YumYum247
  • YumYum247
7
Physics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
YumYum247
  • 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
YumYum247
  • YumYum247
@aaronq
YumYum247
  • YumYum247
@MrNood can you check my work?!?!?!?!?!

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YumYum247
  • YumYum247
@Astrophysics
Astrophysics
  • 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.
YumYum247
  • YumYum247
50 thousand
Astrophysics
  • Astrophysics
\[10\log \frac{ I_{rc} }{ I_0 }-10\log \frac{ I{nc} }{ I_0 }\]
YumYum247
  • YumYum247
what is I/I0
YumYum247
  • YumYum247
so in this case, I = 110dB/60dB?!?!?!?!?!
Astrophysics
  • Astrophysics
\[I_0 = 10^{-12}W/m^2\]
YumYum247
  • YumYum247
threshold......oh i c
Astrophysics
  • Astrophysics
\[\large 110-60 = 10\log \frac{ I_{rc} }{ I_0 }-10\log \frac{ I_{nc} }{ I_0 }\] this is what we should have
Astrophysics
  • Astrophysics
Now solve for \[\frac{ I_{rc} }{ I_{nc} }\]
YumYum247
  • 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....
Astrophysics
  • Astrophysics
What do you mean?
Astrophysics
  • Astrophysics
You want to compare the intensities, you're given the decibel readings
Astrophysics
  • Astrophysics
So you should have a ratio, no units.
YumYum247
  • YumYum247
|dw:1436314773756:dw|
Astrophysics
  • Astrophysics
Mhm?
Astrophysics
  • Astrophysics
You need to know your logarithm rules
YumYum247
  • YumYum247
|dw:1436314848076:dw|
YumYum247
  • YumYum247
60
YumYum247
  • YumYum247
|dw:1436314962935:dw|
Astrophysics
  • 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 \[50 = 10\left( 10 \log \frac{ I_{rc} }{ I_0 } \times \frac{ I_0 }{ I_{nc} }\right) \implies 50 = 10 \log \frac{ I_{rc} }{ I_{nc} }\]
YumYum247
  • YumYum247
so......wait
YumYum247
  • YumYum247
spit it out Irish!!! :"D
YumYum247
  • YumYum247
i don't know what do do here....
IrishBoy123
  • IrishBoy123
lol!! small typo in last bit but usual brilliant job from @Astrophysics as the song goes: "this is how you do it"
YumYum247
  • YumYum247
since |dw:1436315441718:dw|
IrishBoy123
  • IrishBoy123
look at everything @Astrophysics has posted and ignore everything else in this thread that is *the path*
Astrophysics
  • 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!
YumYum247
  • YumYum247
we want to know how mch is 110dB louder than 60db...
Astrophysics
  • 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
YumYum247
  • YumYum247
Haaannn????????????????O_O
Astrophysics
  • Astrophysics
So now we have \[5 = \log \frac{ I_{rc} }{ I_{nc} } \implies 10^5 = \frac{ I_{rc} }{ I_{nc} }\]
Astrophysics
  • Astrophysics
That's intense :o
Astrophysics
  • Astrophysics
Please ask if you don't follow something @YumYum247
YumYum247
  • YumYum247
i don't get how did you get from there to here?!?!?!?
YumYum247
  • YumYum247
can you help me learn the way i did it?????please
YumYum247
  • YumYum247
|dw:1436315726234:dw|
Astrophysics
  • Astrophysics
I don't know what that is lol
Astrophysics
  • Astrophysics
It's best if you do the algebra first then the calculations after
YumYum247
  • YumYum247
|dw:1436315751186:dw|
YumYum247
  • YumYum247
|dw:1436315795001:dw|
YumYum247
  • YumYum247
|dw:1436315857550:dw|
Astrophysics
  • Astrophysics
No, that doesn't even make sense
YumYum247
  • YumYum247
Awwwwnnnnnnn!!! :"(
Astrophysics
  • Astrophysics
Haha, well lets go over it again I suppose
YumYum247
  • YumYum247
ok we have 110dB and 60dB, we want to know how much 110dB is louder than 60dB
Astrophysics
  • 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.
YumYum247
  • YumYum247
ok
Astrophysics
  • 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?
YumYum247
  • YumYum247
yep.
Astrophysics
  • 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 \[\beta = 10 \log \frac{ I }{ I_0 }\]
YumYum247
  • YumYum247
B as beta?
Astrophysics
  • Astrophysics
yes
YumYum247
  • YumYum247
aight...go on!!! :")
Astrophysics
  • Astrophysics
So we represent the decibels we're given as \[\beta _{rc} = 110 dB~~~\text{and}~~~\beta_{nc} = 60 dB\] good?
YumYum247
  • YumYum247
good.
Astrophysics
  • Astrophysics
Ok, so we have \[\beta _{rc} =10\log \frac{ I_{rc} }{ I_0 }~~~\text{and}~~~\beta_{nc} = 10\log \frac{ I_{nc} }{ I_0 }\]
YumYum247
  • YumYum247
ok
YumYum247
  • YumYum247
so we ply in the numbers into those two separete formulas to get our numbers?!!?!?!?
Astrophysics
  • Astrophysics
Now since we want to compare the intensities, we need to take the difference of the decibel readings, where we have \[\beta_{rc} - \beta _{nc} = 10\log \frac{ I_{rc} }{ I_{0} } - 10 \log \frac{ I_{nc} }{ I_0 }\]
Astrophysics
  • Astrophysics
Our goal here is to get the ratio \[\frac{ I_{rc} }{ I_{nc} }\]
YumYum247
  • YumYum247
ok but can you just say βnc = 10logIrcI0 and plug in the numbers to get the difference.
Astrophysics
  • 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.
YumYum247
  • YumYum247
Oh ok i got it.....but it's too long.....
YumYum247
  • YumYum247
i wanted to do all that in 2 steps :"(
YumYum247
  • YumYum247
Thank you very very much!!!! :") and thank yo Irish!!! :)
Astrophysics
  • Astrophysics
Np :)
IrishBoy123
  • IrishBoy123
:"( :"| :") :")) :"))) life as it should be!! good night folks!!
YumYum247
  • YumYum247
stay yum yum!!!
Astrophysics
  • Astrophysics
Haha, I agree! You can see the struggle in the faces and then eventually turning into happiness :)
YumYum247
  • YumYum247
(logIrcI0×I0Inc) how did you get that.....?!?!?!?! Since 10logIrcI0−10logIncI0
YumYum247
  • YumYum247
how did yo trn two subtracting emtities into multiplication??????
Astrophysics
  • 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
YumYum247
  • YumYum247
ok. thanks!!
YumYum247
  • YumYum247
ok so do yo remember the question i asked you yesterday.....about triangle and rectangle.......was it right the way i did it???
YumYum247
  • YumYum247
YumYum247
  • YumYum247
is this really the remainder???????????????????|dw:1436317748361:dw|

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