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### problem44_12

Course: PHYS 218, Spring 2008
School: Texas A&M
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eBR eB 3.97 107 s. b) R 3.12 107 m s. c) For threem m figure precision, the relativistic form of the kinetic must energy be used, ( 1 )mc 2 eV ( 1 )mc 2 , so eV ( 1 )mc 2 , so V 5.11 106 V. e 44.12: a) 2 f

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Coursehero >> Texas >> Texas A&M >> PHYS 218

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eBR eB 3.97 107 s. b) R 3.12 107 m s. c) For threem m figure precision, the relativistic form of the kinetic must energy be used, ( 1 )mc 2 eV ( 1 )mc 2 , so eV ( 1 )mc 2 , so V 5.11 106 V. e 44.12: a) 2 f
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Texas A&M - PHYS - 218
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
Texas A&M - PHYS - 218
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)
Texas A&M - PHYS - 218
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
Texas A&M - PHYS - 218
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.
Texas A&M - PHYS - 218
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)
Texas A&M - PHYS - 218
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)
Texas A&M - PHYS - 218
44.20: From Table (44.2), (mme2mv )c 2105.2 MeV.
Texas A&M - PHYS - 218
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.
Texas A&M - PHYS - 218
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
Texas A&M - PHYS - 218
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
Texas A&M - PHYS - 218
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
Texas A&M - PHYS - 218
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
Texas A&M - PHYS - 218
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
Texas A&M - PHYS - 218
44.26: a)The particle has Q 1 (as its label suggests) and S 3. Its appears as a &quot;hole&quot;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
Texas A&M - PHYS - 218
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;
Texas A&M - MEEN - 221
Texas A&M - MEEN - 221
Texas A&M - MEEN - 221
Texas A&M - MEEN - 221
Texas A&M - MEEN - 221
Iowa State - EE - 442/448
Allan R. Hambley, Electrical Engineering: Principles and Applications, Third Edition, ISBN 0-13-147046-9 2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
Iowa State - EE - 442/448
Allan R. Hambley, Electrical Engineering: Principles and Applications, Third Edition, ISBN 0-13-147046-9 2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
Iowa State - EE - 442/448
Allan R. Hambley, Electrical Engineering: Principles and Applications, Third Edition, ISBN 0-13-147046-9 2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
Iowa State - EE - 442/448
Allan R. Hambley, Electrical Engineering: Principles and Applications, Third Edition, ISBN 0-13-147046-9 2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
Iowa State - EE - 442/448
Allan R. Hambley, Electrical Engineering: Principles and Applications, Third Edition, ISBN 0-13-147046-9 2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
Iowa State - EE - 442/448
Allan R. Hambley, Electrical Engineering: Principles and Applications, Third Edition, ISBN 0-13-147046-9 2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
Texas A&M - CVEN - 305
Texas A&M - CVEN - 305
Cincinnati - ENGIN - 375
Cincinnati - ENGIN - 375
Cincinnati - ENGIN - 375
Cincinnati - ENGIN - 375
Cincinnati - ENGIN - 375
Cincinnati - ENGIN - 375
Cincinnati - ENGIN - 375
Wisconsin - MATH - 222
CHAPTER 1 PRELIMINARIES1.1 REAL NUMBERS AND THE REAL LINE 1. Executing long division, 2. Executing long division,&quot; 9 &quot; 11oe 0.1,2 9oe 0.2,2 113 9oe 0.3,3 118 9oe 0.8,9 119 9oe 0.911 11oe 0.09,oe 0.18,oe 0.27,oe 0.81,
Wisconsin - MATH - 222
CHAPTER 2 LIMITS AND CONTINUITY2.1 RATES OF CHANGE AND LIMITS 1. (a) Does not exist. As x approaches 1 from the right, g(x) approaches 0. As x approaches 1 from the left, g(x) approaches 1. There is no single number L that all the values g(x) get ar
Wisconsin - MATH - 222
CHAPTER 3 DIFFERENTIATION3.1 THE DERIVATIVE OF A FUNCTION 1. Step 1: f(x) oe 4 c x# and f(x b h) oe 4 c (x b h)# Step 2: oe c2x c h Step 3: f w (x) oe lim (c2x c h) oe c2x; f w (c\$) oe 6, f w (0) oe 0, f w (1) oe c2h!# # # # # \$ # # # #f(x b h)
Wisconsin - MATH - 222
CHAPTER 4 APPLICATIONS OF DERIVATIVES4.1 EXTREME VALUES OF FUNCTIONS 1. An absolute minimum at x oe c# , an absolute maximum at x oe b. Theorem 1 guarantees the existence of such extreme values because h is continuous on [a b]. 2. An absolute minimu
