中南大学大学物理双语版答案(共31页).doc

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1、精选优质文档-倾情为你奉上Problem 1. Answers: 1. ; ; 7.13.() 2. 3. a-e, b-d, c-f. 4. d: , , , , 5. (a), (Answer)(b) , , (Answer) (Answer)(c) , (Answer)6. Solution: From the definition of acceleration for a straight line motion , and the given condition , we have .Apply chain rule to dv/dt, the equation can be re

2、written as Separating the variables gives Take definite integration for both sides of the equation with initial conditions, we have ,or (Answer)Problem 2. Answers: 1. 13.0 m/s2, 5.7m/s, 7.5 m/s2, 2. , . Solution: At initial moment when the ball is just kicked out: , . In order the ball not to hit th

3、e rock, , ; (Answer)For the vertical motion: , , (Answer) 3. d. 4. d.5. Solution: at any moment the speed of the projectile is Tangential acceleration:, Radial acceleration:, From , we have the radius of curvature is given by (Answer)6. Solution: Problem 3. Answers: 1. Solution: Fig. 3-1 (a), (b) 2.

4、 (a) Mg, (b) 3. d. 4. e 5. Solution: from the given condition and with , at 20 s, , we have (a),With 40.0 s, we can get (b) 2.50 m/s, Solution: (c) From , we have Therefore , (Answer)which means the acceleration of the boat is proportional to the speed at any time. 6. Solution: From Newtons second l

5、aw, we have By the definition of acceleration, this equation can be rewritten as Separating the variables obtains Take the definition integral with the initial conditions, we hve Then . (Answer)Problem 4. Answers: 1.(a) 21J, (b) , q =19.4 2. . 3. c. 4. d 5. (a) =, (b) (c) , 6. Solution: (a) In the p

6、rocess of the pendulum swinging down, only the gravity does work , mechanic energy is conserved: When the sphere is released from a certain height, in order to make the ball will return to this height after the string strikes the peg:, . (Answer)Solution: (b) If the pendulum is to swing in a complet

7、e circle centered on the peg, at the top of the path, the tension T on the cord must not be zero. , Because So From the conservation of mechanical energy, we have Inserting into above equation, obtains (Answer)Problem 5. Answers: 1.(a) 2.(a) 0.284, (b) 114.6 fJ, 45.4 fJ. Solution: (a) Momentum and k

8、inetic energy are conserved: (1), So, (2) From Eq.1, , insert Eq.2 , . (Answer) (Answer) (Answer) 3. a. , , 4. a Solution: For m: in the elastic collision process, , in the inelastic collision: , 5. Solution: , , 6. proof problem Solution: , , Problem 6. Answers: 1. 144 rad. Solution: , ， , (Answer)

9、 Fig. 6-2 2. (a), Solution: (b) 3. c. Fig. 6-4 4. a. Solution: The initial angular acceleration of the rod and the initial translational acceleration of the right end of the rod: , , Fig.6-55. 21.5 N Solution: For the flywheel: , , (Answer) Fig. 6-66. Solution: from the conservation of mechanical en

10、ergy: , Problem 7. Answers: 1.(a) , Solution: (a), , , (b) . 2. (a) Mvd, (b) , (c) . Solution: (a) Angular momentum is conserved:Fig. 7-2; (b) (c) Fig. 7-5 3. d. 4. a 5.(a) , (b) Solution: (a) In the process of collision, angular momentum and mechanical energy are conserved: (1) Or (2) Form Eq.1, we

11、 have (3) (4) From Eq.(3), we have (5) Eq.(4) + Eq.(5) (Answer) Inserting w into Eq.(4), obtains (Answer) Solution: (b) , Fig. 7-6 6. Solution: (a) angular momentum is conserved (Answer) (b) Mechanical energy is not conserved because the clay and the solid cylinder undergo an inelastic collision. In

