The earplug can reduce the sound level to about 18 decibels (dB). What percentage reduction is this intensity?

Physics

2 answers:

zlopas [31]3 years ago

8 0

Answer:

1 x 10 -10 whisper at 1m distance.

Explanation:

Properly fitted ear plugs an reduce noise form 15-30db. Although they are better for low frequency

marishachu [46]3 years ago

8 0

Answer:

The change in intensity is 63%.

Explanation:

intensity level = 18 db

Let the intensity is I.

Io = 10^(-12) W/m^2

Use the formula of intensity

$dB = 10 log\left ( \frac{I}{Io} \right )\\\\18 = 10 log\left ( \frac{I}{Io} \right )\\\\1.8 = log\left ( \frac{I}{Io} \right )\\\\\left ( \frac{I}{Io} \right )=63.1$

So, the change in intensity is 63%.

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3 years ago

in a solar system far, far away the sun's intensity is 200 w/m2 for an inner planet located a distance r away. what is the sun's

GarryVolchara [31]

The sun's intensity for an outer planet located at a distance 6r from the sun is 5.55 W/m². The result is obtained by using the inverse square law formula.

<h3>What is the Inverse Square Law formula?</h3>

The Inverse Square Law formula describes the intensity of light is inversely proportional to the square of the distance. It can be expressed as

$\frac{I_{1} }{I_{2} } = \frac{d_{2}^{2} }{d_{1}^{2}}$

Where

I₁ = Intensity at distance 1 (W/m²)
I₂ = Intensity at distance 2 (W/m²)
d₁ = distance 1 from a light source (m)
d₂ = distance 2 from a light source (m)

Given the case the sun's intensity is 200 W/m² for an inner planet at the distance r. If an outer planet is at a distance 6r, what is the sun's intensity?

By using the inverse square law formula, the sun's intensity for an outer planet is

$\frac{I_{1} }{I_{2} } = \frac{d_{2}^{2} }{d_{1}^{2}}$

$\frac{200 }{I_{2} } = \frac{(6r)^{2} }{r^{2}}$