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VashaNatasha [74]
3 years ago
6

When a sound wave moves past a point inair, what happens to the density of air at thispoint?1.There is no change in the density

of air.2.The density of air increases and thendecreases as the sound wave passes.correct3.There is no air after the sound wavepasses
Physics
1 answer:
Westkost [7]3 years ago
8 0

Answer:

2.The density of air increases and thendecreases as the sound wave passes.

Explanation:

Sound waves are mechanical waves, which consist of oscillation of the particles in the medium where the wave is transmitted through.

Sound waves are also longitudinal waves, which means that the direction of oscillations of the particles of the medium occurs in a direction parallel to the direction of motion of the wave (so, essentially back and forth).

Due to the nature of longitudinal waves, they create alternating regions of the medium where the density of particles are higher and lower. The former are called compressions, while the latter are called rarefactions.

Therefore, when a sound wave travels through the air, the density of one region of air continuously changes: compression first (high density), rarefaction then (lower density), then compression again, etc..

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Talja [164]

Answer:

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Explanation:

3 0
2 years ago
13.An airplane starts from rest at the end of a runway and accelerates at a constant rate. In the first second, the airplane tra
Murrr4er [49]

Answer:

The answer is 3.33m

Explanation:

The acceleration "a" is constant.

Acceleration is the variation of velocity over time,

\frac{dv}{dt} = a.

solving the last equation

\int_{v_0}^v dv = a\int_0^t dt \rightarrow v-v_0 = at,

where v_0=0 because the airplane starts from rest.

Once again, velocity is the variation of distance over time.

\frac{dx}{dt} = at \rightarrow \int_{x_0}^x dx = a\int_0^t t\ dt

then

x- x_0 = \frac{1}{2}at^2

where x_0=0 if we consider  the end of the runway as the initial point (this step is for simplicity but you can let it expressed, it's going to cancel anyway).

If x=1.11\ m at t=1s, then

a = \frac{2x}{t^2} = 2.22\ m/s^2

and the final expression for the distance is

x = 1.11 t^2.

If t = 2s, x = 4.44 m. Which means thad the additional distance is

x(2s) - x(1s) = 4.44 - 1.11 = 3.33\ m

8 0
4 years ago
If you planned to bike to a park that was five miles away, what average speed would you have to maintain to arrive in about 15 m
AfilCa [17]

The average speed you need to maintain is 0.33 miles per minute.

The average speed is the speed that you maintain throughout the journey. It is obtained as; Total distance covered/Total time taken

If we have the following information from the question;

Distance covered = 5 miles

Time taken = 15 mins

Then it follows that;

Average speed = Distance/time

Average speed= 5 miles/15mins

Average speed = 0.33 miles per minute.

Learn more: brainly.com/question/17661499

5 0
3 years ago
A slow moving river of ice
Komok [63]

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3 0
3 years ago
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The road from city A to city B is described by a car with Vm 40 km / h. When the car turns (from B to A) the average speed is 60
True [87]

Answer:

i think the answer is 20.......

7 0
2 years ago
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