{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

# chap37 - 37.1 If O sees simultaneous flashes then O will...

This preview shows pages 1–4. Sign up to view the full content.

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!

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
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.
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.

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

### Page1 / 22

chap37 - 37.1 If O sees simultaneous flashes then O will...

This preview shows document pages 1 - 4. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online