At a concert by The Who in 1976, the sound level 50 meters from the stage registered 120 decibels. The threshold of pain for the human ear is 90 decibles. 1. calculate the sound intensity of the Who concert and the sound intensity of the threshold of pain.

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At a concert by The Who in 1976, the sound level 50 meters from the stage registered 120 decibels. The threshold of pain for the human ear is 90 decibles. 1. calculate the sound intensity of the Who concert and the sound intensity of the threshold of pain.

Mathematics
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Intensity is inversely proportional to the square of the distance, \[I=\frac{k}{r^2}\]k is some constant. You can compare intensities to distance as,\[\frac{I_0}{I_{pain}}=\frac{r_{pain}^2}{r_0^2} \rightarrow r_{pain}=\sqrt{\frac{I_0}{I_{pain}}}r_0=\frac{100}{\sqrt{3}}m\]
is that the answere
can you show me

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that's the answer. pain dist. ~ 57.7m
ok and what about the sound intensity of the concert
and this time put it in scientific notation
You need to be using the decibel formula or something. http://hyperphysics.phy-astr.gsu.edu/hbase/sound/intens.html You should post new questions in the question box.
whait what I dont get that
i did not understand so that is no the answer you gave me the first on
The answer for the distance is wrong. I didn't read decibels; I took it as intensity. You have to convert decibels to intensity using the formula\[D=10\log_{10}(10^{12}I)\]
You should post this question again so someone else can take it. I'm not round for much longer.
ok but ou ar to good in math
If you dont mind me asking what grade are you in
University.
wow
Are you doing a timed test or something?
im in a community collesge
college
is this an online assignment?
yes why
one more question befroe leaving
when is it due? i can look at it later...that's all. when i'm rushed me, i make mistakes.
can you go to the other question ou answerd for me the when about the websites
here i will post it to you
i can't find it. if you have the link, post it.
the number of websites is growing by 15% each month. In June 2005, there were 4000 websites. Write a formula for the number of websites as a function of months since june 2005.38 minutes ago N(t)=4000(1.15) t Now with that question how long will it take fr the number of sites to reach 8000? Give the answer in exact value and decimal form
8000=4000(1.15)^t so 2=1.15^t \[t=\frac{\log 2}{\log 1.15} \approx 4.96\]months
and 16000
how long will it take fr the number of sites to reach 16000
16000=4000(1.15)^t so 4=(1.15)^t\[\log 4 = t \log 1.15 \rightarrow t=\frac{\log 4}{\log 1.15} \approx 9.92\]months
thanks
ok
you are thhe bomb
The Who question:using the formula I gave you,\[120=10\log_{10}(10^{12}I) \rightarrow I=1\]i.e. Intensity is 1 Watt/m^2 at 50 meters. To find out the distance for 90 decibels,\[90=10\log _{10}(10^{12}I_{pain}) \rightarrow I_{pain}=10^{-10}\]Now, again, intensity is inversely proportional to the square of the distance, so\[\frac{I_0}{I_{pain}}=\frac{k/r_0^2}{k/r_{pain}^2}=\frac{r_{pain}^2}{r_0^2}\]so that\[r^2_{pain}=r_0^2\frac{I_0}{I_{pain}}=50^2 \times \frac{1}{10^{-3}}=2.5 \times 10^6m^2 \]so the pain distance is\[r_{pain}=\sqrt{2.5 \times 10^6}\approx1581m\]I don't know what level of significance you want for significant figures, so I can't do that part.
ok
you still there
you still there
are you looking for toxicsugar?
yeah, i was going to say
I = P / 4pi r^2
Yeah...I skipped it. I think he just wants answers, though.
:s
yeah
im just googling this, learning it
ah
apparently it is difficult to change db to hertz
is it?
how can you
ok how many hertz is 90 db
I don't know. I haven't really looked into it. I just saw a couple of things doing a search. It seems to depend on the context.
yeah but you solved this problem well. you know your physics
i have question. two astronauts are pulling on each other with 20 N on a rope. what is the tension in the rope (no gravity here)
Sorry, was away. It' 20N.
how did you get that
give me a minute. i need coffee.
sure
hey
sry, back
Tension isn't a force, which is what screws people up. Mathematically, it's a rank 2 tensor - a function that takes a vector and spits out another. The input is a direction vector and the output is the force exerted on that part of the rope you're looking at, in the direction of the vector you plugged in. We know the direction up yields 20N, and direction down yields 20N...we *say* the tension in the rope is 20N.
