西尔斯大学物理双语版答案
大学物理西尔斯
1. Atomic Spectra of Hydrogen
1) Atomic Spectra ： emission spectrum：the spectrum formed during the process of radiation of electromagnetic waves.
absorption spectrum： the spectrum formed during the process of absorption of electromagnetic waves with certain wavelengths.
1) Ground-state level : Every atom has a lowest energy level that includes the minimum internal energy state that the atom can have.
2) Excited level :
Limit of wavelength of Balmer series
n , B 364.6nm
1890:
1
R(
1
2
2
1
2
)
n
n 3, 4,5,
R＝ 4
1.097
373
7 1
10 m
B
➢ Atomic Spectra of Hydrogen
Lyman series:
UV
~ R( 1 1 )
Q1：There must be something with positive charge in the atom. How does the mass of the atom distribute? Q2：How does the charges of the atom distribute?
西尔斯大学物理
Example 3-6 A mass m is set at rest on a ramp; there is friction between the mass and the ramp. The ramp is slowly raised from a horizontal orientation. By analyzing the forces on the mass, find the critical ramp angle c at which the mass will start to slide.
** Applying of Newton’s Laws
Example 3-4 The figure represents a stock car of mass m = 1600 kg traveling at a constant speed v = 20 m/s around a flat, circular track of radius R = 190 m. For what value of s between the track and the tires of the car will the car be on the verge of sliding off the track?
l
11
University Physics
** Applying of Newton’s Laws
Solution
List the components of Newton’s second law for The radial component
l
T
v
mg
v2 T mg sin man m l Tangential component dv mg cos mat m dt
大学物理双语2012-2013-1月A答案及评分标准
标准答案及评分标准一．Choice(20分，每题4分)1. a2. b3. c4. d5. a二．Blanks （20分）6. 0.5 （1分） 3 (1分) 1/2 (1分)π/6（2分） 3π or 9.42 (3pts)7. 1.26 （3分）8. 2.26°(3分)9. 3.56×10-28 (3分)10. 1.14eV (3分)三．Questions(10分)11. (5pts)The relativity principle: The laws of physics must be the same in all inertial reference frames. 一切物理规律在惯性系中相同。
(2分)The constancy of the speed of light: The speed of light in vacuum has the same value c in all inertial frames, regardless of the velocity of the observer or the velocity of the source emitting the light. 真空中的光速在任何惯性系中都是c ，与光源或观察者的运动无关。
（3分）12. (5pts)The maximum kinetic energy of photoelectrons is independent of light intensity. (2分)No electrons are emitted if the incident light frequency falls below some cutoff frequency fc , whose value is characteristic of the material being illuminated, regardless of the light intensity.（3分）四. Problems （50分）13(10pts) (1) The average transitional kinetic energy 21101.623-⨯==kT K t J ………………… 3pts (2) the rms speed s m MRT v rms /4803≈= ………………… 3pts (3) the internal energy RT i E 2= ………………… 1pts For O 2, i = 5, ………………… 1pts the internal energy 60912≈=RT i E J ………………… 2pts 14(10pts)(1) 1→2: Isothermal expansion )/ln(12V V nRT Q H H = ………………… 2pts 3→4: Isothermal compression )/ln(34V V nRT Q L L = ………………… 2pts (2) 2→3: Adiabatic expansion 1312--=γγV T V T L H ………………… 1pts4 1: Adiabatic compression. 