Through a window in Carl's spaceship, passing al 0.5c, you watch Carl doing an important physics calculation. By your watch it takes him 1 min. How much time did Carl spend on his calculation?

Short Answer

Expert verified

The time spends by Carl in spaceship for calculation is0.87min.

Step by step solution

01

Identification of given information

The given data can be listed below as:

  • The speed of spaceship is, v=0.5c.
  • The time measured by watch is, .
02

Usage of time dilation concept

The problem is based on the concept of time dilation. The spaceship for this problem is moving at half the speed of light.

03

Determination of the formula of the time dilation for spaceship

The formula to calculate the time dilation for spaceship is:

t2=t11-v2c2

04

Determination of the time dilation for spaceship

Substitute all the values in the above equation.

1min=t11-(0.5c)2c2t1=0.87min

Therefore, the time spend by Carl in spaceship for calculation is0.87min.

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Most popular questions from this chapter

Exercise 117 gives the speed u of an object accelerated under a constant force. Show that the distance it travels is given by

x=mc2f2[1+Ftmc2-1]

In Example 2.5, we noted that Anna could go wherever she wished in as little time as desired by going fast enough to length-contract the distance to an arbitrarily small value. This overlooks a physiological limitation. Accelerations greater than about 30gare fatal, and there are serious concerns about the effects of prolonged accelerations greater than 1g. Here we see how far a person could go under a constant acceleration of 1g, producing a comfortable artificial gravity.

(a) Though traveller Anna accelerates, Bob, being on near-inertial Earth, is a reliable observer and will see less time go by on Anna's clock (dt')than on his own (dt). Thus, dt'=(1y)dt, where u is Anna's instantaneous speed relative to Bob. Using the result of Exercise 117(c), with g replacing Fm, substitute for u, then integrate to show that t=cgsinhgt'c.

(b) How much time goes by for observers on Earth as they “see” Anna age 20 years?

(c) Using the result of Exercise 119, show that when Anna has aged a time t', she is a distance from Earth (according to Earth observers) of x=c2g(coshgt'c-1).

(d) If Anna accelerates away from Earth while aging 20 years and then slows to a stop while aging another 20. How far away from Earth will she end up and how much time will have passed on Earth?

Classically, the net work done on an initially stationary object equals the final kinetic energy of the object. Verify that this also holds relativistically. Consider only one-dimension motion. It will be helpful to use the expression for p as a function of u in the following:

W=Fdx=dpdtdx=dxdtdp=udp

In Example 2.5, we noted that Anna could go wherever she wished in as little time as desired by going fast enough to length-contract the distance to an arbitrarily small value. This overlooks a physiological limitation. Accelerations greater than about 30g are fatal, and there are serious concerns about the effects of prolonged accelerations greater than 1g. Here we see how far a person could go under a constant acceleration of 1g, producing a comfortable artificial gravity.

(a) Though traveller Anna accelerates, Bob, being on near-inertial Earth, is a reliable observer and will see less time go by on Anna's clock (dt') than on his own (dt). Thus,, whereuis Anna's instantaneous speed relative to Bob. Using the result of Exercise 117(c), withgreplacingF/m, substitute for u, then integrate to show that

(b) How much time goes by for observers on Earth as they “see” Anna age 20 years?

(c) Using the result of Exercise 119, show that when Anna has aged a timet’, she is a distance from Earth (according to Earth observers) of

(d) If Anna accelerates away from Earth while aging 20 years and then slows to a stop while aging another 20. How far away from Earth will she end up and how much time will have passed on Earth?

Is it possible for the momentum of an object to be mc. If not. why not? If so, under what condition?

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