## YumYum247 one year ago 7

1. YumYum247

i say 5 million times since 110dB = 1X10^-1 w/m^2 whereas the 60dB = 1 X 10^-6w/m^2 difference = 1X10^-1 w/m^2 All Divided by 1 X 10^-6w/m^2 and you get 5 and 5 = 5,000,000w/m^2

2. YumYum247

@aaronq

3. YumYum247

@MrNood can you check my work?!?!?!?!?!

4. YumYum247

@Astrophysics

5. Astrophysics

Well as before let me list the formula first $\beta = 10 \log \frac{ I }{ I_0 }$ so we want the intensity of the rock concert compared to the sound of a normal conversation which is $\frac{ I_{rc} }{ I_{nc} }$ So we want $\beta_{rc}-\beta _{nc}$ where rc represents rock concert, and nc is normal conversation.

6. YumYum247

50 thousand

7. Astrophysics

$10\log \frac{ I_{rc} }{ I_0 }-10\log \frac{ I{nc} }{ I_0 }$

8. YumYum247

what is I/I0

9. YumYum247

so in this case, I = 110dB/60dB?!?!?!?!?!

10. Astrophysics

$I_0 = 10^{-12}W/m^2$

11. YumYum247

threshold......oh i c

12. Astrophysics

$\large 110-60 = 10\log \frac{ I_{rc} }{ I_0 }-10\log \frac{ I_{nc} }{ I_0 }$ this is what we should have

13. Astrophysics

Now solve for $\frac{ I_{rc} }{ I_{nc} }$

14. YumYum247

ok wait let me solve this way, then maybe i'll make my own smaller and shorter way to do the same qestion......cuz that's too long....

15. Astrophysics

What do you mean?

16. Astrophysics

You want to compare the intensities, you're given the decibel readings

17. Astrophysics

So you should have a ratio, no units.

18. YumYum247

|dw:1436314773756:dw|

19. Astrophysics

Mhm?

20. Astrophysics

You need to know your logarithm rules

21. YumYum247

|dw:1436314848076:dw|

22. YumYum247

60

23. YumYum247

|dw:1436314962935:dw|

24. Astrophysics

$110-60 = 10\log \frac{ I_{rc} }{ I_0 }-10\log \frac{ I_{nc} }{ I_0 } \implies 50 = 10\log \frac{ I_{rc} }{ I_0 }-10\log \frac{ I_{nc} }{ I_0 }$ right? Now we want to solve for the ratio $50 = 10\left( 10 \log \frac{ I_{rc} }{ I_0 } \times \frac{ I_0 }{ I_{nc} }\right) \implies 50 = 10 \log \frac{ I_{rc} }{ I_{nc} }$

25. YumYum247

so......wait

26. YumYum247

spit it out Irish!!! :"D

27. YumYum247

i don't know what do do here....

28. IrishBoy123

lol!! small typo in last bit but usual brilliant job from @Astrophysics as the song goes: "this is how you do it"

29. YumYum247

since |dw:1436315441718:dw|

30. IrishBoy123

look at everything @Astrophysics has posted and ignore everything else in this thread that is *the path*

31. Astrophysics

Oh haha, I see the type, thanks for pointing it out @IrishBoy123 and thanks for the comment that means a lot coming from you!

32. YumYum247

we want to know how mch is 110dB louder than 60db...

33. Astrophysics

$50 = 10\left( \log \frac{ I_{rc} }{ I_0 } \times \frac{ I_0 }{ I_{nc} }\right) \implies 50 = 10 \log \frac{ I_{rc} }{ I_{nc} }$ this is it

34. YumYum247

Haaannn????????????????O_O

35. Astrophysics

So now we have $5 = \log \frac{ I_{rc} }{ I_{nc} } \implies 10^5 = \frac{ I_{rc} }{ I_{nc} }$

36. Astrophysics

That's intense :o

37. Astrophysics

38. YumYum247

i don't get how did you get from there to here?!?!?!?

