Solution manual for modern physics 6th edition by tipler

Solution Manual for Modern Physics 6th Edition by Tipler Full file at https://TestbankDirect.eu/Solution-Manual-for-Modern-Physics-6th-Edition-by-Tipler Chapter – Relativity I 1-1 (a) Speed of the droid relative to Hoth, according to Galilean relativity, uHoth , is uHoth  uspaceship  udroid  2.3 108 m / s  2.1108 m / s  4.4 108 m / s (b) No, since the droid is moving faster than light speed relative to Hoth 1-2 L  2.74 10 m  (a) t    1.83 104 s c 3.00 10 m / s (b) From Equation 1-6 the correction  t  2L v  c c2  t  1.83 104 s 104   1.83 1012 s c km / s  1.3 105 c 299, 796 km / s No, the relativistic correction of order 10-8 is three orders of magnitude smaller than (c) From experimental measurements  the experimental uncertainty 0.4 fringe  1.0 fringe  v2  1.0  29.9 km / s   2.22 103  v  47.1 km / s 0.4 1-3  29.8km / s  1-4 (a) This is an exact analog of Example 1-1 with L = 12.5 m, c = 130 mph, and v = 20  v km / s  mph Calling the plane flying perpendicular to the wind plane #1 and the one flying parallel to the wind plane #2, plane #1 win will by Δt where Lv 12.5mi  20mi / h  t    0.0023 h  8.2s c 130mi / h  (b) Pilot #1 must use a heading   sin 1  20 /130   8.8 relative to his course on both legs Pilot #2 must use a heading of 0 relative to the course on both legs Full file at https://TestbankDirect.eu/Solution-Manual-for-Modern-Physics-6th-Edition-by-Tipler Solution Manual for Modern Physics 6th Edition by Tipler Full file Chapter at https://TestbankDirect.eu/Solution-Manual-for-Modern-Physics-6th-Edition-by-Tipler – Relativity I 1-5 (a) In this case, the situation is analogous to Example 1-1 with L  108 m, v  104 m / s, and c  108 m / s If the flash occurs at t = 0, the interior is dark until t =2s at which time a bright circle of light reflected from the circumference of the great circle plane perpendicular to the direction of motion reaches the center, the circle splits in two, one moving toward the front and the other moving toward the rear, their radii decreasing to just a point when they reach the axis 10-8 s after arrival of the first reflected light ring Then the interior is dark again (b) In the frame of the seated observer, the spherical wave expands outward at c in all directions The interior is dark until t = 2s at which time the spherical wave (that reflected from the inner surface at t = 1s) returns to the center showing the entire inner surface of the sphere in reflected light, following which the interior is dark again 1-6 Yes, you will see your image and it will look as it does now The reason is the second postulate: All observers have the same light speed In particular, you and the mirror are in the same frame Light reflects from you to the mirror at speed c relative to you and the mirror and reflects from the mirror back to you also at speed c, independent of your motion 1-7 N  Lv (Equation 1-10) Where λ = 590 nm, L = 11 m, and ΔN = 0.01 fringe c2 v2  N  c   0.01 fringe   590 109 m  3.00 108 m / s  11m  2L v  4.91103 m / s  km / s 1-8 (a) No Results depends on the relative motion of the frames (b) No Results will depend on the speed of the proton relative to the frames (This answer anticipates a discussion in Chapter If by “mass”, the “rest mass” is implied, then the answer is “yes”, because that is a fundamental property