Chapter37 - 1. From the time dilation equation t = t0...

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1. From the time dilation equation t = γ t 0 (where t 0 is the proper time interval, γβ =− 11 2 / , and β = v / c ), we obtain F H G I K J 1 0 2 t t . The proper time interval is measured by a clock at rest relative to the muon. Specifically, t 0 = 2.2000 µ s. We are also told that Earth observers (measuring the decays of moving muons) find t = 16.000 s. Therefore, 2 2.2000 s 1 0.99050. 16.000 s ⎛⎞ = ⎜⎟ ⎝⎠
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2. (a) We find β from γβ =− 11 2 /: () 2 2 1 1 0.14037076. 1.0100000 β γ = (b) Similarly, 2 1 10.000000 0.99498744. = (c) In this case, 2 1 100.00000 0.99995000. = (d) The result is 2 1 1000.0000 0.99999950. =
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3. In the laboratory, it travels a distance d = 0.00105 m = vt , where v = 0.992 c and t is the time measured on the laboratory clocks. We can use Eq. 37-7 to relate t to the proper lifetime of the particle t 0 : () 2 2 0 0 2 1 1 0.992 0.992 1/ t vd tt t cc vc ⎛⎞ =⇒ = = ⎜⎟ ⎝⎠ which yields t 0 = 4.46 × 10 –13 s = 0.446 ps.
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4. From the value of t in the graph when β = 0, we infer than t o in Eq. 37-9 is 8.0 s. Thus, that equation (which describes the curve in Fig. 37-23) becomes 0 22 8.0 s 1(/) 1 t t vc β ∆= = −− . If we set β = 0.98 in this expression, we obtain approximately 40 s for t .
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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. ≈×
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6. Due to the time-dilation effect, the time between initial and final ages for the daughter is longer than the four years experienced by her father: t f daughter t i daughter = γ (4.000 y) where γ is Lorentz factor (Eq. 37-8). Letting T denote the age of the father, then the conditions of the problem require T i = t i daughter + 20.00 y , T f = t f daughter – 20.00 y . Since T f T i = 4.000 y, then these three equations combine to give a single condition from which γ can be determined (and consequently v ): 44 = 4 γ γ = 11 β = 2 30 11 = 0.9959.
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should be admitted that this is a fairly subtle question which has occasionally precipitated debates among professional physicists. 7. (a) The round-trip (discounting the time needed to “turn around”) should be one year according to the clock you are carrying (this is your proper time interval t 0 ) and 1000 years according to the clocks on Earth which measure t . We solve Eq. 37-7 for β : 2 2 0 1y 1 1 0.99999950. 1000y t t ⎛⎞ =− = ⎜⎟ ⎝⎠ (b) The equations do not show a dependence on acceleration (or on the direction of the velocity vector), which suggests that a circular journey (with its constant magnitude centripetal acceleration) would give the same result (if the speed is the same) as the one described in the problem. A more careful argument can be given to support this, but it
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8. Only the “component” of the length in the x direction contracts, so its y component stays sin30 (1.0 m)(0.50) 0.50m yy == ° = = AAA while its x component becomes 22 1 (1.0 m)(cos30 ) 1 (0.90) 0.38m. xx β =− = ° = AA Therefore, using the Pythagorean theorem, the length measured from S' is () () 2 2 (0.38 m) (0.50 m) 0.63m.
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Chapter37 - 1. From the time dilation equation t = t0...

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