns.APIEL is a trademark owned by the College Entrance Examination Board. PSAT/NMSQT is a registered 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 TestingService.1997M1. A nonlinear spring is compressed horizontally. The spring exerts a force that obeys the equation F(x) = Ax½, where x is the distance from equilibrium that the spring is compressed and A is a constant. A physics student records data on the force exerted by the spring as it is compressed and plots the two graphs below, which include the data and the student's best-fit curves.a. From one or both of the given graphs, determine A. Be sure to show your work and specify theunits.b. i. Determine an expression for the work done in compressing the spring a distance x.ii. Explain in a few sentences how you could use one or both of the graphs to estimate a numerical answer to part (b)i for a given value of x.c. The spring is mounted horizontally on a countertop that is 1.3 m high so that its equilibrium positionis just at the edge of the countertop. The spring is compressed so that it stores 0.2 J of energy and is then used to launch a ball of mass 0.10 kg horizontally from the countertop. Neglecting friction,determine the horizontal distance d from the edge of the countertop to the point where the hall strikes the floor1997M2. An open-top railroad car (initially empty and of mass M o) rolls with negligible friction along a straight horizontal track and passes under the spout of a sand conveyor. When the car is under the conveyor, sand is dispensed from the conveyor in a narrow stream at a steady rate ∆M/∆t = C and falls vertically from an average height h above the floor of the railroad car. The car has initial speed v o and sand is filling it from time t = 0 to t = T. Express your answers to the following in terms of the given quantities and g.a. Determine the mass M of the car plus the sand that it catches as a function of time t for 0 < t < T.b. Determine the speed v of the car as a function of time t for 0 < t < T.c. i. Determine the initial kinetic energy K i of the empty car.ii. Determine the final kinetic energy K f of the car and its load.iii. Is kinetic energy conserved? Explain why or why not.d. Determine expressions for the normal force exerted on the car by the tracks at the following times.i. Before t = 0ii. For 0 < t < Tiii. After t = T1997M3. A solid cylinder with mass M, radius R, and rotational inertia ½MR2 rolls without slipping down the inclined plane shown above. The cylinder starts from rest at a height H. The inclined plane makes an angle θ with the horizontal. Express all solutions in terms of M, R, H, θ, and g.a. Determine the translational speed of the cylinder when it reaches the bottom of the inclined plane.b. On the figure below, draw and label the forces acting on the cylinder as it rolls down the inclinedplane Your arrow should begin at the point of application of each force.c. Show that the acceleration of the center of mass of the cylinder while it is rolling down the inclinedplane is (2/3)g sinθ.d. Determine the minimum coefficient of friction between the cylinder and the inclined plane that isrequired for the cylinder to roll without slipping.e. The coefficient of friction μ is now made less than the value determined in part (d), so that thecylinder both rotates and slips.i. Indicate whether the translational speed of the cylinder at the bottom of the inclined plane is greaterthan, less than, or equal to the translational speed calculated in part (a). Justify your answer.ii. Indicate whether the total kinetic energy of the cylinder at the bottom of the inclined plane is greater than, less than, or equal to the total kinetic energy for the previous case of rolling withoutslipping. Justify your answer.1997E1. A technician uses the circuit shown above to test prototypes of a new battery design. The switch is closed, and the technician records the current for a period of time. The curve that best fits the results is shown in the graph below.The equation for this curve is I = I o e -kt where t is the time elapsed from the instant the switch is closed and I oand k are constants.a. i.o across the resistorimmediately after the switch is closed.ii. Would the open circuit voltage of the fresh battery have been less than, greater than, or equal to the value in part i ? Justify your answer.b. Determine the value of k from this best-fit curve. Show your work and be sure to include units in your answer.c. Determine the following in terms of R, I o , k, and t.i. The power delivered to the resistor at time t = 0ii. The power delivered to the resistor as a function of time tiii. The total energy delivered to the resistor from t = 0 until the current is reduced to zero1997E2. A nonconducting sphere with center C and radius a has a spherically symmetric electric charge density. The total charge of the object is Q > 0.a. Determine the magnitude and direction of the electric field at point P, which is a distance R > a to theright of the sphere's center.b. Determine the flux of the electric field through the spherical surface centered at C and passingthrough P.A point particle of charge -Q is now placed a distance R below point P. as shown above.c. Determine the magnitude and direction of the electric field at point P.d. Now consider four point charges, q1, q2, q3, and q4, that lie in the plane of the page as shown in thediagram above. Imagine a three-dimensional closed surface whose cross section in the plane of the page is indicated.i. Which of these charges contribute to the net electric flux through the surface?ii. Which of these charges contribute to the electric field at point P1 ?iii. Are your answers to i and ii the same or are they different? Explain why this is so.e. If the net charge enclosed by a surface is zero, does this mean that the field is zero at all points on thesurface? Justify your answer.f. If the field is zero at all points on a surface, does this mean there is no net charge enclosed by thesurface? Justify your answer.1997E3. A long, straight wire lies on a table and carries a constant current I 0, as shown above.a. Using Ampere's law, derive an expression for the magnitude B of the magnetic field at a perpendicular distance r from the wire.A rectangular loop of wire of length l , width w, and resistance R is placed on the table a distance s from thewire, as shown below. b. What is the direction of the magnetic field passing through the rectangular loop relative to thecoordinate axes shown above on the right?c. Show that the total magnetic flux φm through the rectangular loop is Il s w s02μπln()+The rectangular loop is now moved along the tabletop directly away from the wire at a constant speed v = |ds/dt ∣as shown above.d. What is the direction of the current induced in the loop? Briefly explain your reasoning.e. What is the direction of the net magnetic force exerted by the wire on the moving loop relative to the coordinate axes shown above on the right? Briefly explain your reasoning.f. Determine the current induced in the loop. Express your answer in terms of the given quantities and fundamental constants.。

