Chapter 44
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Chapter 44

Course Number: PHYS 2306, Fall 2007

College/University: Virginia Tech

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44.1: m a) K 9 . 109 mc 2 1 1 v 2 1 c 2 0 . 1547 mc 2 10 31 kg, so K 1 . 27 10 14 J b) The total energy of each electron or positron is E K mc 2 1 .1547 mc 2 14 9 . 46 10 J. The total energy of the electron and positron is converted into the total energy of the two photons. The initial momentum of the system in the lab frame is zero (since the equal-mass particles have equal speeds in opposite...

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K 9 44.1: m a) . 109 mc 2 1 1 v 2 1 c 2 0 . 1547 mc 2 10 31 kg, so K 1 . 27 10 14 J b) The total energy of each electron or positron is E K mc 2 1 .1547 mc 2 14 9 . 46 10 J. The total energy of the electron and positron is converted into the total energy of the two photons. The initial momentum of the system in the lab frame is zero (since the equal-mass particles have equal speeds in opposite directions), so the final momentum must also be zero. The photons must have equal wavelengths and must be traveling in opposite directions. Equal means equal energy, so each photon has energy 9 . 46 10 14 J. c) E hc so hc E hc ( 9 . 46 10 14 J ) 2 . 10 pm The wavelength calculated in Example 44.1 is 2.43 pm. When the particles also have kinetic energy, the energy of each photon is greater, so its wavelength is less. 44.2: The total energy of the positron is E K mc 2 5 . 00 MeV 2 0.511 MeV 5.51 MeV. 2 We can calculate the speed of the positron from Eq. 37.38 E mc 1 2 2 2 v c 1 mc E 2 1 0 . 511 MeV 5.51 MeV 0 . 996 . v c 44.3: Each photon gets half of the energy of the pion E 1 2 f E h c f 3 . 00 1 .7 m c 2 1 2 ( 270 m e ) c 10 7 2 1 2 ( 270 ) ( 0 . 511 MeV) 10 19 69 MeV 10 22 ( 6 .9 eV ) (1 . 6 10 34 J eV ) 1 .7 Hz ( 6 . 63 10 10 8 J s) 14 m s Hz 22 1 . 8 10 m gamma ray. 44.4: a) hc E hc m c 2 h m c ( 6 . 626 (207) (9.11 10 14 31 10 34 J s) 10 m s ) 8 kg) (3.00 1 . 17 10 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 would be created with non-zero kinetic energy. m 270 m e 207 m e 63 m e 44.5: a) m m E 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. 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 wavelength 7 . 02 10 16 m. 44.7: E ( m ) c 2 ( 400 kg 400 kg ) ( 3 . 00 10 8 m s ) 2 7 . 20 10 19 J. 12 1 C 0n 44.8: 4 He 9 Be 2 4 6 We take the masses for these reactants from Table 43.2, and use Eq. 43.23 Q 1 0 ( 4 . 002603 u 10 5 7 3 4 2 9.012182 u 12.000000 reaction. u 1.008665 u) (931.5 MeV u ) 5 . 701 MeV. This is an exoergic 44.9: n 1 B Li He m (0 n m ( 3 Li m 7 10 5 B) He) 1.008665 7.016004 u u 10.012937 4.002603 u u 11.021602 11.018607 u u 4 2 0.002995 u; (0.002995 u) (931.5 MeV u ) 2 . 79 MeV The mass decreases so energy is released and the reaction is exoergic. 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 E a 226 MeV. 44.11: a) B B qB B m m q 10 1 . 60 2 mf q 27 2 ( 2 . 01 u) (1.66 kg u ) (9.00 19 10 6 Hz) 10 C 1 . 18 T b) K q B R 2m 3.42 2 2 2 (1 . 60 6 10 19 C) (1.18 T) (0.32 m) 10 27 2 2 2 5 . 47 10 13 J 2 ( 2 . 01 u ) (1 . 