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3 years ago

How do sound waves move in air

Physics

2 answers:

Ira Lisetskai [31]3 years ago

7 0

Answer:

emitted as energy waves

Explanation:

will stay in motion until absorbed.

dusya [7]3 years ago

5 0

Sound waves traveling through air are indeed longitudinal waves with compressions and rarefactions. As sound passes through air (or any fluid medium), the particles of air do not vibrate in a transverse manner.

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The pilot of an aircraft wishes to fly due west in a 33.9 km/h wind blowing toward the south. The speed of the aircraft in the a

diamong [38]

Answer: $\theta =9.96^{\circ}$ North of west

Explanation:

Given

Plane wishes to fly in west

but wind with speed 33.9 km/h towards south obstructing its path

so plane must fly at an angle of \theta w.r.t west such that it final velocity is towards west

Plane absolute speed=195 km/h

To fly towards west velocity in Y direction should be zero

thus $195sin\theta =33.9$

$\theta =9.96^{\circ}$

so Plane should head towards $9.96^{\circ}$ North of west in order to fly in west.

So plane

actual velocity is

$v=-195cos9.96\hat{i}+195sin9.96\hat{j}$

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3 years ago

An electromagnetic radio wave is received by a transmitter before it is converted to a sound wave. The radio wave has a waveleng

Dennis_Churaev [7]

Answer:

Explanation:

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3 years ago

Two electrodes, separated by a distance d, in a vacuum are maintained at a constant potential difference. An electron, accelerat

Alja [10]

Answer:

Explanation:

Given that, the distance between the electrode is d.

The electron kinetic energy is Ek when the electrode are at distance "d" apart.

So, we want to find the K.E when that are at d/3 distance apart.

K.E = ½mv²

Note: the mass doesn't change, it is only the velocity that change.

Also,

K.E = Work done by the electron

K.E = F × d

K.E = W = ma × d

Let assume that if is constant acceleration

Then, m and a is constant,

Then,

K.E is directly proportional to d

So, as d increase K.E increase and as d decreases K.E decreases.

So,

K.E_1 / d_1 = K.E_2 / d_2

K.E_1 = E_k

d_1 = d

d_2 = d/3

K.E_2 = K.E_1 / d_1 × d_2

K.E_2 = E_k × ⅓d / d

Then,

K.E_2 = ⅓E_k

So, the new kinetic energy is one third of the E_k

7 0

3 years ago

A car begins at rest (0 velocity), 5 seconds later it is travelling at 20 meters/per second. What was the acceleration of the ve

klasskru [66]

use the formula

v= u+ at

v is final velocity , u is initial velocity , a is acceleration and t is time

put the values

20 = 0+ a×5

a = 4 m/s²

4 0

2 years ago

Read 2 more answers

The concentration of Biochemical Oxygen Demand (BOD) in a river just downstream of a wastewater treatment plant’s effluent pipe

shtirl [24]

Answer:

The BOD concentration 50 km downstream when the velocity of the river is 15 km/day is 63.5 mg/L

Explanation:

Let the initial concentration of the BOD = C₀

Concentration of BOD at any time or point = C

dC/dt = - KC

∫ dC/C = -k ∫ dt

Integrating the left hand side from C₀ to C and the right hand side from 0 to t

In (C/C₀) = -kt + b (b = constant of integration)

At t = 0, C = C₀

In 1 = 0 + b

b = 0

In (C/C₀) = - kt

(C/C₀) = e⁻ᵏᵗ

C = C₀ e⁻ᵏᵗ

C₀ = 75 mg/L

k = 0.05 /day

C = 75 e⁻⁰•⁰⁵ᵗ

So, we need the BOD concentration 50 km downstream when the velocity of the river is 15 km/day

We calculate how many days it takes the river to reach 50 km downstream

Velocity = (displacement/time)

15 = 50/t

t = 50/15 = 3.3333 days

So, we need the C that corresponds to t = 3.3333 days

C = 75 e⁻⁰•⁰⁵ᵗ

0.05 t = 0.05 × 3.333 = 0.167

C = 75 e⁻⁰•¹⁶⁷

C = 63.5 mg/L

5 0

3 years ago