Question #96eae

1 Answer
anor277
Sep 3, 2017

Well, we know that #c=nuxxlambda#..........

Explanation:

And so #lambda("wavelength")=c/nu#...and so we put the numbers into this quotient.....

#lambda=(3.00xx10^8*m*s^-1)/(3.64xx10^7*s^-1)#

(Note #"Hz"-="frequency"# and has units of #s^-1#)

And so we get.....#lambda=8.24*m#....i.e. we are in the radiowave section.

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