66 10 eV 3.42 MeV 2 ( 5 . 47 (2.01 u) (1.66 eB m 3 . 97 10 7 kg u ) and v 2K m 10 10 13 J) kg u ) R eBR m 27 1 . 81 10 m s . 7 7 44.12: a) 2 f s. b ) 3 . 12 10 m s. c ) For three- figure precision, the relativistic form of the kinetic energy must be used, eV ( 1 )mc , so eV 2 ( 1 )mc , so V 2 ( 1 )mc e 2 5 .11 10 V. 6 44.13: a) E a2 Em 2 mc ( E m 2 mc ) 2 2 Ea 2 2 mc 2 mc 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 ( 2 He) So E m 4 4.00 u (16 . 0 GeV) (4.00 u) (0.9315 GeV u ) 2 3.73 GeV. 3 . 73 GeV 30.6 GeV. 2(3.73 GeV) b) For colliding beams of equal mass, each has half the available energy, so each has 8.0 GeV. 44.14: a) 1000 10 MeV 3 1065 . 8 , so v 0 . 999999559 c . 938.3 MeV b) Nonrelativistic: eB m 3 . 83 10 8 rad s . Relativistic: eB 1 m 3 . 59 10 rad s . 2 5 44.15: a) With E m So E m mc , E m 2 Ea 2 2 Eq. (44.11). 2 mc [ 2 ( 38 . 7 GeV)] 2 ( 0 . 938 GeV) 3190 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 total of 77.4 GeV. 44.16: The available energy E a must be ( m 2 m ) c 2 , so Eq. (44.10) becomes 0 ( m 0 Et 2m p ) c (m 0 2 4 2m pc (E t 2 2 2 2 m p c ), or 2 2 2m p ) c 2m p 2m pc ( 547 . 3 MeV 2(938.3 MeV)) 2 2 ( 938 . 3 MeV) 2 1254 MeV. 2(938.3 MeV) 44.17: Section 44.3 says m ( Z 0 ) E 91 . 2 10 eV 0 9 91 . 2 GeV c . 8 1.461 10 J; m E c 2 1 . 63 10 25 kg m (Z ) m ( p) 97 . 2 44.18: a) We shall assume that the kinetic energy of the can set the value of the photon's energy equal to Q. Q (1193 6 0 is negligible. In that case we 1116 ) MeV 18 77 MeV E photon . b) The momentum of this photon is p E photon c (77 10 eV) (1.60 10 8 10 J eV ) 4.1 10 20 kg m s 0 (3.00 m s) To justify our original assumption, we can calculate the kinetic energy of a this value of momentum K p 0 that has 2 E 2 2 (77 MeV) 2 2.7 MeV 0 Q 77 MeV. 2m 2 mc 2(1116 MeV) Thus, we can ignore the momentum of the 44.19: m M( ) 2 without introducing a large error. mp m 0 . Using Table (44.3): 938.3 MeV 2 E ( m )c 1189 MeV 135.0 MeV 116 MeV. 44.20: From Table (44.2), ( m me 2 m v )c 105 . 2 MeV. 44.21: Conservation of lepton number. a) e ve v Lu : 1 1, L e : 0 so lepton numbers are not conserved. e ve v Le : 0 1 1 b) L : 1 1 1 1 so lepton numbers are conserved. e . Lepton numbers are not conserved since just one lepton is c) produced from zero original leptons. p e e Le : 0 1 1, so the lepton numbers are conserved. d) n 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 baryon number 1, and the pions have baryon number 0. d) Conserved: The initial and final baryon numbers are 1+1 = 1+1+0. 44.23: Conservation of strangeness: v . Strangeness is not conserved since there is just one strange a) K particle, in the initial states. b) n K p cannot be conserved. c) K K d) p K 0 0 . Again there is just one strange particle so strangeness 0 0 0 S: 1 1 S :0 1 1 0 , so strangeness is conserved. 0 , so strangeness is conserved. 44.24: a) Using the values of the constants from Appendix F, e 4 2 c 0 7 . 29660475 10 3 1 137 . 050044 , or 1 137 to three figures. b) From Section 38.5 v1 e 2 2 0 h e 2 but notice this is just 4 0c c as claimed rewriting as h 2 . 