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kari74 [83]
3 years ago
9

Contrasting the Doppler effect with shock waves, the one that requires the faster source is?

Physics
2 answers:
MAXImum [283]3 years ago
6 0

I think shock waves require more speed they travel at the speed of sound

WITCHER [35]3 years ago
5 0

Answer:

Shock waves

Explanation:

The Doppler effect is well noticed at moments where the speed of the source is going slower than the speed of the waves. E.g. the change in apparent frequency of the sound of a train horn. As the train nears the observer, the blast of its horn is observed at a high pitch and as the train accelerates further, the blast of its horn is observed at a low pitch.

A shock wave is observed if the source is going at an equal speed as or faster than the wave can go. When the origin of sound moves at an equal speed as sound, it results in the source being at the front edge of the waves that it forms at all times.

The attached image shows a snapshots in time of diverse wavefronts forned by an aircraft that is accelerating at an equal speed as sound. The circular lines represent compressional wavefronts of the sound waves. Notice that these circles are bunched up at the front of the aircraft. This phenomenon is known as a shock wave. Shock waves are also produced if the aircraft moves faster than the speed of sound. If a moving source of sound moves faster than sound, the source will always be ahead of the waves that it produces. The diagram at the right depicts snapshots in time of a variety of wavefronts produced by an aircraft that is moving faster than sound. Note that the circular compressional wavefronts fall behind the faster moving aircraft (in actuality, these circles would be spheres).

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What is the difference between mass and weight?
I am Lyosha [343]
Mass is the actual amount of material contained in a body and is measured by kg, gm, etc. Whereas weight is the force extorted by the gravity on that object mg. Note that mass is independent of everything but weight is different on earth, moon, etc.

5 0
3 years ago
Spot the dog is running after a ball.In 22m,he accelerates at a constant rate,and his velocity increases from 5.0m/s to 11m/s. W
MAXImum [283]

Here it is given that speed of the dog will increase from 5 m/s to 11 m/s in order to cover the distance of 22 m

so here we can use kinematics to find the acceleration

v_f = 11 m/s

v_i = 5 m/s

d = 22 m

now we will have

v_f^2 - v_i^2 = 2 a d

11^2 - 5^2 = 2(a)(22)

a = \frac{11^2 - 5^2}{44}

a = 2.18 m/s^2

so here it will accelerate with a = 2.18 m/s^2

7 0
2 years ago
A rocket of initial mass 115 kg (including all the contents) has an engine that produces a constant vertical force (the thrust)
sweet-ann [11.9K]

Answer:

Part (i) the initial acceleration of the rocket is 6.98 m/s²

Part(ii) the floor pushes on the power supply at 120m altitude by a force of 31.68 N

Explanation:

Part (i) the initial acceleration of the rocket.

For the rocket to accelerate, the force applied to it must overcome gravitational force due to its own weight.

F_{Net} = M(a+g)\\\\1930 = 115(a+9.8)\\\\a +9.8 =\frac{1930}{115} \\\\a +9.8 = 16.78\\\\a = 16.78-9.8\\\\a = 6.98 \frac{m}{s^2}

Part(ii) how hard the floor pushes on the power supply at 120 m altitude

At 120 m height, the acceleration of the rocket is 6.98 m/s², which is the same as the power supply.

given force on power supply;

F = 18.5 N

Applying Newton's second law of motion, the mass of the power supply = 18.5/9.8

= 1.888 kg

The force on power supply at this altitude = m(a+g)

                                           = 1.888(6.98 +9.8)

                                           = 1.888(16.78)

                                           = 31.68 N

Therefore, the floor pushes on the power supply at 120 m altitude by a force of 31.68 N

4 0
3 years ago
The energy absorbed or released during a reaction in which a substance is produced is called the
sweet [91]

the answer is bond energy but I am not pretty sure

7 0
3 years ago
A car accelerates uniformly in a straight line
julia-pushkina [17]

The car travels a distance <em>d</em> from rest with acceleration <em>a</em> after time <em>t</em> of

<em>d</em> = 1/2 <em>a</em> <em>t</em>²

It covers 69 m with 2.8 m/s² acceleration, so that

69 m = 1/2 (2.8 m/s²) <em>t</em>²

<em>t</em>² = 2 (69 m) / (2.8 m/s²)

<em>t</em> ≈ 7.02 s

where we take the positive square root because we're talking about time *after* the car begins accelerating.

8 0
3 years ago
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