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Mathematics 18 Online
OpenStudy (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.

OpenStudy (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,\[\frac{I_0}{I_{pain}}=\frac{r_{pain}^2}{r_0^2} \rightarrow r_{pain}=\sqrt{\frac{I_0}{I_{pain}}}r_0=\frac{100}{\sqrt{3}}m\]

OpenStudy (toxicsugar22):

is that the answere

OpenStudy (toxicsugar22):

can you show me

OpenStudy (anonymous):

that's the answer. pain dist. ~ 57.7m

OpenStudy (toxicsugar22):

ok and what about the sound intensity of the concert

OpenStudy (toxicsugar22):

and this time put it in scientific notation

OpenStudy (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.

OpenStudy (toxicsugar22):

whait what I dont get that

OpenStudy (toxicsugar22):

i did not understand so that is no the answer you gave me the first on

OpenStudy (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\[D=10\log_{10}(10^{12}I)\]

OpenStudy (anonymous):

You should post this question again so someone else can take it. I'm not round for much longer.

OpenStudy (toxicsugar22):

ok but ou ar to good in math

OpenStudy (toxicsugar22):

If you dont mind me asking what grade are you in

OpenStudy (anonymous):

University.

OpenStudy (toxicsugar22):

wow

OpenStudy (anonymous):

Are you doing a timed test or something?

OpenStudy (toxicsugar22):

im in a community collesge

OpenStudy (toxicsugar22):

college

OpenStudy (anonymous):

is this an online assignment?

OpenStudy (toxicsugar22):

yes why

OpenStudy (toxicsugar22):

one more question befroe leaving

OpenStudy (anonymous):

when is it due? i can look at it later...that's all. when i'm rushed me, i make mistakes.

OpenStudy (toxicsugar22):

can you go to the other question ou answerd for me the when about the websites

OpenStudy (toxicsugar22):

here i will post it to you

OpenStudy (anonymous):

i can't find it. if you have the link, post it.

OpenStudy (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

OpenStudy (anonymous):

8000=4000(1.15)^t so 2=1.15^t \[t=\frac{\log 2}{\log 1.15} \approx 4.96\]months

OpenStudy (toxicsugar22):

and 16000

OpenStudy (toxicsugar22):

how long will it take fr the number of sites to reach 16000

OpenStudy (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

OpenStudy (toxicsugar22):

thanks

OpenStudy (anonymous):

ok

OpenStudy (toxicsugar22):

you are thhe bomb

OpenStudy (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.

OpenStudy (toxicsugar22):

ok

OpenStudy (anonymous):

you still there

OpenStudy (anonymous):

you still there

OpenStudy (anonymous):

are you looking for toxicsugar?

OpenStudy (anonymous):

yeah, i was going to say

OpenStudy (anonymous):

I = P / 4pi r^2

OpenStudy (anonymous):

Yeah...I skipped it. I think he just wants answers, though.

OpenStudy (anonymous):

:s

OpenStudy (anonymous):

yeah

OpenStudy (anonymous):

im just googling this, learning it

OpenStudy (anonymous):

ah

OpenStudy (anonymous):

apparently it is difficult to change db to hertz

OpenStudy (anonymous):

is it?

OpenStudy (anonymous):

how can you

OpenStudy (anonymous):

ok how many hertz is 90 db

OpenStudy (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.

OpenStudy (anonymous):

yeah but you solved this problem well. you know your physics

OpenStudy (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)

OpenStudy (anonymous):

Sorry, was away. It' 20N.

OpenStudy (anonymous):

how did you get that

OpenStudy (anonymous):

give me a minute. i need coffee.

OpenStudy (anonymous):

sure

OpenStudy (anonymous):

hey

OpenStudy (anonymous):

sry, back

OpenStudy (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.

OpenStudy (anonymous):

hmmm

OpenStudy (anonymous):

what if one of the astronauts died, and the live astronaut pulls on the dead one 20 N

OpenStudy (anonymous):

It's a third law of motion thing.

OpenStudy (anonymous):

is the tension still 20 N

OpenStudy (anonymous):

There'd be no tension.

OpenStudy (anonymous):

there is, because the dead astronaut has weight

OpenStudy (anonymous):

err, inertia i mean

OpenStudy (anonymous):

Oh...I thought he floated away :)

OpenStudy (anonymous):

no, he just died when the moment came to pull on it

OpenStudy (anonymous):

lol, are you just coming up with this?

OpenStudy (anonymous):

If he died and is still attached, the tension will be whatever force the live astronaut applies to accelerate him or her.

OpenStudy (anonymous):

The force is 'communicated' through the rope by the third law.

OpenStudy (anonymous):

so its 20 N in both cases

OpenStudy (anonymous):

If that's what the live one is pulling the dead one with.

OpenStudy (anonymous):

just found this - http://physics.stackexchange.com/questions/1220/rope-tension-question it might help

OpenStudy (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

OpenStudy (anonymous):

so there is a net force of 10 N on the rope. so why doesnt it accelerate 10/1 = a

OpenStudy (anonymous):

10 m/s^2

OpenStudy (anonymous):

I'm not sure what you mean. If there is a net force on the rope, it will accelerate.

OpenStudy (anonymous):

You have to remember that most questions in physics on this matter assume a massless rope...

OpenStudy (anonymous):

right but 10 m/s^2 is crazy fast , it will shoot off

OpenStudy (anonymous):

It's okay if it's fast.

OpenStudy (anonymous):

but the masses are accelerating VERY slow

OpenStudy (anonymous):

I'm not sure what you mean.

OpenStudy (anonymous):

say the guy has a mass of 50 kg

OpenStudy (anonymous):

Oh...I see...the tension would fall off since the rope wouldn't be taught.

OpenStudy (anonymous):

Actually, no...the acceleration would remain the same assuming they're all accelerating at 10m/s^2

OpenStudy (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

OpenStudy (anonymous):

so 30 - T = 50*a T - 20 = 40 *a

OpenStudy (anonymous):

The tension's 20N

OpenStudy (anonymous):

Do a free body diagram for each astronaut

OpenStudy (anonymous):

The net force on each is 20N to the right.

OpenStudy (anonymous):

i changed my problem

OpenStudy (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

OpenStudy (anonymous):

The rope communicates that force.

OpenStudy (anonymous):

but the tension is not 20

OpenStudy (anonymous):

most people think that, but if you draw the force body diagram for the guys you have

OpenStudy (anonymous):

the tension is actually 24.444

OpenStudy (anonymous):

lets call m1 = 50 kg, and m2 = 40 kg

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