44.25: f 2 (J m) (J s) (m s ) 1 c 1 and thus f 2 c is dimensionless. (Recall f 2 has units of energy times distance.) 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 it turns out, all the other particles on this lattice had been discovered already and it was this "hole" and mass regularity that led to an accurate prediction of the properties of the ! b) See diagram. Use quark charges u 2 3 1 3 1 3 2 3 ,d 1 3 , and s 1 3 as a guide. 44.27: a) uds : Q e 0; B S C 1 3 0 1 3 0 1 3 ( 1) 1; 1 0. 2 b) cu : 0 0 0 Q 2 0; 0; e B S C 3 1 3 0 1 0 0 3 1 3 0; 1. Q c) ddd: 3 3(0 ) 1 3 0 0 1 3 0; C 2 3 1; B 3(0 ) 3 1 3 0. 1 3 1; e S Q d) 1; B 0 ( 1) 1 3 1. 0; dc : e S 0; C 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 u s . b) The particle has an s antiquark, and for a baryon number of 1 the particle must consist of three antiquarks. For a net charge of e, the quark content must be d d s . c) S 2 means that there are two s quarks, and for baryon number 1 there must be one more quark. For a charge of 0 the third quark must be a u quark and the quark content is uss. 44.29: a) The antiparticle must consist of the antiquarks so: n udd . b) So n udd is not its own antiparticle c c so c c so the is its own antiparticle. c) 5906 MeV 44.30: ( m 2 m ) c 2 ( 9460 MeV 2(1777 MeV)) 44.4 for masses). 0 1 44.31: In decay, 1 p e 0 n ve 1 1 1 1 (see Sections 44.3 and in decay a u quark changes to a d quark. 44.32: a) Using the definition of z from Example 44.9 we have that p uud , n 1 0 udd , so 1 z 1 ( 0 0 s) 0 s . Now we use Eq. 44.13 to obtain 1 z c c v v 1 1 v c v c 1 1 . b) Solving the above equation for (1 (1 z) z) 2 2 we obtain 1 1 .5 1 1 .5 2 2 1 1 0 . 3846 . Thus, v 0 . 3846 c 1 . 15 10 8 m s . c) We can use Eq. 44.15 to find the distance to the given galaxy, r v H0 (1 . 15 ( 7 . 1 10 H 0r 4 10 8 m s) 1 . 6 10 3 Mpc 1 . 04 10 5 ( m s ) Mpc) km s . 44.33: a) v b) 0 s ( 20 ( km s ) Mly )( 5210 Mly ) c c v v 3 . 0 10 km s 3 . 0 10 km s 5 5 1 . 04 1 . 04 10 km s 10 km s 8 5 5 1 . 44 . 44.34: From Eq. (44.15), r c H0 3 . 00 10 m s 1 . 5 10 4 Mly . b) This distance 20 ( km s ) Mly represents looking back in time so far that the light has not been able to reach us. 44.35: a) v ( 0 s ) ( 0 s ) 2 2 1 1 7 c 658 . 5 590 . 0 658 . 5 590 . 0 2 2 1 1 ( 2 . 998 1 0 m s) 8 3 . 280 10 m s. 10 m s 4 7 b) r v H0 3.280 2.0 1640 Mly. 10 m Mly 44.36: Squaring both sides of Eq. (44.13) and multiplying by 2 2 c v gives 0 ( c v ) s ( c v ), and solving this for v gives Eq. (44.14). 44.37: a) m M ( 1 H ) 1 electron masses. m E ( m )c 2 M (1 H ) 2 M ( 2 He ) where atomic masses are used to balance u 4 2 3 1 . 00 7825 u 3 2 2.014102 3 3.16029 u 5.898 10 3 u ( 5 . 89 8 10 u) (931.5 Me V u ) 5.494 MeV . b) m mn M ( He ) M ( He) 3.016029 u 4.002603 u 1 . 008 6649 u 0 . 022 091 u E 4 12 ( m )c 2 ( 0 . 02209 1 u) (931.5 Me V u) 10 3 20.58 MeV. 44.38: 3 m ( He ) m ( C ) 44.39: m m e m p m n m E 0 . 00054 86 u ( m )c m 4 He 2 7 . 80 u, or 7.27 MeV. m ve 0, m v e so assuming 1.007276 10 4 u 1.008665 u 8.40 10 4 u 2 ( 8 .40 7 . 69 8 u) (931.5 Me V u) 3 0.783 MeV and is endoergic. 44.40: m 12 6 C m 16 O 10 u, or 7.16 MeV, an exoergic reaction. 10 3 44.41: For blackbody radiation m T (1 . 