hw08_sol - Physics133HW8Spring2009 1.[HRW37.05] 5. We solve...

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Physics 133 HW8 Spring 2009 1. [HRW 37.05] 2. [HRW 37.08] 3. [HRW 37.14] 5. We solve the time dilation equation for the time elapsed (as measured by Earth observers): ! ! t t " # 0 2 1 0 9990 ( . ) where ! t 0 = 120 y. This yields ! t = 2684 y 3 2.68 10 y. $% 8. Only the “component” of the length in the x direction contracts, so its y component stays sin 30 (1.0 m)(0.50) 0.50m yy ! " " # " " !!! while its x component becomes 22 1 (1.0 m)(cos30 ) 1 (0.90) 0.38m. xx $ ! " % " # % " !! Therefore, using the Pythagorean theorem, the length measured from S' is &’ &’ 2 2 (0.38 m) (0.50 m) 0.63m. xy ! ! ! " ( " ( " ! ! ! 14. (a) We solve Eq. 37-13 for v and then plug in: 2 2 0 1 1 1 0.866. 2 L L ! "# $ % $ % $ &’ () (b) The Lorentz factor in this case is *+ 2 1 2.00 1/ vc , $$ % .
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Physics 133 HW8 Spring 2009 4. [HRW 37.15] 5. [HRW 37.76] 22 11 7.09. 1 1 (0.99) ! " # # # $$ Thus, % t 0 = (26.26 y)/(7.09) = 3.705 y. 15. (a) The speed of the traveler is v = 0.99 c , which may be equivalently expressed as 0.99 ly/y. Let d be the distance traveled. Then, the time for the trip as measured in the frame of Earth is % t = d / v = (26 ly)/(0.99 ly/y) = 26.26 y. (b) The signal, presumed to be a radio wave, travels with speed c and so takes 26.0 y to reach Earth. The total time elapsed, in the frame of Earth, is 26.26 y + 26.0 y = 52.26 y . (c) The proper time interval is measured by a clock in the spaceship, so % t 0 = % t / . Now 76. (a) The relative contraction is 2 1 2 2 2 0 8 00 12 (1 ) || 1 1 1 630m/s 1 1 1 1 2 2 2 3.00 10 m/s 2.21 10 .
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hw08_sol - Physics133HW8Spring2009 1.[HRW37.05] 5. We solve...

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