Chapter 04, Problem 061 When a large star becomes a supernova, its core may be c

Question

Chapter 04, Problem 061 When a large star becomes a supernova, its core may be compressed so tightly that it becomes a neutron star, with a radius of about 21.0 km (about the size of a typical city). If a neutron star rotates once every 0.540 seconds, (a) what is the speed of a particle on the star's equator and (b) what is the magnitude of the particle's centripetal acceleration? (c) If the neutron star rotates faster, do the answers to (a) and (b) increase, decrease, or remain the same? (a) Number (b) Number SHOW HINT LINK TO TEXT LINK TO SAMPLE PROBLE LINK TO SAMPLE PROBLEM LINK TO SAMPLE PROBLE MATH HELP MATH HELP MATH HELP MATH HELP lacy Policy I 2000-2017 lohn Miley & Sons, Inc. All Rights Reserved. A Division of lohn Wiley a sons, Inc. Version 4.22.3.4 o e a 9

Explanation / Answer

a)distance covered

d = 2 pi r

d = 2 x 3.14 x 21000 = 131880 m

t = 0.54 s

v = d/t = 131880/0.54 = 2.44 x 10^5 m/s

Hence, v = 2.44 x 10^5 m/s

b)a(c) = v^2/R

a(c) = (2.44 x 10^5)^2/21000 = 2.84 x 10^6 m/s^2

Hence, a(c) = 2.84 x 10^6 m/s^2

c)Increases, since t will decrease and velocity will increase so acceleration will also increase

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