of protons.) Full file at https://TestbankDirect.eu/Solution-Manual-for-Modern-Physics-6th-Edition-by-Tipler Solution Manual for Modern Physics 6th Edition by Tipler Full file at https://TestbankDirect.eu/Solution-Manual-for-Modern-Physics-6th-Edition-by-Tipler Chapter – Relativity I (Problem 1-8 continued) (c) Yes This is guaranteed by the 2nd postulate (d) No The result depends on the relative motion of the frames (e) No The result depends on the speeds involved (f) Yes Result is independent of motion (g) Yes The charge is an intrinsic property of the electron, a fundamental constant 1-9 The wave from the front travels 500 m at speed c + (150/3.6) m/s and the wave from the rear travels at c – (150/3.6) m/s As seen in Figure 1-14, the travel time is longer for the wave from the rear t  tr  t f  500m 500m  3.00 10 m / s  150 / 3.6  m / s 3.00  10 m / s  150 / 3.6  m / s  108  150 / 3.6  108  150 / 3.6         500   108   150 / 3.6   108   150 / 3.6 2     500 150 / 3.6   10   4.63  1013 s * 1-10 A * B * v C While the wavefront is expanding to the position shown, the original positions of A, B, and C have moved to the * marks, according to the observer in S (a) According to an S  observer, the wavefronts arrive simultaneously at A and B (b) According to an S observer, the wavefronts not arrive at A and C simultaneously (c) The wavefront arrives at A first, according to the S observer, an amount Δt before arrival at C  , where Full file at https://TestbankDirect.eu/Solution-Manual-for-Modern-Physics-6th-Edition-by-Tipler Solution Manual for Modern Physics 6th Edition by Tipler Full file Chapter at https://TestbankDirect.eu/Solution-Manual-for-Modern-Physics-6th-Edition-by-Tipler – Relativity I (Problem 1-10 continued) t  BC  BA  cv cv since BC   BA  L, Thus c  v  c  v   2v  t  L   L  2  2   c v  c  v  1-11 β   1/ 1    1/ 0.2 0.4 0.6 0.8 0.85 0.90 0.925 0.950 0.975 0.985 0.990 0.995 1-12 vx   t1    t1  20  c   1.0206 1.0911 1.2500 1.6667 1.8983 2.2942 2.6318 3.2026 4.5004 5.7953 7.0888 10.0125 vx   t2    t2  20  c   ( from Equation 1-19) vx vx   (a) t2  t1    t2  20  t1  20     t2  t1  c c   (b) The quantities x1 and x2 in Equation 1-19 are each equal to x0 , but x1 and x2 in Equation 1-18 are different and unknown 1-13 (a)   1/ 1  v / c  1/  1/ 1   0.85c  / c    1/  1.898 x    x  vt   1.898 75m   0.85c   2.0  105 s    9.537  103 m y  y  18m z  z  4.0m t     t  vx / c   1.898  2.0 105 s   0.85c  75m  / c   3.756  105 s Full file at https://TestbankDirect.eu/Solution-Manual-for-Modern-Physics-6th-Edition-by-Tipler Solution Manual for Modern Physics 6th Edition by Tipler Full file at https://TestbankDirect.eu/Solution-Manual-for-Modern-Physics-6th-Edition-by-Tipler Chapter – Relativity I (Problem 1-13 continued) (b) x    x  vt    1.898  9.537 103 m   0.85c   3.756 10 5 s    75.8m difference is due to rounding of  , x, and t  y  y  18m z  z  4.0m t    t   vx / c   1.898 3.756 105 s   0.85c   9.537 103 m  / c   2.0 10 5 s 1-14 To show that Δt = (refer to Figure 1-8 and Example 1-1) t1  L c v 2  L c v 2  2L c v 2  2L c  v2 / c2 t2 , because length parallel to motion is shortened, is given by: t2  L  v / c L  v / c 2 Lc  v / c   cv cv c 1  v / c  t2  2L c   v2 / c2  v2 / c2   2L  t1 c  v2 / c2 Therefore, t2 – t1 = and no fringe shift is expected 1-15 (a) Let frame S be the rest frame of Earth and frame S  