062 10 3 2 . 90 7 m K , so m 1 T1 m 2 T2 m1 m) 2.728 K 3000 K 9.66 10 m. 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 ) ( L T) 1 G / c 2 3 are L. 2 2 1/2 E M T 2 L T T 2 3 L J s) (6.673 L 10 11 3 b) G c 3 2 (6.626 10 34 N m kg ) 2 1 2 2 (3.00 10 m s) 8 1.616 10 35 m. 44.43: a) E a 2 ( 7 TeV) 14 TeV b) Fixed target; equal mass particles, Em Ea 2 2 mc 10 11 2 (1.4 10 7 MeV) 2 938.3MeV 2 mc 1.04 2(938.3 MeV) MeV 1.04 10 TeV. 5 44.44: K mpc 2 hc ,K hc mpc 2 6 52 MeV. 44.45: The available energy must be the sum of the final rest masses: (at least) Ea 2mec 2 m 0 c 2 2 ( 0.511 MeV) 136.0 MeV. 135.0 MeV For alike target and beam particles: E m MeV 1 . 81 10 MeV. So K 4 Ea e 2 2 2 2mec mec mec 1 . 81 2 (136.0 MeV) 2(0.511 MeV) 10 4 2 0 . 511 (1 . 81 10 4 MeV ) MeV . 44.46: In Eq.(44.9), Ea (m 0 m K 0 ) c , and with M 2 mp , m 2 2 m and E m (m )c 2 K, K Ea 2 (m c ) 2 2 2 (mpc ) (m 2 )c 2 2mpc (1193 MeV 497.7 MeV) (139.6 MeV) 2 (938.3 MeV) 2 139.6MeV 2(938.3 MeV) 904 MeV. 44.47: The available energy must be at least the sum of the final rest masses. Ea (m 2 0 )c 2 (m K )c 2 2 (m K )c 2 1116 MeV 2 ( 493 . 7 MeV ) 2103 MeV . E a 2 2(m p )c E 2 K (( m p ) c ) 2 (( m K ) c ) . 2 So E K EK Ea 2 (( m p ) c ) 2 2 (( m K ) c ) 2 2 2 2 ( 2103 ) 2 ( 938 . 3 ) 2 ( 493 . 7 ) 2 MeV 2(m p )c 1759 MeV 2 ( 938 . 3 ) K. ( m K )c So the threshold energy K = 1759 MeV 493.7 MeV=1265 MeV. 44.48: a) The decay products must be neutral, so the only possible combinations are 0 0 0 or 0 0 2 b) m 3m 0 142 . 3 Me V c , (m so the kinetic energy of the 0 0 mesons is 142.3 MeV. For the other reaction, K m 0 m m )c 2 133 . 1 MeV . 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 MeV 0.511 MeV = 139.1 MeV. b) Conservation of momentum leads to the neutrinos carrying away most of the energy. 44.50: E (4.4 (1.054 6 10 34 J s) 10 19 1.5 J/eV) 2 10 22 s. 10 eV) (1.6 2 44.51: a) E ( m )c ( m p )c 2 ( m K )c (m K )c 2 1019.4 MeV 32.0 MeV. 2(493.7 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 been produced in addition to the kaons. K or K . c) Conservation of strangeness will not allow 44.52: a) The baryon number is 0, the charge is e , the strangeness is 1, all lepton numbers are zero, and the particle is K . b) The baryon number is 0, the charge is e , the strangeness is 0, all lepton numbers are zero, and the particle is . c) The baryon numbers is 1, the charge is 0, the strangeness is zero, all lepton numbers are 0, and the particle is an antineutron. d) The baryon number is 0, the charge is e , the strangeness is 0, the muonic lepton number is 1, all other lepton numbers are 0, and the particle is . t E m c 2 44.53: t 7 . 6 10 21 s E 5 1 . 054 10 34 21 J s s 1 . 39 10 14 J 87 keV 7 . 6 10 0 . 087 MeV 3097 MeV 2 . 8 10 . 44.54: a) The number of protons in a kilogram is (1 . 00 kg ) 6.023 10 23 molecules 3 mol ( 2 protons molecule) 6.7 10 . 25 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 decay is m p c 2 938 . 3 MeV 1 . 503 10 10 J , and so the energy deposited in a year, per kilogram, is ( 6 . 7 10 25 ) 7 . 0 10 3 ln (2) 1.0 10 18 (1 y ) (1 . 50 y 10 10 J) Gy 0 . 