12、 this process, some kinetic energy must be lost.: ， Problem 9. Answers: 1. (a) 3.07 MeV（Or: 6mc2, 4.9210-13 J）, (b) 0.986c. Solution: (a), , For an electron, Total energy: Solution: (b) , 2. 2mc2 = 20.511=1.02 MeV. (1.6410-13 J ) 3. c. Solution: 4. d Solution: , .5. 1.63103 MeV/c. Solution: , , , Fo

13、r a proton, 6. (a) 5.3710-11 J = 335 MeV (5.3610-11 J) (b) 1.3310-9 J = 8.31 GeV (1.3310-9 J) Solution: (a) ; (b) 7. Solution: , , , , Problem 10. Answers: 1. 1.21022 for 37C. Solution: , 2. 385 K, 7.9710-21J, molar mass of the gas. Solution: , , M is the molar mass of the gas. 3. b. 4. a. 5. (a) 2/

14、3v0, (b) N/3, (c) 1.22v0, (d) 1.31v0. Solution: (a) , , ; (b) (c) (d) (Answer)6. Solution: (a) .(b) , Molar mass: , oxygen gas.(c) . (d) Problem 8. Answers: 1. 0.92c , 2. (a) 2.210-6 s, (b) 653 m. Solution: , , 3. b. (the constancy of the speed of light) 4. b. As the figure shows 5. (a) L0 = 17.4m ,

15、 (b) q = 3.3 Fig. 8-5 Solution: (a) Solution: (b) , 6. (a) 2.50108 m/s, (b) 4.97 m, (c) - 1.3310-8 s.Solution: (a) From the given condition, From the Lorentz transformation, Solution: (b) From the Lorentz transformation Solution (c) From the Lorentz transformation, . In S, red flash occurs first.Pro

16、blem 11. Answers: 1. (a) 3.5103 J, (b) 2.5103 J , (c) -1000 J. Solution: Work done by the gas: Work done on the gas: 2. 227K, Solutio: , 3. b . As the figure shows. 4. e. Solution: For a free expansion: , 5. Solution: (a) Fig. 11-5 (b) , . (c) , 6. Solution: From the definition for molar specific he

17、at at constant pressure, we have From the first Law of thermal dynamics, we have , Here, Combining theses equations, obtains ,From the ideal gas law, for the constant pressure process with one mole of ideal gas, we have Then Therefore Problem 12. Answer: 1. 3.28 J/K Solution: The entropy change of t

18、he Universe due to the energy transfer by radiation from the Sun to the Earth is . 2. 57.2 J/K.Solution: Suppose the process can be replaced by a reversible isothermal process, then .3. d. 4. c. Solution: free expansion is an irreversible process which occurs in an isolated system. 5. (a) -0.390nR,

19、(b) -0.545nR, Solution: (a) (b) 6. (a) 4500 J, (b) - 4986 J, (c) 9486 JSolution: (a) , The area under the T-S curveFig. 12-6(b) (c) Problem 13. Answer: 1. 1.86 2. 1/3, 2/3. , ， . Fig. 13-53. b. 4. d. 5. (a) 4.10103 J, (b) 1.42104 J, (c) 1.01104 J, (d) 28.9% Solution: (a) (b) A B: C A: . (c) B C: (d)

20、 (e) , the efficiency of the cycle is much lower than that of a Carnot engine operating between the same temperature extremes.6. Solution: (a) The work done by the gas from state a to c along abc is given by In the constant volume process along b to c, Wbc = 0. Therefore, (Answer) Solution: (b) the

21、change in internal energy form state b to c is, Using the ideal gas law to this constant volume process, obtaining Fig. 13-6 Then (Answer) From the definition of change in entropy, we have Using the first law of thermodynamics to this constant volume process with dW= 0, obtaining Substituting this i

22、nto the expression of entropy change, we have = (Answer) Solution: (c) Because both internal energy and entropy are state properties, for one complete cycle, (Answer)Problem 14. Answers: 1. 40.9 N/m. , 2. . Fig. 14-2Solution: For block B: , For block P: If block B is no to slip, , , , 3. c. , 4. c.

