大学物理双语(下)试题补考
西尔斯大学物理双语版答案
1. S: 2kv dtdva -==2kv dxdvv dt dx dx dv -==k d x v dvxx vv -=⎰⎰)(ln00x x k v v--= )(00x x k e v v --= (answer)2. S: j t i t dt rd v )3cos 15()3sin 15(+-== jt i t dtv d a )3sin 45()3cos 45(-+-==()()j t i t j t i t v r)3cos 15()3sin 15()3sin 5()3cos 5(+-⋅+=⋅j j t t i i t t⋅⋅+⋅⋅-=)3c o s 3s i n 75()3sin 3cos 75( 0= (proved c)3. S: dtdv v m k m f a =-==dt mkv dv t t v v -=⎰⎰0)(0t mkv t v -=0)(ln t m ke v t v -=0)( (answer) D: t m k e v dtdxv -==0dt e v dx t m k tt x -⎰⎰=00)(0kmv x e kmv ekmv t x t m k t t mk 0max 00),1()(=-=-=--4. S: )()32(j y d i dx j i x r d f dw+⋅+=⋅=dy xdx dw w fi32+==⎰⎰dy xdx 323342⎰⎰--+== -6 J (answer)5. S: 23230.60.4)0.30.4(t t t t t dtddt d +-=+-==θω, t t t dtddt d 60.6)30.60.4(2+-=+-==ωα 0.40300.60.4)0(2=⨯+⨯-=ω (answer of a)0.28)0.4(30.40.60.4)0.4(2=⨯+⨯-=ω rad/s (answer of a ) 60.266)0.2(=⨯+-=α rad/s 2 (answer of b )t t 60.6)(+-=α is time varying not a constant (answer of c) 6. S: ω20031222ML L v m L mv +⋅= MLmv ML L mv 4343020==ω (answer a))c o s 1(2)31(21m a x 22θω-=LMg ML ]1631[cos 2221maxgLM v m -=-θ (answer b) 7. G: m =1.0g, M =0.50kg, L =0.60m, I rod =0.0602m kg ⋅,s rod /5.4=ωR:I sys , v 0S: I sys =I rod +(M+m)L2=0.060+(0.50+0.0010)×0.602= 0.24 2m kg ⋅(answer)the system ’s angular momentum about rotating axis is conservative in the collision.sysI L mv ω=0s m mL I v sys/108.160.00010.024.05.430⨯=⨯⨯==ω (answer )D: The bullet momentum 0v m p=(before impact), its angular momentumabout rotating axis can be expressed as L mv 0(a scalar) 8. S:γ==00.800x xt v c -∆==0811800.600 3.0010t t γ∆=∆=⨯⨯ 9. S: 202202)(mc E cp E E γγ==+=222c p m c m c m c =10. S: 0i n t =-=∆n e t n e t W Q E n e t n e t W Q = 1(3010)(4.0 1.0)2=-- J 30= (answer)11. S: from nRT PV =and K T A 300= we can get:KT K T C B 100300== (answer of a)Change of internal energy between A and B:0)(23int =-=∆A B T T k n E (answer of b)The net work of the cycle:))(100300()13(2121m N AC BC W ⋅-⋅-=⋅=J 200= (answer of c) From the first law : W E Q +∆=int we can derive:the net heat of the whole cycle is J W Q 200== (answer)12. S: 131)(320===⎰⎰∞F v Av dv Av dv v p F33FvA =(answer of a ) F F v a v g v Av dv vAv v F4341420===⎰13. G: T 1=T 2=T , m 1, p 1, v rms,1, m 2, p 2=2p 1, v avg,2 = 2v rms,1 R: m 1 / m 2 S: v avg,2 =1.602m kTv rms,1 = 1.731m kTv avg,2 = 2v rms,167.4)60.173.12(221=⨯=m m (answer) 14. S: dE int =dQ – dWd Q = dE int + dW = n C v dT+pdV VdVnR T dT nC dV T p T dT nC T dQ dS v v +=+==if i f v VV v T T V V nR T T nC V dVnR T dT nC ds S f i filnln +=+==∆⎰⎰⎰ 15. S: dA E q θεcos 0⎰=212100)0.60100(1085.8⨯-⨯⨯=- C 61054.3-⨯= 16. S: 2041)(r Qr E πε=(R < r <∞) dr rQ dr r E udV dU 2022208421πεπε=⋅== RQ r dr Q udV U R0220288πεπε===⎰⎰∞(answer) RQ r dr Q U r r Rεπεεπε02202*88==⎰∞(answer ) 18. S: in the shell of r – r + drdr r R r dV r dq 204)/1()(πρρ-==)34(31)/(4)(4303200r Rr dr R r r dq r q r-=-==⎰⎰πρπρfrom the shell theorems , within the spherical symmetry distribution )34(12)(41)(20020r Rr Rr r q r E -==ερπε (answer of b)R r r R Rdr dE 320)64(12*00=⇒=-=ερ 00200*max 9])32(3324[12)(ερερRR R R R r E E =-⨯== 19. S: j yV i x V V gradV y x E∂∂-∂∂-=-∇=-=),( )0.20.2(y x x VE x +-=∂∂-= x yV E y 0.2-=∂∂-= )/(480.2)0.20.2()0.2,0.2(m V j i j x i y x E--=-+-=20. S: Q in = - q , Q out = q (answer ) 1010241241)0(R qq V q πεπε==104)0(R qV in πε-=204)0(R q V o u t πε=)0()0()0()0(out in q V V V V ++= )11(4210R R q +=πε21. S: from the planar symmetry and superposition principle, Emust in normal direction of the plates and 1σ,2σ,3σ,4σ must be const. Fromcharge conservationA Q S =+)(21σσ ⇒ SQ A=+21σσ (1) B Q S =+)(43σσ ⇒ SQ B=+43σσ (2) Apply Gauss ’ law in the closed surface shown in Fig. 032=+σσ (3)within the metal, 0=p Ewhich leads to002222432104030201=-++⇒=-++σσσσεσεσεσεσFrom(1), (2), (3), (4) yield:⎪⎪⎩⎪⎪⎨⎧-=-=+==S Q Q SQ Q B AB A 223241σσσσ (answer of a) (6 points) 004030201122222εεσεσεσεσS Q Q E BA p -=--+= (1 point) 004030201222222εεσεσεσεσS Q Q E BA p +=+++=(1 point) (answer of b) d S Q Q d E d E V BA p AB 012ε-==⋅= (2 points) (answer of c)27.33()(32)18w F x dx x dx J ==+=⎰⎰；at x=3m212W mV =， 6/V m s =。
大学物理II补考卷
大学物理II补考卷一、选择题（每题4分，共40分）A. 速度B. 力C. 位移D. 加速度2. 在LC振荡电路中，振荡电流的最大值出现在电容器完全放电时，此时电路中的能量主要储存在（）。
A. 电场能B. 磁场能C. 动能D. 电势能A. 日食B. 彩虹C. 薄膜干涉D. 光的衍射4. 磁感应强度B与磁场力F的关系是（）。
A. B = F / ILB. B = F / qvC. B = F / (IL)D. B = F / (qv)5. 一个电子以速度v垂直射入磁感应强度为B的匀强磁场，其运动轨迹是（）。
A. 直线B. 抛物线C. 椭圆D. 圆6. 在恒力F作用下，物体从静止开始沿直线运动，若物体在t时间内的位移为s，则物体在t+Δt时间内的位移为（）。
A. s + FΔtB. s + F(t+Δt)C. s + FtD. s + FΔt²/27. 一根均匀的细杆，其长度为L，质量为m，求杆的重心位置（）。
A. L/4B. L/2C. 3L/4D. LA. 速度B. 加速度C. 力D. 位移9. 一个单摆在位移为A/2处释放，其最大速度出现在（）。
A. 位移为A/2处B. 位移为A处C. 位移为0处D. 无法确定A. 电场强度与电荷量成正比B. 磁感应强度与电流成正比C. 磁通量与磁感应强度成正比D. 电阻与导体长度成正比二、填空题（每题4分，共40分）1. 在真空中，光速c = ______ m/s。
2. 一个电容器的电容C = ______ F，若其两板间的电压为U，则储存的电场能量为______。
3. 根据电磁感应定律，闭合回路中的感应电动势E = ______。
4. 一个平面简谐波的波动方程为y = Asin(2π/λ)x 2π/T)t，其中A为振幅，λ为波长，T为周期，波速v = ______。
