【AP物理C】【真题】2004年力学解答题答案
2004年全国高考卷2物理试题及答案(精编清晰)

2004年普通高等学校招生全国统一考试(甘肃、青海、宁夏、贵州、新疆等地)14.在核反应方程式kX Xe Sr n U ++→+1365490381023592中( )A .X 是中子,k=9B .X 是中子,k=10C .X 是质子,k=9D .X 是质子,k=1015.如图所示,在x ≤0的区域内存在匀强磁场,磁场的方向垂直于x y 平面(纸面)向里。
具有一定电阻的矩形线框a bcd 位于x y 平面内,线框的a b 边与y 轴重合。
令线框从t=0的时刻起由静止开始沿x 轴正方向做匀加速运动,则线框中的感应电流I (取逆时针方向的电流为正)随时间t 的变化图线I —t 图可能是下图中的哪一个?16.一定质量的理想气体,从某一状态开始,经过系列变化后又回一开始的状态,用W 1表示外界对气体做的功,W 2表示气体对外界做的功,Q 1表示气体吸收的热量,Q 2表示气体放出的热量,则在整个过程中一定有( )A .Q 1—Q 2=W 2—W 1B .Q 1=Q 2C .W 1=W 2D .Q 1>Q 217.图中M 是竖直放置的平面镜,镜离地面的距离可调节。
甲、乙二人站在镜前,乙离镜的距离为甲离镜的距离的2倍,如图所示。
二人略错开,以便甲能看到乙的像。
以l 表示镜的长度,h 表示乙的身高,为使甲能看到镜中乙的全身像,l 的最小值为 ( ) A .h 31B .h 21C .h 43D .h18.已知:一简谐横波在某一时刻的波形图如图所示,图中位于a 、b 两处的质元经过四分之一周期后分别运动到a '、b '处。
某人据此做出如下判断: ①可知波的周期, ②可知波的传播速度, ③可知的波的传播方向, ④可知波的波长。
其中正确的是( )A .①和④B .②和④C .③和④D .②和③19.如图,在倾角为α的固定光滑斜面上,有一用绳子拴着的长木板,木板上站着一只猫。
已知木板的质量是猫的质量的2倍。
2004年全国各地高考物理试题分类详解-力学

2004高考力学题【湘鄂、全国】18.如图所示,四个完全相同的弹簧都处于水平位置,它们的右端受到大小皆为F 的拉力作用,而左端的情况各不相同:①中弹簧的左端固定在墙上。
②中弹簧的左端受大小也为F 的拉力作用。
③中弹簧的左端拴一上物块,物块在光滑的桌面上滑动。
④中弹簧的左端拴一小物块,物块在有摩擦的桌面上滑动。
若认为弹簧的质量都为零,以l 1、l 2、l 3、l 4依次表示四个弹簧的伸长量,则有【D 】A .l 2>l 1 B.l 4>l 3 C.l 1>l 3 D.l 2=l 4【上海】5.物体B 放在物体A 上,A 、B 的上下表面均与斜面平行(如图),当两者以相同的初速度靠惯性沿光滑固定斜面C 向上做匀减速运动时【C 】(A )A 受到B 的摩擦力沿斜面方向向上。
(B )A 受到B 的摩擦力沿斜面方向向下。
(C )A 、B 之间的摩擦力为零。
(D )A 、B 之间是否存在摩擦力取决于A 、B 表面的性质。
【两广】7.用三根轻绳将质量为m 的物块悬挂在空中,如图所示.已知ac 和bc 与竖直方向的夹角分别为030和060,则 ac 绳和bc 绳中的拉力分别为【A 】 A1,2mg B.12mg C1,2mg D.12mg 【新课程】32.三个完全相同的物块1、2、3放在水平桌上,它们与桌面间的动摩擦因数都相同。
现用大小相同的外力F 沿图示方向分别作用在1和2上,用21F 的外力沿水平方向作用在3上,使三者都做加速运动,令a 1、a 2、a 3分别代表物块1、2、3的加速度,则【C 】A .a 1=a 2=a 3B .a 1= a 2,a 2>a 3C .a 1>a 2,a 2<a 3D .a 1>a 2,a 2>a 3【山西】15.如图所示,ad 、bd 、cd 是竖直面内三根固定的光滑细杆,a 、b 、c 、d 位于同一圆周上, a 为圆周的最高点,d为最低点。
每根杆上都套着一个小滑环(图中未画出),三个滑环分别从 a 、b 、c 处释放(初速为0),用t 1、、、t 2、、t 3依次表示m各滑环到达d 所用的时间,则【D 】A .t 1=、t 2、=t 3B .t 1、>、t 2、>t 3C .t 3 > t 1、>t 2、、D .t 1、、<t 2、<t 3【两广】9.一杂技演员,用一只手抛球.他每隔0.40s 抛出一球,接到球便立即把球抛出,已知除抛、接球的时刻外,空中总有四个球,将球的运动看作是竖直方向的运动,球到达的最大高度是(高度从抛球点算起,取210/g m s =)【C 】A . 1.6mB . 2.4mC .3.2mD .4.0m【湘鄂、全国】21.放在水平地面上的一物块,受到方向不变的水平推力F 的作用,F 的大小与时t 的关系和物块速度υ与时间t 的关系如图所示。
