Bộ tài liệu bài tập tham khảo môn vật lý bậc đại học bằng tiếng anh Chapter 37 | 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. a y . . 71 - 2 t yr x 10-6s x 10-6 s. b d vt x 108 ms x 10-6s x 103 m km. 71 -u7c2 1 -u7C2 V2 1 - T A At - At0 1 - 1 - u 7c2 At U 2 At 250 ms 2 4 hrs 3600 2 x 108 ms 2 At -At 0 x 10 -9s. The clock on the plane shows the shorter elapsed time. y . . 71 - 2 y At x 10-6s x 10 4s ms. a A 1 - U2 A 2 At 0 Ï At At 0 At C. 1 42 u c 1 2 2 u . b Ax uAt x 108 ms x 10-7s 126 m. y a At0 A L2 X 108 m s. 0 y y b s x 107m. c At0 7Y s. This is what the racer measures your clock to read at that instant. At your origin you read the original x 108m 3 x 108 ms s. Clearly the observes you and the racer will not agree on the order of events 2 212 I I x 106 m s Z1 -u2 c2 At 1 -I-------------- M 1 yr 1 11 1 x 108 ms J At- At0 x 10 4 yr hrs. 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. a The frame in which the source the searchlight is stationary is the spacecraft s frame so ms is the proper time. b To three figures u c. Solving Eq. for u c in terms of y c 41 - Vr 2 -1 - Ar. Using 1 y At0 A t ms 190 ms gives u c . y 71 - u c 2 71 - 2 x l0 55 km a l ------ km. Y b In muon s frame d uAt x 10-6 s km. d 51 . h c In earth s frame At At0y x 10-6 s x 10-5s d uAt x 10-5 s km d h 390tm km a t x 10 m b y 1 10-4 s. h . 71 - 2 45 km ------- km. h Y s h . t . . c ----------- x 10 5 s and x 10 s so the results agree but the particle s .