5. 一个质量为m的物体在水平地面上受到一个大小为F的恒力作用，若物体与地面间的摩擦系数为μ，则物体所受的摩擦力f =______。
10下双语物理复习题
2010.12.大学物理B（下，双语）复习范围Ch.12 P.317: 9, 12, 26, 45, 77, 87Ch.13 P.348: 11, 21, 22 , 23, 26, 47Ch.15 p.382-: 13, 22, 23, 25Ch.16 p.400: 8. 11. 14. 58Ch.17 p.424: 11,12, 21,27, 29, 37, 29, 41, 48Ch.18 p.451: 6, 15, 30; 67, 68;Ch.30 p.689: 5, 7,8,11, 22, 23, 25, 31, 32, 37, 41, 43, 45Ch.31 P.727: 3, 6, 7; 13, 16, 20, 25, 29, 34, 65,45,47,49,51Ch32 P.758: 3, 4, 8, 9, 10, 21, 26, 62, 35, 39, 43, 65Ch33 P.787: 2,3,6,13,15,19, 29, 37, 39考试时间：2011年1月日，答疑时间：考试前3天答疑地点：物理楼314，207，ljq@Ch12 Oscillations；I-Ex. P.317: 9, 12,26, 45 ,77,8787I-Ch.13 Waves Motion , p348:11,21,22,23,26,4711. (II) S and P waves from an earthquake travel at different speeds and this difference helps in the determination of the earthquake "epicenter" (where the disturbance took place), (a) Assuming typical speeds of 9.0km/s and 5.5km/s for P and S waves, respectively, how far away did the earthquake occur if a particular seismic station detects the arrival of these two types of waves exactly 94 s apart? (b) Is one seismic station sufficient to determine the position of the focus? Explain.21. (II) Consider a point on the string of Example 13-3 that is 1.00 m from the left-hand end. Determine (a) the maximum velocity of this point, and (b) its maximum acceleration, (c) What is its velocity and acceleration at t = 2.0 s?22. (II) Show, for a sinusoidal transverse wave traveling on a string, that the slope of the string at any point x is equal to the ratio of the transverse speed of the particle to the speed of the wave at that point.23. (II) A transverse wave pulse travels to the right along a string with a speed υ = 2.0 m/s. At t = 0 the shape of the pulse is given by the function D = 0.45cos(3.0x + 1.2). (a) Plot D versus x at t=0. (b) Determine a formula for the wave pulse at any time t assuming there are no frictional losses,(c) Plot D(x, t) versus x at t = 1.0 s. (d) Repeat parts (b) and (c) assuming the pulse is traveling to the left.Ch15 Temperature and the Ideal Gas Law; Ex. p.382-: 13,22, 23, 2513.(II) A hot-air balloon achieves its buoyant lift by heating the air inside the balloon which makes it less dense than the air outside. Suppose the volume of a balloon is 1800 m3 and the required lift is 2700N (rough estimate of the weight of the equipment and passenger). Calculate the temperature of the air inside the balloon which will produce the required lift. Assume that the outside air temperature is 0o and that air is an ideal gas under these conditions. What factors limit the maximum altitude attainable by this method for a given load?22．(II) A cubic box of volume 5.1 X 10 2 m 3is filled with air at atmospheric pressure at 20°C. The box is closed and heated to 180°C. What is the net force on each side ofthe box?23. Estimate how many molecules of air are in each 2.0-L breath you inhale that werealso in the last breath Galileo took. [Hint: Assume the atmosphere is about 10 kmhigh and of constant density.]25．Write the ideal gas law in terms of the density of the gas.I-Ch16 Kinetic Theory of Gases; Ex. p.400: 8. 11. 14. 588. Show that the rms speed of molecules in a gas is given by v rms =ρ/3p , where P is the pressure in the gas, and ρis the gas density.Solution: Solution:Solution:Solution:11.(II) Calculate (a) the rms speed of an oxygen molecule at 0°C and (b) determine how many times per second it wouldmove back and forth across a 7.0-m-long room on the average, assuming it made very few collisions with other molecules.14. (I) A group of 25 particles have the following speeds: two have speed 10 m/s, seven have 15 m/s, four have 20 m/s, three have 25 m/s, six have 30 m/s, one has 35 m/s, and two have 40 m/s. Determine (a) the average speed, (b) the rms speed, and (c)the most probable speed.58.At room temperature, it takes approximately 2.45×103J to evaporate 1.00g of water.Estimate the average speed of evaporating molecules. What multiple of v rms (at 20oC) for watermolecules is this? (Assume Eq.16-4 holds.)Ch.17 Heat and the First Law of Thermodynamics; Ex. P.424: 11,12, 21, 27,29,37, 41, 48 (e-p.384)11.The 1.20-kg head of a hammer has a speed of 6.5m/s just before it strikes a nail (Fig. 17-23)and is brought to rest. Estimate the temperature rise of a 14-g iron nail generated by ten suchhammer blows done in quick succession. Assume the nail absorbs all the energy.12. (II) (a) Show that if the specific heat varies as a function of temperature, c(T), the heat neededto raise the temperature of a substance from T 1 to T 2 is given by ⎰=12)(TT dT T mc Q . (b) Suppose c(T) = c o (l + aT) for some substance, where a = 2.0 ×103C 0-1 and T is the Celsius temperature.kT v m K 23212==Determine the heat required to raise the temperature from T1 to T2. (c) What is the mean value of c over the range