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3 years ago

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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The law of universal gravitation states that any two objects in the universe, without exception,

STatiana [176]

A. attract each other.

The Law of Universal Gravitation discusses the phenomenon of gravity. Remember that gravity is the force that keeps us on Earth; the Earth pulls us down, and our bodies pull back. Gravity is the force of attraction, so the correct answer is a).

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3 years ago

An asteroid orbiting the Sun has a mass of 4.00×1016 kg. At a particular instant, it experiences a gravitational force of 3.14×1

Ksivusya [100]

<h2>The asteroid is 4.11 x 10¹¹ m far from Sun</h2>

Explanation:

We have gravitational force

$F=\frac{GMm}{r^2}$

Where G = 6.67 x 10⁻¹¹ N m²/kg²

M = Mass of body 1

M = Mass of body 2

r = Distance between them

Here we have

M = Mass of Sun = 1.99×10³⁰ kg

m = Mass of asteroid = 4.00×10¹⁶ kg

F = 3.14×10¹³ N

Substituting

$F=\frac{GMm}{r^2}\\\\3.14\times 10^{13}=\frac{6.67\times 10^{-11}\times 1.99\times 10^{30}\times 4\times 10^{16}}{r^2}\\\\r=4.11\times 10^{11}m$

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3 years ago

You’re driving down the highway late one night at 20 m/s when a deer steps out onto the road 35 m in front on you. Your reaction

Salsk061 [2.6K]

Answer:

i would juh hit da deer no kap

Explanation:

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3 years ago

The box leaves position x=0x=0 with speed v0v0. The box is slowed by a constant frictional force until it comes to rest at posit

const2013 [10]

Answer:

fr = ½ m v₀²/x

Explanation:

This exercise the body must be on a ramp so that a component of the weight is counteracted by the friction force.

The best way to solve this exercise is to use the energy work theorem

W = ΔK

Where work is defined as the product of force by distance

W = fr x cos 180

The angle is because the friction force opposes the movement

Δk = $K_{f}$ –K₀

ΔK = 0 - ½ m v₀²

We substitute

- fr x = - ½ m v₀²

fr = ½ m v₀²/x

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3 years ago

If a sound travels 343m/s through air and has a frequency of 800Hz, what would the wavelength be of the same sound traveling thr

VARVARA [1.3K]

Answer:

0.625m

Explanation:

Now Velocity of a wave ,V = frequency × wavelength

Wavelength =velocity /frequency

=500/800 =0.625m

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2 years ago