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KIM [24]
3 years ago
8

The sound intensity at the ear of passenger in a car with a damaged muffler is 8.0 × 10-3 W/m2. What is the intensity level of t

his sound in decibels? Use the threshold of hearing (1.0 × 10-12 W/m2) as the refere
Physics
1 answer:
Mama L [17]3 years ago
4 0

Answer:

99 dB

Explanation:

We have given that sound intensity with a damaged muffler =8× 10^{-3}

we have to find the intensity level of the this sound in decibels

for calculating in decibels we have to use the formula

β=10log\frac{I}{I_0}

  =10log\frac{.008}{10^{-12}}

   =10 log8+10 log10^{9}

   =10 log8+90 log10

   =10×0.9030+90

   =99 dB

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Answer:

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Explanation:

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I attach a 4.1 kg block to a spring that obeys Hooke's law and supply 3.8 J of energy to stretch the spring. I release the block
borishaifa [10]

Answer:

The amplitude of the oscillation is 2.82 cm

Explanation:

Given;

mass of attached block, m = 4.1 kg

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period of oscillation, T = 0.13 s

First, determine the spring constant, k;

T = 2\pi \sqrt{\frac{m}{k} }

where;

T is the period oscillation

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k is the spring constant

T = 2\pi \sqrt{\frac{m}{k} } \\\\k = \frac{m*4\pi ^2}{T^2} \\\\k = \frac{4.1*4*(3.142^2)}{(0.13^2)} \\\\k = 9580.088 \ N/m\\\\

Now, determine the amplitude of oscillation, A;

E = \frac{1}{2} kA^2

where;

E is the energy of the spring

k is the spring constant

A is the amplitude of the oscillation

E = \frac{1}{2} kA^2\\\\2E = kA^2\\\\A^2 = \frac{2E}{k} \\\\A = \sqrt{\frac{2E}{k} } \\\\A =  \sqrt{\frac{2*3.8}{9580.088} }\\\\A = 0.0282 \ m\\\\A = 2.82 \ cm

Therefore, the amplitude of the oscillation is 2.82 cm

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