be the spaceship moving at speed v to the right relative to Earth The other spaceship moving to the left relative to Earth at speed u is the “particle” Then v = 0.9c and ux = −0.9c u x   ux  v  ux v / c (Equation 1-22) 0.9c  0.9c 1.8c   0.9945c 1.81   0.9c  0.9c  / c (b) Calculating as above with v  3.0 104 m / s  ux ux  1 3.0  104 m / s  3.0  104 m / s 6.0  104 m / s   6.0  104 m / s 8 4  10  3.0  10 m / s  3.0  10 m / s   3.0 10 m / s  Full file at https://TestbankDirect.eu/Solution-Manual-for-Modern-Physics-6th-Edition-by-Tipler Solution Manual for Modern Physics 6th Edition by Tipler Full file Chapter at https://TestbankDirect.eu/Solution-Manual-for-Modern-Physics-6th-Edition-by-Tipler – Relativity I 1-16 ax  dux dt  where u x  And t     t  vx / c  ux  v  ux v / c (Equation 1-22) (Equation 1-18) dux   ux  v   vdu x / c 1  u x v / c   1  u x v / c  du x 2 1  v     ux  v  du x  1  u x v / c  du x c   1  ux v / c2  dt     dt  vdx / c   v  u  v  du x / dt   1  u x v / c   du x / dt    x  du c ax  x    dt   1  u v / c  x  dux / dt  1  v / c    1  u x v / c ay  du y dt    where u y  ax  1  u x v / c  uy (Equation 1-22)  1  ux v / c  du y   du y /   1  u x v / c    u y /   1  u x v / c  du x 1 2  du  1  u v / c    u v / c  du   1  u v / c  y x y x 2 x 2 duy  du y / dt  1  u x v / c    u y v / c   du x / dt  ay   dt   1  u v / c   1  u v / c  x  x a y 1  u x v / c   ax  u y v / c   1  u x v / c  az is found in the same manner and is given by: az  az 1  ux v / c   ax  u z v / c   1  ux v / c  Full file at https://TestbankDirect.eu/Solution-Manual-for-Modern-Physics-6th-Edition-by-Tipler Solution Manual for Modern Physics 6th Edition by Tipler Full file at https://TestbankDirect.eu/Solution-Manual-for-Modern-Physics-6th-Edition-by-Tipler Chapter – Relativity I 1-17 (a) As seen from the diagram, when the observer in the rocket ( S  ) system sees c∙s tick by on the rocket’s clock, only 0.6 c∙s have ticked by on the laboratory clock ct ct  x 1 0 x (b) When 10 seconds have passed on the rocket’s clock, only seconds have passed on the laboratory clock 1-18 (a) y u x  u y  c y x v x ux  uy  (b) u x  v 0v  v 1  vu x / c u y  1  vu x / c   (Equation 1-23) c c   1    u  ux2  u y2  v  c /   v  c 1  v / c   c Full file at https://TestbankDirect.eu/Solution-Manual-for-Modern-Physics-6th-Edition-by-Tipler Solution Manual for Modern Physics 6th Edition by Tipler Full file Chapter at https://TestbankDirect.eu/Solution-Manual-for-Modern-Physics-6th-Edition-by-Tipler – Relativity I 1-19 By analogy with Equation 1-23, 1-20 (a) ux  ux  v 0.9c  0.9c 1.8c    0.9945c 2  vux / c   0.9c  0.9c  / c 1.81 (b) ux  1.8c /1.81  0.9c  1.8   0.9 1.81 c  3.429 c  0.9997c ux  v   vux / c  1.8c /1.81 0.9c  / c 1.81  1.8  0.9  3.430  (a) (Equation 1-16)  v2 / c2  1  v / c 2   v  1       c  1/ v2 v4  1   c2 c2 (b)       v                 2!  c  v2  1 c2   v / c  1  v / c  1/ 2 2    v      v                     c     2!  