## Summersnow8 2 years ago Male mosquitoes in the mood for mating find female mosquitoes of their own species by listening for the characteristic "buzzing" frequency of the female's wing beats. This frequency is about 620 wing beats per second. How many cycles of oscillation does the radiation from a cesium-133 atom complete during one mosquito wing beat? I know cesium-133 atom is 9,192,631,770 cycles/ second

1. whpalmer4

620 wing beats = 1 sec 9,192,631,770 cycles = 1 sec 9,192,631,770 cycles = 620 wing beats divide both by 620 wing beats 9192631770/620 cycles / wing beat

2. Summersnow8

i tried that but it said it was wrong

3. whpalmer4

what did you get for the answer?

4. Summersnow8

25535088.25

5. whpalmer4

well, that IS wrong :-) try doing it again.

6. whpalmer4

9192631770/620 =

7. Summersnow8

1414251.042

8. whpalmer4

no, that's not right either. have you pressed the clear or all clear button on your calculator before starting the problem?

9. Summersnow8

gahhhh

10. whpalmer4

correct answer should be 1.48268 * 10^7, or digits to that effect...

11. Summersnow8

14142510.42

12. whpalmer4

you aren't entering the same numbers as I am.

13. Summersnow8

okay what are you entering

14. whpalmer4

9 1 9 2 6 3 1 7 7 0 / 6 2 0 =

15. Summersnow8

bahahaha, okay, i was doing 650, damn thing... so do sif figs count? 1.4 x 10 ^7

16. Summersnow8

sig*

17. whpalmer4

I can't really advise you on whether or not they want you to round to an appropriate number of significant figures. If doing it on paper, I would write something like "\(1.48268 * 10^7\) cycles which is \(1.5 *10^7\) cycles to 2 significant figures" [note that it is 1.48... so rounds to 1.5, not 1.4]

18. whpalmer4

because the problem says "about 620 beats per second" perhaps we should go with the two significant figures suggested by "620".

19. Summersnow8

okay thanks :) lifesaver

20. whpalmer4

you're welcome!