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2 years ago

When a sound wave travels through air, how is this air affected by its passage

1 answer:

6 0

Sound is a mechanical wave <span>that results from the back and forth vibration of the particles of the medium through which the sound wave is moving. If a sound wave is moving from left to right through air, then particles of air will be displaced both rightward and leftward as the energy of the sound wave passes through it. The motion of the particles is parallel (and anti-parallel) to the direction of the energy transport. This is what characterizes sound waves in air as longitudinal waves.</span><span>
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When a rocket is traveling toward a mountain at 100 m/s, the sound waves from this rocket's engine approach the mountain at spee

Charra [1.4K]

Answer:

The correct option is C

Explanation:

From the question we are told that

The initial speed of the rocket is $v_i = 100 m/s$

The speed of the rocket engine sound is $V$

The final speed of the rocket is $v_f = 200 \ m/s$

The speed of the sound at $v_f$ would still remain V this because the speed of sound wave is constant and is not dependent on the speed of the observer(The mountain ) or the speed of the source (The rocket ).

A clear example when lightning strikes you will first see (that is because it travels at the speed of light which is greater than the speed of sound) but it would take some time before you hear the sound of the lightning

Here we see that the speed of the lightning(speed of sound) does not affect the speed of the sound it generates

3 0

3 years ago

My Notes Ask Your Teacher 6. Six blocks with different masses, m, each start from rest at the top of smooth, frictionless inclin

Tresset [83]

Answer:

Kf > Ka = Kb > Kc > Kd > Ke

Explanation:

We can apply

E₀ = E₁

where

E₀: Mechanical energy at the beginning of the motion (top of the incline)

E₁: Mechanical energy at the end (bottom of the incline)

then

K₀ + U₀ = K₁ + U₁

If v₀ = 0 ⇒ K₀

and h₁ = 0 ⇒ U₁ = 0

we get

U₀ = K₁

U₀ = m*g*h₀ = K₁

we apply the same equation in each case

a) U₀ = K₁ = m*g*h₀ = 70 Kg*9.81 m/s²*8m = 5493.60 J

b) U₀ = K₁ = m*g*h₀ = 70 Kg*9.81 m/s²*8m = 5493.60 J

c) U₀ = K₁ = m*g*h₀ = 35 Kg*9.81 m/s²*4m = 1373.40 J

d) U₀ = K₁ = m*g*h₀ = 7 Kg*9.81 m/s²*16m = 1098.72 J

e) U₀ = K₁ = m*g*h₀ = 7 Kg*9.81 m/s²*4m = 274.68 J

f) U₀ = K₁ = m*g*h₀ = 105 Kg*9.81 m/s²*6m = 6180.30 J

finally, we can say that

Kf > Ka = Kb > Kc > Kd > Ke

8 0

2 years ago

I'm new to brainy, but here is my question.

skad [1K]

680 is the right answer

8 0

3 years ago

Read 2 more answers

Using Newton's 2nd Law of Motion Formula (F=MA) answer the following.

vladimir1956 [14]

Explanation:

1200 is your answer for this question

6 0

2 years ago

A system of ideal gas at 22°C undergoes an ischoric process with an internal energy decrease of 4.30 × 10 3 4.30×103 J to a fina

Komok [63]

Answer:

The approximate change in entropy is -14.72 J/K.

Explanation:

Given that,

Temperature = 22°C

Internal energy $U=4.30\times10^{3}\ J$

Final temperature = 16°C

We need to calculate the approximate change in entropy

Using formula of the entropy

$\Delta S=\dfrac{\Delta U}{T}$

Where, $\Delta U$ = internal energy

T = average temperature

Put the value in to the formula

$\Delta S=\dfrac{-4.30\times10^{3}}{\dfrac{22+273+16+273}{2}}$

$\Delta S=-14.72\ J/K$

Hence, The approximate change in entropy is -14.72 J/K.

5 0

3 years ago