c   1 (c) v2 v4   c2 c2  1  t  t  v2 c2 v2 v4   c2 c4   1  1 1-21  1   1   v2 v4  c2 c4 v2 c2 (Equation 1-26) t  t  t   t  v2 v2    1      t  t  c2 c2 t  t  v  2c t  2 t  t    v  c 2  t    1/  c   0.01 1/  0.14c Full file at https://TestbankDirect.eu/Solution-Manual-for-Modern-Physics-6th-Edition-by-Tipler Solution Manual for Modern Physics 6th Edition by Tipler Full file at https://TestbankDirect.eu/Solution-Manual-for-Modern-Physics-6th-Edition-by-Tipler Chapter – Relativity I S 1-22 Orion (#2) v = 1000 km/h Lyra (#1) S Earth 2.5 103 c y 2.5 103 c y (a) Note that   1/  v / c  and c y  c  3.15 107 s  From Equation 1-27: t  t2  t1  , since the novas are simultaneous in system S (Earth) Therefore, in S  (the aircraft) v t   t2  t1    x2  x1  c 10 m / h  2.5 103  2.5 103   c   3.15 107 s    3600s / h  c  1.46 105 s  40.5h (b) Since t  is positive, t2  t1 ; therefore, the nova in Lyra is detected on the aircraft before the nova in Orion 1-23 (a) L  Lp /  (Equation 1-28) =Lp  v / c  1.0m 1   0.6c  / c    1/  0.80m (b) t  L / v  0.80m / 0.6c  4.4 109 s (c) The projection OA on the x axis is L The length OB on the ct axis yields t ct  ct back of meterstick passes x = x B meterstick AA x Full file at https://TestbankDirect.eu/Solution-Manual-for-Modern-Physics-6th-Edition-by-Tipler Solution Manual for Modern Physics 6th Edition by Tipler Full file Chapter at https://TestbankDirect.eu/Solution-Manual-for-Modern-Physics-6th-Edition-by-Tipler – Relativity I t  t  t   1-24 (a)  (Equation 1-26)  v2 / c2 2.6  108 s 1   0.9c  / c    1/  2.6 108 s 0.19  5.96  108 s (b) s  vt   0.9   3.0 108 m / s  6.0 108 s   16.1m (c) s  vt   0.9   3.0 108 m / s  2.6 108 s   7.0m (d)  s    ct    x  2 (Equation 1-31)  c  6.0  108    16.1m   324  259  65  s  7.8m 1-25 From Equation 1-28, L  Lp /   Lp  v / c where L  85m and Lp  100m  v / c  L / Lp  85 /100 Squaring  v / c  85 /100  2  v  1   85 /100   c  0.2775c and v  0.527c  1.58  108 m / s   1-26 (a) In the spaceship the length L = the proper length Lp; therefore, ts  Lp c  Lp c  Lp c (b) In the laboratory frame the length is contracted to L  Lp /  and the round trip time is tL   L L L   c  v c  v c 1  v / c  Lp /  c 1  v / c   Lp  v / c c   v2 / c2   Lp c  v2 / c2 (c) Yes The time ts measured in the spaceship is the proper time interval τ From time dilation (Equation 1-26) the time interval in the laboratory tL   ; therefore, tL  Lp  v2 / c2 c which agrees with (b) 10 Full file at https://TestbankDirect.eu/Solution-Manual-for-Modern-Physics-6th-Edition-by-Tipler Solution Manual for Modern Physics 6th Edition by Tipler Full file Chapter at https://TestbankDirect.eu/Solution-Manual-for-Modern-Physics-6th-Edition-by-Tipler – Relativity I 1-39 y y S (other) c S  (Earth) v θ x x (a) cx  c cos   v c y  c sin  tan    c y cx  c sin  sin   c cos   v cos   v / c (b) If   90 , tan    1  1 v/c   90     45 e.g., if v  0.5c,    63 1-40 (a) Time t for information to reach front of rod is given by: ct  Lp   vt  t  Lp  (c  v ) Distance information travels in time t: c (c  v)(c  v) / c c  v2 / c2 cv ct   Lp  Lp  Lp  (c  v ) (c  v ) cv (c  v ) cLp Since (c  v) (c  v)  for v > 1, the distance the information must travel to reach the front of the rod is  Lp ; therefore, the rod has extended beyond its proper length (b)  1 v / c (ct  Lp )  1 Lp 1 v / c 16 Full file at https://TestbankDirect.eu/Solution-Manual-for-Modern-Physics-6th-Edition-by-Tipler Solution Manual for Modern Physics 6th Edition by Tipler Full file at https://TestbankDirect.eu/Solution-Manual-for-Modern-Physics-6th-Edition-by-Tipler Chapter – Relativity I Δ 0.11 0.29 0.57 0.73 1.17 2.00 2.50 3.36 5.25 8.95 v/c 0.10 0.25 0.40 0.50 0.65 0.80 0.85 0.90 0.95 0.98 Coefficient of extension vs v/c 10 Delta 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 v/c (c) As v  c,    , the maximum length of the rod   also 1-41 (a) Alpha Centauri is c∙y away, so the traveler went L    8c  y  in y, or 8c  y  v / c  v  y   v / c  v  / 8c    /  v / c      3 / 4  2 3 / 4       1/ 1  0.5625  v  0.8c (b) t  t0    y  and   1/    1.667 t  1.667  y   10 y or 4y older than the other traveler 17 Full file at https://TestbankDirect.eu/Solution-Manual-for-Modern-Physics-6th-Edition-by-Tipler Solution Manual for Modern Physics 6th Edition by Tipler Full file Chapter at https://TestbankDirect.eu/Solution-Manual-for-Modern-Physics-6th-Edition-by-Tipler – Relativity I (Problem 1-41 continued) (c) ct  10 ct (c∙y) x Alpha Centauri Earth 10 x 1-42 Orbit circumference  4.0 107 m Satellite speed v  4.0 107 m /  90  60s /   7.41103 m / s t  t0  tdiff 1  t  t /   tdiff  t 1  /    t    (Problem 1-20) 2  tdiff   3.16  107 s  1 /   7.41  103 / 3.0  108   0.0096s  9.6ms 1-43 (a) t   t  v c2 x (Equation 1-20) For events to be simultaneous in S , t   v v 1.5 103 m   c c 6 v   10 s  10 m / s  s /1.5 103 m t  x  106 s   1.2 108 m / s  0.4c (b) Yes 18 Full file at https://TestbankDirect.eu/Solution-Manual-for-Modern-Physics-6th-Edition-by-Tipler Solution Manual for Modern Physics 6th Edition by Tipler Full file at https://TestbankDirect.eu/Solution-Manual-for-Modern-Physics-6th-Edition-by-Tipler Chapter – Relativity I (Problem 1-43 continued) ct  ct (c) 1000 x 500 B 500 1000 1500 x A (d)  s    x    ct  2 (Equation 1-33)  1.5  103     108 m / s   106 s    2.25  106  3.6  105  1.89  106 m2 s  1370m L  s  1370m 2 1-44 (a)   1/    1/   0.92   2.55 (b)   2.6 108 s t  lab     2.55  2.6 108 s   6.63 108 s (c) N  t   N0 et /  (Equation 1-29) L    L0    0.92   50m   19.6m Where L is the distance in the pion system At 0.92c, the time to cover 19.6m is: t  19.6m / 0.92c  7.0 108 s So, for N0 = 50,000 pions initially, at the end of 50m in the lab, N   5.0 104  e7.0/ 2.6  3,390 (d) 47 19 Full file at https://TestbankDirect.eu/Solution-Manual-for-Modern-Physics-6th-Edition-by-Tipler Solution Manual for Modern Physics 6th Edition by Tipler Full file Chapter at https://TestbankDirect.eu/Solution-Manual-for-Modern-Physics-6th-Edition-by-Tipler – Relativity I 1-45  v2  L  Lp  L  Lp  Lp 1/    Lp 1  1/    Lp   (See Problem 1-20) 2c  For Lp  11m and v  104 m / s L  11 0.5 108   5.5 108 m "Shrinkage"  5.5 108 m  550 atomic diameter 1010 m / atomic diameter 1-46 (a) ct  ct 1500 B 1000 500 A 0 500 1000 1500 2000 x (b) Slope of ct  axis  2.08  1/  , so   0.48 and v  1.44 108 m / s (c) ct    ct and   1/   so ct     ct ct  877m t   1.5  877  / c  4.39 s For ct   1000m and   0.48 (d) t  t   1.14t   t   5 s /1.14  4.39 s 1-47 (a) L  Lp /   Lp  u / c  100m   0.85  52.7m (b) u  u u 0.85c  0.85c 1.70c    0.987c 2  uu / c 1.72   0.85 (c) L  Lp /    Lp  u2 / c2  100m  1.70 /1.72   16.1m (d) As viewed from Earth, the ships pass in the time required for one ship to move its L 52.7m  2.1107 s own contracted length t   u 0.85  3.00 10 m / s 20 Full file at https://TestbankDirect.eu/Solution-Manual-for-Modern-Physics-6th-Edition-by-Tipler Solution Manual for Modern Physics 6th Edition by Tipler Full file at https://TestbankDirect.eu/Solution-Manual-for-Modern-Physics-6th-Edition-by-Tipler Chapter – Relativity I (Problem 1-47 continued) 0.987c (e) 100 m 16 m 1-48 In Doppler radar, the frequency received at the (approaching) aircraft is