toxicsugar22
  • toxicsugar22
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
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
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,\[\frac{I_0}{I_{pain}}=\frac{r_{pain}^2}{r_0^2} \rightarrow r_{pain}=\sqrt{\frac{I_0}{I_{pain}}}r_0=\frac{100}{\sqrt{3}}m\]
toxicsugar22
  • toxicsugar22
is that the answere
toxicsugar22
  • toxicsugar22
can you show me

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anonymous
  • anonymous
that's the answer. pain dist. ~ 57.7m
toxicsugar22
  • toxicsugar22
ok and what about the sound intensity of the concert
toxicsugar22
  • toxicsugar22
and this time put it in scientific notation
anonymous
  • anonymous
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.
toxicsugar22
  • toxicsugar22
whait what I dont get that
toxicsugar22
  • toxicsugar22
i did not understand so that is no the answer you gave me the first on
anonymous
  • anonymous
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\[D=10\log_{10}(10^{12}I)\]
anonymous
  • anonymous
You should post this question again so someone else can take it. I'm not round for much longer.
toxicsugar22
  • toxicsugar22
ok but ou ar to good in math
toxicsugar22
  • toxicsugar22
If you dont mind me asking what grade are you in
anonymous
  • anonymous
University.
toxicsugar22
  • toxicsugar22
wow
anonymous
  • anonymous
Are you doing a timed test or something?
toxicsugar22
  • toxicsugar22
im in a community collesge
toxicsugar22
  • toxicsugar22
college
anonymous
  • anonymous
is this an online assignment?
toxicsugar22
  • toxicsugar22
yes why
toxicsugar22
  • toxicsugar22
one more question befroe leaving
anonymous
  • anonymous
when is it due? i can look at it later...that's all. when i'm rushed me, i make mistakes.
toxicsugar22
  • toxicsugar22
can you go to the other question ou answerd for me the when about the websites
toxicsugar22
  • toxicsugar22
here i will post it to you
anonymous
  • anonymous
i can't find it. if you have the link, post it.
toxicsugar22
  • toxicsugar22
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
anonymous
  • anonymous
8000=4000(1.15)^t so 2=1.15^t \[t=\frac{\log 2}{\log 1.15} \approx 4.96\]months
toxicsugar22
  • toxicsugar22
and 16000
toxicsugar22
  • toxicsugar22
how long will it take fr the number of sites to reach 16000
anonymous
  • anonymous
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
toxicsugar22
  • toxicsugar22
thanks
anonymous
  • anonymous
ok
toxicsugar22
  • toxicsugar22
you are thhe bomb
anonymous
  • anonymous
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.
toxicsugar22
  • toxicsugar22
ok
anonymous
  • anonymous
you still there
anonymous
  • anonymous
you still there
anonymous
  • anonymous
are you looking for toxicsugar?
anonymous
  • anonymous
yeah, i was going to say
anonymous
  • anonymous
I = P / 4pi r^2
anonymous
  • anonymous
Yeah...I skipped it. I think he just wants answers, though.
anonymous
  • anonymous
:s
anonymous
  • anonymous
yeah
anonymous
  • anonymous
im just googling this, learning it
anonymous
  • anonymous
ah
anonymous
  • anonymous
apparently it is difficult to change db to hertz
anonymous
  • anonymous
is it?
anonymous
  • anonymous
how can you
anonymous
  • anonymous
ok how many hertz is 90 db
anonymous
  • anonymous
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.
anonymous
  • anonymous
yeah but you solved this problem well. you know your physics
anonymous
  • anonymous
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)
anonymous
  • anonymous
Sorry, was away. It' 20N.
anonymous
  • anonymous
how did you get that
anonymous
  • anonymous
give me a minute. i need coffee.
anonymous
  • anonymous
sure
anonymous
  • anonymous
hey
anonymous
  • anonymous
sry, back
anonymous
  • anonymous
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.
anonymous
  • anonymous
hmmm
anonymous
  • anonymous
what if one of the astronauts died, and the live astronaut pulls on the dead one 20 N
anonymous
  • anonymous
It's a third law of motion thing.
anonymous
  • anonymous
is the tension still 20 N
anonymous
  • anonymous
There'd be no tension.
anonymous
  • anonymous
there is, because the dead astronaut has weight
anonymous
  • anonymous
err, inertia i mean
anonymous
  • anonymous
Oh...I thought he floated away :)
anonymous
  • anonymous
no, he just died when the moment came to pull on it
anonymous
  • anonymous
lol, are you just coming up with this?
