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Course: PSYCH 830:101, Spring 2009School: Rutgers
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Illustrations To Text Accompany Meriam/Kraige Statics 6e Appendix D: Useful Tables. D_01tbl D_02tbl D_03tbl D_03tbl_cont D_04_tbl_cont2 D_04_tbl_cont3 D_04tbl D_04tbl_cont
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To Text Accompany Meriam/Kraige Statics 6e Appendix D: Useful Tables.
D_01tbl
D_02tbl
D_03tbl
D_03tbl_cont
D_04_tbl_cont2
D_04_tbl_cont3
D_04tbl
D_04tbl_cont
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Rutgers - PSYCH - 830:101
Text IllustrationsTo Accompany Meriam/Kraige Statics 6e Chapter 1 : Introduction to Statics.01_0101_0101_0201_0201_0301_0301_0401_0401_0501_0601_0701_0701_0801_08P_01_02P_01_02P_01_09P_01_09P_01_10
Rutgers - PSYCH - 830:101
Text IllustrationsTo Accompany Meriam/Kraige Statics 6e Chapter 2 : Force Systems.02_0102_0202_0302_0402_0502_0602_0702_0802_0902_0902_1002_1102_1102_1202_1202_1302_1402_1502_1602_1702_1802_1902_1902_
Rutgers - PSYCH - 830:101
Text IllustrationsTo Accompany Meriam/Kraige Statics 6e Chapter 3 : Equilibrium.03_0103_01_cont03_0203_0303_0403_0503_0603_0703_0803_0903_1003_A03_B03_CP_03_01P_03_02P_03_02P_03_03P_03_04P_03_05P_03_06P_0
Rutgers - PSYCH - 830:101
Text IllustrationsTo Accompany Meriam/Kraige Statics 6e Chapter 4 : Structures.04_0104_0104_0204_0304_0404_0504_0604_0704_0804_0904_0904_1004_1004_1104_1204_1304_1404_1404_1504_16P_04_04P_04_04P_04_06
Rutgers - PSYCH - 830:101
Text IllustrationsTo Accompany Meriam/Kraige Statics 6e Chapter 5 : Distributed Forces.05_0105_0205_0205_0305_0405_0505_0605_0705_0805_0905_1005_1105_1105_1205_1205_1305_1405_1505_1605_1705_1805_1905_19
Rutgers - PSYCH - 830:101
Text IllustrationsTo Accompany Meriam/Kraige Statics 6e Chapter 6 : Friction.06_0106_0106_0206_0206_0306_0306_0406_0406_0506_0506_0606_0606_0706_0706_0806_0806_0906_0906_1006_1006_1106_1106_1206_12P_06_01P_06_01P_06_02P_06_02
Rutgers - PSYCH - 830:101
Text IllustrationsTo Accompany Meriam/Kraige Statics 6e Chapter 7 : Virtual Work.07_0107_0107_0207_0207_0307_0307_0407_0407_0507_0507_0607_0607_0707_0707_0807_0807_0907_0907_1007_1007_1107_1107_1207_1
Rutgers - PHYSICS - 750:116
44.54: a) The number of protons in a kilogram is 6.023 1023 molecules mol (1.00 kg ) (2 protons molecule) 6.7 1025. 3 18.0 10 kg mol Note that only the protons in the hydrogen atoms are considered as possible sources of proton decay. The energy per d
Rutgers - PHYSICS - 750:116
HR H , presumed to be the same for R dr dR all points on the surface. b) For constant , HR Hr. c) See part (a), dt dt dR dt dR H0 . d) The equation H 0 R is a differential equation, the solution to R dt which, for constant H 0 is R(t ) R0e H 0t , whe
UC Irvine - BIOL - 05400
Rutgers - PSYCH - 830:101
44.1:ma) Kmc 2311 1 v c2 210.1547mc 2149.109 10kg, so K1.27 10Jb) The total energy of each electron or positron is E K mc2 1.1547mc2 9.46 10 14 J. The total energy of the electron and positron is converted into the total energ
Rutgers - PHYSICS - 750:116
44.2: The total energy of the positron is E K mc 2 5.00 MeV 0.511 MeV 5.51 MeV. We can calculate the speed of the positron from Eq. 37.38Emc 2 1v2 c2v c1mc 2 E210.511 MeV 5.51 MeV20.996.
Rutgers - PHYSICS - 750:116
44.3: Each photon gets half of the energy of the pion 1 1 1 E m c2 (270 me )c 2 (270)(0.511 MeV) 69 MeV 2 2 2 E (6.9 107 eV)(1.6 10 19 J eV) f 1.7 1022 Hz 34 h (6.63 10 J s)c f3.00 108 m s 1.7 1022 Hz1.8 10 14 m gamma ray.
