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Inessa05 [86]
3 years ago
13

What is the sound intensity of a whisper at a distance of 2.0 m, in W/m²? What is the corresponding sound intensity level in dB?

Physics
1 answer:
Rufina [12.5K]3 years ago
4 0

To solve this problem we apply the concepts related to the Intensity, which is defined as the Power given by the object on unit area. Mathematically this is,

I = \frac{P}{A}

Where,

I = Intensity

P = Power

A = Area

Considering that a whisper is around to 1*10^{-10}W, then we have that

I = \frac{1*10^{-10}}{2^2}

I = 2.5*10^{-11} W/m^2

Intensity level in

\beta = 10log (\frac{I}{I_0})

Where,

I_0 = Threshold intensity level

\beta = 10log (\frac{2.5*10^{-11}}{1*10^{-12}})

\beta = 14dB

Therefore the corresponding sound intensity level  is 14dB

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What is an Amplitude
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A 50 kg wooden crate is slid across a flat surface for 60 meters by a force of 90.0 N. If the coefficient of sliding friction is
Alika [10]

The final speed of the crate is 6.3 m/s

Explanation:

Please note that there is a typo in the text of the question: the coefficient of sliding friction is 0.15, not 9.15.

First of all, we need to find the force of friction acting on the crate. This is given by:

F_f = \mu mg

where:

\mu=0.15 is the coefficient of friction

m = 50 kg is the mass of the crate

g=9.8 m/s^2 is the acceleration of gravity

Substituting,

F_f = (0.15)(50)(9.8)=73.5 N

Now we can find the acceleration of the crate by using Newton's second law:

F-F_f = ma

where

F = 90.0 N is the force applied forward on the crate

F_f = 73.5 N is the force of friction, acting backward

a is the acceleration of the crate

Solving for a,

a=\frac{F-F_f}{m}=\frac{90.0-73.5}{50}=0.33 m/s^2 in the forward direction.

Finally, we can find the final speed of the crate by using suvat equations:

v^2-u^2 = 2as

where:

v is the final speed

u = 0 is the initial speed

a=0.33 m/s^2 is the acceleration

s = 60 m is the distance covered

Solving for v,

v=\sqrt{u^2+2as}=\sqrt{0+2(0.33)(60)}=6.3 m/s

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6 0
4 years ago
A ray of light (f = 5.09 x 1014 Hz) is incident on the boundary between air and an unknown material X at an angle of incidence o
11Alexandr11 [23.1K]

Answer:

A) Material X is flint glass

B) Speed of light in the X material is 1.807 x 10^{8} m/s

C) Angle of refraction of light in medium X is 29.57^{o}

Explanation:

Given data:

frequency of the light f = 5.09 x 10^{14} Hz

angle of incidence Θ_{i} = 55^{o}

index of refraction of material n_{2} = 1.66

A) To find material X

Given the index of refraction is 1.66 and hence the material is the flint glass

B) To calculate the speed of light in the material.

We know that the relation between index of refraction (n), velocity of light (c = 3 x 10^{8} m/s) and velocity of light is given by the equation:

n = c/v

Hence,

Speed of light in the X material v = c/n

                                                          = \frac{3 x 10^{8} }{1.66}

                                                          = 1.807 x 10^{8} m/s

C) To calculate angle of refraction of light in medium X

We know that the Snell's law states that

                       n_{1} sin Θ_{i}  = n_{2} sin Θ_{r}

n_{1} = incident index

n_{2}  = refracted index

Θ_{i} = incident angle

Θ_{r} = refracted angle

In given problem, n_{1} = 1 since medium is air

Substituting the known values, we get

1 x  sin 55^{o}  = 1.66 x sin Θ_{r}

sin Θ_{r} = sin 55^{o} /1.66

           = 0.4935

Hence, angle of refraction of light in medium X Θ_{r} = sin^{-1} (0.4935)

                                                                                       = 29.57^{o}

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3 years ago
Which of the following is not a requirement of practical fuel
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Nothing on that  list of choices is a requirement.
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4 years ago
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