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

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

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