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Lemuel physics 2 13

# Lemuel physics 2 13 - Wong Lemuel Homework 13 Due 11:00 pm...

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Wong, Lemuel – Homework 13 – Due: Apr 30 2007, 11:00 pm – Inst: Matt Mackie 1 This print-out should have 13 questions. Multiple-choice questions may continue on the next column or page – fnd all choices beFore answering. The due time is Central time. NOTE the due time is 11pm central, which is MIDNIGHT our time. Also, you may need to use E = mc 2 , and that electon- volts are converted to joules using 1eV = 1 . 602 × 10 - 19 J. 001 (part 1 oF 1) 10 points Use the relativistic coordinate transFormation ( x,y,z,t ) ←- ( x 0 ,y 0 ,z 0 ,t 0 ) shown and given below where the latter Frame S 0 , ( x 0 ,y 0 ,z 0 ,t 0 ), has a velocity v in the positive direction rela- tive to the Frame S , ( x,y,z,t ). y x z S y' x' z' S' v The relativistic coordinate transFormation is x = x 0 x 0 = x y = y 0 y 0 = y z = γ ( z 0 + v t 0 ) z 0 = γ ( z - v t ) t = γ µ t 0 + β c z 0 t 0 = γ µ t - β c z . Given T 0 = 39 s, c = 2 . 99792 × 10 8 m / s, v = 2 . 69813 × 10 8 m / s, β v c , and γ r 1 1 - β 2 , what is the observed time interval T = t 2 - t 1 in the stationary reFerence Frame given T 0 = t 0 2 - t 0 1 in the moving reFerence Frame? Correct answer: 89 . 4721 s. Explanation: Given : β = v c = 2 . 69813 × 10 8 m / s 2 . 99792 × 10 8 m / s = 0 . 9 γ = r 1 1 - β 2 = s 1 1 - (0 . 9) 2 = 2 . 29416 In the moving reFerence Frame S 0 suppose there is an oscillator with a period oF T 0 = t 0 2 - t 0 1 . In the stationary reFerence Frame S an observer measures the period to be T = t 2 - t 1 . The oscillator is resting ( z 0 constant) in the moving reFerence Frame S 0 . ThereFore, we can exclude the z 0 -coordinate, by trans- Forming From the moving reFerence Frame; i.e. , z 0 2 = z 0 1 = z 0 , into the stationary reFerence Frame. Applying the equation t = γ µ t 0 - β c z 0 , we have T = t 2 - t 1 = γ µ t 0 2 - β c z 0 - γ µ t 0 1 - β c z 0 = γ ( t 0 2 - t 0 1 ) = γ T 0 = 1 p 1 - β 2 T 0 . ThereFore T = γ T 0 = (2 . 29416) (39 s) = 89 . 4721 s . Thus an observer in Frame S measures a longer period than is measured in S 0 . IF the period is that oF a clock, the observer in S would say it is running too slowly, For while he has measured a time oF one minute, the clock in S 0 has not yet reached one minute. This time dilation is with the length con- traction: iF a rod oF length L is at rest in S , its length as measured From S 0 is L 0 = 1 γ L ; iF the oscillator is at rest in S , its period as measured From S 0 is T 0 = γ T . keywords:

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Wong, Lemuel – Homework 13 – Due: Apr 30 2007, 11:00 pm – Inst: Matt Mackie 2 002 (part 1 of 1) 10 points Use the relativistic coordinate transformation ( x,y,z,t ) ←- ( x 0 ,y 0 ,z 0 ,t 0 ) shown below where the latter frame S 0 , ( x 0 ,y 0 ,z 0 ,t 0 ), has a velocity v in the positive direction relative to the frame S , ( x,y,z,t ). y
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Lemuel physics 2 13 - Wong Lemuel Homework 13 Due 11:00 pm...

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