70 rad. b) For an RBE of unity, the equivalent dose is (1) (0.70 rad) = 0.70 rem. 44.55: a) E ( m )c 2 (m )c 2 (m 0 )c 2 (m )c 2 1321 MeV 1116 MeV 139.6 MeV E 65 MeV. 0 b) Using (nonrelativistic) conservation of momentum and energy: P m m 0v 0 0 Pf m 0 v E v 0 v 0. m Also K K 0 from part (a). 2 So K K 1 2 0 m v E m K 1 m 0 0 m 0v 2 0 K 0 1 m m 0 2 m 65 MeV 1116 Me V 139 . 6 MeV 57 . 8 MeV. 7 . 2 MeV 1 m K 65 0 1 7 . 2 MeV So the fractions of energy carried off by the particles are 0.89 for the . dR dt dR dt R dr dt HR R dR dt 7 .2 65 0 . 11 for the 0 and 44.56: a) For this model, all points on the surface. H0 dR dt R . HR , so H , presumed to be the same for HR Hr . b) For constant , dR dt R0e H 0t c) See part (a), d) The equation H 0 R is a differential equation, the solution to , where R 0 which, for constant H 0 is R(t ) is the value of R at t d dt ln ( R ) 0 . This equation may be solved by separation of variables, as dR dt R H 0 and integrating both sides with respect to time. e) A constant H 0 would mean a constant critical density, which is inconsistent with uniform expansion. 44.57: From Pr.(44.56): r So So dR dt 1 dR R dt dv d R R 1 dr dt v d d r . d dt 0. r H 0r. 1 dr dt r d 2 since dr dt dR dt dt 1 dr r dt r dR R dt 1 dr R dt 0 d d 1 dR R dt Now dR dt dR dt H0 K K where K is a constant. R K t since 1 t d dt 1 dR R dt Kt K . 0. So the current value of the Hubble constant is universe. 1 T where T is the present age of the 44.58: a) For mass m, in Eq. ( 37 . 23 ) u v cm , v v 0 , and so v m v0 1 v cm 2 . v 0 v cm c For mass M , u v cm , v 0 , so v M v cm . b) The condition for no net momentum in the center of mass frame is m m v m M M v M 0 , where m and M correspond to the velocities found in part (a). The algebra reduces to m m ( 0 ) 0 M , where v0 0 , v cm c c 0 , and the condition for no net momentum becomes m( M ) 0 M 0 M , or 1 m 0 M m 0 m M 1 2 0 , and v cm m M mv 0 1 (v0 / c ) 2 . c) Substitution of the above expression into the expressions for the velocities found in part (a) gives the relatively simple forms vm v0 0 M m M 0 ,v M v0 0 m m 0 M m 2 mM 0 . 2 . M After some more algebra, m m m m m M 2 0 2 , M 0 2 m 0 2 , from which 0 M 2 mM m 2 M 2 mM MM M This last expression, multiplied by c 2 , is the available energy E a in the center of mass frame, so that Ea 2 (m 2 2 M 2 2 2 2 mM 0 )c 2 2 2 4 2 2 (mc ) ( mc ) 2 (Mc ) 2 ( 2 Mc )(m 0 c ) 2 Mc E m , 2 ( Mc ) which is Eq. (44.9). 44.59: 0 n 0 a) E ( m )c 2 (m 0 )c 2 ( m n )c ( m 2 0 )c 2 1116 MeV 41 . 4 MeV. 939 . 6 MeV 135 . 0 MeV b) Using conservation of momentum and kinetic energy; we know that the momentum of the neutron and pion must have the same magnitude, p n p Kn En mnc 2 2 (m nc ) ( pnc) K K 2 2 2 ( pnc) 2 2 mnc 2 (m nc ) (m nc ) K Kn K 2 2 2 mnc 2 2 2 2 2m c K (m nc ) E 2 2 mnc K 2 2 2m c K 2 2 2 mnc 2 2 2 E. 2 EK 2m nc K . 2 2 ( mnc ) 2 2 2m c K 2 (m n c ) K 2 2 Em c Collecting terms we find : K ( 2 m c 2 K K ( 41 . 4 MeV ) 2 (135 . 0 MeV ) 35 . 62 MeV. 2 2E 2m nc ) E 2 2 Em n c 2 ( 41 . 4 MeV )( 939 . 6 MeV ) 2 ( 41 . 4 MeV ) 2 ( 939 . 6 MeV ) . So the fractional energy carried by the pion is 35 . 62 41 . 4 0 . 86 , and that of the neutron is 0.14.

