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toxicsugar22
 5 years ago
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.
toxicsugar22
 5 years ago
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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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Intensity 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
 5 years ago
Best ResponseYou've already chosen the best response.0is that the answere

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0that's the answer. pain dist. ~ 57.7m

toxicsugar22
 5 years ago
Best ResponseYou've already chosen the best response.0ok and what about the sound intensity of the concert

toxicsugar22
 5 years ago
Best ResponseYou've already chosen the best response.0and this time put it in scientific notation

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0You need to be using the decibel formula or something. http://hyperphysics.phyastr.gsu.edu/hbase/sound/intens.html You should post new questions in the question box.

toxicsugar22
 5 years ago
Best ResponseYou've already chosen the best response.0whait what I dont get that

toxicsugar22
 5 years ago
Best ResponseYou've already chosen the best response.0i did not understand so that is no the answer you gave me the first on

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0The 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
 5 years ago
Best ResponseYou've already chosen the best response.0You should post this question again so someone else can take it. I'm not round for much longer.

toxicsugar22
 5 years ago
Best ResponseYou've already chosen the best response.0ok but ou ar to good in math

toxicsugar22
 5 years ago
Best ResponseYou've already chosen the best response.0If you dont mind me asking what grade are you in

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Are you doing a timed test or something?

toxicsugar22
 5 years ago
Best ResponseYou've already chosen the best response.0im in a community collesge

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0is this an online assignment?

toxicsugar22
 5 years ago
Best ResponseYou've already chosen the best response.0one more question befroe leaving

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0when is it due? i can look at it later...that's all. when i'm rushed me, i make mistakes.

toxicsugar22
 5 years ago
Best ResponseYou've already chosen the best response.0can you go to the other question ou answerd for me the when about the websites

toxicsugar22
 5 years ago
Best ResponseYou've already chosen the best response.0here i will post it to you

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i can't find it. if you have the link, post it.

toxicsugar22
 5 years ago
Best ResponseYou've already chosen the best response.0the 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
 5 years ago
Best ResponseYou've already chosen the best response.08000=4000(1.15)^t so 2=1.15^t \[t=\frac{\log 2}{\log 1.15} \approx 4.96\]months

toxicsugar22
 5 years ago
Best ResponseYou've already chosen the best response.0how long will it take fr the number of sites to reach 16000

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.016000=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

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0The 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.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0are you looking for toxicsugar?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0yeah, i was going to say

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Yeah...I skipped it. I think he just wants answers, though.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0im just googling this, learning it

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0apparently it is difficult to change db to hertz

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ok how many hertz is 90 db

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I 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
 5 years ago
Best ResponseYou've already chosen the best response.0yeah but you solved this problem well. you know your physics

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i 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
 5 years ago
Best ResponseYou've already chosen the best response.0Sorry, was away. It' 20N.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0give me a minute. i need coffee.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Tension 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
 5 years ago
Best ResponseYou've already chosen the best response.0what if one of the astronauts died, and the live astronaut pulls on the dead one 20 N

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0It's a third law of motion thing.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0is the tension still 20 N

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0There'd be no tension.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0there is, because the dead astronaut has weight

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Oh...I thought he floated away :)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0no, he just died when the moment came to pull on it

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0lol, are you just coming up with this?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0If he died and is still attached, the tension will be whatever force the live astronaut applies to accelerate him or her.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0The force is 'communicated' through the rope by the third law.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so its 20 N in both cases

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0If that's what the live one is pulling the dead one with.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0just found this  http://physics.stackexchange.com/questions/1220/ropetensionquestion it might help

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0how 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
 5 years ago
Best ResponseYou've already chosen the best response.0so there is a net force of 10 N on the rope. so why doesnt it accelerate 10/1 = a

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I'm not sure what you mean. If there is a net force on the rope, it will accelerate.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0You have to remember that most questions in physics on this matter assume a massless rope...

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0right but 10 m/s^2 is crazy fast , it will shoot off

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0It's okay if it's fast.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0but the masses are accelerating VERY slow

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I'm not sure what you mean.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0say the guy has a mass of 50 kg

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Oh...I see...the tension would fall off since the rope wouldn't be taught.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Actually, no...the acceleration would remain the same assuming they're all accelerating at 10m/s^2

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0the 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
 5 years ago
Best ResponseYou've already chosen the best response.0so 30  T = 50*a T  20 = 40 *a

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Do a free body diagram for each astronaut

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0The net force on each is 20N to the right.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0the 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
 5 years ago
Best ResponseYou've already chosen the best response.0The rope communicates that force.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0but the tension is not 20

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0most people think that, but if you draw the force body diagram for the guys you have

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0the tension is actually 24.444

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0lets call m1 = 50 kg, and m2 = 40 kg

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0the forces on m1 is 30 N to the right, and has tension to the left. so 30  T = 50* a

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0The concept of tension remains the same, except when answering the second question, I got slack and didn't bother to calculate things.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.030  T = 50 * a T  10 = 40 * a

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Yes, so long as we assume the rope is fully taught.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I agree with those equations.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0oh so 18.88888 N in tension

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so if the rope has 1kg say, it is accelerating 18.8888 ms^2

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Actually, I think these equations are for massless rope.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0suppose that it is massless. and the rope has two forces on it, at either end, so F1  F2 = 0*a, so F1 = F2

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0but the forces at the end are unequal

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Why do you get F1F2=0? The object's accelerating.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0For it to remain taught and communicate the force, it must be accelerating with the system.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0but yuo said massless

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Sorry, I'm being distracted.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0if 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
 5 years ago
Best ResponseYou've already chosen the best response.0does that make sense?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Oh, I thought you left.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Problems are coming in because of the massless thing.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0In reality, the rope will have mass.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so the rope exerts a force of 18.888 on either end

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0The rope carries the force through tension.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I'm trying to come up with a physical example.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Sorry, I was preoccupied. Set the problem up like the attachment that's coming.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0The 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\[\frac{m}{L}(Lx)\]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\[F_2+\frac{m}{L}(Lx)aT(x)=\left( m+\frac{F_1}{a}+\frac{F_2}{a} \right)a\]where we're dealing with magnitudes (directions of forces are contained in the signs). Solving for T(x), we have,\[T(x)=F_2+\frac{m}{L}(Lx)a\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Now, since acceleration is constant throughout, at the point x, we have by equating forces, \[F_2maF_2=Ma\]where M is the total mass of the system. Hence,\[a=\frac{F_1F_2}{M+m}\]so that\[T(x)=F_2+\frac{m}{L}(Lx)\frac{F_1F_2}{M+m}\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0...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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