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Sol8 - UAHuntsville Spring 2010 PH 113 Homework Solutions...

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UAHuntsville Spring, 2010 PH 113: Homework Solutions No. 8 Solution 47: (a) From the Lorentz transformation equations, we find ( ) ( ) 2 [400 m (1.00 s)] 400 m (299.8 m) 1 x x v t x c t c γ γ β γ β μ β β ʹஒ Δ = Δ − Δ = Δ − Δ = = (b) A plot of ' x Δ as a function of β with 0 0.01 β < < is shown below: (c) A plot of ' x Δ as a function of β with 0.1 1 β < < is shown below: (d) To find the minimum, we can take a derivative of Δ x ʹஒ with respect to β , simplify, and then set equal to zero:
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2 3/ 2 2 0 (1 ) 1 d x d x c t x c t d d β β β β β β Λি Ξ৯ ʹஒ Δ Δ − Δ Δ − Δ Μ৏ Ο৿ = = = Μ৏ Ο৿ Νয় Πਏ This yields 8 6 (2.998 10 m/s)(1.00 10 s) 0.7495 0.750 400 m c t x β Δ × × = = = Δ (e) Substituting this value of β into the part (a) expression yields Δ x ʹஒ = 264.8 m 265 m for its minimum value. Solution 48: (a) In the messenger’s rest system (called S m ), the velocity of the armada is v v v vv c c c c c c c m m ' / . . ( . )( . ) / . . = = = 1 080 0 95 1 080 0 95 0 625 2 2 The length of the armada as measured in S m is 2 0 1 (1.0ly) 1 ( 0.625) 0.781 ly .
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