T1 to T2for part (b), expressed in terms of c0, the specific heat at 0°C?21.Consider the following two-step process. Heat is allowed to flow out of an ideal gas at constant volume so that its pressure drops from 2.2atm to 1.5atm. Then the gas expands at constant pressure, from a volume of 6.8L to 10.0L, where the temperature reaches its original value. See Fig. 17-24. Calculate (a) the total work done by the gas in the process, (b) the change in internal energy of the gas in the process, and (c) the total heat flow into or out of the gas.27.When a gas is taken from a to c along the curved path in Fig. 17-26, the work done by the gas is W=-35J and the heat added to the gas is Q= -63J. Along path abc, the work done is W=-48J. (a) What is Q for path abc? (b) If P c = P b/2, what is W for path cda? (c) What is Q for path cda? (d) What is U A-U C(e) If Ud-Uc=5J, what is Q for path da?29.37.The specific heat at constant volume of a particular gas is 0.182kcal/kg.K at room temperature, and its molecular mass is 34. (a) What is its specific heat at constant pressure? (b) What do you think is the molecular structure of the gas?41. A 1.00-mol sample of an ideal diatomic gas at a pressure of 1.00atm and temperature of 490K undergoes a process in which its pressure increases linearly with temperature. The final temperature and pressure are 720 K and 1.60atm. Determine (a) the change in internal energy, (b) the work done by the gas, and (c) the heat added to the gas. (Assume 5 active degrees of freedom.)48. A 1.00-mol sample of an ideal monatomic gas, originally at a pressure of 1.00atm, undergoesa three-step process: (1) it is expanded adiabatically from T1= 550K to T2= 389K; (2) it is compressed at constant pressure until its temperature reaches T3 ; (3) it then returns to its original pressure and temperature by a constant-volume process, (a) Plot these processes on a PV diagram,(b) Determine T3 . (c) Calculate the change in internal energy, the work done by the gas, and the heat added to the gas for each process, and (d) for the complete cycle.I-Ch.18 The Second Law of Thermodynamics; Ex. P.451(e-p405):6, 15, 30; Ch17- 29, 67, 68*6.Figure 18-15 is a PV diagram for a reversible heat engine in which 1.0mol of argon, a nearly ideal monatomic gas, is initially at STP( point a). Points b and c are on an isotherm at T =423K. Process ab is at constant volume. Process ac at constant pressure. (a) Is the path of the cycle carried out clockwise or counterclockwise? (b) What is the efficiency of engine?15.Assume that a hiker needs 4000kcal of energy to supply a day's worth of metabolism. Estimate the maximum height the person can climb in one day. using only this amount of energy. As a rough prediction, treat the person as an isolated heat engine, operating between the internal temperature of 37℃(98.6℉) and the ambient air temperature of 20℃.30.(a) Given that the coefficient of performance of a refrigerators defined as show that for an ideal (Carnot)refrigerator, (b) write the CP in term of the efficiency e of the reversible heat engine obtained by running the refrigerator backward. (c) What is the coefficient of performance for an ideal refrigerator that maintains a freezer compartment at-16℃when the condenser’s temperature is 22℃?67.The Stirling cycle, shown in Fig.18-17, is useful to describe external combustion engines as well as solar-power systems. Find the efficiency of the cycle in terms of the parameters shown, assuming a monatomic gas as the working substance. The processes ab and cd are isothermal whereas bc and da are at constant volume. How does it compare to the Carnot efficiency?68.One mole of an ideal monatomic gas at STP first undergoes an isothermal expansion so that the volume at b is 2.5 times the volume at a (Fig, 18-18). Next, heat is extracted at a constant volume so that the pressure drops. The gas is then compressed adiabatically back to the original state, (a) Calculate the pressures at b and c. (b) Determine the temperature at c. (c) Determine the work done, heat input or extracted, and the change in entropy for each process, (d) What is theefficiency of this cycle?Figure 18-155, 7,8,11, 22, 23, 25, 31,32, 37, 41, 43, 45(dark)24=m (dark)2λm t d ==3012024118 2λλ==+=+=∆h L .292422-118 2λλ548==+=+-=∆y h L (dark) )(212222λλ+=+m t n 121=-m m (dark)2)12(022λ+=+m t nCh31 Diffraction and Polarization,I-p.727：3,6,7,13,16, 20,25,31,32,37,41,43,453.Light of wavelength 550nm falls on slit that is 3.5×10-3mm wide. Estimate how far from thecentral maximum the first diffraction maximum fringe is if the screen 10.0m away?6.(a) For a given wavelength λ, what is maximum slit width for which there will be nodiffraction minima?