Question

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.

Answer 1

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

Answer 2

is that the answere

Answer 3

can you show me

Answer 4

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

Answer 5

ok and what about the sound intensity of the concert

Answer 6

and this time put it in scientific notation

Answer 7

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.

Answer 8

whait what I dont get that

Answer 9

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

Answer 10

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

Answer 11

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

Answer 12

ok but ou ar to good in math

Answer 13

If you dont mind me asking what grade are you in

Answer 14

University.

Answer 15

wow

Answer 16

Are you doing a timed test or something?

Answer 17

im in a community collesge

Answer 18

college

Answer 19

is this an online assignment?

Answer 20

yes why

Answer 21

one more question befroe leaving

Answer 22

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

Answer 23

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

Answer 24

here i will post it to you

Answer 25

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

Answer 26

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

Answer 27

8000=4000(1.15)^t so 2=1.15^t

\[t=\frac{\log 2}{\log 1.15} \approx 4.96\]months

Answer 28

and 16000

Answer 29

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

Answer 30

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

Answer 31

thanks

Answer 32

ok

Answer 33

you are thhe bomb

Answer 34

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.

Answer 35

ok

Answer 36

you still there

Answer 37

you still there

Answer 38

are you looking for toxicsugar?

Answer 39

yeah, i was going to say

Answer 40

I = P / 4pi r^2

Answer 41

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

Answer 42

:s

Answer 43

yeah

Answer 44

im just googling this, learning it

Answer 45

ah

Answer 46

apparently it is difficult to change db to hertz

Answer 47

is it?

Answer 48

how can you

Answer 49

ok how many hertz is 90 db

Answer 50

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.

Answer 51

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

Answer 52

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)

Answer 53

Sorry, was away. It' 20N.

Answer 54

how did you get that

Answer 55

give me a minute. i need coffee.

Answer 56

sure

Answer 57

hey

Answer 58

sry, back

Answer 59

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.

Answer 60

hmmm

Answer 61

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

Answer 62

It's a third law of motion thing.

Answer 63

is the tension still 20 N

Answer 64

There'd be no tension.

Answer 65

there is, because the dead astronaut has weight

Answer 66

err, inertia i mean

Answer 67

Oh...I thought he floated away :)

Answer 68

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

Answer 69

lol, are you just coming up with this?

Answer 70

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

Answer 71

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

Answer 72

so its 20 N in both cases

Answer 73

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

Answer 74

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

it might help

Answer 75

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

Answer 76

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

Answer 77

10 m/s^2

Answer 78

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

Answer 79

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

Answer 80

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

Answer 81

It's okay if it's fast.

Answer 82

but the masses are accelerating VERY slow

Answer 83

I'm not sure what you mean.

Answer 84

say the guy has a mass of 50 kg

Answer 85

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

Answer 86

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

Answer 87

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

Answer 88

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

Answer 89

The tension's 20N

Answer 90

Do a free body diagram for each astronaut

Answer 91

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

Answer 92

i changed my problem

Answer 93

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

Answer 94

The rope communicates that force.

Answer 95

but the tension is not 20

Answer 96

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

Answer 97

the tension is actually 24.444

Answer 98

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