【AP物理C】【真题】解答题2008力学

AP® Physics C: Mechanics2008 Free-Response QuestionsThe 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 5,000 schools, colleges, universities, and other educational organizations. Each year, the College Board serves seven 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.© 2008 The College Board. All rights reserved. College Board, Advanced Placement Program, AP, AP Central, SAT, and the acorn logo are registered trademarks of the College Board. PSAT/NMSQT is a registered trademark of the College Board and National Merit Scholarship Corporation.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: .PHYSICS C: MECHANICSSECTION IITime-45 minutes3 QuestionsDirections: Answer all three questions. The suggested time is about 15 minutes for answering each of the questions, which are worth 15 points each. The parts within a question may not have equal weight. Show allyour work in the pink booklet in the spaces provided after each part, NOT in this green insert.2008M1. A skier of mass M is skiing down a frictionless hill that makes an angle θ with the horizontal, as shown in the diagram. The skier starts from rest at time t = 0 and is subject to a velocity-dependent drag force due to air resistance of the form F = –bv, where v is the velocity of the skier and b is a positive constant. Express all algebraic answers in terms of M, b, θ, and fundamental constants.(a) On the dot below that represents the skier, draw a free-body diagram indicating and labeling all of the forces that act on the skier while the skier descends the hill.(b) Write a differential equation that can be used to solve for the velocity of the skier as a function of time.(c) Determine an expression for the terminal velocity v T of the skier.(d) Solve the differential equation in part (b) to determine the velocity of the skier as a function of time, showing all your steps.(e) On the axes below, sketch a graph of the acceleration a of the skier as a function of time t, and indicate the initial value of a. Take downhill as positive.taO2008M2. The horizontal uniform rod shown above has length 0.60 m and mass 2.0 kg. The left end of the rod is attached to a vertical support by a frictionless hinge that allows the rod to swing up or down. The right end of the rod is supported by a cord that makes an angle of 30° with the rod. A spring scale of negligible mass measures the tension in the cord. A 0.50 kg block is also attached to the right end of the rod.(a) On the diagram below, draw and label vectors to represent all the forces acting on the rod. Show each forcevector originating at its point of application.(b)Calculate the reading on the spring scale.ML2, where M is the mass of the rod and L is its length.The rotational inertia of a rod about its center is 112Calculate the rotational inertia of the rod-block system about the hinge.(d)If the cord that supports the rod is cut near the end of the rod, calculate the initial angular acceleration of therod-block system about the hinge.2008M3. In an experiment to determine the spring constant of an elastic cord of length 0.60 m, a student hangs the cord from a rod as represented above and then attaches a variety of weights to the cord. For each weight, the student allows the weight to hang in equilibrium and then measures the entire length of the cord. The data are recorded in the table below:(a)Use the data to plot a graph of weight versus length on the axes below. Sketch a best-fit straight line throughthe data.the cord.The student now attaches an object of unknown mass m to the cord and holds the object adjacent to the point at which the top of the cord is tied to the rod, as represented above. When the object is released from rest, it falls1.5 m before stopping and turning around. Assume that air resistance is negligible.(c)Calculate the value of the unknown mass m of the object.(d)i. Calculate how far down the object has fallen at the moment it attains its maximum speed.ii.Explain why this is the point at which the object has its maximum speed.iii.Calculate the maximum speed of the object.。

