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avanturin [10]
3 years ago
6

A bell is rung. What best describes the density of air around the bell? The air density does not change. The air density increas

es and then returns to normal levels. The air density decreases and then returns to normal levels. The air density increases and decreases repeatedly before returning to normal.
Physics
2 answers:
dezoksy [38]3 years ago
7 0

I think the closest possible answer to this question is The air density increases and decreases repeatedly before returning to normal.Thank you for your question. Please don't hesitate to ask in Brainly your queries. 
Yakvenalex [24]3 years ago
4 0

The correct answer to the question is : D). The air density increases and decreases repeatedly before returning to normal

EXPLANATION:

Before coming into any conclusion, first we have to understand sound wave.

A sound wave is a longitudinal wave in which the vibration of particles are parallel to the direction of wave propagation. It moves in the form of compression and rarefaction.

Compression is the high pressure region where particles are closely aggregated. Rarefaction is the region of low pressure where particles of the medium are not so close to each other just like compression.

Hence, the correct statement  will be that the air density increases and decreases repeatedly before returning to normal .

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3 years ago
Which type of mirror produces images that are always upright and at the same distance from the mirror as the object is?
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3 years ago
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A 50 kg runner runs up a flight of stairs. The runner starts out covering 3 steps every second. At the end the runner stops. Thi
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To solve the problem it is necessary to take into account the concepts of kinematic equations of motion and the work done by a body.

In the case of work, we know that it is defined by,

W = F * d

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d = Distance

The distance in this case is a composition between number of steps and the height. Then,

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On the other hand we have the speed changes, depending on the displacement and acceleration (omitting time)

V_f^2-V_i^2 = 2a\Delta X

Where,

V_f = Final velocity

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a = Acceleration

\Delta X = Displacement

PART A) For the particular case of work we know then that,

W = F*d

W = m*g*(h*N)

W = 50*9.8*(0.3*30)

W = 4.41kJ

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V_f^2-V_i^2 = 2a\Delta X

Here,

V_i = \frac{0.3*3}{1} = 0.90m/s\rightarrow3 steps in one second

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Replacing,

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Re-arrange for a,

a = -\frac{0.9^2}{2*30*0.3}

a = -45*10^{-3}m/s^2

At this point we can calculate the time, which is,

t = \frac{\Delta V}{a}

t = \frac{0-0.9}{-45*10^{-3}}

t = 20s

With time and work we can finally calculate the power

P = \frac{W}{t} = \frac{4.41}{20}

P = 0.2205kW

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Here,

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We know that,

The time period for the spring is calculated with the equation:

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