Rutgers - PHYSICS - 750:116
44.4: a) hc Ehc m c 2h m c(6.626 10 34 J s) (207)(9.11 10 31 kg) (3.00 108 m s)1.17 10 14 m 0.0117 pm. In this case, the muons are created at rest (no kinetic energy). b) Shorter wavelengths would mean higher photon energy, and the muons wo
Rutgers - PHYSICS - 750:116
44.5: a)mmm270 me207 me63 meE 63(0.511 MeV) 32 MeV. b) A positive muon has less mass than a positive pion, so if the decay from muon to pion was to happen, you could always find a frame where energy was not conserved. This cannot occur.
Rutgers - PHYSICS - 750:116
44.6: a) The energy will be the proton rest energy, 938.3 MeV, corresponding to a frequency of 2.27 10 23 Hz and a wavelength of 1.32 10 15 m. b) The energy of each photon will be 938.3 MeV 830 MeV 1768 MeV, with frequency 42.8 10 22 Hz and wavelengt
Rutgers - PHYSICS - 750:116
1 44.8: 4 He 9 Be 12 C 0 n 2 4 6 We take the masses for these reactants from Table 43.2, and use Eq. 43.23 Q (4.002603 u 9.012182 u 12.000000 u 1.008665 u) (931.5 MeV u )5.701 MeV. This is an exoergic reaction.
Rutgers - PHYSICS - 750:116
44.9:1 0n10 5B4 27 3Li4 2He4.002603u 11.018607 um( n m( Li7 31 010 5B) 1.008665 u 10.012937 u 11.021602u He) 7.016004 um 0.002995 u; (0.002995u) (931.5 MeV u ) 2.79 MeV The mass decreases so energy is released and the rea
Rutgers - PHYSICS - 750:116
44.10: a) The energy is so high that the total energy of each particle is half of the available energy, 50 GeV. b) Equation (44.11) is applicable, and Ea 226 MeV.
Rutgers - PHYSICS - 750:116
44.11:a) BqB mBm 2mf q q2 (2.01 u) (1.66 10 27 kg u )(9.00 106 Hz) 1.60 10 19 C B 1.18 Tq 2 B 2 R 2 (1.60 1019 C) 2 (1.18 T) 2 (0.32 m) 2 b) K 5.47 1013 J 27 2m 2(2.01 u )(1.66 10 kg u ) 3.42 106 eV 3.42 MeV and v 2K 2(
Rutgers - PHYSICS - 750:116
eB eBR 3.97 107 s. b) R 3.12 107 m s. c) For threem m figure precision, the relativistic form of the kinetic energy must be used, ( 1 )mc 2 eV ( 1 )mc 2 , so eV ( 1 )mc 2 , so V 5.11 106 V. e44.12: a) 2 f
Rutgers - PHYSICS - 750:116
44.13: a) Ea22mc 2 ( Emmc 2 )Ea2 Em mc 2 2 2mc The mass of the alpha particle is that of a 4 He atomic mass, minus two electron masses. 2 But to 3 significant figures this is just M ( 4 He) 4.00 u (4.00 u) (0.9315 GeV u ) 3.73 GeV. 2(16.0 GeV)
Rutgers - PHYSICS - 750:116
1000 103 MeV 1065.8, so v 938.3 MeV b) Nonrelativistic: eB 3.83 108 rad s . m Relativistic: eB 1 3.59 105 rad s . m 44.14: a) 0.999999559 . c
Rutgers - PHYSICS - 750:116
44.15: a) With Em So Emmc , Em2Ea2 Eq. (44.11). 2mc 2[2(38.7 GeV)] 2 3190 GeV. 2(0.938 GeV) b) For colliding beams the available energy E a is that of both beams. So two proton beams colliding would each need energy of 38.7 GeV to give a tota
Rutgers - PHYSICS - 750:116
44.16: The available energy E a must be (m02m )c 2 , so Eq. (44.10) becomes( m 0 Et2m p ) 2 c 4 ( m 02m p c 2 ( E t 2m p c 22m p c 2 ), or2m p ) 2 c 2 2m p(547.3 MeV 2(938.3 MeV) 2 2(938.3 MeV)2(938.3 MeV) 1254 MeV.
