Chapter 37 - 37.1 If O sees simultaneous flashes then O...

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37.1: If O sees simultaneous flashes then O will see the ) A A( flash first since O would believe that the A flash must have traveled longer to reach O , and hence started first. 37.2: a) . 29 . 2 ) 9 . 0 ( 1 1 γ 2 = - = s. 10 5.05 s) 10 20 . 2 ( ) 29 . 2 ( γ 6 6 - - × = × = = τ t b) km. 1.36 m 10 1.36 s) 10 05 . 5 ( ) s m 10 00 . 3 ( ) 900 . 0 ( 3 6 8 = × = × × = = - vt d 37.3: + - - = - 2 2 2 1 2 2 2 2 2 1 ) 1 ( 1 c u c u c u s. 10 00 . 5 ) ( ) s m 10 2(3.00 (3600) hrs) 4 ( ) s m 250 ( 2 ) )( 1 1 ( ) ( 9 0 2 8 2 2 2 2 2 0 - × = - × = = - - = - t t t c u t c u t t The clock on the plane shows the shorter elapsed time. 37.4: . 79 . 4 ) 978 . 0 ( 1 1 γ 2 = - = ms. 0.395 s 10 3.95 s) 10 4 . 82 ( ) 79 . 4 ( γ -4 6 = × = × = - t 37.5: a) 2 0 2 2 2 2 0 1 1 = - - = t t c u c u t t 2 2 0 42 6 . 2 1 1 - = - = c t t c u . 998 . 0 c u = b) m. 126 s) 10 (4.2 ) s m 10 00 . 3 ( ) 998 . 0 ( 7 8 = × × = = - t u x 37.6: 667 . 1 γ = a) s. 300 . 0 ) 800 . 0 ( γ m 10 20 . 1 γ 8 0 = × = = c t t b) m. 10 7.20 ) (0.800 s) 300 . 0 ( 7 × = c c) s. 180 . 0 γ s 300 . 0 0 = = t (This is what the racer measures your clock to read at that instant.) At your origin you read the original s. 5 . 0 ) s m 10 (3 (0.800) m 10 20 . 1 8 8 = × × Clearly the observes (you and the racer) will not agree on the order of events!
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37.7: yr) (1 s m 10 3.00 s m 10 4.80 1 1 2 8 6 2 2 0 × × - = - = t c u t hrs. 1.12 yr ) 10 28 . 1 ( ) ( 4 0 = × = - - t t The least time elapses on the rocket’s clock because it had to be in two inertial frames whereas the earth was only in one. 37.8: a) The frame in which the source (the searchlight) is stationary is the spacecraft’s frame, so 12.0 ms is the proper time. b) To three figures, . c u = Solving Eq. (37.7) for c u in terms of , γ . γ 2 1 1 ) γ 1 ( 1 2 2 - - = c u Using . 998 . 0 gives ms 190 ms 0 . 12 γ 1 0 = = = c u t t 37.9: 00 . 6 ) 9860 . 0 ( 1 1 ) ( 1 1 γ 2 2 = - = - = c u a) km. 17 . 9 6.00 km 55 γ 0 = = = l l b) In muon’s frame: %. 1 . 7 071 . 0 17 . 9 651 . 0 % km. 0.651 s) 10 20 . 2 )( 9860 . 0 ( 6 = = = = = × = = - h d c t u d c) In earth’s frame: %. 1 . 7 km 55.0 km 90 . 3 % km 3.90 s) 10 )(1.32 (0.9860 s 10 1.32 s)(6.00) 10 2 . 2 ( γ 5 5 6 0 = = = = × = = × = × = = - - - h d c t u d t t 37.10: a) s. 10 1.51 0.99540 m 10 4.50 4 4 - × = × = c t km. 4.31 10.44 km 45 γ 10.44 (0.9954) 1 1 γ ) 2 = = = = - = h h b c) s; 10 1.44 γ and s, 10 1.44 0.99540 5 5 - - × = × = t c h so the results agree but the particle’s lifetime is dilated in the frame of the earth.
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37.11: a) m 3600 0 = l m. 3568 m)(0.991) (3600 ) s m 10 00 . 3 ( ) s m 10 00 . 4 ( 1 ) m 3600 ( 1 2 8 2 7 0 2 2 0 = = × × - = - = l c u l l ) b . s 10 9.00 s m 10 4.00 m 3600 5 7 0 0 - × = × = = u l t c) . s 10 8.92 s m 10 4.00 m 3568 5 7 - × = × = = u l t 37.12: . s m 10 2.86 952 . 0 ) γ 1 ( 1 so , 3048 . 0 1 γ 8 2 × = = - = = c c u 37.13: 2 2 0 2 2 0 1 1 c u l l c u l l - = - = m. 92.5 0.600 1 m 74.0 2 0 = - = c c l 37.14: Multiplying the last equation of (37.21) by u and adding to the first to eliminate t gives , 1 1 2 2 x c u x t u x γ = - γ = + and multiplying the first by 2 c u and adding to the last to eliminate x gives , γ 1 1 γ 2 2 2 t c u t x c u t = - = + ), ( γ and ) ( γ so 2 c x u t t t u x x + = + = which is indeed the same as Eq. (37.21) with the primed coordinates replacing the unprimed, and a change of sign of u.
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