shifted by approximately f / f0  v / c Another frequency shift in the same direction occurs at the receiver, so the total shift f / f0  2v / c v   c /  8 107   120m / s 1-49 #1 #2 For star #1: v  32km / s  3.2 104 m / s cm period  115d L c  c  3.2 104 c+ c− Earth c  c  3.2  104 Simultaneous images of star#1 in opposition will appear at Earth when L is at least as large as: 21 Full file at https://TestbankDirect.eu/Solution-Manual-for-Modern-Physics-6th-Edition-by-Tipler Solution Manual for Modern Physics 6th Edition by Tipler Full file Chapter at https://TestbankDirect.eu/Solution-Manual-for-Modern-Physics-6th-Edition-by-Tipler – Relativity I (Problem 1-49 continued) L L  57.4d  c  3.2 10 c  3.2  104 c  3.2  104 L L c  3.2  104  75.5 d  24 h / d  3600 s / h   c  3.2  104 c  3.2  104 c  3.2  104 c  3.2  104  c  3.2 10  L   57.5d  24  3600   c  3.2 10  L c   3.2 10  c   3.2  10  4 2 2  c  3.2 104    c  3.2  104   L  c   3.2  104    57.5  24  3600      c   3.2  104 2   57.5  24  3600    L  6.4 104 L  6.99  1018 m  739c y 1-50 t2  t1    t2  t1   v  xb  xa  (Equation 1-20) c2 (a) t2  t1    t2  t1    v / c   xb  xa    0.5 1.0 y   v / c   2.0 1.0  c y Thus, 0.5   v / c   v  0.5c in the  x direction (b) t     t  vx / c  Using the first event to calculate t  (because t  is the same for both events), t   /   0.5  1 y   0.5c 1c y  / c   1.155 1.5  1.7 y   (c)  s    x    ct   1c y    0.5c y   0.75  c y  2 2  s  0.866c y (d) The interval is spacelike (e) L  s  0.866c y 22 Full file at https://TestbankDirect.eu/Solution-Manual-for-Modern-Physics-6th-Edition-by-Tipler Solution Manual for Modern Physics 6th Edition by Tipler Full file at https://TestbankDirect.eu/Solution-Manual-for-Modern-Physics-6th-Edition-by-Tipler Chapter – Relativity I 1-51 (a) ct (c∙y) 1.7y 1.0 x x (c∙y) -2.0 -1.0 1.0 2.0 Because events are simultaneous in S  , line between and is parallel to x axis Its slope is 0.5   v  0.5c (b) From diagram t   1.7 y 1-52 xb  xr    xb  xr   v  tb  tr  (1) tb  tr    tb  tr   v  xb  xr  / c  (2) Where xb  xr  2400m tb  tr  5 s xb  xr  2400m tb  tr  5 s Dividing (1) by (2) and inserting the values, 2400  v  106  400   2400  v  2400  / 106 c  2400  10 6 v 6 5 10 10  v  2400  / c   2400 2  6 v   10   4800  v  2.69 108 m / s in  x direction 6  10 c  1-53 ux  0.85c cos50 u y  0.85c sin 50   0.72   1/    1.441 v  0.72c 23 Full file at https://TestbankDirect.eu/Solution-Manual-for-Modern-Physics-6th-Edition-by-Tipler Solution Manual for Modern Physics 6th Edition by Tipler Full file Chapter at https://TestbankDirect.eu/Solution-Manual-for-Modern-Physics-6th-Edition-by-Tipler – Relativity I (Problem 1-53 continued) ux  ux  v  vux / c ux  0.85c cos 50  0.72c 0.1736c   0.286c   0.72c   0.85c cos 50 / c   0.3934 uy  uy  uy  1  vux / c  0.85c sin 50 1.441 1   0.72c   0.85c cos50 / c    0.745c u  ux2  uy2  0.798c tan    uy / ux  0.745 /  0.286    111 with respect to the  x axis 1-54 This is easier to in the xy and xy planes Let the center of the meterstick, which is parallel to the x-axis and x  y  x  y  at t  t   moves upward with speed vy in S, at The right hand end of the stick, e.g., will not be at t   in S  because the clocks in S  are not synchronized with those in S In S  the components of the sticks velocity are: uy  ux  uy  1  vu x / c   vy  because u y  v y and u x  ux  v  v because u x   vu x / c When the center