Rutgers - PHYSICS - 750:116
44.17: Section 44.3 says m( Z0 ) 91.2 GeV c 2 .E91.2 109 eV 1.461 10 8 J; m 97.2E c21.63 10 25 kgm(Z0 ) m(p)
Rutgers - PHYSICS - 750:116
44.18: a) We shall assume that the kinetic energy of the 0 is negligible. In that case we can set the value of the photon's energy equal to Q. Q (1193 1116) MeV 77 MeV Ephoton . b) The momentum of this photon is Ephoton (77 106 eV) (1.60 10 18 J eV)
Rutgers - PHYSICS - 750:116
44.19: m M ( ) m p m 0 . Using Table (44.3): E (m)c 2 1189 MeV 938.3 MeV 135.0 MeV 116 MeV.
Rutgers - PHYSICS - 750:116
44.21: Conservation of lepton number. a) e ve v Lu : 1 1, Le : 0 so lepton numbers are not conserved. b) e ve v Le : 0 1 1 L : 1 1 so lepton numbers are conserved.1 1e . Lepton numbers are not conserved since just one lepton is c) produced from
Rutgers - PHYSICS - 750:116
44.22: a) Conserved: Both the neutron and proton have baryon number 1, and the electron and neutrino have baryon number 0. b) Not conserved: The initial baryon number is 1 +1 = 2 and the final baryon number is 1. c) Not conserved: The proton has bary
Rutgers - PHYSICS - 750:116
44.23: Conservation of strangeness: a) K v . Strangeness is not conserved since there is just one strange particle, in the initial states. b) n K p 0 . Again there is just one strange particle so strangeness cannot be conserved. c) K K
Rutgers - PHYSICS - 750:116
44.24: a) Using the values of the constants from Appendix F, e2 1 7.29660475 10 3 , 4 0 c 137.050044 or 1 137 to three figures. b) From Section 38.5 e2 v1 2 0 h but notice this is juste2 h c as claimed rewriting as . 4 0c 2
Rutgers - PHYSICS - 750:116
44.25:f2 cand thus(J m) 1 (J s)(m s 1 )f2 is dimensionless. (Recall f 2 has units of energy times distance.) c
Rutgers - PHYSICS - 750:116
44.26: a)The particle has Q 1 (as its label suggests) and S 3. Its appears as a "hole"in an otherwise regular lattice in the S Q plane. The mass difference between each S row is around 145 MeV (or so). This puts the mass at about the right spot. As
Rutgers - PHYSICS - 750:116
44.27: a)uds :Q eb)2 1 1 0; 3 3 3 1 1 1 B 1; 3 3 3 S 0 0 ( 1) 1 C 0 0 0 0. Q 2 2 cu : 0; e 3 3 1 1 B 0; 3 3 S 0 0 0; C 1 0 1.Q 1 1 3 1; B 3 e 3 3 S 3(0) 0; C 3(0) 0. Q 1 e 3 S 0 0 2 3 0; C 1; B 0 ( 1) 1 3 1. 1;c)ddd:d)dc :1 30;
Rutgers - PHYSICS - 750:116
44.28: a) S 1 indicates the presence of one s antiquark and no s quark. To have baryon number 0 there can be only one other quark, and to have net charge + e that quark must be a u, and the quark content is us . b) The particle has an s antiquark, an
Rutgers - PHYSICS - 750:116
44.29: a) The antiparticle must consist of the antiquarks so: n u dd . b) So n udd is not its own antiparticle cc so c c so the is its own antiparticle. c)
Rutgers - PHYSICS - 750:116
44.30: (m 2m )c 2 44.4 for masses).(9460 MeV 2(1777 MeV)5906 MeV (see Sections 44.3 and
Rutgers - PHYSICS - 750:116
44.31: In1 1decay, 1 p 11 00 1e1 0n vepuud , nudd , so indecay a u quark changes to a d quark.
Rutgers - PHYSICS - 750:116
44.32: a) Using the definition of z from Example 44.9 we have that ( 0 s ) 0 1 z 1 . 0 s Now we use Eq. 44.13 to obtain 1 v c v 1 c 1 z . v c v 1 c 1 b) Solving the above equation for we obtain(1 z ) 2 1 1.5 2 1 0.3846. (1 z ) 2 1 1.5 2 1 Thus, v
Rutgers - PHYSICS - 750:116
44.33: a) v b)H0rc v c v(20 (km s) Mly)(5210 Mly) 1.04 105 km s .3.0 105 km s 1.04 105 km s 3.0 105 km s 1.04 105 km s 1.44.0 s
Rutgers - PHYSICS - 750:116
3.00 108 m s c 1.5 104 Mly. b) This distance H 0 20(km s) Mly represents looking back in time so far that the light has not been able to reach us.44.34: From Eq. (44.15), r
Rutgers - PHYSICS - 750:116
44.35: a) v( 0 s ) 2 1 c ( 0 s ) 2 13.280 107 m s.658.5 590.0 658.5 590.02 21 (2.998 108 m s) 1b) rv H03.280 107 m s 2.0 104 m Mly1640 Mly.
