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Nonamiya [84]
4 years ago
9

The loudness l of a sound, measured in decibels, is given by l=10log10r, where r is the sound's relative intensity. suppose one

person talks with a relative intensity of 10^6 or 60 decibels. how much louder would 100 people be, talking at the same intensity?
Physics
1 answer:
marishachu [46]4 years ago
5 0

Answer

given,                              

I is the loudness of sound

I = 10 Log₁₀ r                  

r is relative intensity                    

at when relative intensity is 10⁶        

I = 60 dB                                                  

how much louder when 100 people would be talking together

I = 10 Log₁₀ r                

I = 10 Log₁₀ (10⁶ x 100)  

I = 10 Log₁₀ (10⁸)                

I = 80 dB                      

hence, the intensity will be increased by (80 dB -60 dB) 20 dB when 100 people start talking together.

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Answer:

2.25 %

Explanation:

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at the range of mean plus or minus two standard deviation, P([\mu -2\sigma \leq X \leq \mu+2\sigma])\approx 95.5%

at the range of mean plus or minus three standard deviation, P([\mu - 3\sigma\leq X \leq \mu+3\sigma])\approx 99.7%

So, note that  they are asking about the probability that it is greater than 0.32, that is the mean (0.3) plus two times the standard deviation (0.1) (P(X \leq \mu+2\sigma))  

So we know that the 95.5% is between \mu - 2\sigma = 0.3 -2*0.1 = 0.28 and \mu + 2\sigma = 0.3 +2*0.1 = 0.32, hence approximately the 4.5% (100%-95.5%) is greater than 0.32 or less than  0.28. But half (4.5%/2=2.25%) is greater than 0.32 and the other half is less than 0.28.

So P(X \leq \mu+2\sigma) \approx 2.25%

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3 years ago
Which of the following is the correct unit for time when calculating power in watts?
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A silica-rich igneous rock that has large crystals and makes up much of the continental crust is ________
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(a) Calculate the electric field strength near a 10.0 cm diameter conducting sphere that has 1.00 C of excess charge on it. (b)
Romashka-Z-Leto [24]

Electric field strength near the surface of sphere is given by

E = \frac{kQ}{r^2}

Now we have

Q = 1 C

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now electric field is given by

E = \frac{9*10^9 * 1}{0.05^2}

E = 3.6 * 10^{12} N/C

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4 0
3 years ago
7. A 1300 kg car starts at rest and rolls down a hill from a height of 10.0 m. It then moves across a level (frictionless) surfa
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Answer:

0.505 m

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From the question,

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Also given: m = 1300 kg, e = 1.0×10⁶ N/m =1000000 N/m

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