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1. Each atom has a mass of m = M/NA, where M is the molar mass and NA is the Avogadro constant. The molar mass of arsenic is 74.9 g/mol or 74.9 103 kg/mol. 7.50 1024 arsenic atoms have a total mass of (7.50 1024) (74.9 103 kg/mol)/(6.02 1023 mol
SUNY Buffalo - PHY - 107-207
1. (a) Since the gas is ideal, its pressure p is given in terms of the number of moles n, the volume V, and the temperature T by p = nRT/V. The work done by the gas during the isothermal expansion isW=V2 V1p dV = n RTV2 V1dV V = n RT ln 2 . V
SUNY Buffalo - PHY - 107-207
1. (a) With a understood to mean the magnitude of acceleration, Newton's second and third laws lead to m2 a2 = m1a1c6.3 10 kghc7.0 m s h = 4.9 10 m =-7 2 2-79.0 m s2kg.(b) The magnitude of the (only) force on particle 1 isq q q F = m1
SUNY Buffalo - PHY - 107-207
1. We note that the symbol q2 is used in the problem statement to mean the absolute value of the negative charge which resides on the larger shell. The following sketch is for q1 = q2 .The following two sketches are for the cases q1 > q2 (left figu
SUNY Buffalo - PHY - 107-207
1. The vector area A and the electric field E are shown on the diagram below. The angle between them is 180 35 = 145, so the electric flux through the area is = E A = EA cos = (1800 N C ) 3.2 10-3 m cos145 = -1.5 10-2 N m 2 C.2()2. We u
SUNY Buffalo - PHY - 107-207
1. (a) An Ampere is a Coulomb per second, so84 A h = 84FG HCh sIJ FG 3600 s IJ = 3.0 10 K H hK5C.(b) The change in potential energy is U = qV = (3.0 105 C)(12 V) = 3.6 106 J.2. The magnitude is U = eV = 1.2 109 eV = 1.2 GeV.3. T
SUNY Buffalo - PHY - 107-207
1. Charge flows until the potential difference across the capacitor is the same as the potential difference across the battery. The charge on the capacitor is then q = CV, and this is the same as the total charge that has passed through the battery.
SUNY Buffalo - PHY - 107-207
1. (a) The charge that passes through any cross section is the product of the current and time. Since 4.0 min = (4.0 min)(60 s/min) = 240 s, q = it = (5.0 A)(240 s) = 1.2 103 C. (b) The number of electrons N is given by q = Ne, where e is the magnitu
SUNY Buffalo - PHY - 107-207
1. (a) The cost is (100 W 8.0 h/2.0 W h) ($0.80) = $3.2 102. (b) The cost is (100 W 8.0 h/103 W h) ($0.06) = $0.048 = 4.8 cents.2. The chemical energy of the battery is reduced by E = q, where q is the charge that passes through in time t = 6.
