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YumYum247
 one year ago
7
YumYum247
 one year ago
7

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YumYum247
 one year ago
Best ResponseYou've already chosen the best response.0i 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
 one year ago
Best ResponseYou've already chosen the best response.0@MrNood can you check my work?!?!?!?!?!

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.4Well 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.

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.4\[10\log \frac{ I_{rc} }{ I_0 }10\log \frac{ I{nc} }{ I_0 }\]

YumYum247
 one year ago
Best ResponseYou've already chosen the best response.0so in this case, I = 110dB/60dB?!?!?!?!?!

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.4\[I_0 = 10^{12}W/m^2\]

YumYum247
 one year ago
Best ResponseYou've already chosen the best response.0threshold......oh i c

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.4\[\large 11060 = 10\log \frac{ I_{rc} }{ I_0 }10\log \frac{ I_{nc} }{ I_0 }\] this is what we should have

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.4Now solve for \[\frac{ I_{rc} }{ I_{nc} }\]

YumYum247
 one year ago
Best ResponseYou've already chosen the best response.0ok 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
 one year ago
Best ResponseYou've already chosen the best response.4What do you mean?

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.4You want to compare the intensities, you're given the decibel readings

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.4So you should have a ratio, no units.

YumYum247
 one year ago
Best ResponseYou've already chosen the best response.0dw:1436314773756:dw

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.4You need to know your logarithm rules

YumYum247
 one year ago
Best ResponseYou've already chosen the best response.0dw:1436314848076:dw

YumYum247
 one year ago
Best ResponseYou've already chosen the best response.0dw:1436314962935:dw

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.4\[11060 = 10\log \frac{ I_{rc} }{ I_0 }10\log \frac{ I_{nc} }{ I_0 } \implies 50 = 10\log \frac{ I_{rc} }{ I_0 }10\log \frac{ I_{nc} }{ I_0 } \] 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
 one year ago
Best ResponseYou've already chosen the best response.0spit it out Irish!!! :"D

YumYum247
 one year ago
Best ResponseYou've already chosen the best response.0i don't know what do do here....

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.1lol!! small typo in last bit but usual brilliant job from @Astrophysics as the song goes: "this is how you do it"

YumYum247
 one year ago
Best ResponseYou've already chosen the best response.0since dw:1436315441718:dw

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.1look at everything @Astrophysics has posted and ignore everything else in this thread that is *the path*

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.4Oh haha, I see the type, thanks for pointing it out @IrishBoy123 and thanks for the comment that means a lot coming from you!

YumYum247
 one year ago
Best ResponseYou've already chosen the best response.0we want to know how mch is 110dB louder than 60db...

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.4\[50 = 10\left( \log \frac{ I_{rc} }{ I_0 } \times \frac{ I_0 }{ I_{nc} }\right) \implies 50 = 10 \log \frac{ I_{rc} }{ I_{nc} }\] this is it

YumYum247
 one year ago
Best ResponseYou've already chosen the best response.0Haaannn????????????????O_O

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.4So now we have \[5 = \log \frac{ I_{rc} }{ I_{nc} } \implies 10^5 = \frac{ I_{rc} }{ I_{nc} }\]

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.4That's intense :o

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.4Please ask if you don't follow something @YumYum247

YumYum247
 one year ago
Best ResponseYou've already chosen the best response.0i don't get how did you get from there to here?!?!?!?

YumYum247
 one year ago
Best ResponseYou've already chosen the best response.0can you help me learn the way i did it?????please

YumYum247
 one year ago
Best ResponseYou've already chosen the best response.0dw:1436315726234:dw

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.4I don't know what that is lol

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.4It's best if you do the algebra first then the calculations after

YumYum247
 one year ago
Best ResponseYou've already chosen the best response.0dw:1436315751186:dw

YumYum247
 one year ago
Best ResponseYou've already chosen the best response.0dw:1436315795001:dw

YumYum247
 one year ago
Best ResponseYou've already chosen the best response.0dw:1436315857550:dw

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.4No, that doesn't even make sense

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.4Haha, well lets go over it again I suppose

YumYum247
 one year ago
Best ResponseYou've already chosen the best response.0ok we have 110dB and 60dB, we want to know how much 110dB is louder than 60dB

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.4Ok 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 unitless number that tells us how much bigger one thing is than the other.

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.4So 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?

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.4Alright, 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
 one year ago
Best ResponseYou've already chosen the best response.0aight...go on!!! :")

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.4So we represent the decibels we're given as \[\beta _{rc} = 110 dB~~~\text{and}~~~\beta_{nc} = 60 dB\] good?

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.4Ok, so we have \[\beta _{rc} =10\log \frac{ I_{rc} }{ I_0 }~~~\text{and}~~~\beta_{nc} = 10\log \frac{ I_{nc} }{ I_0 }\]

YumYum247
 one year ago
Best ResponseYou've already chosen the best response.0so we ply in the numbers into those two separete formulas to get our numbers?!!?!?!?

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.4Now 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
 one year ago
Best ResponseYou've already chosen the best response.4Our goal here is to get the ratio \[\frac{ I_{rc} }{ I_{nc} }\]

YumYum247
 one year ago
Best ResponseYou've already chosen the best response.0ok but can you just say βnc = 10logIrcI0 and plug in the numbers to get the difference.

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.4No, 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
 one year ago
Best ResponseYou've already chosen the best response.0Oh ok i got it.....but it's too long.....

YumYum247
 one year ago
Best ResponseYou've already chosen the best response.0i wanted to do all that in 2 steps :"(

YumYum247
 one year ago
Best ResponseYou've already chosen the best response.0Thank you very very much!!!! :") and thank yo Irish!!! :)

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.1:"( :" :") :")) :"))) life as it should be!! good night folks!!

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.4Haha, I agree! You can see the struggle in the faces and then eventually turning into happiness :)

YumYum247
 one year ago
Best ResponseYou've already chosen the best response.0(logIrcI0×I0Inc) how did you get that.....?!?!?!?! Since 10logIrcI0−10logIncI0

YumYum247
 one year ago
Best ResponseYou've already chosen the best response.0how did yo trn two subtracting emtities into multiplication??????

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.4Oh 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
 one year ago
Best ResponseYou've already chosen the best response.0ok so do yo remember the question i asked you yesterday.....about triangle and rectangle.......was it right the way i did it???

YumYum247
 one year ago
Best ResponseYou've already chosen the best response.0is this really the remainder???????????????????dw:1436317748361:dw
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