Wisconsin - MATH - 222
CHAPTER 5 INTEGRATION5.1 ESTIMATING WITH FINITE SUMS 1. faxb oe x# Since f is increasing on ! &quot;, we use left endpoints to obtain lower sums and right endpoints to obtain upper sums.(a) ~x oe (b) ~x oe (c) ~x oe (d) ~x oe 2. faxb oe x\$&quot;c! # &quot;c! %
Texas A&M - CVEN - 306
Cornell - MATH - 2940
ISM: Linear AlgebraSection 1.1Chapter 1 1.11. x + 2y x + 2y = 1 -y 2x + 3y = 1 -2 1st equation x + 2y = 1 -2 2nd equation x y=1 y 2. 4x + 3y 7x + 5y3 x + 4y 1 -4y 3 x + 4y = 2 4 =3 7x + 5y=1 = -1 (-1)= -1 , so that (x, y) = (-1, 1). =
Carnegie Mellon - MATH - 21-260
- CHAPTER 1. -Chapter OneSection 1.1 1.For C &quot;&amp; , the slopes are negative, and hence the solutions decrease. For C &quot;&amp; , the slopes are positive, and hence the solutions increase. The equilibrium solution appears to be Ca&gt;b oe &quot;&amp; , to which all
Carnegie Mellon - MATH - 21-260
- CHAPTER 2. -Chapter TwoSection 2.1 1a+ba,b Based on the direction field, all solutions seem to converge to a specific increasing function. a- b The integrating factor is .a&gt;b oe /\$&gt; , and hence Ca&gt;b oe &gt;\$ &quot;* /#&gt; - /\$&gt; It follows that all s
Carnegie Mellon - MATH - 21-260
- CHAPTER 3. -Chapter ThreeSection 3.1 1. Let C oe /&lt;&gt; , so that C w oe &lt; /&lt;&gt; and C ww oe &lt; /&lt;&gt; . Direct substitution into the differential equation yields a&lt;# #&lt; \$b/&lt;&gt; oe ! . Canceling the exponential, the characteristic equation is &lt;# #&lt; \$ o
Carnegie Mellon - MATH - 21-260
- CHAPTER 4. -Chapter FourSection 4.1 1. The differential equation is in standard form. Its coefficients, as well as the function 1a&gt;b oe &gt; , are continuous everywhere. Hence solutions are valid on the entire real line. 3. Writing the equation in
Carnegie Mellon - MATH - 21-260
- CHAPTER 5. -Chapter FiveSection 5.1 1. Apply the ratio test : lim aB \$b8&quot; k a B \$b 8 kHence the series converges absolutely for kB \$k &quot; . The radius of convergence is 3 oe &quot; . The series diverges for B oe # and B oe % , since the n-th ter
Carnegie Mellon - MATH - 21-260
- CHAPTER 6. -Chapter SixSection 6.1 3.The function 0 a&gt;b is continuous. 4.The function 0 a&gt;b has a jump discontinuity at &gt; oe &quot; . 7. Integration is a linear operation. It follows that (E !-9=2 ,&gt; /=&gt; .&gt; oe&quot; E ,&gt; =&gt; &quot; E ,&gt; =&gt; ( / / .&gt;
Carnegie Mellon - MATH - 21-260
- CHAPTER 7. -Chapter SevenSection 7.1 1. Introduce the variables B&quot; oe ? and B# oe ? w . It follows that B&quot;w oe B# and B#w oe ? ww oe #? !&amp; ? w . In terms of the new variables, we obtain the system of two first order ODEs B&quot;w oe B# B#w oe #B&quot;
Carnegie Mellon - MATH - 21-260
- CHAPTER 8. -Chapter EightSection 8.1 2. The Euler formula for this problem is C8&quot; oe C8 2^&amp; &gt;8 \$C8 , C8&quot; oe C8 &amp;82# \$2 C8 ,in which &gt;8 oe &gt;! 82 Since &gt;! oe ! , we can also writea+b. Euler method with 2 oe !&amp; &gt;8 C8 8oe# !&quot; &quot;&amp;*)! 8oe% !
Carnegie Mellon - MATH - 21-260
- CHAPTER 9. -Chapter NineSection 9.1 2a+b Setting x oe 0 /&lt;&gt; results in the algebraic equations OE &amp;&lt; \$For a nonzero solution, we must have ./&gt;aA &lt; Ib oe &lt;# ' &lt; ) oe ! . The roots of the characteristic equation are &lt;&quot; oe # and &lt;# oe % . For
Carnegie Mellon - MATH - 21-260
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.
Carnegie Mellon - MATH - 21-260
1.2:0.473 L1000 cm3 1L1in 2.54 cm328.9 in 3 .
Carnegie Mellon - MATH - 21-260
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
Carnegie Mellon - MATH - 21-260
1.4:g 11.3 cm31 kg 1000 g100 cm 1m31.13 10 4kg . m3
Carnegie Mellon - MATH - 21-260
1.5:327 in 32.54 cm in31 L 1000 cm 35.36 L.
Carnegie Mellon - MATH - 21-260
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
Carnegie Mellon - MATH - 21-260
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.
Carnegie Mellon - MATH - 21-260
1.8:180,000furlongs fortnight1 mile 8 furlongs1 fortnight 14 day1 day 24 h67mi . h
Carnegie Mellon - MATH - 21-260
1.9:15.0km L1 mi 1.609 km3.788 L 1 gal35.3mi . gal
Syracuse - CEN - 231
CHAPTER TWO2.1 (a)= 18144 10 9 ms . 1 wk 1 d 1 h 1 s 38.1 ft / s 0.0006214 mi 3600 s (b) = 25.98 mi / h 26.0 mi / h 3.2808 ft 1 h 3 wk 7d 24 h 3600 s 1000 ms(c)554 m 4 1d 1h d kg 24 h 60 min1 kg 108 cm 4 = 3.85 10 4 cm 4 / min g 1000 g 1