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.

Related questions
See all questions in Calculations with wavelength and frequency
Impact of this question
1681 views around the world
You can reuse this answer
Creative Commons License