ch01-p014 - 14. We denote the pulsar rotation rate f (for...

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(a) Multiplying f by the time-interval 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 1 rotation 604800 s 388238218.4 1.55780644887275 10 s N §· == ¨¸ × ©¹ which should now be rounded to 3.88 × 10 8 rotations since the time-interval 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 part (a) takes the form N = ft , or 6 3 1rotation
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This note was uploaded on 09/03/2009 for the course PHYS 1200 taught by Professor Stoler during the Spring '06 term at Rensselaer Polytechnic Institute.

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