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Vera_Pavlovna [14]
3 years ago
12

A sound wave moving through the air causes particles in the air to move. If air particles are moving, why doesn’t this motion cr

eate a breeze?
Physics
1 answer:
mylen [45]3 years ago
3 0

Answer:

The motion of air particle caused by sound wave is vibrational and cannot create breeze because breeze involves translational motion (permanent displacement) of the air particles.

Explanation:

This sound wave results from the back and forth vibration of the particles of air through which the sound wave is moving. The sound wave causes disturbance within the particles of air by transfering energy from one particle to the adjacent particles of air without causing permanent displacement of the particles of air. The motion between these particles of air caused by sound wave is vibrational, while breeze involves displacement of particles of air through translational motion. Hence, breeze is a moving mass of air.

Thus, the motion of air particle caused by sound wave is vibrational and cannot create breeze because breeze involves translational motion (permanent displacement) of the air particles.

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Marina86 [1]
Answer A is incorrect
A crest is just one point. It is not the distance between 2 crests.

B  is incorrect
A trough is just 1 point. It is not the distance between 2 troughs.

C is incorrect.
the amplitude measures the height of a crest from the middle of the wave to the crest (or trough).

D is the correct answer. That is the distance between 2 crests or 2 troughs 
8 0
3 years ago
Which two quantities below can be used to calculate average speed?
defon
Change of position and time
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If we're interested in knowing the rate at which light energy is received by a unit of area on a particular surface, we're reall
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Illluminance is the measurement of photo-metric power. That means, illuminance is the rate of photo-metric flux that is received by a surface per area. It is usually expressed as a unit of W/m^2. Thus, from the choices, the answer we're looking for is illuminance.


6 0
3 years ago
(b) Fig. 1.1 shows two airports A and C. north SE с sea land क WE not to scale Fig. 1.1 An aircraft flies due north from A for a
GalinKa [24]

The addition of vectors and the uniform motion allows to find the answers for the questions about distance and time are:

  • The distance to go between airports A and C  is 373.6 10³ m
  • The time to go from airport A to B is 2117 s

Vectors are quantities that have modulus and direction, so their addition must be done using vector algebra.

In this case the plane flies towards the North a distance of y = 360 10³ m at an average speed of v = 170 m / s, when arriving at airport B it turns towards the East and travels from x = 100 10³ m, until' it the distance reaches the airport C

Let's use the Pythagoras theorem to find the distance traveled

               R = Ra x² + y²

               R =   10³

               R = 373.6 10³ m

They indicate the average speed for which we can use the uniform motion ratio

               v = \frac{\Delta y }{t}

                t = \frac{\Delta y}{v}

They ask for the time in in from airport A to B, we calculate

                t = 360 10 ^ 3/170

                t = 2.117 10³ s

In conclusion we use the addition of vectors the uniform motion we can find the answer for the question of distance and time are:

  • The distance to go between airports A and C B is 373.6 10³ m
  • The time to go from airport A to B is 2117 s

Learn more here: brainly.com/question/15074838

5 0
3 years ago
In Hooke's law, Fspring=kΔx, what does the ∆x stand for
monitta

Answer:

Change in Displacement

Explanation:

delta/triangle = change

x = displacement

formula (if needed): final x - initial x

7 0
2 years ago
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