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Sollutions Assignment S 06 1

# Sollutions Assignment S 06 1 - Solutions Assignment 1...

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Solutions Assignment 1, Chapter 2: 4, 6, 11, 20, 23, 24, 27 3. Using the vector triangle shown, the speed of the light coming toward the mirror is Qc 2 —v and the same onthe oc return trip. Therefore the total time is distance 2(2 = speed = t/c2 Notice that smO= soO= siir 1 (!‘\ c c) 032km/hl_ 199 and 4. As in Problem 3, sinS = vj/v 2 , so S = sin(vi/v 2 ) = sin [ Ij = = J(0.94 km/h) 2 (0.32 km/h) 2 = 0.88 km/h. Ad * 6. Let ii = the number of fringes shifted, then i.= —i-. Smce Ad = c(At’ At), we have (At’ At) 2 (i + £2) 1? A c 2 A Solving for v and noting that (i + 2 = 22 m, = /__nA = (3.00 x 108 m/s) J (0.005) (589 10 m) = 3.47 km/s 22 in V i+C2 *1Li,3flienv cwefindl 6 2 —1,so i—3d = x 3d = x vt = t Ox/c t Ox/c t _32 x’ + 3d’ _sx’+/Id’=x’+vt’ 52 = I’ + Ox’/c t’ + flx’/c 52

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12. a) Conversion 100 km/h = 27.77 rn/s so 6 = v/c = (27 77 m/s) (3 00 108 mfs) = 9.3 x 10 b) fi = v/c = (290 m/s)/ (3.00 x 108 m/s) = 9.7 x 1O c)u=2.3vj=(2.3-330m/s) and 6 = v/c = (2.3 .330 m/s) / (3.00 x 108 rn/a) 2.5 x 10 d) Conversion 27,000 km/h = 7500 in/s so = t,/c = (7500m/s)/ (3.00 x 10 8 m/s) = 2.5 x 10 e) (25cm) / (2ns) = 1.25 x 108 rn/s so fl u/c = (1.28 x 108 m/s)/ (3.00 x io m/s) = 042 f) (1 x 1O 14 m ) / (0.3.5 x 1022s) = 2.857 x 108 rn/s 6 = v c = (2.857 x 108 m/s) (3.00 x 108 m/s) = 0.95 20. The round-trip distance is d = 40 ly.
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Sollutions Assignment S 06 1 - Solutions Assignment 1...

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