of the stick is located as noted above, the right end in S  will be at: x    x  vt   0.5 because t  The S  clock there will read: t     t  vx / c   0.5 v / c Because t = Therefore, when t   at the center, the right end is at xy given by: x  0.5 and    tan 1 For   0.65 v y  0.5 v    c  v  0.5 v  y  tan 1 y   / 0.5  tan 1  v y v / c    x   c  y   uy t       0.494v y / c  24 Full file at https://TestbankDirect.eu/Solution-Manual-for-Modern-Physics-6th-Edition-by-Tipler Solution Manual for Modern Physics 6th Edition by Tipler Full file at https://TestbankDirect.eu/Solution-Manual-for-Modern-Physics-6th-Edition-by-Tipler Chapter – Relativity I 1-55  x  y  z   ct     x  vt     y 2  z 2  c  t   vx   / c  2  x2    c 2 v / c   y 2  z 2  t 2   v  c 2   x t   2 v  2vc 2 / c   x2  y 2  z 2   ct  1-56 The solution to this problem is essentially the same as Problem 1-53, with the manhole taking the place of the meterstick and with the addition of the meterstick moving to the right along the x-axis Following from Problem 1-53, the manhole is titled up on the right and so the meterstick passes through it; there is no collision y manhole x meterstick 1-57 (a) t2    t2  vx2 / c  and t1    t1  vx1 / c  t2  t1   t2  t1  v  x2  x1  / c    T  vD / c  (b) For simultaneity in S , t2  t1, or T  vD / c2  v / c  cT / D Because v / c  1, cT / D is also this is always positive because v/c < Thus, t2  t1   T  vD / c  is always positive (d) Assume T  D / c with c  c Then  c v T  vD / c   D / c    vD / c    D / c      c c  This changes sign at v / c  c / c which is still smaller than For any larger v still smaller than c) t2  t1   T  vD / c   or t1  t2 25 Full file at https://TestbankDirect.eu/Solution-Manual-for-Modern-Physics-6th-Edition-by-Tipler Solution Manual for Modern Physics 6th Edition by Tipler Full file Chapter at https://TestbankDirect.eu/Solution-Manual-for-Modern-Physics-6th-Edition-by-Tipler – Relativity I 1-58  v  0.6c 1   0.6   1.25 (a) The clock in S reads   60min  75min when the S  clock reads 60 and the first signal from S  is sent At that time, the S  observer is at v  75min  0.6c  75min  45c The signal travels for 45 to reach the S observer and arrives at 75 + 45 = 120 on the S clock (b) The observer in S sends his first signal at 60 and its subsequent wavefront is found at x  c  t  60  The S  observer is at x  vt  0.6ct and receives the wavefront when these x positions coincide, i.e., when c  t  60   0.6ct 0.4ct  60c t   60c  / 0.4c  150 x  0.6c    90c The confirmation signal sent by the S  observer is sent at that time and place, taking 90 to reach the observer in S It arrives at 150 + 90 = 240 (c) Observer in S: Sends first signal Receives first signal Receives confirmation 60 120 240 The S  observer makes identical observations 1-59 Clock at r moves with speed u  r , so time dilation at that clock’s location is: t0  t  tr  t0  u / c  t0  r 2 / c   c, tr  t0 1  r 2 / c      t0 1  r 2 / c   t0 t  t0 r 2   And, r   t0 t0 2C Or, for r 26 Full file at https://TestbankDirect.eu/Solution-Manual-for-Modern-Physics-6th-Edition-by-Tipler Solution Manual for Modern Physics 6th Edition by Tipler Full file at https://TestbankDirect.eu/Solution-Manual-for-Modern-Physics-6th-Edition-by-Tipler Chapter – Relativity I 1-60 vB ux  ux  v  ux v / c uy  (a) For vBA : v  vB , vAx  0, vAy  vA So, vAx  vAy  B 1  vAx v / c   v