Please solve the problem correctly. Thank you! A 5.57-kg object oscillates back and forth at the end of a spring whose spring constant is 58.9 N/m. An observer is traveling at a speedof 2.24 x 10 m/s relative to the fixed end of the spring. What does this observer measure for the period of oscillation?NumberUnits

To determine the period of oscillation as measured by the observer, we need to account for the…

Answered: A 5.57-kg object oscillates back and…

Physics for Scientists and Engineers: Foundations and Connections

1st Edition

ISBN:9781133939146

Author:Katz, Debora M.

Publisher:Katz, Debora M.

Chapter1: Getting Started

Section: Chapter Questions

Problem 30PQ

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A car traveling at 35.0 m/s takes 26.0 minutes to travel a certain distance according to the drivers clock in the car. How long does the trip take according to an observer at rest on Earth? Hint: The following approximation is helpful: [1x]121+12xforx1.
Unreasonable Results (a) Find the value of for the following situation. An astronaut measures the length of her spaceship to be 25.0 m, while an Earth-bound observer measures it to be 100 m. (b) What is unreasonable about this result? (c) Which assumptions are unreasonable or inconsistent?
An Earth satellite used in the Global Positioning System moves in a circular orbit with period 11 h 58 min. (a) Determine the radius of its orbit. (b) Determine its speed. (c) The satellite contains an oscillator producing the principal nonmilitary GPS signal. Its frequency is 1 575.42 MHz in the reference frame of the satellite. When it is received on the Earths surface, what is the fractional change in this frequency due to time dilation, as described by special relativity? (d) The gravitational blueshift of the frequency according to general relativity is a separate effect. The magnitude of that fractional change is given by ff=Ugmc2 where Ug/m is the change in gravitational potential energy per unit mass between the two points at which the signal is observed. Calculate this fractional change in frequency. (e) What is the overall fractional change in frequency? Superposed on both of these relativistic effects is a Doppler shift that is generally much larger. It can be a redshift or a blueshift, depending on the motion of a particular satellite relative to a GPS receiver (Fig. P1.39).
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A car traveling at 35.0 m/s takes 26.0 minutes to travel a certain distance according to the drivers clock in the car. How long does the trip take according to an observer at rest on Earth? Hint: The following approximation is helpful: [1x]121+12xforx1.
A spacecraft zooms past the Earth with a constant velocity. An observer on the Earth measures that an undamaged clock on the spacecraft is ticking at one-third the rate of an identical clock on the Earth. What does an observer on the spacecraft measure about the Earth-based clock's ticking rate? (a) It runs more than three times faster than his own clock. (b) It runs three times faster than his own. (c) It runs at the same rate as his own. (d) It runs at one-third the rate of his own. (e) It runs at less than one-third the rate of his own.
A spacecraft zooms past the Earth with a constant velocity. An observer on the Earth measures that an undamaged clock on the spacecraft is ticking at one-third the rate of an identical clock on the Earth. What does an observer on the spacecraft measure about the Earth-based clocks ticking rate? (a) It runs more than three times faster than his own clock. (b) It runs three times faster than his own. (c) It runs at the same rate as his own. (d) It runs at one-third the rate of his own. (e) It runs at less than one-third the rate of his own.
The original “clock” used to define the length of the second was the daily rotation of Earth about its axis. Why has this clock been replaced by one based on the oscillation period of light waves emitted by atoms like cesium and rubidium?
A muon formed high in the Earths atmosphere is measured by an observer on the Earths surface to travel at speed = 0.990c for a distance of 4.60 km before it decays into an electron, a neutrino, and an antineutrino (c+v+v). (a) For what time interval does the muon live as measured in its reference frame? (b) How far does the Earth travel as measured in the frame of the muon?
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A 5.57-kg object oscillates back and forth at the end of a spring whose spring constant is 58.9 N/m. An observer is traveling at a speed of 2.24 x 10 m/s relative to the fixed end of the spring. What does this observer measure for the period of oscillation? Number Units