Concept explainers
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
- As seen by spectators in the grandstand, how much faster than Jo is the light beam?
- As seen by Jo, how much faster is the light beam than he is? Explain your answer clearly.
- 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.
- 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?

Want to see the full answer?
Check out a sample textbook solution
Chapter S2 Solutions
COSMIC PERSPECTIVE LL FD
- 20. Two small conducting spheres are placed on top of insulating pads. The 3.7 × 10-10 C sphere is fixed whie the 3.0 × 107 C sphere, initially at rest, is free to move. The mass of each sphere is 0.09 kg. If the spheres are initially 0.10 m apart, how fast will the sphere be moving when they are 1.5 m apart?arrow_forwardpls help on allarrow_forwardpls help on thesearrow_forward
- pls help on all asked questions kindlyarrow_forwardpls help on all asked questions kindlyarrow_forward19. Mount Everest, Earth's highest mountain above sea level, has a peak of 8849 m above sea level. Assume that sea level defines the height of Earth's surface. (re = 6.38 × 106 m, ME = 5.98 × 1024 kg, G = 6.67 × 10 -11 Nm²/kg²) a. Calculate the strength of Earth's gravitational field at a point at the peak of Mount Everest. b. What is the ratio of the strength of Earth's gravitational field at a point 644416m below the surface of the Earth to a point at the top of Mount Everest? C. A tourist watching the sunrise on top of Mount Everest observes a satellite orbiting Earth at an altitude 3580 km above his position. Determine the speed of the satellite.arrow_forward
- pls help on allarrow_forwardpls help on allarrow_forward6. As the distance between two charges decreases, the magnitude of the electric potential energy of the two-charge system: a) Always increases b) Always decreases c) Increases if the charges have the same sign, decreases if they have the opposite signs d) Increases if the charges have the opposite sign, decreases if they have the same sign 7. To analyze the motion of an elastic collision between two charged particles we use conservation of & a) Energy, Velocity b) Momentum, Force c) Mass, Momentum d) Energy, Momentum e) Kinetic Energy, Potential Energyarrow_forward
- Physics for Scientists and Engineers: Foundations...PhysicsISBN:9781133939146Author:Katz, Debora M.Publisher:Cengage LearningPrinciples of Physics: A Calculus-Based TextPhysicsISBN:9781133104261Author:Raymond A. Serway, John W. JewettPublisher:Cengage LearningUniversity Physics Volume 1PhysicsISBN:9781938168277Author:William Moebs, Samuel J. Ling, Jeff SannyPublisher:OpenStax - Rice University
- Physics for Scientists and Engineers, Technology ...PhysicsISBN:9781305116399Author:Raymond A. Serway, John W. JewettPublisher:Cengage LearningAstronomyPhysicsISBN:9781938168284Author:Andrew Fraknoi; David Morrison; Sidney C. WolffPublisher:OpenStaxGlencoe Physics: Principles and Problems, Student...PhysicsISBN:9780078807213Author:Paul W. ZitzewitzPublisher:Glencoe/McGraw-Hill





