## 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.

1. 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,$\frac{I_0}{I_{pain}}=\frac{r_{pain}^2}{r_0^2} \rightarrow r_{pain}=\sqrt{\frac{I_0}{I_{pain}}}r_0=\frac{100}{\sqrt{3}}m$

2. toxicsugar22

3. toxicsugar22

can you show me

4. anonymous

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

5. toxicsugar22

ok and what about the sound intensity of the concert

6. toxicsugar22

and this time put it in scientific notation

7. 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.

8. toxicsugar22

whait what I dont get that

9. toxicsugar22

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

10. 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$D=10\log_{10}(10^{12}I)$

11. anonymous

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

12. toxicsugar22

ok but ou ar to good in math

13. toxicsugar22

14. anonymous

University.

15. toxicsugar22

wow

16. anonymous

Are you doing a timed test or something?

17. toxicsugar22

im in a community collesge

18. toxicsugar22

college

19. anonymous

is this an online assignment?

20. toxicsugar22

yes why

21. toxicsugar22

one more question befroe leaving

22. anonymous

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

23. toxicsugar22

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

24. toxicsugar22

here i will post it to you

25. anonymous

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

26. 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

27. anonymous

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

28. toxicsugar22

and 16000

29. toxicsugar22

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

30. 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

31. toxicsugar22

thanks

32. anonymous

ok

33. toxicsugar22

you are thhe bomb

34. 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.

35. toxicsugar22

ok

36. anonymous

you still there

37. anonymous

you still there

38. anonymous

are you looking for toxicsugar?

39. anonymous

yeah, i was going to say

40. anonymous

I = P / 4pi r^2

41. anonymous

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

42. anonymous

:s

43. anonymous

yeah

44. anonymous

im just googling this, learning it

45. anonymous

ah

46. anonymous

apparently it is difficult to change db to hertz

47. anonymous

is it?

48. anonymous

how can you

49. anonymous

ok how many hertz is 90 db

50. 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.

51. anonymous

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

52. 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)

53. anonymous

Sorry, was away. It' 20N.

54. anonymous

how did you get that

55. anonymous

give me a minute. i need coffee.

56. anonymous

sure

57. anonymous

hey

58. anonymous

sry, back

59. 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.

60. anonymous

hmmm

61. anonymous

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

62. anonymous

It's a third law of motion thing.

63. anonymous

is the tension still 20 N

64. anonymous

There'd be no tension.

65. anonymous

there is, because the dead astronaut has weight

66. anonymous

err, inertia i mean

67. anonymous

Oh...I thought he floated away :)

68. anonymous

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

69. anonymous

lol, are you just coming up with this?

70. anonymous

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

71. anonymous

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

72. anonymous

so its 20 N in both cases

73. anonymous

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

74. anonymous

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

75. 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

76. anonymous

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

77. anonymous

10 m/s^2

78. anonymous

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

79. anonymous

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

80. anonymous

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

81. anonymous

It's okay if it's fast.

82. anonymous

but the masses are accelerating VERY slow

83. anonymous

I'm not sure what you mean.

84. anonymous

say the guy has a mass of 50 kg

85. anonymous

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

86. anonymous

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

87. 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

88. anonymous

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

89. anonymous

The tension's 20N

90. anonymous

Do a free body diagram for each astronaut

91. anonymous

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

92. anonymous

i changed my problem

93. 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

94. anonymous

The rope communicates that force.

95. anonymous

but the tension is not 20

96. anonymous

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

97. anonymous

the tension is actually 24.444

98. anonymous

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

99. anonymous

the forces on m1 is 30 N to the right, and has tension to the left. so 30 - T = 50* a

100. anonymous

I get 18 + 8/9.

101. anonymous

The concept of tension remains the same, except when answering the second question, I got slack and didn't bother to calculate things.

102. anonymous

so do you agree that

103. anonymous

30 - T = 50 * a T - 10 = 40 * a

104. anonymous

Yes, so long as we assume the rope is fully taught.

105. anonymous

I agree with those equations.

106. anonymous

oh so 18.88888 N in tension

107. anonymous

so if the rope has 1kg say, it is accelerating 18.8888 ms^-2

108. anonymous

Actually, I think these equations are for massless rope.

109. anonymous

ok

110. 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

111. anonymous

but the forces at the end are unequal

112. anonymous

Why do you get F1-F2=0? The object's accelerating.

113. anonymous

For it to remain taught and communicate the force, it must be accelerating with the system.

114. anonymous

but yuo said massless

115. anonymous

so m = 0

116. anonymous

F1 - F2 = 0*a

117. anonymous

Sorry, I'm being distracted.

118. anonymous

thats ok

119. 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

120. anonymous

does that make sense?

121. anonymous

Oh, I thought you left.

122. anonymous

nope

123. anonymous

Problems are coming in because of the massless thing.

124. anonymous

In reality, the rope will have mass.

125. anonymous

so the rope exerts a force of 18.888 on either end

126. anonymous

The rope carries the force through tension.

127. anonymous

I'm trying to come up with a physical example.

128. anonymous

Sorry, I was preoccupied. Set the problem up like the attachment that's coming.

129. anonymous

130. 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,$T(x)=F_2+\frac{m}{L}(L-x)a$

131. 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$T(x)=F_2+\frac{m}{L}(L-x)\frac{F_1-F_2}{M+m}$

132. 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.