1411--=γγV T V T L H ………………… 1pts Therfore, 4312V V V V = ………………… 1pts HL H L T T Q Q = ………………… 1pts The efficiency of Carnot engine is HL H L T T Q Q -=-=11η ………………… 2pts 15 (17pts)(1) The light path difference isθδs i nd = ………………… 2pts For bright fringesλθδm d ==s i n………………… 2pts The separation between the 2nd-order bright fringe and the 3rd-order bright fringe is423105.523tan tan -⨯==-≈-=∆d D d D d D D D x λλλθθm ………………… 3pts (2) The light path difference is()L n d 1sin -+=θδ ……………………………………… 3ptsFor bright fringes()λθδm L n d =-+=1s i n …………………………… 1pts At the central, point θ=0 ……………………………………… 1ptsThe central point on the viewing screen is now occupied by the seventh bright slid fringe (m =7) ()λδ710sin =-+=L n d ……………………………………… 1pts 61064.6)1/(7-⨯=-=n L λm ……………………………………… 2pts(3) No ……………………………………… 2pts16 (13pts)(1) The period measured by the observers on the spaceship is the propertime. ……………1pts126.020==km T πs ……………4pts (2) The mass and the spaceship is moving at a very high speed relative to the Earth. The effect of special theory of relativity must be taken into account. ………2ptsThe period measured by the observers on the Earth is not the propertime. ……………2ptsAccording to the special theory of relativity,209.0/1220=-=c v T T s ……………4pts。
西尔斯物理学
129. Superposition: n. 重叠，重合，叠合。合成。 130. Utmost: n. 极限,最大可能,极力。adj. 极度的,最远的. 131. Puck: n. 恶作剧的小妖精; 冰球. 132. Sustain: 维持,持续。 133. Ketchup: 调味番茄汁。 134. Contact force: 触点压力 135. Calibrate: v.校准。 136. Platinum: n.白金，铂。 137. Iridium: n. 铱。 138. Dimensionally: adv.尺寸上的。量纲。 139. Handled: adj.把手的。手柄的。 140. Pose: n.姿势，姿态。v.（使）摆好姿势，形成，因其，造成。 141. Equilibrium: n.平衡，平静。 142. Trivial: adj; 琐细的，价值不高，微不足道的。 143. Furthermore: adv.此外，而且。 144. Cutting-blade: 切割刀具。 145. Aligned:vi.排列。vt.使结盟，使成一行。 146. Toboggan: n.平底雪橇。V. 乘撬滑下，急剧滑下。 147. Iceboat: n.冰上滑行船，碎冰船。 148. Analytical:adj:分析的，解析的。
189. Individual: n. 个人，个体。adj.个别的，单独的，个人的。 190. Infinitesimal: adj. 无穷小的，极小的，无限小的；n. 极小量，无限小。 191. Rigorously: adj. 严厉的，残酷地. 192. Evaluate: vt. 评估，估价，求…值的；v. 评价。 193. Obnoxious: adj. 不愉快的，讨厌的。 194. Swing: v. 摇摆，摆动，回转，旋转；n. 秋千，摇摆，摆动. 195. Quotient: n. 商，份额。 196. Deposit: n. 堆积物，沉淀物，存款，押金，保证金，存放物；vt. 存放， 堆积；vi.沉淀。 197. Hammerhead: n.锤头；adj.锤头状的，鲁钝的。 198. Momentarily: adv. 即刻。 199. Slant: v. 倾斜，歪向；n. 倾斜。 200. Elongation: n. 延长。 201. Coil: v.盘绕，卷。 202. Bounce: v. 反跳，弹起；n. 跳起，弹回。 203. Suspension: n. 吊，悬挂，暂停，中止，延迟。 204. Flex: vt. 弯曲，伸缩，折曲；n. vi 折曲，弯曲，电线，松紧带。 205. Strategy: n. 策略。 206. Wrinkle: n. 皱纹；v. 使皱。
英文版大学物里答案
习题11-1 P 点相对于原点的位矢26=-+p r i j m , P 点到Q 点的位移42∆=-r i j m, 求Q 点相对于原点的位矢并画图.解：设Q 点相对与原点的位矢为Q r ，则：24=+∆=+Q p r r r i j1-2一质点作直线运动，它的运动方程是2ct bt x -=, b , c 是常数. (1) 求此质点的速度和加速度函数；(2) 作出x t -，t υ-和a t -图解：这是一个一维的问题.速度 (2)dx ct b dtυ==-+， 加速度 2d a c dtυ==-. 图略.1-3物体按照29.4t x =的规律运动，x 的单位为米，t 的单位为秒. (1) 计算下列各时间段内的平均速度：1s 到1.1s,1s 到1.01s,1s 到1.001s; (2) 求1s 末的瞬时速度；(3) 解释上述结果解：这也是一个一维的问题.(1) 平均速度 x tυ∆=∆. 1s 到1.1s 内： 224.9 1.1 4.911.11x t υ∆⨯-⨯==∆-＝10.29 （m/s ）, 1s 到1.01s 内：224.9 1.01 4.911.011x t υ∆⨯-⨯==∆-＝9.849（m/s ）, 1s 到1.001s 内：224.9 1.001 4.911.0011x t υ∆⨯-⨯==∆-＝9.8049（m/s ）. (2) 速度 9.8dx t dtυ==. 1-4一质点以110m s -⋅的恒定速率向东运动. 当它刚到达距出发点为d 的一点时，立即以120m s -⋅的恒定速率返回原处. 问: 质点在全过程中的平均速度和平均速率为多少？解：取出发点为原点，向东为x 轴正方向. 从原点到x ＝d 处，作匀速直线运动，时间 11s t υ∆∆=＝d/10.从x ＝d 处返回原点作匀速直线运动，时间22st υ∆∆=＝d/20 (全过程中，平均速率 12s d d t t t υ∆+===∆∆+∆13.3 （m/s ） 返回原处时，位移x ∆＝0，平均速度x tυ∆=∆＝0. 1－5 矿井里的升降机由井底从静止开始匀加速上升，经过3s 速度达到13m s -⋅，然后以这个速度匀速上升6s ，最后减速上升经过3s 后到达井口时刚好停止. (1) 求矿井深度；(2) 作出x t -，t υ-和a t -图.解：（1）以井底为原点，向上为x 轴正向.在0—3s 内，升降机作匀加速直线运动:210112x t a t υ∆=+ (1) 2210112a x υυ=+∆. (2)其中00υ=. 由（1）、（2）两式得：1x ∆＝4.5(m).在3—9s 内，升降机以1υ＝3m/s 作匀加速直线运动,21x t υ∆=＝18（m/s ） (3)在9—12s 内，升降机作匀减速直线运动231212x t a t υ∆=- (4) 2221232a x υυ=-∆, (5) 其中20υ=. 由（4）和（5）两式得3x ∆＝4.5(m)矿井深度 123H x x x =∆+∆+∆＝4.5＋18＋4.5＝27(m).1-6湖中有一小船，岸上有人用一根跨过定滑轮的绳子拉船靠岸。