2004年全国各地高考物理试题分类详解-力学

2004高考力学题【湘鄂、全国】18.如图所示,四个完全相同的弹簧都处于水平位置,它们的右端受到大小皆为F 的拉力作用,而左端的情况各不相同:①中弹簧的左端固定在墙上。
②中弹簧的左端受大小也为F 的拉力作用。
③中弹簧的左端拴一上物块,物块在光滑的桌面上滑动。
④中弹簧的左端拴一小物块,物块在有摩擦的桌面上滑动。
若认为弹簧的质量都为零,以l 1、l 2、l 3、l 4依次表示四个弹簧的伸长量,则有【D 】A .l 2>l 1 B.l 4>l 3 C.l 1>l 3 D.l 2=l 4【上海】5.物体B 放在物体A 上,A 、B 的上下表面均与斜面平行(如图),当两者以相同的初速度靠惯性沿光滑固定斜面C 向上做匀减速运动时【C 】(A )A 受到B 的摩擦力沿斜面方向向上。
(B )A 受到B 的摩擦力沿斜面方向向下。
(C )A 、B 之间的摩擦力为零。
(D )A 、B 之间是否存在摩擦力取决于A 、B 表面的性质。
【两广】7.用三根轻绳将质量为m 的物块悬挂在空中,如图所示.已知ac 和bc 与竖直方向的夹角分别为030和060,则 ac 绳和bc 绳中的拉力分别为【A 】 A1,2mg B.12mg C1,2mg D.12mg 【新课程】32.三个完全相同的物块1、2、3放在水平桌上,它们与桌面间的动摩擦因数都相同。
现用大小相同的外力F 沿图示方向分别作用在1和2上,用21F 的外力沿水平方向作用在3上,使三者都做加速运动,令a 1、a 2、a 3分别代表物块1、2、3的加速度,则【C 】A .a 1=a 2=a 3B .a 1= a 2,a 2>a 3C .a 1>a 2,a 2<a 3D .a 1>a 2,a 2>a 3【山西】15.如图所示,ad 、bd 、cd 是竖直面内三根固定的光滑细杆,a 、b 、c 、d 位于同一圆周上, a 为圆周的最高点,d为最低点。
每根杆上都套着一个小滑环(图中未画出),三个滑环分别从 a 、b 、c 处释放(初速为0),用t 1、、、t 2、、t 3依次表示m各滑环到达d 所用的时间,则【D 】A .t 1=、t 2、=t 3B .t 1、>、t 2、>t 3C .t 3 > t 1、>t 2、、D .t 1、、<t 2、<t 3【两广】9.一杂技演员,用一只手抛球.他每隔0.40s 抛出一球,接到球便立即把球抛出,已知除抛、接球的时刻外,空中总有四个球,将球的运动看作是竖直方向的运动,球到达的最大高度是(高度从抛球点算起,取210/g m s =)【C 】A . 1.6mB . 2.4mC .3.2mD .4.0m【湘鄂、全国】21.放在水平地面上的一物块,受到方向不变的水平推力F 的作用,F 的大小与时t 的关系和物块速度υ与时间t 的关系如图所示。
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

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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.。
2004年物理高考试题

2004年全国普通高等学校招生考试理科综合能力测试第Ⅰ卷(选择题 共126分)本卷共21题,每题6分,共126分。
以下数据可供解题时参考:原子量;C 17 N 14 O 16 Na 23 Mg 24 P 31 Cl 35.5 K 39 Ca40 Fe56l .下列关于光合作用强度的叙述,正确的是A .叶片从幼到老光合作用强度不变B .森林或农田中植株上部叶片和下部叶片光合作用强度有差异C .光合作用强度是由基因决定的,因此是固定不变的D .在相同光照条件下,各种植物的光合作用强度相同2.某生物的体细胞染色体数为2n 。
该生物减数分裂的第二次分裂与有丝分裂相同之处是A .分裂开始前,都进行染色体的复制B .分裂开始时,每个细胞中的染色体数都是2nC .分裂过程中,每条染色体的着丝点都分裂成为两个D .分裂结束后,每个子细胞的染色体数都是n3.用一定量的甲状腺激素连续饲喂正常成年小白鼠4周,与对照组比较,实验组小白鼠表现为A .耗氧量增加、神经系统的兴奋性降低B .耗氧量增加、神经系统的兴奋性增强C .耗氧量减少、神经系统的兴奋性降低D .耗氧量减少、神经系统的兴奋性增强4.下列属于生态系统食物同特征的是A .一种生物只能被另一种生物捕食B .食物链的环节数是无限的C .一种生物可能属于不同的营养级D .食物网上的生物之间都是捕食关系5.用动物细胞工程技术获取单克隆抗体,下列实验步骤中错误的是)A .