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rodikova [14]
3 years ago
11

Suppose a large housefly 3.0 m away from you makes sound with an intensity level of 40.0 dB. What would be the sound intensity l

evel if there 1000 identical flies at that same distance?
Physics
1 answer:
melisa1 [442]3 years ago
5 0

Answer:

Explanation:

Let the intensity of the noise be represented by I

Given that

40dB = 10 log 10 ⁡ ( I /I•) ........ 1

I• is the lowest or threshold intensity of sound made.

I represents the intensity of the sound/ noise

The intensity of noise of 1000flies will be

β = 10 log 10 ⁡(1000I/I•)

Open up the bracket

β = 10 log 10(1000)+ 10 log 10(I/I•)

10 log 10(10^3)+10 log 10(I/I•)

3×10(10 log 10) +10 log 10(I/I•)

Recall, 10 log 10 = 1

30×1 + 10 log 10(I/I•).........2

Put equation 1 into 2

β =30+40

= 70db

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I'm going to assume the friction force is pointed down the bank.  If I get a negative answer, that'll just mean it's actually pointed up the bank.

Sum of the forces in the radial direction (+x):

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N sin θ + F cos θ = m v² / r

Sum of the forces in the y direction:

∑F = ma

N cos θ - F sin θ - W = 0

To solve the system of equations for F, first solve for N and substitute.

N = (W + F sin θ) / cos θ

Substituting:

((W + F sin θ) / cos θ) sin θ + F cos θ = m v² / r

(W + F sin θ) tan θ + F cos θ = m v² / r

W tan θ + F sin θ tan θ + F cos θ = m v² / r

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W tan θ + F sec θ = m v² / r

F sec θ = m v² / r - W tan θ

F = m v² cos θ / r - W sin θ

F = m (v² cos θ / r - g sin θ)

Given that m = 1900 kg, θ = 11°, v = 27.2 m/s, and r = 55 m:

F = 1900 ((27.2)² cos 11 / 55 - 9.8 sin 11)

F = 21577 N

Rounding to two sig-figs, you need at least 22000 N of friction force.

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