Fundamentals Of Physics (6ºed)\chap01

Fundamentals Of Physics (6ºed)\chap01 - N = ft 1 × 10 6...

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18. We denote the pulsar rotation rate f (for frequency). f = 1 rotation 1 . 55780644887275 × 10 3 s (a) Multiplying f by the time-interval t =7 . 00 days (which is equivalent to 604800 s, if we ignore signifcant fgure considerations for a moment), we obtain the number of rotations: N = µ 1 rotation 1 . 55780644887275 × 10 3 s (604800 s) = 388238218 . 4 which should now be rounded to 3 . 88 × 10 8 rotations since the time-interval was speciFed in the problem to three signiFcant Fgures. (b) We note that the problem speciFes 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 part (a) takes the form
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Unformatted text preview: N = ft 1 × 10 6 = µ 1 rotation 1 . 55780644887275 × 10 − 3 s ¶ t which yields the result t = 1557 . 80644887275 s (though students who do this calculation on their calculator might not obtain those last several digits). (c) Careful reading of the problem shows that the time-uncertainty per revolution is ± 3 × 10 − 17 s. We therefore expect that as a result of one million revolutions, the uncertainty should be ( ± 3 × 10 − 17 )(1 × 10 6 ) = ± 3 × 10 − 11 s....
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