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BARSIC [14]
3 years ago
12

In which of the following situations is the Doppler Effect absent?

Physics
1 answer:
Orlov [11]3 years ago
8 0

Answer:

Correct option is

C

The listener is moving towards the source of sound

The Doppler effect (or the Doppler shift) is the change in frequency or wavelength of a wave (or another periodic event) for an observer moving relative to its source.

So if both listener and source of sound are stationary or are moving with the same velocity that is relative velocity is zero, there won't be any change in frequency.

Thus, in this case, the effect will arise when a listener is moving towards the source of the sound.

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A 1.5 kg ball pushed with a force of 13.5 N accelerates to the left. What is the acceleration of the ball?​
GarryVolchara [31]

Explanation:

F= ma

13.5 =1.5a

1.5a/1.5 =13.5/1.5

a= 9 m/s^2

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3 years ago
a machine gun fires 10 rounds per second the speed of the bullets is 300 m/s. what is the distance in the air between the flying
vovangra [49]
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Read 2 more answers
What are (a) the charge and (b) the charge density on the surface of a conducting sphere of radius 0.20 m whose potential is 240
Lady bird [3.3K]

Answer:

(a) charge q=5.33 nC

(b) charge density σ=10.62 nC/m²

Explanation:

Given data

radius r=0.20 m

potential V=240 V

coulombs constant k=9×10⁹Nm²/C²

To find

(a) charge q

(b) charge density σ

Solution

For (a) charge q

As

V_{potential}=kq/r\\ q=rV_{potential}/k\\q=\frac{(0.20)(240)}{9*10^{9} }\\ q=5.333*10^{-9}C\\or\\ q=5.33nC

For (b) charge density

As charge density σ is given as:

σ=q/(4πR²)

σ=(5.333×10⁻⁹) / (4π×(0.20)²)

σ=10.62 nC/m²

3 0
3 years ago
When responding to sound, the human eardrum vibrates about its equilibrium position. Suppose an eardrum is vibrating with an amp
Paraphin [41]

Answer:

796.18 Hz

Explanation:

Applying,

Maximum velocity = Amplitude×Angular velocity

Therefore,

V' = A(2πf)............... Equation 1

Where V' = maximum velocity of the eardrum, A = Amplitude of vibration of the eardrum, f = frequency of the eardrum vibration, π = pie

make f the subject of the equation

f = V'/2πA................ Equation 2

From the question,

Given: V' = 3.6×10⁻³ m/s, A' = 7.2×10⁻⁷ m,

Constant: 3.14.

Substitute these values into equation 2

f = 3.6×10⁻³/( 7.2×10⁻⁷×2×3.14)

f = 796.18 Hz

6 0
3 years ago
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