(b) What is the maximum slit width so that no visible light exhibits adiffraction minimum?13.If a double-slit pattern contain exactly seven fringes in the central diffraction peak, whatcan you say about the slit width and separation?16 How many fringes are contained in the central diffraction peak fora double-slit pattern if (a) d=2.00a,(b) d=12.0a, (c) d=4.50a,(d) d=7.20a20.Two stars 10light-years away are barely resolved by a 90cm (mirror diameter) telescope. How far apart are the star? Assume λ=550nm and the resolution is limited by diffraction65. (a)How far away can a human eye distinguish two car headlights 2.0m apart? Consider only diffraction effects and assume an eye diameter of 5.0mm and a wavelength of 550nm. (b)What is minimum angular separation an eye could resolve when viewing two stsrs, considering only diffraction effects? In reality, it is 1’ of arc. Why is it not equal to your answer in (b)?25.A grating has 6600lines/cm. How many spectral orders can be seen when it is illuminated by white light?47.At what angle should the axes of two Polaroids be placed so as to reduce the intensity of the incident unpolarized light to (a) 1/3, (b) 1/10?34.Missing orders occur for a diffraction grating when a diffraction minimum coincides with an interference maximum. Let a be the width of each slit and d the separation of slits and show (a ) that if d=2a, all even orders (m=2,4,6,…) are missing. (b ) Show that there will be missing orders whenever d /a =m 1/m 2 ,where m 1 and m 2 are integer. (c ) Discuss the case d =a , the limit in which the space between slits becomes negligible.45. What is the Brewster ’s angle for an air-glass(n=1.56) surface?49.What would Br ewster’s angle be for reflections off the surface of water for light coming from beneath the surface? Compare to the angle for total internal reflection, and to Brewster’s anglefrom above the surface.51.Unpolarized light passed through five successive Polaroid sheets each of whose axis makes a 450 angle with the previous one. What is the intensity of the transmitted beam?29. Show that the second-and third-order spectra of white light produced by a diffraction grating always overlap. What is wavelengths overlap?Ch32.Special Theory of Relativity P758:3,4,8,9,10,21,26,62,35,39,43,6510. A friend of your travels by you in her fast sports car at a speed of 0.660c. It is measured in your frame to be 4.80m long and 1.25m high. (a) What will be its length and height at rest? (b) How many seconds would you say elapsed on your friend’s watch when 20.0s passed on you? (c) How fast did you appear to be traveling according to your friend? (d) How many seconds would she say elapsed on your watch when she saw 20.0s pass on her?21. A stick of length Lo, at rest in reference frame S, makes an angle ө with the x axis. In reference frame S'. which moves to the right with velocity v = vi with respect to S, determine (a) the length L of the stick, and (b) the angle ө it makes with the x' axis.26.A particle of rest mass m travels at a speed v=0.20c. At what speed will its momentum be doubled?62.The nearest star to Earth is Proxima Centauri, 4.3 light-years away, (a) At what constant velocity must a spacecraft travel from Earth if it is to reach the star in 4.0 years, as measured by travelers on the spacecraft? (b) How long does the trip take according to Earth observers?35.Show that when the kinetic energy of a particle equals its rest energy, the speed of the particle is about 0.866c39.What is the momentum of a 750-MeV proton (that is, its kinetic energy is 750MeV)?43.Two identical particles of rest mass m approach each other at equal and opposite speeds , v. The collision is completely inelastic and results in a single particle at rest. What is the rest mass of the new particle? How much energy was lost in the collision? How much kinetic energy is lost in this collision?65.(a) What is the speed of an electron whose kinetic energy is 10,000 times its rest energy? Such speeds are reached in the Stanford Linear Accelerator, SLAC. (b) If the electrons travel in the lab through a tube 3.0 km long (as at SLAC), how long is this tube in the electrons’ reference frame?Ch33 Early Quantum Theory and Models of the AtomP.787: 2,3,6,13,15,19,29, 37, 392.(a)What is the temperature if the peak of blackbody spectrum is at 25.0nm?(b) What is the wavelength at the peak of blackbody spectrum if the body is at temperature of 2800K?3. An HCI molecule vibrates with a natural frequency of 8.1×1013Hz. What is the difference in energy (in joules and electron volts) between possible values of the oscillation energy?6.Estimate the peak wavelength of light issuing from the pupil of the human eye (which approximates a blackbody assuming normal body temperature.13.What is the longest wavelength of light that will emit electrons from a metal whose work function is 4.3×10-19J?15. In a photoelectric-effect experiment it is observed that no current flows unless the wavelength is less than 570nm. (a) What is the wok function of this material? (b) What is the stopping voltage required if light of wavelength 400nm is used?19.A certain type of film is sensitive only to light whose wavelength is less than 660nm.What is the energy (eV;kcal/mol) needed for the chemical reaction to occur which causes the film to change?。