23、, , Fig. 14-5, , .5. (a) xm = 2.00 cm, (b) T = 4.00 s, (c) , (d) , (e) , (f) . 6. proof problem. Fig. 14-6 Solution: (a) the net force acting on the ball is given by(b) Comparing the force acting on the ball to the Hookes law , obtains , the ball moves in SHM, with angular frequency (Answer) Fig. 15

24、-1Problem 15. Answers: 1. 3.5 s. Solution: For a torsion pendulum, , , , 2. 0.944 kgm2.Solution: For a physical pendulum, , here h is the distance between the com and the rotation axis. 3. d. , *4. a. Solution: when or resonance occurs, here f is the nature frequency of the system. The oscillation o

25、f the system is due to the spring property of the diving board, so it can be treated as a torsion pendulum: , , 5. .Solution:Method one:Method two: , The mechanical energy is conserved: , , 6. Solution:Method one, Fig. 15-6For small angle, such as, , Method twoFor small angle, such as, , ,Problem 16

26、. Answers: 1. 0.319 m. Solution: , the speed of the wave: 2. (a) 0.25m,(b) 40.0 rad/s,(c) 0.300 rad/m, (d) 20.9 m, (e) 133m/s, (f) positive x direction.Solution: Standard SH wave function: , , 3. d. Standard traveling wave function: 4. c. , 5. (a) 31.4 rad/s, (b) 314 rad/m, (c) , (d) 3.77 m/s, (e)11

27、8 m/s2.Solution: (a) ; Fig. 16-5 (b) (c) (d) (e) Fig. 16-66. Solution: wave speed for a transverse wave in a string is given by , The time interval required for a transverse wave to travel from one end of the string to the other is (Answer)Problem 17. Answers: 1. (a) 2.0 Hz, (b) 3.38 m/s. Solution:(

28、a) Hz Fig. 17-2 (b) , vd = 3.38 m/s *2. 1029 m/s, 58.3 s. Solution: (a) From the definition of March number, (b) In the time interval t, the sound travels a distance h, , 3. d. 4. c. 5. (a) 308 m/s, (a)163 N, (b) 660 Hz. Solution: (a) For the fundamental mode, (b) the wave speed: 163 N. (c) , .6. (a

29、) As the figure shows(b) 2 cm /s,(c) , with y and x in cm and t in s.(d)- 2.5 cm/sSolution: From the curve: , ,(a) Transverse velocity , .At t = 0, (b) (c) (d) Fig. 18-1Problem 18. Answers: 1. 1500 nm, 5p Solution:, , ; 2. Because the half-wave loss in the reflection, at point O is dark, interferenc

30、e pattern only Fig. 18-2locates above O, and reverse the positions of dark and bright fringes.3. c. , 4. d. , 5. 632 nmSolution: For m = 1, (Answer)6. Solution: , , Problem 19. Answers: 1. 0.50 cm，2. 0.221 mm. Fig. 19-23. d. Solution: 4. d. Solution: For , optical path length difference Fig. 19-4 co

31、rresponds to the second minimum.5. 8.7 mm 6. 1.32 Fig. 19-5 Solution: For m = 10, ,When a liquid is introduced into the apparatus, (Answer)Problem 20. Answers: 1. 5l/2, 5 Solution: In single slit diffraction, bright fringes correspond to the path length difference for the two rays emitted from the t

32、op and the bottom of the slit Fig. 20-5, for the second order bright fringe, , , 5 half-wave zones.2. 650 (red) , 430 (blue),In a diffraction of single slit, half angular width for the bright fringe, at A, smaller wavelength is zero intensity, 650 nm is non-zero Fig. 20-2intensity. At B, 430 nm is non-zero intensity. 3. b. “No diffraction minima are observed” means the first minimum appears at infinity Or , If , , the first minimum will appear on the viewing screen.4. b. , 5. 25 cmSolution: For the third-order minimum: , The position of the third