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16. We denote the pulsar rotation rate
f
(for frequency).
3
1rotation
1.55780644887275
10
s
f
−
=
×
(a) Multiplying
f
by the timeinterval
t
= 7.00 days (which is equivalent to 604800 s, if
we ignore
significant figure
considerations for a moment), we obtain the number of
rotations:
()
3
604800 s
388238218.4
1.55780644887275
10
s
N
−
§·
==
¨¸
×
©¹
which should now be rounded to 3.88
×
10
8
rotations since the timeinterval was
specified in the problem to three significant figures.
(b) We note that the problem specifies the
exact
number of pulsar revolutions (one
million). In this case, our unknown is
t
, and an equation similar to the one we set up in
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This note was uploaded on 09/09/2008 for the course PHYS 100 taught by Professor Shylnov during the Spring '08 term at Illinois Tech.
 Spring '08
 shylnov
 Physics

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