WheypLusC Chapter 25, Prob x ps://edugen.wileyplus.com/edugen/student/mainfr.uni

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WheypLusC Chapter 25, Prob x ps://edugen.wileyplus.com/edugen/student/mainfr.uni ent FULL SCREEN PRINTER VERSION BACK NEXT Chapter 24, Problem 37 A distant galaxy is simultaneously rotating and receding from the earth. As the drawing shows, the galactic center is receding from the earth at a relative speed of uG-2.20 x 106 m/s. Relative to the center, the tangential speed is 0.370 x 10 m/s for locations A and B, which are equidistant from the center. When the frequencies of the light coming from regions A and B are measured on earth, they are not the same and each is different from the emitted frequency of 7,130 x 10 Find the measured frequency for the light from (a) region A and (b) region B. (Give your answer to 4 significant digits. Use 2.998 × 10-m/s as the speed of light.) Galay Earth (a) Number Units Units (b) Number Version 4.24.5.1 All Rights Reserved. A Division of PolicyI .2000-2018.John Wiley&Sons;

Explanation / Answer

Given :-

uG = 2.20 x 10^6 m/s

vT = 0.370 x 10^6 m/s

fs = 7.130 x 10^14 Hz

Speed at location A is

VA = uG - vT

VA = (2.20 x 10^6 m/s) - ( 0.370 x 10^6 m/s)

VA = 1.83 x 10^6 m/s

Speed at location B is

VB = uG + vT

VB = (2.20 x 10^6 m/s) + (0.370 x 10^6 m/s)

VB = 2.57 x 10^6 m/s

a)

The measured frequency at A is

fA = fs ( 1 - VA /c)

fA = (7.310 x 10^14 Hz) ( 1 -   (1.83 x 10^6 m/s) / ( 3 x 10^8 m/s) )

fA = 7.299 x 10^14Hz

b)

frequency at B is

fB = fs ( 1 - VB /c)

fB = (7.310 x 10^14 Hz) ( 1 -   (2.57 x 10^6 m/s) /( 3 x 10^8 m/s) )

f = 7.2474 x 10^14 Hz

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