EBK THE COSMIC PERSPECTIVE
EBK THE COSMIC PERSPECTIVE
9th Edition
ISBN: 9780135161760
Author: Voit
Publisher: VST
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Chapter S2, Problem 57EAP

Racing a Light Beam II. Following his humiliation in the first race against the light beam (Problem 56), Jo went into hiding for the next 2 years. By that time, most people had forgotten about both him and the money they had wasted on the pay-per-view event. However, Jo was secretly in training during this time. He worked out hard and tested new performance-enhancing substances. One day, he emerged from hiding and called another press conference. “I’m ready for a rematch,” he announced. Sponsors were few this time and spectators scarce in the huge Olympic stadium where Jo and the flashlight lined up at the starting line. But those who were there will never forget what they saw, although it all happened very quickly. Jo blasted out of the starting block at 99.9% of the speed of light. The light beam, emitted from the flashlight, took off at the speed of light. The light beam won again—but barely! After the race, TV commentators searched for Jo, but he seemed to be hiding again. Finally, they found him in a corner of the locker room, sulking under a towel. “What’s wrong? You did great!” said the commentators. Jo looked back sadly, saying, “Two years of training and experiments, for nothing!” Let’s investigate what happened.

  1. As seen by spectators in the grandstand, how much faster than Jo is the light beam?
  2. As seen by Jo, how much faster is the light beam than he is? Explain your answer clearly.
  3. Using your results from parts a and b, explain why Jo can say that he was beaten just as badly as before, while the spectators can think he gave the light beam a good race.
  4. Although Jo was disappointed by his performance against the light beam, he did experience one pleasant surprise: The 100-meter course seemed short to him. In Jo’s reference frame during the race, how long was the 100-meter course?

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A cylinder with a piston contains 0.153 mol of nitrogen at a pressure of 1.83×105 Pa and a temperature of 290 K. The nitrogen may be treated as an ideal gas. The gas is first compressed isobarically to half its original volume. It then expands adiabatically back to its original volume, and finally it is heated isochorically to its original pressure. Part A Compute the temperature at the beginning of the adiabatic expansion. Express your answer in kelvins. ΕΠΙ ΑΣΦ T₁ = ? K Submit Request Answer Part B Compute the temperature at the end of the adiabatic expansion. Express your answer in kelvins. Π ΑΣΦ T₂ = Submit Request Answer Part C Compute the minimum pressure. Express your answer in pascals. ΕΠΙ ΑΣΦ P = Submit Request Answer ? ? K Pa
Learning Goal: To understand the meaning and the basic applications of pV diagrams for an ideal gas. As you know, the parameters of an ideal gas are described by the equation pV = nRT, where p is the pressure of the gas, V is the volume of the gas, n is the number of moles, R is the universal gas constant, and T is the absolute temperature of the gas. It follows that, for a portion of an ideal gas, pV = constant. Τ One can see that, if the amount of gas remains constant, it is impossible to change just one parameter of the gas: At least one more parameter would also change. For instance, if the pressure of the gas is changed, we can be sure that either the volume or the temperature of the gas (or, maybe, both!) would also change. To explore these changes, it is often convenient to draw a graph showing one parameter as a function of the other. Although there are many choices of axes, the most common one is a plot of pressure as a function of volume: a pV diagram. In this problem, you…
Learning Goal: To understand the meaning and the basic applications of pV diagrams for an ideal gas. As you know, the parameters of an ideal gas are described by the equation pV = nRT, where p is the pressure of the gas, V is the volume of the gas, n is the number of moles, R is the universal gas constant, and T is the absolute temperature of the gas. It follows that, for a portion of an ideal gas, pV = constant. T One can see that, if the amount of gas remains constant, it is impossible to change just one parameter of the gas: At least one more parameter would also change. For instance, if the pressure of the gas is changed, we can be sure that either the volume or the temperature of the gas (or, maybe, both!) would also change. To explore these changes, it is often convenient to draw a graph showing one parameter as a function of the other. Although there are many choices of axes, the most common one is a plot of pressure as a function of volume: a pV diagram. In this problem, you…

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EBK THE COSMIC PERSPECTIVE

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