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YumYum247

  • one year ago

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  1. YumYum247
    • one year ago
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    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
    • one year ago
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    @aaronq

  3. YumYum247
    • one year ago
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    @MrNood can you check my work?!?!?!?!?!

  4. YumYum247
    • one year ago
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    @Astrophysics

  5. Astrophysics
    • one year ago
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    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
    • one year ago
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    50 thousand

  7. Astrophysics
    • one year ago
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    \[10\log \frac{ I_{rc} }{ I_0 }-10\log \frac{ I{nc} }{ I_0 }\]

  8. YumYum247
    • one year ago
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    what is I/I0

  9. YumYum247
    • one year ago
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    so in this case, I = 110dB/60dB?!?!?!?!?!

  10. Astrophysics
    • one year ago
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    \[I_0 = 10^{-12}W/m^2\]

  11. YumYum247
    • one year ago
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    threshold......oh i c

  12. Astrophysics
    • one year ago
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    \[\large 110-60 = 10\log \frac{ I_{rc} }{ I_0 }-10\log \frac{ I_{nc} }{ I_0 }\] this is what we should have

  13. Astrophysics
    • one year ago
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    Now solve for \[\frac{ I_{rc} }{ I_{nc} }\]

  14. YumYum247
    • one year ago
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    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
    • one year ago
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    What do you mean?

  16. Astrophysics
    • one year ago
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    You want to compare the intensities, you're given the decibel readings

  17. Astrophysics
    • one year ago
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    So you should have a ratio, no units.

  18. YumYum247
    • one year ago
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    |dw:1436314773756:dw|

  19. Astrophysics
    • one year ago
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    Mhm?

  20. Astrophysics
    • one year ago
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    You need to know your logarithm rules

  21. YumYum247
    • one year ago
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    |dw:1436314848076:dw|

  22. YumYum247
    • one year ago
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    60

  23. YumYum247
    • one year ago
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    |dw:1436314962935:dw|

  24. Astrophysics
    • one year ago
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    \[110-60 = 10\log \frac{ I_{rc} }{ I_0 }-10\log \frac{ I_{nc} }{ I_0 } \implies 50 = 10\log \frac{ I_{rc} }{ I_0 }-10\log \frac{ I_{nc} }{ I_0 } \] 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
    • one year ago
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    so......wait

  26. YumYum247
    • one year ago
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    spit it out Irish!!! :"D

  27. YumYum247
    • one year ago
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    i don't know what do do here....

  28. IrishBoy123
    • one year ago
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    lol!! small typo in last bit but usual brilliant job from @Astrophysics as the song goes: "this is how you do it"

  29. YumYum247
    • one year ago
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    since |dw:1436315441718:dw|

  30. IrishBoy123
    • one year ago
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    look at everything @Astrophysics has posted and ignore everything else in this thread that is *the path*

  31. Astrophysics
    • one year ago
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    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
    • one year ago
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    we want to know how mch is 110dB louder than 60db...

  33. Astrophysics
    • one year ago
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    \[50 = 10\left( \log \frac{ I_{rc} }{ I_0 } \times \frac{ I_0 }{ I_{nc} }\right) \implies 50 = 10 \log \frac{ I_{rc} }{ I_{nc} }\] this is it

  34. YumYum247
    • one year ago
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    Haaannn????????????????O_O

  35. Astrophysics
    • one year ago
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    So now we have \[5 = \log \frac{ I_{rc} }{ I_{nc} } \implies 10^5 = \frac{ I_{rc} }{ I_{nc} }\]

  36. Astrophysics
    • one year ago
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    That's intense :o

  37. Astrophysics
    • one year ago
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    Please ask if you don't follow something @YumYum247

  38. YumYum247
    • one year ago
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    i don't get how did you get from there to here?!?!?!?

  39. YumYum247
    • one year ago
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    can you help me learn the way i did it?????please

  40. YumYum247
    • one year ago
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    |dw:1436315726234:dw|

  41. Astrophysics
    • one year ago
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    I don't know what that is lol

  42. Astrophysics
    • one year ago
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    It's best if you do the algebra first then the calculations after

  43. YumYum247
    • one year ago
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    |dw:1436315751186:dw|

  44. YumYum247
    • one year ago
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    |dw:1436315795001:dw|

  45. YumYum247
    • one year ago
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    |dw:1436315857550:dw|

  46. Astrophysics
    • one year ago
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    No, that doesn't even make sense

  47. YumYum247
    • one year ago
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    Awwwwnnnnnnn!!! :"(

  48. Astrophysics
    • one year ago
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    Haha, well lets go over it again I suppose

  49. YumYum247
    • one year ago
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    ok we have 110dB and 60dB, we want to know how much 110dB is louder than 60dB

  50. Astrophysics
    • one year ago
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    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
    • one year ago
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    ok

  52. Astrophysics
    • one year ago
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    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
    • one year ago
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    yep.

  54. Astrophysics
    • one year ago
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    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
    • one year ago
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    B as beta?

  56. Astrophysics
    • one year ago
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    yes

  57. YumYum247
    • one year ago
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    aight...go on!!! :")

  58. Astrophysics
    • one year ago
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    So we represent the decibels we're given as \[\beta _{rc} = 110 dB~~~\text{and}~~~\beta_{nc} = 60 dB\] good?

  59. YumYum247
    • one year ago
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    good.

  60. Astrophysics
    • one year ago
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    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
    • one year ago
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    ok

  62. YumYum247
    • one year ago
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    so we ply in the numbers into those two separete formulas to get our numbers?!!?!?!?

  63. Astrophysics
    • one year ago
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    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
    • one year ago
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    Our goal here is to get the ratio \[\frac{ I_{rc} }{ I_{nc} }\]

  65. YumYum247
    • one year ago
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    ok but can you just say βnc = 10logIrcI0 and plug in the numbers to get the difference.

  66. Astrophysics
    • one year ago
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    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
    • one year ago
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    Oh ok i got it.....but it's too long.....

  68. YumYum247
    • one year ago
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    i wanted to do all that in 2 steps :"(

  69. YumYum247
    • one year ago
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    Thank you very very much!!!! :") and thank yo Irish!!! :)

  70. Astrophysics
    • one year ago
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    Np :)

  71. IrishBoy123
    • one year ago
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    :"( :"| :") :")) :"))) life as it should be!! good night folks!!

  72. YumYum247
    • one year ago
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    stay yum yum!!!

  73. Astrophysics
    • one year ago
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    Haha, I agree! You can see the struggle in the faces and then eventually turning into happiness :)

  74. YumYum247
    • one year ago
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    (logIrcI0×I0Inc) how did you get that.....?!?!?!?! Since 10logIrcI0−10logIncI0

  75. YumYum247
    • one year ago
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    how did yo trn two subtracting emtities into multiplication??????

  76. Astrophysics
    • one year ago
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    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
    • one year ago
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    ok. thanks!!

  78. YumYum247
    • one year ago
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    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
    • one year ago
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  80. YumYum247
    • one year ago
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    is this really the remainder???????????????????|dw:1436317748361:dw|

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