hmmm
what if one of the astronauts died, and the live astronaut pulls on the dead one 20 N
It's a third law of motion thing.
is the tension still 20 N
There'd be no tension.
there is, because the dead astronaut has weight
err, inertia i mean
Oh...I thought he floated away :)
no, he just died when the moment came to pull on it
lol, are you just coming up with this?
If he died and is still attached, the tension will be whatever force the live astronaut applies to accelerate him or her.
The force is 'communicated' through the rope by the third law.
so its 20 N in both cases
If that's what the live one is pulling the dead one with.
just found this - http://physics.stackexchange.com/questions/1220/rope-tension-question it might help
how come, say there is mass in the rope, 1kg and one guy pulls 30 N and the other pulls 20 N in opposite direction
so there is a net force of 10 N on the rope. so why doesnt it accelerate 10/1 = a
10 m/s^2
I'm not sure what you mean. If there is a net force on the rope, it will accelerate.
You have to remember that most questions in physics on this matter assume a massless rope...
right but 10 m/s^2 is crazy fast , it will shoot off
It's okay if it's fast.
but the masses are accelerating VERY slow
I'm not sure what you mean.
say the guy has a mass of 50 kg
Oh...I see...the tension would fall off since the rope wouldn't be taught.
Actually, no...the acceleration would remain the same assuming they're all accelerating at 10m/s^2
the guy on the right is 50 kg, guy on left is 40 kg, the 50 kg guy is applying 30 N, and the 40 kg guy is applying 10 N , on a rope
so 30 - T = 50*a T - 20 = 40 *a
The tension's 20N
Do a free body diagram for each astronaut
The net force on each is 20N to the right.
i changed my problem
the guy on the right is 50 kg, guy on left is 40 kg, the 50 kg guy is applying 30 N, and the 40 kg guy is applying 10 N , on a rope
The rope communicates that force.
but the tension is not 20
most people think that, but if you draw the force body diagram for the guys you have
the tension is actually 24.444
lets call m1 = 50 kg, and m2 = 40 kg
the forces on m1 is 30 N to the right, and has tension to the left. so 30 - T = 50* a
I get 18 + 8/9.
The concept of tension remains the same, except when answering the second question, I got slack and didn't bother to calculate things.
so do you agree that
30 - T = 50 * a T - 10 = 40 * a
Yes, so long as we assume the rope is fully taught.
I agree with those equations.
oh so 18.88888 N in tension
so if the rope has 1kg say, it is accelerating 18.8888 ms^-2
Actually, I think these equations are for massless rope.
ok
suppose that it is massless. and the rope has two forces on it, at either end, so F1 - F2 = 0*a, so F1 = F2
but the forces at the end are unequal
Why do you get F1-F2=0? The object's accelerating.
For it to remain taught and communicate the force, it must be accelerating with the system.
but yuo said massless
so m = 0
F1 - F2 = 0*a
Sorry, I'm being distracted.
thats ok
if the tension is 18.8888 in our example, then 18.888 is the force that is exerted on the rope by both masses being pulled
does that make sense?
Oh, I thought you left.
nope
Problems are coming in because of the massless thing.
In reality, the rope will have mass.
so the rope exerts a force of 18.888 on either end
The rope carries the force through tension.
I'm trying to come up with a physical example.
Sorry, I was preoccupied. Set the problem up like the attachment that's coming.
The rope is assumed to have a uniform density, so mass per unit length will be constant. At the point x, the rope will have mass *to the right* of x of\[\frac{m}{L}(L-x)\]This mass, times the acceleration of the system (since the rope is taught) *plus* the force F_2, will equal the tension to the left at the point x. So we'd have\[F_2+\frac{m}{L}(L-x)a-T(x)=\left( m+\frac{F_1}{a}+\frac{F_2}{a} \right)a\]where we're dealing with magnitudes (directions of forces are contained in the signs). Solving for T(x), we have,\[T(x)=F_2+\frac{m}{L}(L-x)a\]
Now, since acceleration is constant throughout, at the point x, we have by equating forces, \[F_2-ma-F_2=Ma\]where M is the total mass of the system. Hence,\[a=\frac{F_1-F_2}{M+m}\]so that\[T(x)=F_2+\frac{m}{L}(L-x)\frac{F_1-F_2}{M+m}\]
...and if there's a mistake in there, I wouldn't be surprised. I rarely translate what's on my pad to this site intact.

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