西尔斯当代大学物理双语PPT课件
.
10
1.6 Vector Algebra
a Vector quantity can be expressed by a
• Every physical theory has a range of validity. • No theory has ever regarded as the final truth.
1.2 Idealized Models A model is a simplified version of a physical
Chapter 1 Physical Quantities
and Vectors
.
1
1.1 The nature of physics
• Physics is an experimental science. • Physicists observe the phenomena of nature and try to find patterns and principles that relate these phenomena. These patterns are called physical theories or physical laws, or principles.
ball move in a vacuum.
We ignore the earth’s rotation.
point masses
We make the weight constant.
.
5
1.3 Standards and Units
Physics is an experimental science. Experiments require measurements.
(双语版)大学物理学Ch15-Ch17习题
v ≈ v2
Ch.17-12(p.424) (a) Show that if the specific heat varies as a function of temperature, c(T) , the heat needed to raise the temperature of a T substance from T1 to T2 is given by Q = mc(T)dT ∫T (b) Suppose c(T)=co(l+ at) for some substance, where a=2.0×103C0-1 × and T is the Celsius temperature. 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 T2 for part (b), expressed in terms of c0 , the specific heat at 0°C? °
1 2
Solution:
Ch.17-41(p.426) 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.)
西尔斯物理第23章课后习题答案
Chapter 23 Exercises23-3 a) For surface S 1, the electric flux is ()()221CL j L k D j C i B S E -=-⋅-+-=⋅=Φ .For surface S 2, the electric flux is ()()222DL k L k D j C i B S E -=⋅-+-=⋅=Φ .For surface S 3, the electric flux is ()()223CL j L k D j C i B S E =⋅-+-=⋅=Φ .For surface S 4, the electric flux is ()()224DL k L k D j C i B S E =-⋅-+-=⋅=Φ.For surface S 5, the electric flux is ()()225BL i L k D j C i B S E -=⋅-+-=⋅=Φ.For surface S 6, the electric flux is ()()226BL i L k D j C i B S E =-⋅-+-=⋅=Φb) The total electric flux is 0654321=Φ+Φ+Φ+Φ+Φ+Φ=Φ.23-6 a) Because the S1 encloses the charge q1, the electric flux through the S1 isb) Because the S2 encloses the charge q2, the electric flux through the S2 isc) Because the S3 encloses the charge q1and q2, the electric flux through the S3 isd) Because the S4 encloses the charge q1 and q3, the electric flux through the S4 ise) Because the S5 encloses the charge q1 ,q2 and q3, the electric flux through the S5 is23-8 a) When the spherical surface ’s radius is 0.500m, the charges q1 and q2 aren ’t closed by this surface, so the electric flux is zero. b) When the spherical surface ’s radius is 1.50m, the second charge q2 is closed by this surface, so the electric flux is)/(1045.01085.81000.426126011C m N q ∙⨯=⨯⨯==Φ---ε)/(1088.01085.81080.726126022C m N q ∙⨯-=⨯⨯-==Φ---ε)/(1043.01085.81080.71000.42612660213C m N q q ∙⨯-=⨯⨯-⨯=+=Φ----ε)/(1072.01085.81040.21000.42612660314C m N q q ∙⨯=⨯⨯+⨯=+=Φ----ε)/(1016.01085.81040.21080.71000.4261266603215C m N q q q ∙⨯-=⨯⨯+⨯-⨯=++=Φ-----εc) When the spherical surface ’s radius is 2.50m, these two charges enclose by the surface, so the electric flux is23-24 a) When r <a, the Gaussian surface doesn’t enclose any electric charges, so the magnitude of electric field is zero.When a <r <b, the Gaussian surface encloses the charge q, so the magnitude of electric field is When b <r <c, because the hollow sphere is a conductor, the electric field is zero inside it. When r>c, the Gaussian surface encloses the charge q, so the magnitude of electric field ise)b)c m N q 2612602/1068.01085.81000.6⨯-=⨯⨯-==Φ--εc m N q q 261266012/10025.01085.81000.61000.4⨯-=⨯⨯-⨯=+=Φ---ε204r qπε=Φ204r q πε=Φc) The negative charge –q is on the inner surface of the hollow sphere. d) The positive charge +q is on the outer surface of the hollow sphere.23-27 a) when r <a, the surface isn ’t closed any charge, so the electric field is E = 0 b) When a <r <b, the surface doesn ’t enclose any charge, so the electric field is E = 0 c) When b <r <c, the surface does enclose the charge +2q, so the electric field isd) When c <r <d, the surface doesn ’t enclose any net charge, so the electric flux is E = 0. e) When r > d, the surface does enclose the charge +2q and +4q, so the electric flux isThe total charge on the inner surface of the small shell is zero. The total charge on the outer surface of the small shell is +2q. The total charge on the inner surface of the large shell is -2q. The total charge on the outer surface of the large shell is +6q.23-28. a) when r <a, the surface doesn ’t closed any charge, so the electric field is E = 0. b) When a <r <b, the surface doesn ’t enclose any charge, so the electric field is E = 0. c) When b <r <c, the surface does enclose the charge +2q, so the electric field isd) When c <r <d, the surface doesn ’t enclose any net charge, so the electric flux is E = 0. e) When r > d, the surface does enclose the charge +2q and -2q, so the electric flux isO a b c d2042r q E πε=202046442r q r q q E πεπε=+=2042r q E πε=042220=-=rq q E πεThe total charge on the inner surface of the small shell is zero. The total charge on the outer surface of the small shell is +2q. The total charge on the inner surface of the large shell is -2q. The total charge on the outer surface of the large shell is zero.23-29. a) when r <a, the surface doesn ’t closed any charge, so the electric field is E = 0. b) When a <r <b, the surface doesn ’t enclose any charge, so the electric field is E = 0. c) When b <r <c, the surface does enclose the charge +2q, so the electric field ise) When r > d, the surface does enclose the charge +2q and -4q, so the electric flux isThe total charge on the inner surface of the small shell is zero. The total charge on the outer surface of the small shell is +2q. The total charge on the inner surface of the large shell is -2q. The total charge on the outer surface of the large shell is -2q.23-33 a) Between the cylinders, using the Gauss ’s law,the electric field isb) At any point outside the outer cylinder, the electric field isc)O a b c2c2042r q E πε=202042442r q r q q E πεπε-=-=⎰=⋅sq s d E 0ε 02ελπl l r E =⋅⋅r E 02πελ=r E 02πελ=。
- 1、下载文档前请自行甄别文档内容的完整性，平台不提供额外的编辑、内容补充、找答案等附加服务。
- 2、"仅部分预览"的文档,不可在线预览部分如存在完整性等问题,可反馈申请退款(可完整预览的文档不适用该条件!)。
- 3、如文档侵犯您的权益，请联系客服反馈,我们会尽快为您处理(人工客服工作时间：9:00-18:30)。
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 =。