29-9: Picture the Problem: Due to time dilation, as a proper time of 1.00 seconds pass on the ship, 1.00 minute passes on Earth. Strategy:The two events to be considered in this case are successive ticks of the clock that is in your spaceship. From your point of view, these events occur at the same place, so you observe the proper time interval t0and an observer on Earth measures the dilated time interval t. Solve equation 29-2 for the speed of the ship, where and 01.0000 st1.0000 m = 60.000 s.t Solution: 1. Solve equation 29-2 for the speed: 002211ttvcvctt 2.Insert the two time intervals:21.0000 s10.9998660.000 svccInsight:If 1.0000 hour had passed according to an Earth observer, instead of 1.0000 minute, the speed would have to be much closer to the speed of light. The speed would have been 0.9999999614c. 29-26: Picture the Problem: The image shows a rectangular painting as viewed in its rest frame. The painting travels at sufficient speed that a stationary observer views the painting as square. Strategy:The height of the painting will not contract because that dimension is perpendicular to the motion. Therefore, use equation 29-3 to calculate the painting’s speed at which its length will be measured as 80.5 cm.Solution:1. Solve equation 29-3 for the speed v: 22220011LLvcvcLL2.Substitute numerical values:2180.5 cm 124 cm0.761vccInsight:It is not possible to make the painting square by moving the painting in the vertical direction, because length contraction would only shorten that dimension. 29-41: Picture the Problem: The image shows two ships (1 and 2) heading toward Earth (3) at speeds andrelative to Earth. 130.80vc230.80v cStrategy: We want to calculate the speed of the one ship relative to the other. Use equation 29-4 to calculatethe speed of Ship A relative to Ship B. 12,vSolution:1.Multiply both sides of equation 29-4 by the denominator: 122313212232131223122311vvvv vcvv vcvv2.Isolate the unknown variable and solve: 12v2122313132313231222231310.800.80110.800.80vv vcvvccvvvv vcccc
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