39. YumYum247

can you help me learn the way i did it?????please

40. YumYum247

|dw:1436315726234:dw|

41. Astrophysics

I don't know what that is lol

42. Astrophysics

It's best if you do the algebra first then the calculations after

43. YumYum247

|dw:1436315751186:dw|

44. YumYum247

|dw:1436315795001:dw|

45. YumYum247

|dw:1436315857550:dw|

46. Astrophysics

No, that doesn't even make sense

47. YumYum247

Awwwwnnnnnnn!!! :"(

48. Astrophysics

Haha, well lets go over it again I suppose

49. YumYum247

ok we have 110dB and 60dB, we want to know how much 110dB is louder than 60dB

50. Astrophysics

Ok so your question is asking how much more intense is one sound over the other, to figure this out we have to make a ratio, the ratio is a unit-less number that tells us how much bigger one thing is than the other.

51. YumYum247

ok

52. Astrophysics

So we label the intensities we are looking for as $I_{rc} = \text{rock concert}$ and $I_{nc} = \text{normal conversation}$ and we want to compare how much louder is the rock concert than the normal conversation. So we are looking for $\frac{ I_{rc} }{ I_{nc} }$ so far so good?

53. YumYum247

yep.

54. Astrophysics

Alright, now notice we don't have either of the intensities, but we are given the decibels which is represented as $\beta$ in the following formula $\beta = 10 \log \frac{ I }{ I_0 }$

55. YumYum247

B as beta?

56. Astrophysics

yes

57. YumYum247

aight...go on!!! :")

58. Astrophysics

So we represent the decibels we're given as $\beta _{rc} = 110 dB~~~\text{and}~~~\beta_{nc} = 60 dB$ good?

59. YumYum247

good.

60. Astrophysics

Ok, so we have $\beta _{rc} =10\log \frac{ I_{rc} }{ I_0 }~~~\text{and}~~~\beta_{nc} = 10\log \frac{ I_{nc} }{ I_0 }$

61. YumYum247

ok

62. YumYum247

so we ply in the numbers into those two separete formulas to get our numbers?!!?!?!?

63. Astrophysics

Now since we want to compare the intensities, we need to take the difference of the decibel readings, where we have $\beta_{rc} - \beta _{nc} = 10\log \frac{ I_{rc} }{ I_{0} } - 10 \log \frac{ I_{nc} }{ I_0 }$

64. Astrophysics

Our goal here is to get the ratio $\frac{ I_{rc} }{ I_{nc} }$

65. YumYum247

ok but can you just say βnc = 10logIrcI0 and plug in the numbers to get the difference.

66. Astrophysics

No, because you don't have the intensities in the first place, that's what we're looking for, otherwise you're just taking the difference of the decibels which isn't really useful.

67. YumYum247

Oh ok i got it.....but it's too long.....

68. YumYum247

i wanted to do all that in 2 steps :"(

69. YumYum247

Thank you very very much!!!! :") and thank yo Irish!!! :)

70. Astrophysics

Np :)

71. IrishBoy123

:"( :"| :") :")) :"))) life as it should be!! good night folks!!

72. YumYum247

stay yum yum!!!

73. Astrophysics

Haha, I agree! You can see the struggle in the faces and then eventually turning into happiness :)

74. YumYum247

(logIrcI0×I0Inc) how did you get that.....?!?!?!?! Since 10logIrcI0−10logIncI0

75. YumYum247

how did yo trn two subtracting emtities into multiplication??????

76. Astrophysics

Oh logarithm rules $\log_ax - \log_a y = \log_a \left( \frac{ x }{ y } \right)$ and then since were dividing we flip the second fraction and multiply

77. YumYum247

ok. thanks!!

78. YumYum247

ok so do yo remember the question i asked you yesterday.....about triangle and rectangle.......was it right the way i did it???

79. YumYum247

80. YumYum247

is this really the remainder???????????????????|dw:1436317748361:dw|