沈阳药科大学2009年药学(英语)大学物理补考试题
College Physics Test（B）(July,2009)适用对象：2007级药学（英语）专业及2008级基础药学理科基地班专业级班姓名学号题号一二三四总分1 2 3 4 5 6 7满分20 10 10 10 10 10 10 10 5 5 100 得分评阅人І. Put correct answer in to the following blank. （2 scores for eachblank, total score: 20）1.The rod carries a uniformlydistributed charge Q .Determinethe potential V(y) for pointsalong the y (y>0) axis outsidethe rod . （1）2.A uniform rod of mass M and length l can pivot freely (i.e., weignore friction) about a hingewhich is at the end of rod. Therod is held horizontally and thenreleased. At the moment ofrelease, the linear acceleration ofthe tip of the rod （2）. Determine the angular velocity of the rod when it reaches the vertial position. （3）3. A parallel-plate capacitor has fixed charges +Q and –Q. The separation of the plates is then doubled. (a)By what factor does the energy stored in the electric field change? （4）;(b)How much work must be done if the separation of the plates is doubled from d to 2d? （5）(The area of each plate is A)4.The graph of displacementversus time for a small massat the end of spring is shownin figure. Att=0 ,x=0.43cm .Write theequation for displacement xas a function of time.(6)5. A section of an infinitely long cylindrical rod with a uniform + . Let us find an expression for the magnitude of the electric field at a distance r from the axis of the rod.(7)6. Red light illuminates a double-slit system with d=0.2mm. The first-order maximum occurs at x=3.3mm on a screen 1.0m from the7.A puddle of water has a thin film of gasoline floating on it. A beam of light is shining perpendicular on the film. If the wavelength of light incident on the film is 560 nm and the indices of refraction ofgasoline and water are 1.40 and 1.33, respectively, what is the minimum thickness of the film to see a bright reflection? (9) 8. A very fine thread is placed between twoglass plates on one side and the otherside is touching to form a wedge. A beamof monochromatic light of wavelength600 nm illuminates the wedge and 15bright fringes are observed. What is thethickness of the thread? (10)II．Choose the best answer to complete each of the following question. (fill in the blank form with correct answers) （1 score for each question, total score: 10）1 2 3 4 5 6 7 8 9 101. A uniform thin rod of mass M(M=1kg) and length L (L=0.6m) is5. (total score: 10)A conducting sphere with positive chargeｑradius R1，surrounded by a conducting shell without charge originally，inner radius R2 and external radius R3.The dielectric material of dielectric constant k fills all the space between the sphere and the shell. Find (1) electric field in space? (3 scores)（2）potential of inner sphere U1？(3 scores)（3）bound charge per unit area in the surface with inner radius R2 (1 score)（4）what is the electric energy in the dielectric space? (3 scores)k。
四川农业大学《大学物理(专科)》1603补考作业试题及答案
一、单选题（共20道试题，共40分）1. 下面哪一种说法是正确的CA. 运动物体的加速度越大,速度越大B. 作直线运动的物体,加速度越来越小,速度也越来越小C. 切向加速度为正值时,质点运动加快D. 加速度为零时,速度也为零2. 在静电场中，有关静电场电场强度与电势之间的关系，下列说法中正确的是（D ）A. 场强大的地方，电势一定高B. 场强为零处，电势也为零C. 场强相等处，电势一定相等D. 场强为零处，电势不一定为零。
3. 对于沿曲线运动的物体，以下几种说法中哪一种是正确的（B）A. 切向加速度必不为零B. 法向加速度必不为零（拐点处除外）C. 法向加速度必为零D. 若物体作匀速率运动，其总加速度必为零。
4. 下列说法中哪个是正确的DA. 合力一定大于分力B. 物体速率不变，所受合力为零C. 速度很大的物体，运动状态不易改变D. 质量越大的物体，运动状态越不易改变5. 下列说法中哪一个是对的AA. 物体的动量不变，动能也不变B. 物体的动能不变，动量也不变C. 物体的动量变化，动能也一定变化D. 物体的动能变化，动量却不一定变化6. 以下说法正确的是（A）A. 若安培环路上B处处为零，则该环路内电流代数和为零B. 若安培环路内无电流，则该环路上的B必定处处为零C. 若安培环路上B处处不为零，则该环路内必有电流D. 若安培环路内有电流，则该环路上B必定处处不为零7. 以下说法正确的是AA. 若安培环路上B处处为零，则该环路内电流代数和为零B. 若安培环路内无电流，则该环路上的B必定处处为零C. 若安培环路上B处处不为零，则该环路内必有电流D. 若安培环路内有电流，则该环路上B必定处处不为零8. 对于理想气体，经历以下过程，其内能一定增大的是（D ）A. 先等温膨胀，后等容降压B. 先等压膨胀，后等容降压C. 等温降压过程D. 等压膨胀过程9. 在系统不受外力作用的非弹性碰撞过程中CA. 动能和动量都守恒B. 动能和动量都不守恒C. 动能不守恒,动量守恒D. 动能守恒,动量不守恒10. 一半径为R的均匀带电球壳，在其球心O处放置一点电荷q，该电荷受的电场力为零，若该电荷偏离球心O ，则该电荷q的受力情况为（A ）A. 变大B. 变小C. 不变D. 无法判断11. 对于理想气体，经历以下过程，其内能一定增大的是DA. 先等温膨胀，后等容降压B. 先等压膨胀，后等容降压C. 等温降压过程D. 等压膨胀过程12. 有一半径为R的单匝圆线圈，通以电流I，若将该线圈弯成匝数N=4的平面圆线圈，导线长度不变，并通以同样的电流，则线圈中心的磁感应强度是原来的（C ）A. 4倍B. 8倍C. 16倍D. 32倍13. 对一个质点系来说，指点系的机械能守恒条件为CA. 合外力为零B. 合外力不作功C. 外力和非保守力都不作功D. 外力和保守力都不作功14. 下列说法中错误的是（D ）A. 系统经历一正循环后，系统的熵没有变化B. 一切与热现象有关的宏观过程都是不可逆的C. 对于孤立系统或绝热系统内所发生的过程，系统的熵永不减少D. 系统经历一正循环后，系统和外界都没有变化15. 下列哪一种说法是正确的DA. 变化着的电场所产生的磁场，一定随时间而变化B. 变化着的磁场所产生的电场，一定随时间而变化C. 有电流就有磁场，没有电流就一定没有磁场D. 变化着的电场所产生的磁场，不一定随时间而变化16. 以下说法正确的是（A ）A. 若高斯面上E处处为零，则该面内电荷代数和为零B. 若高斯面内无电荷，则该面上的E必定处处为零C. 若高斯面上E处处不为零，则该面内必有电荷D. 若高斯面内有电荷，则该面上E必定处处不为零17. 根据热力学第二定律判断下列哪种说法是正确的CA. 热量能从高温物体传到低温物体，但不能从低温物体传到高温物体B. 功可以完全变成为热，但热不能全部变为热C. 气体能自由膨胀，但不能自由收缩D. 有规则运动的能量能够变为无规则运动的能量，但无规则运动的能量不能变为有规则运动的能量18. 对于单匝线圈取自感系数的定义式为L=Q/I。