SUNY Buffalo - PHY - 107-207
1. (a) Eq. 28-3 leads to 6.50 10-17 N FB v= = = 4.00 105 m s . -19 -3 eB sin 160 10 C 2.60 10 T sin 23.0 .chch(b) The kinetic energy of the proton isK=2 1 2 1 mv = 167 10-27 kg 4.00 105 m s = 134 10-16 J. . . 2 2chchThis is
SUNY Buffalo - PHY - 107-207
1. (a) The magnitude of the magnetic field due to the current in the wire, at a point a distance r from the wire, is given byB= 0i2r.With r = 20 ft = 6.10 m, we havec4 10 B=hb 2 b6.10 mg-7T m A 100 Ag = 3.3 10-6T = 3.3 T.(b
SUNY Buffalo - PHY - 107-207
1. The amplitude of the induced emf in the loop is m = A 0 ni0 = (6.8 10-6 m 2 )(4 10 -7 T m A)(85400 / m)(1.28 A)(212 rad/s)= 1.98 10-4 V.2. (a) =d B d = 6.0t 2 + 7.0t = 12t + 7.0 = 12 2.0 + 7.0 = 31 mV. dt dtchb g(b) Appealing to
SUNY Buffalo - PHY - 107-207
1. (a) All the energy in the circuit resides in the capacitor when it has its maximum charge. The current is then zero. If Q is the maximum charge on the capacitor, then the total energy is2.90 10-6 C Q2 U= = = 117 10-6 J. . -6 2C 2 3.60 10 Fc
SUNY Buffalo - PHY - 107-207
1. (a) The flux through the top is +(0.30 T)r2 where r = 0.020 m. The flux through the bottom is +0.70 mWb as given in the problem statement. Since the net flux must be zero then the flux through the sides must be negative and exactly cancel the tota
Maryland - ANTH - 220
Subfields of anthropology: cultural, anthropological linguistics, archaeology, biological Subfields of bioanthropology: human biology, primatology, paleoanthropology What do bioanthropologists study: human biology human biodiversity and microevoluti
SUNY Buffalo - PHY - 107-207
1. In air, light travels at roughly c = 3.0 108 m/s. Therefore, for t = 1.0 ns, we have a distance of d = ct = (3.0 108 m / s) (1.0 10-9 s) = 0.30 m.2. (a) From Fig. 33-2 we find the smaller wavelength in question to be about 515 nm, (b) and the
SUNY Buffalo - PHY - 107-207
1. The image is 10 cm behind the mirror and you are 30 cm in front of the mirror. You must focus your eyes for a distance of 10 cm + 30 cm = 40 cm.2. The bird is a distance d2 in front of the mirror; the plane of its image is that same distance d2
Maryland - ANTH - 220
Microevolution: evolution of populations over short periods of time because of observable causes Heterozygous: has different alleles, producing gametes of different genotypes; good because possess a selective advantage (homozygous is opposite) Forces
SUNY Buffalo - PHY - 107-207
1. Comparing the light speeds in sapphire and diamond, we obtainv = vs - vd = cFG 1 - 1 IJ Hn n K F 1 - 1 IJ = 4.55 10 m s. = c2.998 10 m sh G H 177 2.42 K .s d 8 72. (a) The frequency of yellow sodium light is c 2.998 108 m s f = = = 5.09
SUNY Buffalo - PHY - 107-207
1. The condition for a minimum of a single-slit diffraction pattern isa sin = mwhere a is the slit width, is the wavelength, and m is an integer. The angle is measured from the forward direction, so for the situation described in the problem, i
SUNY Buffalo - PHY - 107-207
1. From the time dilation equation t = t0 (where t0 is the proper time interval, = 1 / 1 - 2 , and = v/c), we obtain = 1-FG t IJ . H t K2 0The proper time interval is measured by a clock at rest relative to the muon. Specifically, t0 = 2.2
SUNY Buffalo - PHY - 107-207
1. (a) Let E = 1240 eVnm/min = 0.6 eV to get = 2.1 103 nm = 2.1 m. (b) It is in the infrared region.2. The energy of a photon is given by E = hf, where h is the Planck constant and f is the frequency. The wavelength is related to the frequency b
SUNY Buffalo - PHY - 107-207
1. According to Eq. 39-4 En L 2. As a consequence, the new energy level E'n satisfiesEn L = En LFG IJ = FG L IJ H K H L K-22=1 , 2which gives L = 2 L. Thus, the ratio is L / L = 2 = 1.41.2. (a) The ground-state energy is( 6.63 10
SUNY Buffalo - PHY - 107-207
1. (a) For a given value of the principal quantum number n, the orbital quantum number ranges from 0 to n 1. For n = 3, there are three possible values: 0, 1, and 2. (b) For a given value of , the magnetic quantum number m ranges from - to + . For =
SUNY Buffalo - PHY - 107-207
1. The number of atoms per unit volume is given by n = d / M , where d is the mass density of copper and M is the mass of a single copper atom. Since each atom contributes one conduction electron, n is also the number of conduction electrons per unit