A where  B  B vBA  vAx2  vAy2  vB2   vA /  B    tan  BA  vB2 / c 2 vAy vA /  B v   A vAx  vB  B vB vBy  A 1  vBx v / c vAB  vAx  v  vB  vAx v / c   (b) For vAB : v  vA , vBy  vB , vBx  So, vBx   vBy  1  ux v / c  VA Space station vAy  uy   vB A  v A    vB /  A    tan  AB where  A  vBx  v  vA  vBx v / c 1  vA2 / c 2 vB v  B  A  v A   A vA (c) The situations are not symmetric B viewed from A moves in the +y direction, and A viewed from B moves in the –y direction, so tan  A   tan  B  45 only if vA  vB and  A   B  1, which requires vA  vB  27 Full file at https://TestbankDirect.eu/Solution-Manual-for-Modern-Physics-6th-Edition-by-Tipler Solution Manual for Modern Physics 6th Edition by Tipler Full file Chapter at https://TestbankDirect.eu/Solution-Manual-for-Modern-Physics-6th-Edition-by-Tipler – Relativity I 1-61 1-62 (a) Apparent time A  B  T /  t A  tB and apparent time B  A  T /  t A  tB where tA = light travel time from point A to Earth and tB = light travel time from point B to Earth A B  T L L T 2vL     2 cv cv c v B A T L L T 2vL     2 cv cv c v (b) Star will appear at A and B simultaneously when tB  T /  t A or when the period is: L  4vL  L T  t B  t A      2  c  v c  v  c  v 1-63 The angle of u  with the x axis is: uy uy 1  vux / c tan     ux ux  v  vu   1  2x  c   tan    uy   ux  v    u sin  sin     u cos   v    cos   v / u  28 Full file at https://TestbankDirect.eu/Solution-Manual-for-Modern-Physics-6th-Edition-by-Tipler Solution Manual for Modern Physics 6th Edition by Tipler Full file at https://TestbankDirect.eu/Solution-Manual-for-Modern-Physics-6th-Edition-by-Tipler Chapter – Relativity I #1 1-64 d  vt cos  θ d xEarth  vt sin  #2 xEarth L L to Earth (a) From position #2 light reaches Earth at time t2  L / c From position #1 light reaches L d Earth at time t1    t c c L L d      t  c c c  vt cos    t c  t 1   cos   t Earth  t2  t1  t Earth t Earth (b) Bapp  vapp c  xEarth vt sin   sin    ctEarth ct 1   cos     cos  (c) For   0.75 and vapp  c  app  1, the result in part (b) becomes 1 0.75sin   sin   cos   1/ 0.75  0.75cos  Using the trigonometric identity p sin   q cos   r sin   A where r  p2  q2 sin A  p / r cos A  q / r sin   cos   sin   45   1/ 0.75 sin   45   1/ 0.75   45  70.6   25.6 29 Full file at https://TestbankDirect.eu/Solution-Manual-for-Modern-Physics-6th-Edition-by-Tipler Solution Manual for Modern Physics 6th Edition by Tipler Full file at https://TestbankDirect.eu/Solution-Manual-for-Modern-Physics-6th-Edition-by-Tipler Full file at https://TestbankDirect.eu/Solution-Manual-for-Modern-Physics-6th-Edition-by-Tipler ... https://TestbankDirect.eu /Solution- Manual- for- Modern- Physics- 6th- Edition- by- Tipler Solution Manual for Modern Physics 6th Edition by Tipler Full file at https://TestbankDirect.eu /Solution- Manual- for- Modern- Physics- 6th- Edition- by- Tipler. .. https://TestbankDirect.eu /Solution- Manual- for- Modern- Physics- 6th- Edition- by- Tipler Solution Manual for Modern Physics 6th Edition by Tipler Full file at https://TestbankDirect.eu /Solution- Manual- for- Modern- Physics- 6th- Edition- by- Tipler. .. https://TestbankDirect.eu /Solution- Manual- for- Modern- Physics- 6th- Edition- by- Tipler Solution Manual for Modern Physics 6th Edition by Tipler Full file at https://TestbankDirect.eu /Solution- Manual- for- Modern- Physics- 6th- Edition- by- Tipler