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problem44_41
Course: PHYSICS 750:116, Spring 2009School: Rutgers
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For 44.41: blackbody radiation mT 2.90 10 3 m K , so m1 T1 T2 m2 m1 (1.062 10 3 m) 2.728 K 3000 K 9.66 10 7 m.
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For 44.41: blackbody radiation mT
2.90 10 3 m K , so m1 T1
T2
m2 m1
(1.062 10 3 m)
2.728 K 3000 K
9.66 10 7 m.
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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
Rutgers - PHYSICS - 750:116
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) Th
Rutgers - PHYSICS - 750:116
44.53:E m c 2t7.6 10 21 sE t1.054 10 34 J s 1.39 10 14 J 87 keV 21 7.6 10 s0.087 MeV 3097 MeV2.8 10 5.
Rutgers - PHYSICS - 750:116
44.55: a) E( m)c 2(m )c 2( m 0 )c 2(m )c 21321MeV 1116 MeV 139.6 MeV E 65 MeV. b) Using (nonrelativistic) conservation of momentum and energy: P 0m 0v00Pfm0vvm m0v 0.Also KK0E from part (a).1 m v2 2 K0So K1
Rutgers - PHYSICS - 750:116
44.57: From Pr.(44.56): r R R Sor.dR 1 dr r d 1 dr d 2 since 0. dt dt dt dt dt 1 dR 1 dr 1 dr dr 1 dR v So r H 0 r. R dt R dt r dt dt R dt dv d r dR d dR 0 Now d d R dt d dt dR K where K is a constant. dt dR K
Rutgers - PHYSICS - 750:116
v0 vcm . 1 v0 vcm c 2 vcm , v 0, so v M vcm . b) The condition for no net momentum in the For mass M , u center of mass frame is mm vm MM vM 0, where m and M correspond to the velocities )0 M , where found in part (a). The algebra reduces to m m ( 0
Rutgers - PHYSICS - 750:116
44.59: 0 n 0 a) E (m)c 2 (m0 )c 2 (mn )c 2 (m 0 )c 2 1116 MeV 939.6 MeV 135.0 MeV 41.4 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 pK n E
Rutgers - PHYSICS - 750:116
2.54 cm in . 1 km 10 5 cm 1.61 km 1.1: 1 mi 5280 ft mi 12 in. ft Although rounded to three figures, this conversion is exact because the given conversion from inches to centimeters defines the inch.
Rutgers - PHYSICS - 750:116
1.3: The time required for light to travel any distance in a vacuum is the distance divided by the speed of light; 103 m 3.33 10 6 s 3.33 103 ns. 8 3.00 10 m s
Rutgers - PHYSICS - 750:116
1.6:1 m31000 L 1 m3 2111.9 bottles1 gal 128 oz. 3.788 L 1 gal 2112 bottles1 bottle . 16 oz.The daily consumption must then be 1 yr bottles 2.11 103 yr 365.24 da5.78bottles . da
Rutgers - PHYSICS - 750:116
1.7:1450 mi hr 1.61 km mi 2330 km hr . 2330 km hr 10 3 m km 1 hr 3600 s 648 m s.
Rutgers - PHYSICS - 750:116
1.8:180,000furlongs fortnight1 mile 8 furlongs1 fortnight 14 day1 day 24 h67mi . h
Rutgers - PHYSICS - 750:116
1.10: a) 60mi hr ft s21h 3600 s 30.48 cm 1ft5280 ft 1 mi 1m 100 cm88ft s m s2b) 329.8g c) 1.0 3 cm100 cm 1m31 kg 1000 g103kg m3
Rutgers - PHYSICS - 750:116
1.11: The density is mass per unit volume, so the volume is mass divided by density. V 60 10 3 g 19 .5 g cm3 3077 cm3 4 3 Use the formula for the volume of a sphere, V r , 3 1/ 3 9.0 cm to calculate r : r 3V 4
Rutgers - PHYSICS - 750:116
1.13: a)10 m 1.1 10 3%. 3 890 10 mb) Since the distance was given as 890 km, the total distance should be 890,000 meters. To report the total distance as 890,010 meters, the distance should be given as 890.01 km.
Rutgers - PHYSICS - 750:116
5.98 mm 72 mm 2 (two significant figures). 1.14: a) 12 mm .98 m m b) 512 m m = 0.50 (also two significant figures). c) 36 mm (to the nearest millimeter). d) 6 mm. e) 2.0.
Rutgers - PHYSICS - 750:116
1.15: a) If a meter stick can measure to the nearest millimeter, the error will be about 0.13%.b) If the chemical balance can measure to the nearest milligram, the error will be about 8.3 10 3%. c) If a handheld stopwatch (as opposed to electric timi
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1.16: The area is 9.69 0.07 cm2, where the extreme values in the piece's length and width are used to find the uncertainty in the area. The fractional uncertainty in the cm2 area is 0.07 cm2 = 0.72%, and the fractional uncertainties in the length and
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