将抗原注入小鼠体内,获得能产生抗体的B 淋巴细胞B .用纤维素酶处理B 淋巴细胞与小鼠骨髓瘤细胞C .用聚乙二醇作诱导剂,促使能产生抗体的B 淋巴细胞与小鼠骨髓瘤细胞融合D .筛选杂交瘤细胞,并从中选出能产生所需抗体的细胞群,培养后提取单克隆抗体6.在pH =l 含+2Ba 离子的溶液中,还能大量存在的离子是A .-2AlOB .-ClOC .-ClD .-24SO7.物质的量浓度相同的下列溶液中,符合按pH 由小到川匝序排列的是A .Na 2CO 3 NaHCO 3 NaCl NH 4ClB .Na 2CO 3 NaHCO 3 NH 4Cl NaClC .(NH 4)2SO 4 NH 4Cl NaNO 3 Na 2SD .NH 4Cl (NH 4)2SO 4 Na 2S NaNO 38.已知(l ))g (O 21)g (H 22+ =H 2O (g ) △H 1=a kJ ·1mol -(2))g (O )g (H 222+ =2H 2O (g ) △H 2=b kJ ·1mol -(3))g (O 21)g (H 22+=H 2O (l ) △H 3=c kJ ·1mol -(4))g (O )g (H 222+ =2H 2O (l ) △H 4=d kJ ·1mol -下列关系式中正确的是A . a <c <0B .b >d >0C .2a =b <0D .2c =d >09.将0.l mol ·1L -醋酸溶液加水稀释,下列说法正确的是A .溶液中c (H +)和c (-OH )都减小B .溶液中c (H +)增大C .醋酸电离平衡向左移动D .溶液的pH 增大10.下列叙述正确的是A .同温同压下,相同体积的物质,它们的物质的量必相等B .任何条件下,等物质的量的乙烯和一氧化碳所含的分子数必相等C .1L 一氧化碳气体一定比1L 氧气的质量小D .等体积、等物质的量浓度的强酸中所含的H +数一定相等11.若1 mol 某气态烃C x H y 完全燃烧,需用3 mol O 2,则A .x = 2,y =2B .x = 2,y =4C .x = 3,y =6D .2=3,y=812.下列分子中,所有原子不可能共处在同一平面上的是A .C 2H 2B .CS 2C .NH 3D .C 6H 613.常温下,下列各组物质不能用一种试剂通过化学反应区别的是A .MnO 2 CuO FeOB .(NH 4)2SO 4 K 2SO 4 NH 4ClC .AgNO 3 KNO 3 Na 2CO 3D .Na 2CO 3 NaHCO 3 K 2CO314.现有1200个氢原子被激发到量子数为4的能级上,若这些受激氢原子最后都回到基态,则在此过程中发出的光子总数是多少?假定处在量子数为n 的激发态的氢原子跃迁到各较低能级的原子数都是处在该激发态能级上的原子总数的1n 1-。
2004年中考物理试题及答案

一、下列各小题均有四个选项,其中只有一个符合题意(共34分,每小题2分)1.在国际单位制中,电功的单位是A库仑 B安培 C焦耳 D伏特2.在通常情况下,下列物体属于绝缘体的是A铜线 B玻璃 C人体 D大地3.下列现象中,属于光的折射现象的是A小孔成像B用放大镜看地图C湖边的景物,在湖中形成的“倒影”D太阳光照射下,地面上出现树的影子4.运动员将足球从后场踢到前场,足球在空中运动的过程中,不计空气阻力,其受力情况是A 只受踢力 B人受重力 C受踢力和重力 D不受力的作用5.下列事悠然中,做功改变物体内能的是A用酒精灯加热烧杯中的水, B冬天室内使用暖气取暖C用锯锯木板,锯条发热 D盆中的热水温度逐渐降低6.下列有关密度的说法,正确的是A一滴水的密度小于一桶水的密度B因为铝比铁轻松,所以铝的密度小于铁的密度C液体的密度一定小于固体的密度D密度是物质的一种特性。
7.运动员跑到终点后,不能立即停下来。
这是因为A失去了惯性 B 运动员具有惯性C运动员不受力的作用 D运动员的惯性大于阻力8.关于家庭电路的下列说法中正确的是A家庭电路的电压是380V,B家庭电路中的用电器都是串联的C用湿手搬动电路中的开关是安全的D保险丝熔断可能是电路的总功率过大造成的9.关于下列装置工作时发生能量转化的说法中,正确的是A电动机将电能转化为机械能 B汽油机将内能转化为电能C干电池将电能转化为化学能 D发电机将电能转化为内能10.在图的四种情境中,人对物体做功的是A . B. C. D.11.在空中喷洒干冰是人工增雨的一种方法,干冰使空气中的水蒸气变成小冰粒,冰粒下降过程中变成雨滴,水蒸气为成冰粒、冰粒变成雨滴的物态变化过程分别属于A凝华、熔化 B 升华、熔化 C升华、液化 D凝华、液化12.根据图2所示的两盏白炽灯的铭牌,可以知道A甲灯一定比乙灯亮B正常发光时,甲灯的电阻大于乙灯的电阻C正常发光时,甲灯的电功率是乙灯电功率的4倍D正常发光时,甲灯消耗的电能是乙灯消耗电能的4倍13.