2019年下学期大学物理C(二)补考试卷
5．如图所示，平板电容器(忽略边缘效应)充电时，沿环路 L1 的磁场
H
L1
强度 H 的环流与沿环路 L2 的磁场强度 H 的环流两者，必有
L2
(A) H dl H dl
(B) H dl H dl
L1
L2
(C) H dl H dl
L1
L2
(D) H dl 0
（A）分布在高斯面内的电荷决定的
（B）分布在高斯面外的电荷决定的
（C）高斯面内电荷代数和决定的 （D）空间所有电荷决定的 4．一孤立导体球壳带有正电荷，若将远处一带电体移至导体球壳附近，则
（A）导体球壳外附近的场强仍与其表面垂直
（B）导体球壳面上的电荷仍为均匀分布
（C）导体球壳的电势仍保持不变
（D）由于静电屏蔽，球壳外的带电体在球壳内产生的场强处处为零
点有点电荷 ，
·qa1
·
b · -q2
此系统的电势能 W=
。
3．（本题 3 分）在加热黑体过程中，单色辐出度极大值对应的波长由 1 变化到 2 ，则
总辐出度变为原来的
倍。
4．（本题 4 分）如图所示，则 L1 B dl =
，
B dl = L2
。
5．（本题 3 分）如图中三条曲线分别为顺磁质、抗磁质和 B
算圆柱形电容器的电容。
E b
3.如图所示，ab 导线与无限长直导线 GE 共面，ab 延长线与 GE 交于 O I2
点成 60角， ab 长 L， a 端距 GE 为 d，若 GE、ab 分别通以电流 I1、I2。求 I1
a
L
ab 在图示位置时所受 GE 产生磁场的作用力大小和方向。
60
O
d G
大学物理II补考卷
大学物理II补考卷（考试时间：90分钟，满分：100分）一、选择题（每题2分，共30分）1. 在简谐振动中，位移与时间的关系是（）A. 正弦函数B. 余弦函数C. 直线函数D. 指数函数2. 关于电容器的电容，下列说法正确的是（）A. 电容与电容器板间的电压成正比B. 电容与电容器板间的电压成反比C. 电容与电容器板间的电介质成正比D. 电容与电容器板间的距离成反比3. 磁场中某点的磁感应强度方向与该点电流元所受磁场力方向的关系是（）A. 平行B. 垂直C. 相互作用D. 无关4. 下列哪种现象属于光的衍射？（）A. 光在空气中的直线传播B. 光照射到物体边缘时产生的影子C. 通过狭缝观察光源时产生的明暗条纹D. 光在两种介质界面上的反射5. 一个电子以垂直于磁场方向的速度进入均匀磁场中，其运动轨迹将是（）A. 直线B. 抛物线C. 圆D. 椭圆6. 根据玻尔理论，氢原子的电子在某一轨道上运动时，其角动量是（）A. 连续的B. 量子化的C. 随机变化的D. 与轨道半径成正比7. 在LC振荡电路中，电容器放电完毕时，电路中的电流为（）A. 最大B. 最小C. 稳定D. 零二、判断题（每题1分，共20分）8. 物体做匀速圆周运动时，其加速度方向始终指向圆心。
（）9. 磁通量的变化率与感应电动势的大小成正比。
（）10. 在干涉现象中，两列波相遇时的相位差会影响干涉条纹的间距。
（）11. 根据量子力学，电子在原子中的运动轨迹是可以精确描述的。
（）12. 物体在静电场中的电势能只与其位置有关。
（）13. 电磁波在真空中的传播速度小于光速。
（）14. 在绝热过程中，理想气体的内能变化等于外界对气体做的功。
（）15. 质点做直线运动时，其速度大小不变但方向可以改变。
（）三、填空题（每空1分，共10分）16. 在简谐振动中，振幅是描述振动强弱的物理量，它等于______的最大值。
17. 一个平行板电容器的电容C由公式C=______决定。
华东理工大学《大学物理》第二学期双语课程期中考试试卷
华东理工大学2014–2015学年第二学期《大学物理》双语课程期中考试试卷2015.4开课学院：理学院专业：中德环境/电子考试形式：闭卷，所需时间120分钟考生姓名：学号：班级任课教师谢海芬Part A.(45points)Each question is worth 3points.Write the letter corresponding to the best answer to the question to the answer sheet below.1.An object moves in a circle of radius R at constant speed with a period t .If you want to change only the period in order to cut the object's acceleration in half,the new period should be:A.4tB.t/2C.2tD.2/t E t22.Two identical stones are dropped from rest and feel no air resistance as they fall.Stone A is dropped from height h ,and stone B is dropped from height 2h .If stone A takes time t to reach the ground,stone B will take time:A.t/2B.4tC.t 2D.2tE.2/t upper stringlower string3.A heavy ball is suspended as shown.A quick jerk on the lower string will break that string but a slow pull on the lower string will break the upper string.The first result occurs because:A.the force is too small to move the ball;B.the ball has inertia;C.action and reaction is operatingD.air friction holds the ball backE.the ball has too much energy 题序Part A Part BPart C 总分123得分评卷人4..Two particles of equal mass undergo a two dimensional elastic collision(not headon),the target particle being initially at rest.The angle between the velocityvectors of the two particles after the collision is:A.less than90°B.greater than90°C.equal to90°D.depends on the speed of the incident particleE.cannot be predicted5.A particle starts from rest and is acted on by a net force that does work at a rate that is proportional to the time t.The speed of the particle is proportional to:(A)1/t(B)t(C)2t(D)t（E))1/t6.Three objects are connected by massless wires over a massless frictionless pulley as shown in Fig..The tension in the wire connecting the10.0-kg and15.0-kg objects is measured to be133N.the tension in wire A is closest to:A.245N;B.280N;C.87.5N;D.517N;E.333N7.The sketch below shows a block of mass m on an inclined plane that makes an angle θwith the horizontal direction.The block is being held stationary by the force of static friction between the block and the plane.The coefficient of static friction between the block and the plane is u s.The magnitude of the net force exerted on the block is:A.u s mg cosθ D.mg cosθB.u s mg sinθ E.mg sinθC.zero8.The pilot of a light airplane with an airspeed of200km/h wants to fly due west. There is a strong wind of120km/h blowing from the north.If the pilot points the nose of the airplane north of west so that her ground track is due west,what will be her groundspeed?A.160km/h.B.120km/hC.80km/hD.180km/hE.It would impossible to fly due west in this situation9.A projectile is fired straight upward from Earth’s surface with a speed that is half the escape speed.If R is the radius of Earth,and air resistance is ignored,the greatestdistance from the center of Earth that the projectile will reach is:A.5R /4;B.4R /3;C.3R /2;D.2R;E.4R10.A particle begins with kinetic energy K 1at point A.It moves along path 1to point B,then along path 2back to point A.It ends with kinetic energy K 2where K 2>K 1.Which statement below can one correctly conclude?A.The net force on the particle was always centripetal.B.The potential energy of the system