Rutgers - PHYSICS - 750:116
44.36: Squaring both sides of Eq. (44.13) and multiplying by 2 c v gives 0 (c v) 2 (c v), and solving this for v gives Eq. (44.14). s
Rutgers - PHYSICS - 750:116
2 44.37: a) m M (1 H) M (1 H) M ( 3 He) where atomic masses are used to balance 1 2 electron masses. m 1.007825 u 2.014102 u 3.16029 u 5.898 10 3 u E ( m)c 2 (5.898 10 3 u) (931.5 Me V u ) 5.494 MeV.b)mmn M ( 3 He) M ( 4 He) 2 2 1.0086649 u 3.0
Rutgers - PHYSICS - 750:116
44.39: mmeEmpmnmve so assuming mve0,8.40 10 4 u 0.783 MeV and is endoergic.m 0.0005486 u 1.007276 u 1.008665u ( m)c 2 ( 8.40 10 4 u) (931.5 Me V u)
Rutgers - PHYSICS - 750:116
44.40: m12 C6m 4 He2m16 O87.69 10 3 u, or 7.16 MeV, an exoergic reaction.
Rutgers - PHYSICS - 750:116
44.41: For blackbody radiation mT2.90 10 3 m K , so m1 T1 m2 T2 m1(1.062 10 3 m)2.728 K 3000 K9.66 10 7 m.
Rutgers - PHYSICS - 750:116
44.42: a) The dimensions of are energy times time, the dimensions of G are energy times time per mass squared, and so the dimensions of(E T) (E L M 2 ) (L T) 31/2G / c3 are2E M34T2 LL TT2 LL.12b)G c312(6.626 10J s)(6.673
Rutgers - PHYSICS - 750:116
44.43: a) E a 2(7 TeV) 14 TeV b) Fixed target; equal mass particles, 2 Ea (1.4 107 MeV) 2 Em mc 2 2(938.3 MeV) 2mc 21.04 1011 MeV 1.04 105 TeV.938.3MeV
Rutgers - PHYSICS - 750:116
44.45: The available energy must be the sum of the final rest masses: (at least) Ea 2mec 2 m 0 c 22(0.511 MeV) 135.0 MeV 136.0 MeV.For alike target and beam particles: EmeEa2 2me c 2me c 2(136.0 MeV) 2 2(0.511MeV)0.511MeV 1.81 104 MeV. S
Rutgers - PHYSICS - 750:116
44.46: In Eq.(44.9), Ea (m 0 mK 0 )c 2 , and with MK Ea2 (m c 2 )2 2mp c2mp , mm and Em(m )c 2K,(mp c 2 ) 2( m )c 2 (938.3 MeV) 2 139.6MeV(1193 MeV 497.7 MeV) 2 (139.6 MeV) 2 2(938.3 MeV) 904 MeV.
Rutgers - PHYSICS - 750:116
44.47: The available energy must be at least the sum of the final rest masses. Ea (m 0 )c 2 (mK )c 2 (mK )c 2 1116 MeV 2(493.7 MeV) 2103 MeV.Ea2(mp )c )2 22(mp )c 2 EK(mK )c ) .2 2So EKEa2(mp )c 2 ) 2(mK )c 2 ) 2(2103) 22(mp )c
Rutgers - PHYSICS - 750:116
44.48: a) The decay products must be neutral, so the only possible combinations are 0 0 0 or 0 b) m03m0142.3Me V c 2 , so the kinetic energy of the (m00mesons is 142.3MeV. For the other reaction, Km0mm )c2133.1 MeV.
Rutgers - PHYSICS - 750:116
44.49: a) If the decays, it must end in an electron and neutrinos. The rest energy of (139.6 MeV) is shared between the electron rest energy (0.511 MeV) and kinetic energy (assuming the neutrino masses are negligible). So the energy released is 139.6
Rutgers - PHYSICS - 750:116
44.50: E(1.054 10 34 J s) (4.4 106 eV) (1.6 10 19 J/eV)1.5 1022s.
Rutgers - PHYSICS - 750:116
44.51: a) E( m)c 2(mp )c 2(mK )c 2(mK )c 21019.4 MeV 2(493.7 MeV) 32.0 MeV. Each kaon gets half the energy so the kinetic energy of the K is 16.0 MeV. b) Since the 0 mass is greater than the energy left over in part (a), it could not have b