/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Q1Q A rod is to move at constant spe... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

A rod is to move at constant speedvalong thexaxis of reference frame S, with the rod’s length parallel to that axis. An observer in frame Sis to measure the lengthLof the rod. Which of the curves in Fig. 37-15 best gives length L(vertical axis of the graph) versus speed parameterβ?

Short Answer

Expert verified

The curvec gives the best graph between length and speed parameter.

Step by step solution

01

Write the given data from the question.

The constant speed of the rod is valong the xaxis.

The length of the rod is L.

The speed parameter is β.

02

Determine the formulas to find the best graph between length and speed parameter.

The expression to calculate the length of the rod in respect to observer in the frameS' is given as follows.

L=L0Y

Here, L0is the proper length of the rod and γis the Lorentz factor.

03

Determine the best graph between length and speed parameter.

Let assume a frame S'which is moving with the velocity of vand for observer in this frame the length of the rod seems to be shorter which is given by,

L=L0γ …… (i)

The expression for the Lorentz factor is given by,

γ=11-β2

Substitute 11-β2forγ into equation (i).

L=L01-β2

As the value of theβ increases, 1-β2, therefore value of length decreases.

Hence the curve cgives the best graph between length and speed parameter.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Question:(a) If m is a particle’s mass, p is its momentum magnitude, and K is its kinetic energy, show that

m=pc2-K22Kc2

(b) For low particle speeds, show that the right side of the equation reduces to m. (c) If a particle has K = 55.0 MeV when p=121 MeV/c, what is the ratio m/me of its mass to the electron mass?

Figure 37-17 shows two clocks in stationary frame S'(they are synchronized in that frame) and one clock in moving frame S. Clocks C1and C'1read zero when they pass each other. When clocks C1and C'2pass each other, (a) which clock has the smaller reading and (b) which clock measures a proper time?

The center of our Milky Way galaxy is about 23000ly away. (a) To eight significant figures, at what constant speed parameter would you need to travel exactly (measured in the Galaxy frame) in exactly 23000ly (measured in your frame)? (b) Measured in your frame and in light-years, what length of the Galaxy would pass by you during the trip?

Question: Temporal separation between two events. Events and occur with the following spacetime coordinates in the reference frames of Fig. 37-25: according to the unprimed frame,andaccording to the primed frame, (xA,tA)and(xB,tB). In the unprimed frame ∆t=tB-tA=1.00μsand Δx=xB-xA=240m. (a) Find an expression for Δt'in terms of the speed parameterβand the given data. Graph Δt' versus βfor the following two ranges of β: (b) 0 to 0.01and 0.1 (c) 0.1 to 1. (d) At what value of βis Δt'minimum and (e) what is that minimum? (f) Can one of these events cause the other? Explain.

(a) What potential difference would accelerate an electron to speed c according to classical physics? (b) With this potential difference, what speed would the electron actually attain?

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.