decreased.C.At least one non-conservative force acted on the particle.D.The kinetic energy of the particle at point B is >K1and <K2.11.The graph below shows the potential energy U for a particle moving on the xaxis.Which of the following five graphs correctly gives the force F exerted on theparticle?12.A small object of mass m ,on the end of a light cord,is held horizontally at a distance r from a fixed support as shown.The object is then released.What is the tension in the cord when theobject is at the lowest point ofits swing?A.m g/2;B.mg ;C.3m g/2;D.2mg ;E.3mg13.The displacement of an object oscillating on a spring is given by x(t)=Acos(ωt +φ).If the object is initially displaced in the negative x direction and given a negative initial velocity,then the phase constant φis between:A.0and π/2rad;B.π/2and πrad;C.πand 3π/2rad;D.3π/2and 2πradE.none of the above (φis exactly 0,π/2,π,or 3π/2rad)14.Block A on the left has mass 1.00kg.Block B on the right has mass 3.00kg.The BPath 2A Path 1blocks are forced together,compressing the spring.Then the system is released from rest on a level,frictionless surface.After the blocks are released,the kinetic energy (KE)of block A isA.1/9the KE of blockB. B.1/3the KE of block B.C.3times the KE of block B.D.9times the KE of block B.E.the same as the KE of block B.15.a mass of 2kg is attached to three springs of spring constants 1N/m,andWhen the mass is at rest the springs are unstretched.For this system:A.2502s rad /.=ω;B.222s rad /=ω;C.2512s rad /.=ω;D.232s rad /=ω;E.212s rad /=ωAnswer sheet of the choices 123456789101112131415Part B Short Answer (20Points),each question is worth 2points1.The position of a particle moving along the x axis depends on the time according to the equation 2x t 4t 6=-+.where x in meters and t in seconds.From t=0s to t=4s.(a)What is the displacement?Answer:_____________________(b)What is the distance?Answer:_____________________2.A block of mass 4kg on a horizontal surface experiences a constant force of 8N in the +X direction and 6N in the –Y direction.It moves from X =2m,Y =2m to X =6m,Y =-1m.How much work does the force do on the block during the move?If the block is moving at the rest ,what is the final speedAnswer:_____________________Answer:_____________________3.3333 r cos(t )i sin(t )j =+where r in meters and t in seconds.When t=1sv =a =ets travel around the sun in elliptical orbits with large eccentricities.when the Comets travel from r 1to r 2,and r are the distance from the center of the sun.,(The sun’s mass is s M ,The gravity constant is S G ,The Comets’s mass is m )a)The change potential energy of the CometsU =b)What is the work done by the sun n moving the comets from r 1to r 2W =5.Imagine that you discover a new type of conservative force.This force F is a function of the position x of an object.Suppose that F(x)=2x -2N where x is measured in meters.(a)Find the change in the potential energy associated with this conservative force for an object that moves from x=1meters to x=2meters.U (b)If the position of zero potential is at x=1meter,How about the potential of the x=4meters?4U Part C.(35points)You must show all your work or state your reasoning in order to receive credit.1.(10points)Two companies,Company A and Company B,sell plastic ball launchers for physics demos.Both launchers cost the same and both are the same size.Both launch a plastic ball of 10grams.The two companies send us specifications of the net force on the ball during launch in graphical form.The respective graphs are shown below and on the next page.Both launchers reach a maximum force of 1.2N and the graphs are composed of straight line segments.We want to pick the launcher that gives the highest final speed to the ball,starting from e principles of physics to show which one we should select.t (s)00.100.080.060.040F (N)0.021.20.80.60.40.21.00F (N)x (cm)03624121.20.80.60.40.21.0Company A Graph of Force versus Time,Company B Graph of Force versus Distance2.(12points)A series of successive vibrations form a positive progressing wave whose snapshot at t=0is shown in the figure.The speed of the wave is u 5m /s(1)write the vibration equation of the particle O(2)Write the wave equation y(x,t )(3)write the vibration equation of the particle at x =λ/2(4)Find the speed of particle B when t =0y(cm)5 u m /soB C 20105－x(cm)10－3.(13points)A car with mass M is setting on the frictionless ground,A particle ball with mass m is released from horizontal position with rest.When the ball arrives at the vertical position.(suppose the length of the string is R)(a)What’s displacement of the car?(b)What is the velocity of ball and the car relative to the ground respectively?(c)What is the tension acted on the ball when the ball is at the vertical position?(d)(additional8points)What is the car’s speed when the ball is rotating through30 degrees relative to the car?m30°M。