在利用蜡烛研究凸透镜成像的实验中,凸透镜的焦距是10cm,点燃蜡烛放在距凸透镜15cm处,在凸透镜另一侧上观察一了蜡烛清晰的像,这个像一定是A倒立、放大的实像 B倒立、缩小的实像C正立、放大的虚像 D正立、放大的实像14.甲、乙两位同学沿同一直线推一辆小车,甲用100N的力向右推,乙用60N的力向左推,则小车受到的合力的大小和方向是A 40N,向右B 40N,向左C 160N向右D 160N ,向左15.甲物体的比热容大于乙物体的比热容,若A甲、乙质量相等,则甲吸收的热量一定多B甲、乙吸收的热量相等,则乙升高的温度一定多C甲、乙质量相等,它们升高的温度相同,则甲吸收的热量一定多D甲、乙质量相等,它们入出的热量也相等,则甲降低的温度一定多16.在图3所示的电路中,电源电压不变,当开关S1、S2都闭合时,电压表V1、V2的示数之比为5:3,当开关S1闭合,S2断开时,电压表V1、V2的示数之比为3:2,在后一种情况下,全部通电时间内,电阻R1与R3产生的热量之比为A 2:3B 1:3C 3:2D 2:117.把同种材料制成的甲、乙两个正方体,放在水平桌面上,甲、乙对桌面的压强分别为P1和P2,如图4所示,把甲、放在乙的上面,则乙对桌面的压强为A P1+P2B P12+P22C(P13+P23)/P22 D(P13+P23)/P12二、下列各小题均有四个选项,其中符合题意的选项均多于一个(共6分,每小题3分,全选对的得3分,选对但不全的得2分,有错选的不得分)18、以下有关磁现象的说法中,正确的是A磁体周围存在着磁场 B磁感线是真实存在的C同名磁极互相吸引 D指南针能指南北是受地磁场的作用19、下列关于力的说法中,正确的是A人推车时,人也受到车给人的推力B两个物体只要互相接触,就一定发生力的作用C用手捏一个空易拉罐,易拉罐变瘪了,表明力可以使物体发生形变D排球运动员扣球使球运动方向发生了改变,表明力可以改变物体的运动状态。
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10 m s 2 is of course also acceptable.
5.
Numerical answers that differ from the published answer due to differences in rounding throughout the question typically receive full credit. The exception is usually when rounding makes a difference in obtaining a reasonable answer. For example, suppose a solution requires subtracting two numbers that should have five significant figures and that differ starting with the fourth digit (e.g. 20.295 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.
1 point
T =
m1 (2 gL ) + m1 g = 2 m1 g + m1 g L T = 3m1 g
3 points
(c)
For any statement of conservation of momentum For a correct application of conservation of momentum, with a clear indication that both masses have a common final velocity m1uB = (m1 + m2 ) uafter m1 uafter = (m1 + m2 ) B For an answer in terms of the required quantities and of the form: m1 (answer from part(a)) (m1 + m2 ) m1 uafter = 2 gL m ( 1 + m2 )
1 point
(b)
