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owl
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Hi there! Any help would be greatly appreciated. Thanks!
If a sound level of 0 dB at 1000 Hz corresponds to a maximum gauge pressure (sound amplitude) of 10^-9 atm, what is the maximum gauge pressure in a 50dB sound at the same frequency? (answer: 3x10^-7 atm)
L (in dB) = 10 log (I/I0)
If it is 0 dB that must mean that I0, ie the reference intensity is 10^-9 atm right?
So ie 0 = 10 log (10^-9/10^-9)
So to get an answer at 50dB... it would be 50 = 10 log ( ??/10^-9)
but then I'm getting 50/10 = log ( ??/10^-9)
10^5 = ?? / 10^-9 (going 10^x both sides to get rid of the log)
which leaves me with ?? = 10^14 which can't be (and isn't) right...?
And the other one...
What is the pressure drop due to the Bernoulli effect as water goes into a 3 cm diameter nozzle from a 9cm diameter fire hose while carrying a flow of 5 x 10^-4 m3s-1? (answer: 247 Pa)
So the bernoulli equation is P1 + 1/2 pv12 = P2 + 1/2 pv22
Leaving out the gravitational parts as horizontal flow is assumed...
(the little p being density = 1000 kgm^3 for water and the big P being pressure)
Well rearranging the bernoulli equation to get P2 - P1 I get
change in P = 1/2 p ( [Q / A1 ]2 - [ Q / A2 ]2 ] ) = 15.44 Pa?
The answer shows up as 247 Pa which I noticed to be approximately 15.44 squared... but I can't seem to come up with anything other than this!
Any help would be really appreciated. Thanks!
Homework Statement
If a sound level of 0 dB at 1000 Hz corresponds to a maximum gauge pressure (sound amplitude) of 10^-9 atm, what is the maximum gauge pressure in a 50dB sound at the same frequency? (answer: 3x10^-7 atm)
Homework Equations
L (in dB) = 10 log (I/I0)
The Attempt at a Solution
If it is 0 dB that must mean that I0, ie the reference intensity is 10^-9 atm right?
So ie 0 = 10 log (10^-9/10^-9)
So to get an answer at 50dB... it would be 50 = 10 log ( ??/10^-9)
but then I'm getting 50/10 = log ( ??/10^-9)
10^5 = ?? / 10^-9 (going 10^x both sides to get rid of the log)
which leaves me with ?? = 10^14 which can't be (and isn't) right...?
And the other one...
Homework Statement
What is the pressure drop due to the Bernoulli effect as water goes into a 3 cm diameter nozzle from a 9cm diameter fire hose while carrying a flow of 5 x 10^-4 m3s-1? (answer: 247 Pa)
Homework Equations
So the bernoulli equation is P1 + 1/2 pv12 = P2 + 1/2 pv22
Leaving out the gravitational parts as horizontal flow is assumed...
(the little p being density = 1000 kgm^3 for water and the big P being pressure)
The Attempt at a Solution
Well rearranging the bernoulli equation to get P2 - P1 I get
change in P = 1/2 p ( [Q / A1 ]2 - [ Q / A2 ]2 ] ) = 15.44 Pa?
The answer shows up as 247 Pa which I noticed to be approximately 15.44 squared... but I can't seem to come up with anything other than this!
Any help would be really appreciated. Thanks!