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- YumYum247

7

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- 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

@aaronq

- YumYum247

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

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## More answers

- YumYum247

@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

50 thousand

- Astrophysics

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

- YumYum247

what is I/I0

- YumYum247

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

- Astrophysics

\[I_0 = 10^{-12}W/m^2\]

- YumYum247

threshold......oh i c

- 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

Now solve for \[\frac{ I_{rc} }{ I_{nc} }\]

- 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

What do you mean?

- Astrophysics

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

- Astrophysics

So you should have a ratio, no units.

- YumYum247

|dw:1436314773756:dw|

- Astrophysics

Mhm?

- Astrophysics

You need to know your logarithm rules

- YumYum247

|dw:1436314848076:dw|

- YumYum247

60

- YumYum247

|dw:1436314962935:dw|

- 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

so......wait

- YumYum247

spit it out Irish!!! :"D

- YumYum247

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

- IrishBoy123

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

- YumYum247

since |dw:1436315441718:dw|

- IrishBoy123

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

- 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

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

- 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

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

- Astrophysics

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

- Astrophysics

That's intense :o

- Astrophysics

Please ask if you don't follow something @YumYum247

- YumYum247

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

- YumYum247

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

- YumYum247

|dw:1436315726234:dw|

- Astrophysics

I don't know what that is lol

- Astrophysics

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

- YumYum247

|dw:1436315751186:dw|

- YumYum247

|dw:1436315795001:dw|

- YumYum247

|dw:1436315857550:dw|

- Astrophysics

No, that doesn't even make sense

- YumYum247

Awwwwnnnnnnn!!! :"(

- Astrophysics

Haha, well lets go over it again I suppose

- YumYum247

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

- 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

ok

- 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

yep.

- 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

B as beta?

- Astrophysics

yes

- YumYum247

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

- Astrophysics

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

- YumYum247

good.

- 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

ok

- YumYum247

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

- 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

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

- YumYum247

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

- 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

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

- YumYum247

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

- YumYum247

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

- Astrophysics

Np :)

- IrishBoy123

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

- YumYum247

stay yum yum!!!

- Astrophysics

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

- YumYum247

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

- YumYum247

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

- 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

ok. thanks!!

- 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

##### 1 Attachment

- YumYum247

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

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