1977年高考物理试卷

选择题题目：关于物体的运动状态与所受合外力的关系，下列说法正确的是：A. 物体做匀速直线运动时，所受合外力一定为零（正确答案）B. 物体做曲线运动时，所受合外力一定变化C. 物体做匀变速直线运动时，所受合外力一定增大D. 物体做匀速圆周运动时，所受合外力方向一定与速度方向相同题目：下列关于电场的描述中，正确的是：A. 电场线总是从正电荷出发，终止于负电荷B. 在电场中，电势降低的方向一定是电场线的方向（正确答案）C. 电场强度为零的地方，电势也一定为零D. 等量同种电荷的电场中，两电荷连线中点处场强为零题目：关于光的传播，下列说法正确的是：A. 光在介质中的传播速度总是大于在真空中的速度B. 光的干涉和衍射现象说明光具有波动性（正确答案）C. 光从一种介质进入另一种介质时，频率一定会改变D. 红光和紫光相比，红光的折射率更大题目：下列关于原子核的描述中，正确的是：A. 原子核由质子和中子组成，质子和中子统称为核子（正确答案）B. 原子核内质子数和中子数一定相等C. 原子核的体积占原子体积的大部分D. 原子核的结合能对原子核的稳定性没有影响题目：关于热力学定律，下列说法正确的是：A. 热量不能自发地从低温物体传到高温物体（正确答案）B. 一定质量的某种理想气体，内能只与温度有关C. 第二类永动机违反了能量守恒定律D. 绝对零度是可以达到的题目：下列关于电磁感应的说法中，正确的是：A. 感应电流的磁场总是阻碍引起感应电流的磁通量的变化（正确答案）B. 导体在磁场中运动时，一定会产生感应电流C. 感应电动势的大小与线圈的匝数无关D. 穿过闭合电路的磁通量发生变化时，电路中不一定有感应电流题目：关于机械波的传播，下列说法正确的是：A. 机械波的传播速度与介质的性质有关（正确答案）B. 横波中质点的振动方向与波的传播方向相同C. 纵波中质点的振动方向与波的传播方向垂直D. 机械波能在真空中传播题目：下列关于电阻的说法中，正确的是：A. 导体的电阻与导体两端的电压成正比B. 导体的电阻与通过导体的电流成反比C. 导体的电阻是导体本身的一种性质，与电流、电压无关（正确答案）D. 不同材料的导体，电阻一定不同。

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AP® Physics C
1977 Scoring Guidelines
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