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2009─2010年 第 二 学期 《 大学物理》双语补考试卷注意：1、本试卷共 6 页； 2、考试时间: 120 分钟 3、姓名、学号必须写在指定地方 4、可以携带计算器Ⅰ. Filling the Blanks(每小题 2 分，共 20 分)1．一空气平行板容器，两板相距为d ，与一电池连接时两板之间相互作用力的大小为F ，在与电池保持连接的情况下，将两板距离拉开到4d ，则两板之间的相互作用力的大小是2. There is a point charge of electric quantity 3Q at the center of a cube, the electric flux through one surface of cube is3．如图1所示，真空中有两个点电荷，带电量分别为2Q 和-Q ，相距2R 。
若以负电荷所在处O 点为中心, 以R 为半径作高斯球面S ，则高斯面上a 、b 两点的电场强度大小分别为 、________________4. As shown in the figure 2, in the vacuum let the metal sphere with radius R be grounded, place a point charge Q with a distance r (r>R) away from the center O of the sphere, the total induced charge on the surface of the metal sphere isa Figure 15. 如图3所示，AOC 为一折成∠形的金属导线（AO =OC = L ），位于xoy 平面上. 磁感应强度为B 的匀强磁场垂直于xoy 平面. 当AOC 以速度v 沿x 轴正向运动时，导线上A 、C 两点间的电势差U AC = ，当以速度v 沿y 轴正向运动时A 、C两点中 点电势高.6．有一由N 匝细导线绕成的平面正三角形线圈，边长为a, 通有电流I ，置于均匀外磁场B 中，当线圈平面的法向与外磁场方向夹角为300时，该线圈所受的磁力矩M m 为________________7． The period of a pendulum(单摆) is measured to be 3.0s in the reference frame of the pendulum. The period when measured by an observer moving at a speed of 0.95c relative to the pendulum is8．把一个静止质量为0m 的粒子，由静止加速到0.6v c =(c 为真空中的光速)需做功为 Ⅱ．Choose the Correct Answer(每小题 3 分，共 30 分)1. 如图4所示，在真空中半径分别为R 和2R 的两个同心球面，其上分别均匀地带有电量+2q 和-2q ，今将一电量为+Q 的带电粒子从内球面处由静止释放，则该粒子到达外球面时的动能为： （ ）(A)02Qq R πε. (B)04Qq Rπε.(C) 08Qq Rπε. (D)038QqRπε.2．In following statements which one is correct ( )2qFigure 4Figure 3(A) The electric potential is definitely zero at the point where the electric field is zero.(B) The electric potential is non-zero at the point where the electric field is not zero.(C) The electric field is definitely zero at the point where the electric potential is zero.(D) The electric field is definitely zero in this area where the electric potential is constant3．有两个长直密绕螺线管，长度及线圈匝数均相同，半径分别为r 1和r 2．管内充满均匀介质，其磁导率分别为μ1和μ2．设r 1∶r 2=1∶2，μ1∶μ2=4∶1，当将两只螺线管串联在电路中通电稳定后，其自感系数之比L 1∶L 2与磁能之比Wm 1∶Wm 2分别为： ( )(A) L 1∶L 2=1∶1，Wm 1∶Wm 2 =1∶1(B) L 1∶L 2=1∶2，Wm 1∶Wm 2 =1∶1 (C) L 1∶L 2=1∶2，Wm 1∶Wm 2 =1∶2 (D) L 1∶L 2=2∶1，Wm 1∶Wm 2 =2∶14. One circular wire loop carries a clockwise current I in a uniform magnetic field B directed to the right as in the figure5. If B increases, what will happen to the net magnetic force and the torque on the current loop, respectively? ( ) (A) increase, increase (B) both keep the same (C) keep the same, decrease (D) keep the same, increase5. 尺寸相同的铁环与铜环所包围的面积中，通以相同变化率的磁通量，则环中： ( ) (A) 感应电动势相同，感应电流相同 (B) 感应电动势不同，感应电流不同 (C) 感应电动势相同，感应电流不同Figure 5(D) 感应电动势不同，感应电流相同6. The segment of wire in Figure 6 carries a current of I, where the radius of the circular arc is R, the magnitudeof the magnetic field at the origin O is ( )(A) R I 80μ (B) R I 40μ (C) RI πμ80 (D) R Iπμ207. 半径分别为4R 和R 的两个金属球，相距很远. 用一根长导线将两球连接，并使它们带电.在忽略导线影响的情况下，两球表面的电荷面密度之比为： ( )(A) 1/2 (B) 4 (C) 1/4 (D) 28. Which of the following statements is NOT true: ( ) (A) No two electric field lines can cross each other(B) The electric field vector is tangent to the electric field line at each point. (C) Magnetic field lines are always closed curves(D) The magnetic fields can be produced by current, so magnetic fields have sources9. 在某地发生两件事，静止位于该地的甲测得时间间隔为3s ，若相对甲以4c/5(c 表示真空中光速)的速率作匀速直线运动的乙测得时间间隔为：(A) 2.4s (B) 4s (C) 3.6s (D) 5s ( )10. A spaceship is measured to be 120.0 m long and 20.0 m in diameter while at rest relative to an observer. If this spaceship now flies by the observer with a speed of 0.99c, the diameter does the observer measure is ( ) (A) 17m (B) 19m (C) 15m (D) 20mⅢ．如图7所示，两个同心的均匀带电球面，内球面半径为R 1，带电量Q 1,外球面半径为R 2，带电Figure 6量为Q 2.设无穷远处为电势零点， 求（1）空间各处电场强度的分布；（2）在内球面内，距中心为r 处的P 点的电势。
(15分)Ⅳ．A rectangular loop of width a and length b islocated near a long wire carrying a current I (Fig. 8). The distance between the wire and the closestside of the loop is c. The wire is parallel to the long side of the loop. Find the total magnetic flux through the loop due to the current in the wire.(10分)Figure 7ⅤA conducting rod of length l moves with velocity v parallel to a long wire carrying a steady current I . Theaxis of the rod is maintained perpendicular to the wire with the near end a distance r away, as shown in Figure 9. Find the magnitude of the emf induced in the rod. (10分)Ⅵ. 地面观测者测得地面上甲、乙两地相聚68.010m ⨯，设测得作匀速直线运动的火车，由甲地到乙地历时2.0s 。
在一列与列车同方向相对地面运行速率0.6u c =的宇宙飞船中观测，试求改列车由甲地到乙地相对地面运行的路程、时间、和速度。
(15分)Figure 92009─2010学年 第二学期《大学物理》(双语)课程补考试卷参考答案及评分标准一、 Fill in the Blanks. (2 ⨯10=20 marks)1. 16F 2.2Q ε 3. 0217Q/(36R )πε， 023Q/(4R )πε4. R Q r-5. sin blv α， A 点6.2IB87. 9.6s8. 204m c二、Choose the Correct Answer. (3⨯10=30 marks)三．（15 marks ）1解：（1）依据高斯定律有=⋅⎰S E d S204r E Q πε= ( 2分)r<R 1 Q=0 E 1=0R 1<r< R 2 Q= Q 1 E 2= Q 1/(4πε0r 2)r> R 2 Q= Q 1 +Q 2 E 3= (Q 1+Q 2 )/(4πε0r 2) ( 5分)方向均沿径向向外。