SUNY Buffalo - PHY - 107-207
1. Our calculation is similar to that shown in Sample Problem 42-1. We set K = 5.30 MeV=U = (1/ 4 0 )( q qCu / rmin ) and solve for the closest separation, rmin:rmin-19 9 q qCu kq qCu ( 2e )( 29 ) (1.60 10 C )( 8.99 10 V m/C ) = = = 4 0 K 4 0 K
SUNY Buffalo - PHY - 107-207
1. If R is the fission rate, then the power output is P = RQ, where Q is the energy released in each fission event. Hence, R = P/Q = (1.0 W)/(200 106 eV)(1.60 10 19 J/eV) = 3.1 1010 fissions/s.2. We note that the sum of superscripts (mass number
UCSD - BILD - 3
Mutations 1) Duplications: retrotransposons: carry genes and insert copies of them into genome -moths and transposons: desaturase performs last step in sex attractant molecule synthesis. The transposable element ezi was inserted into the genome into
UCSD - BILD - 3
Allopatric Speciation -occurs when environment is divided into geographically separate regions and organisms in different regions diverge from common ancestor -drosophila species on Hawaiian islands; flies arrived and evolved in different directions
UCSD - BILD - 3
-the largest increase in life expectancy in the last thirty years was in Asia -the majority of population growth in the future will take place in south Asia, the Middle East, and sub-Saharan Africa -age pyramid for Cambodia= bulge in number of young
UCSD - BILD - 3
Darwin's theory 1. Species are mutable: Observed on Galapagos that species change over time and can give rise to new species over time (Lamarck's theory too). Darwin realized that environment determines what kind of organisms live there. 2. Natural s
UCSD - BILD - 3
Fossils -chimpanzee-like braincase (projecting ridges) and a human-like face: pushes our ancestry back, but molecular evidence suggests that it went extinct Hominid Fossils -Australopithecines went extinct in Africa -Australopithecines gave rise to H
University of Texas - LAT - 506Q
REVISED TEST SCHEDULE! DEPARTMENT OF CLASSICS LAT 506Q Accelerated First Year Latin (33525)FALL 2007Time: M-F 10-11Place: WAG 10Instructor: Professor Ingrid E.M. Edlund-Berry Office: WAG 110; office hours:MTWTh 11-11:30; F 11-12 noon, or by a
Maryland - CLAS - 170
Jason abandons medea and his chil. Medea helps Jason on his quest for the Golden Fleece w/ magic. She kills their kids, Creon and Creon daughter. Pelias, wicked half-uncle sends him on quest.boil uncle to dealth.turning him younger/comparison w/ ilia
San Jose State - GEOG - 001
Brian Kim Geography 001 Homework 3 San Francisco Bay/Sacramento-San Joaquin Delta System The Sacramento-San Joaquin Delta is formed at the flowing together of the south-flowing Sacramento River and the north-flowing San Joaquin River. The delta and S
University of Texas - LIN - 306
Pop quiz dated 09/14/08 correct responses1. non-smoker N nonV smoke N -er2. Dog wood treeN N N dog 3. Worldliness N A world -li -ness N wood N tree
San Jose State - GEOG - 001
The Rock Cycle is a group of changes. Igneous rock can change into sedimentary rock or into metamorphic rock. Sedimentary rock can change into metamorphic rock or into igneous rock. Metamorphic rock can change into igneous or sedimentary rock. Igneou
SUNY Buffalo - PHY - 107-207
1. Conservation of momentum requires that the gamma ray particles move in opposite directions with momenta of the same magnitude. Since the magnitude p of the momentum of a gamma ray particle is related to its energy by p = E/c, the particles have th
San Jose State - GEOG - 001
Geog 001Spring 2008 Gary PereiraHomework 3, due May 5: Minimum three pages of text. Write a report on the San Francisco Bay/Sacramento-San Joaquin Delta system. This should go well beyond a general description as found in Wikipedia. You also must
San Jose State - AAS - 033A
American Colonialism o 1770s-1830s: Arrival of American merchants/missionaries o Destruction of Hawaiian customs and values o Establishment f capitalistic values o Decimation of the Hawaiian people o Sought to transplant/displace tribal cultures with
Wisc La Crosse - HIST - 101
University of Texas - BIO - 309D
BIOLOGY 309D: THE HUMAN BODY 50290, 50295, 50300, 50305 MWF 1:00 2:00 BUR 116 Spring 2008Instructor: Dr. Jessica Wandelt Office: PAI 3.10D Office Hours: TTH 9:30-10:30 and F 10:00-11:00 in PAT 207 or by appointment Email: jewandelt@gmail.com Teach