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kari74 [83]
4 years ago
9

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

Physics
2 answers:
MAXImum [283]4 years ago
6 0

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

WITCHER [35]4 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 temperature of a 3.72 mm cube (e=0.288) that radiates 56.6 W?
blsea [12.9K]

Answer:

The temperature is 2541.799 K

Explanation:

The formula for black body radiation is given by the relation;

Q = eσAT⁴

Where:

Q = Rate of heat transfer 56.6

σ = Stefan-Boltzman constant = 5.67 × 10⁻⁸ W/(m²·k⁴)

A = Surface area of the cube = 6×(3.72 mm)² = 8.3 × 10⁻⁵ m²

e = emissivity = 0.288

T = Temperature

Therefore, we have;

T⁴ = Q/(e×σ×A) = 56.6/(5.67 × 10⁻⁸ × 8.3 × 10⁻⁵ × 0.288) = 4.174 × 10¹⁴ K⁴

T  =  2541.799 K

The temperature = 2541.799 K.

7 0
4 years ago
A real battery with internal resistance 0.460 Ω and emf 9.00 V is used to charge a 56.0-µF capacitor. A 21.0-Ω resistor is put i
Margaret [11]

Answer: 1.176×10^-3 s

Explanation: The time constant formulae for an RC circuit is given below as

t =RC

Where t = time constant , R = magnitude of resistance = 21 ohms , C = capacitance of capacitor = 56 uf = 56×10^-6 F

t = 56×10^-6 × 21

t = 1176×10^-6

t = 1.176×10^-3 s

4 0
3 years ago
Read 2 more answers
What is the period of a wave traveling 5 m/s if its wavelength is 20 m/s
Ilia_Sergeevich [38]

Speed of wave is given as

v = 5 m/s

Wavelength of the wave is given as

\lambda = 20 m

now from the formula of wave time period we can say

speed = \frac{wavelength}{time period}

5 = \frac{20}{T}

T = \frac{20}{5}

T = 4 s

so it will have time period of T = 4 s

7 0
3 years ago
When Trinity pulls on the rope with her weight, Newton's Third Law of Motion tells us that the rope will _____.
Grace [21]
The rope will pull back against you.
5 0
3 years ago
Read 2 more answers
The component of the external magnetic field along the central axis of a 78-turn circular coil of radius 34.0 cm decreases from
grigory [225]

Answer:

Induced current, I = 18.88 A

Explanation:

It is given that,

Number of turns, N = 78

Radius of the circular coil, r = 34 cm = 0.34 m

Magnetic field changes from 2.4 T to 0.4 T in 2 s.

Resistance of the coil, R = 1.5 ohms

We need to find the magnitude of the induced current in the coil. The induced emf is given by :

\epsilon=-N\dfrac{d\phi}{dt}

Where

\dfrac{d\phi}{dt} is the rate of change of magnetic flux,

And \phi=BA

\epsilon=-NA\dfrac{dB}{dt}

\epsilon=-78\times \pi (0.34)^2\dfrac{(0.4-2.4)}{2}

\epsilon=28.32\ V

Using Ohm's law, \epsilon=I\times R

Induced current, I=\dfrac{\epsilon}{R}

I=\dfrac{28.32}{1.5}

I = 18.88 A

So, the magnitude of the induced current in the coil is 18.88 A. Hence, this is the required solution.

5 0
3 years ago
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