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Vikentia [17]
2 years ago
15

It's a frightening idea, but what would be the sound intensity level of 100 physics professors talking simultaneously

Physics
1 answer:
Svetlanka [38]2 years ago
7 0

The sound intensity level of 100 physics professors talking simultaneously is equal to 72 decibel.

<u>Given the following data:</u>

  • Sound intensity = 52 dB.

<h3>What is sound intensity?</h3>

Sound intensity can be defined as a measure of the power of a sound wave per unit area. Thus, sound intensity is a quantity that can be used to measure the energy of sound and its unit of measurement is Watts per square meter.

<h3>How to calculate the sound intensity level.</h3>

Mathematically, sound intensity level is given by this formula:

\beta = 10 log(\frac{I}{I_{ref}} )

<u>Note:</u> The reference value of sound intensity is equal to 1.0 \times 10^{-12}\;W/m^2

Thus, the sound intensity for one (1) professor is given by:

I_1 = 1.0 \times 10^{-12} \times 10^{5.2}\\\\I_1=1.585 \times 10^{-7}\;W/m^2

For 100 professors, the sound intensity is:

I_{100} = 1.585 \times 10^{-7} \times 100\\\\I_{100}=1.585 \times 10^{-5}\;W/m^2

Substituting the parameters into the formula, we have;

\beta = 10 log(\frac{1.585 \times 10^{-5}}{1.0 \times 10^{-12}})\\\\\beta = 10 log(15.58 \times 10^{6})\\\\\beta = 10 \times 7.2\\\\\beta =72\;dB

Read more on sound intensity here: brainly.com/question/17062836

<u>Complete Question:</u>

One physics professor talking produces a sound intensity level of 52 dB. It's a frightening idea, but what would be the sound intensity level of 100 physics professors talking simultaneously?

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Find the hiker’s gravitational potential energy if the cliff is 60m high
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Answer:

Potential energy is U=mgh

Explanation:

The potential energy depends on the mass, the acceleration of gravity g and the height at which the object or person is.

Potential energy  U=mgh

In this case we would need to know the exact mass of the hiker in order to calculate the potential energy.

But we know the values of g and h

g=9.81m/s^2

h=60m

So, the potential energy

U=m(9.81m/s^2)(60m)\\\\U=588.6*m

m is the mass of the hiker, wich is not in the description of the problem.

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Which of the following are found within the electromagnetic spectrum? Check all that apply. sound waves visible light X rays ult
sattari [20]

Answer:

Visible light

X rays

ultraviolet radiation

gamma rays

microwave radiation

Explanation:

Electromagnetic waves consist of oscillating electric and magnetic fields which vibrate in a direction perpendicular to the direction of motion of the wave (transverse wave). Electromagnetic waves have all same speed in a vacuum (c=3.0\cdot 10^8 m/s, known as speed of light) and are classified into 7 different types according to their frequency and wavelength. This classification is called electromagnetic spectrum.

From lowest to highest wavelength, the 7 types are:

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Sound waves, on the contrary, do not belong to the electromagnetic spectrum, since they are another type of wave called mechanical waves (which consist of vibrations of the particles in a medium).

8 0
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A baseball thrown by a pitcher is hit by a batter. At the moment when the ball hits the bat, the force exerted on the bat by the
Slav-nsk [51]

Newton's Laws of motion describe the motion of an object based on the applied force

The correct option that gives the relationships between the forces is the the option;

\underline{F_{ball}}<u> on bat</u> = \underline{F_{bat}}<u> on ball</u>

<em>(Either option A or B without the minus symbol before </em>F_{bat}<em>, likely typographical error)</em>

<em />

Reason:

Force exerted on the bat by the ball = F_{ball}

Force exerted on the ball by the bat = F_{bat}

Given that the batter hits the ball with the bat, the force exerted by the bat on the ball, F_{bat}, is reacted to by the force of the ball acting on the bat, F_{ball}

According to Newton's third law of motion, action and reaction are equal and opposite. Therefore, at the moment when the ball hits the bat, we have;

\underline{F_{ball}}<u> on bat</u> = \underline{F_{bat}}<u> on ball</u>

<em>Where the force of the bat is high, the ball is accelerated to travel at high speed</em>

Learn more Newton's Laws of motion here:

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8 0
3 years ago
In the two-slit experiment, monochromatic light of frequency 5.00 × 1014 Hz passes through a pair of slits separated by 2.20 × 1
asambeis [7]

Explanation:

It is given that,

Frequency of monochromatic light, f=5\times 10^{14}\ Hz

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d\ sin\theta=n\lambda

For third maxima,

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\theta=sin^{-1}(\dfrac{n\lambda}{d})

\theta=sin^{-1}(\dfrac{nc}{fd})  

\theta=sin^{-1}(\dfrac{3\times 3\times 10^8\ m/s}{5\times 10^{14}\ Hz\times 2.2\times 10^{-5}\ m})  

\theta=4.69^{\circ}

(b) For second dark fringe, n = 2

d\ sin\theta=(n+1/2)\lambda

\theta=sin^{-1}(\dfrac{5\lambda}{2d})

\theta=sin^{-1}(\dfrac{5c}{2df})

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\theta=3.90^{\circ}

Hence, this is the required solution.

8 0
3 years ago
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