anonymous
  • anonymous
If he died and is still attached, the tension will be whatever force the live astronaut applies to accelerate him or her.
anonymous
  • anonymous
The force is 'communicated' through the rope by the third law.
anonymous
  • anonymous
so its 20 N in both cases
anonymous
  • anonymous
If that's what the live one is pulling the dead one with.
anonymous
  • anonymous
just found this - http://physics.stackexchange.com/questions/1220/rope-tension-question it might help
anonymous
  • anonymous
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
anonymous
  • anonymous
so there is a net force of 10 N on the rope. so why doesnt it accelerate 10/1 = a
anonymous
  • anonymous
10 m/s^2
anonymous
  • anonymous
I'm not sure what you mean. If there is a net force on the rope, it will accelerate.
anonymous
  • anonymous
You have to remember that most questions in physics on this matter assume a massless rope...
anonymous
  • anonymous
right but 10 m/s^2 is crazy fast , it will shoot off
anonymous
  • anonymous
It's okay if it's fast.
anonymous
  • anonymous
but the masses are accelerating VERY slow
anonymous
  • anonymous
I'm not sure what you mean.
anonymous
  • anonymous
say the guy has a mass of 50 kg
anonymous
  • anonymous
Oh...I see...the tension would fall off since the rope wouldn't be taught.
anonymous
  • anonymous
Actually, no...the acceleration would remain the same assuming they're all accelerating at 10m/s^2
anonymous
  • anonymous
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
anonymous
  • anonymous
so 30 - T = 50*a T - 20 = 40 *a
anonymous
  • anonymous
The tension's 20N
anonymous
  • anonymous
Do a free body diagram for each astronaut
anonymous
  • anonymous
The net force on each is 20N to the right.
anonymous
  • anonymous
i changed my problem
anonymous
  • anonymous
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
anonymous
  • anonymous
The rope communicates that force.
anonymous
  • anonymous
but the tension is not 20
anonymous
  • anonymous
most people think that, but if you draw the force body diagram for the guys you have
anonymous
  • anonymous
the tension is actually 24.444
anonymous
  • anonymous
lets call m1 = 50 kg, and m2 = 40 kg
anonymous
  • anonymous
the forces on m1 is 30 N to the right, and has tension to the left. so 30 - T = 50* a
anonymous
  • anonymous
I get 18 + 8/9.
anonymous
  • anonymous
The concept of tension remains the same, except when answering the second question, I got slack and didn't bother to calculate things.
anonymous
  • anonymous
so do you agree that
anonymous
  • anonymous
30 - T = 50 * a T - 10 = 40 * a
anonymous
  • anonymous
Yes, so long as we assume the rope is fully taught.
anonymous
  • anonymous
I agree with those equations.
anonymous
  • anonymous
oh so 18.88888 N in tension
anonymous
  • anonymous
so if the rope has 1kg say, it is accelerating 18.8888 ms^-2
anonymous
  • anonymous
Actually, I think these equations are for massless rope.
anonymous
  • anonymous
ok
anonymous
  • anonymous
suppose that it is massless. and the rope has two forces on it, at either end, so F1 - F2 = 0*a, so F1 = F2
anonymous
  • anonymous
but the forces at the end are unequal
anonymous
  • anonymous
Why do you get F1-F2=0? The object's accelerating.
anonymous
  • anonymous
For it to remain taught and communicate the force, it must be accelerating with the system.
anonymous
  • anonymous
but yuo said massless
anonymous
  • anonymous
so m = 0
anonymous
  • anonymous
F1 - F2 = 0*a
anonymous
  • anonymous
Sorry, I'm being distracted.
anonymous
  • anonymous
thats ok
anonymous
  • anonymous
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
anonymous
  • anonymous
does that make sense?
anonymous
  • anonymous
Oh, I thought you left.
anonymous
  • anonymous
nope
anonymous
  • anonymous
Problems are coming in because of the massless thing.
anonymous
  • anonymous
In reality, the rope will have mass.
anonymous
  • anonymous
so the rope exerts a force of 18.888 on either end
anonymous
  • anonymous
The rope carries the force through tension.
anonymous
  • anonymous
I'm trying to come up with a physical example.
anonymous
  • anonymous
Sorry, I was preoccupied. Set the problem up like the attachment that's coming.
anonymous
  • anonymous
anonymous
  • anonymous
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,\[T(x)=F_2+\frac{m}{L}(L-x)a\]
anonymous
  • anonymous
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\[T(x)=F_2+\frac{m}{L}(L-x)\frac{F_1-F_2}{M+m}\]
anonymous
  • anonymous
...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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