4 points For any indication that there are two forces acting on the person For an indication that the acceleration of the person is centripetal, i.e. equal to u 2 r or u 2 L For a correct application of Newton’s second law that includes the two forces (tension T and weight) and a non-zero acceleration m u2 T - m1 g = 1 B r 2 mu T = 1 B + m1 g r For substitution of the expression for uB from part (a) and L for the radius 1 point 1 point 1 point
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2
AP PHYSICS C MECHANICS 2004 SCORING GUIDELINES
Question 1 15 points total (a) 2 points For any statement of conservation of energy Taking the zero of potential to be at the height of point B, and setting the kinetic energy at point B equal to the potential energy at point A: 1 m u 2 = m1 gL 2 1 B For the correct answer uB = 2 gL 1 point Distribution of points
AP® Physics C: Mechanics 2004 Scoring Guidelines
The materials included in these files are intended for noncommercial use by AP teachers for course and exam preparation; 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. This permission does not apply to any third-party copyrights contained herein. These materials and any copies made of them may not be resold, and the copyright notices must be retained as they appear here.
2. Generally, double penalty for errors is avoided. For example, if an incorrect answer to part (a) is correctly 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 easily recognized 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 a particular concept is worth one point, and a student’s solution contains the application of that equation to the problem but the student does not write the basic equation, the point is still awarded. 4. The scoring guidelines typically show numerical results using the value g = 9.8 m s 2 , but use of
AP PHYSICS C MECHANICS 2004 SCORING GUIDELINES General Notes about 2004 AP Physics Scoring Guidelines
1.
The solutions contain the most common method(s) of solving the free-response questions, and the allocation of points for these solutions. Other methods of solution also receive appropriate credit for correct work.
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