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3 years ago

What is the name of the perceived change in a sound wave’s frequency due to motion between the observer and the sound source?

Physics

2 answers:

andrew11 [14]3 years ago

6 0

Answer: The answer is Doppler effect.

Explanation:

When there is a relative motion between the source and the observer then there will be change in the frequency of the wave. This phenomenon is called Doppler effect.

For example,when the source and the observer are moving each other then the frequency increases in this case. When the source and the observer are moving away from other then the frequency decreases in this case.

Therefore, Doppler effect is the name of the perceived change in a sound wave's frequency due to motion between the observer and the sound source.

Shkiper50 [21]3 years ago

4 0

Answer: Hello mate!

the name is "doppler effect"

As you may know, the frequency (and hence the wavelength) of the sound determines the pitch.

If you move towards the source of the sound, the wavelenght seems shorter, which means that the frequency is higher, and also the pitch.

In the other case, if you are moving away from the sound, the wavelength seems larger and the frequency smaller, so the pitch is lower.

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3 years ago

g a small smetal sphere, carrying a net charge is held stationarry. what is the speed are 0.4 m apart

Veseljchak [2.6K]

Answer:

The speed of q₂ is $4\sqrt{10}\ m/s$

Explanation:

Given that,

Distance = 0.4 m apart

Suppose, A small metal sphere, carrying a net charge q₁ = −2μC, is held in a stationary position by insulating supports. A second small metal sphere, with a net charge of q₂ = −8μC and mass 1.50g, is projected toward q₁. When the two spheres are 0.800m apart, q₂ is moving toward q₁ with speed 20m/s.

We need to calculate the speed of q₂

Using conservation of energy

$E_{i}=E_{f}$

$\dfrac{1}{2}mv_{i}^2+\dfrac{kq_{1}q_{2}}{r_{i}}=\dfrac{kq_{1}q_{2}}{r_{f}}+\dfrac{1}{2}mv_{f}^2$

$\dfrac{1}{2}m(v_{i}^2-v_{f}^2)=kq_{1}q_{2}(\dfrac{1}{r_{f}}-\dfrac{1}{r_{i}})$

Put the value into the formula

$\dfrac{1}{2}\times1.5\times10^{-3}(20^2-v_{f}^2)=9\times10^{9}\times-2\times10^{-6}\times-8\times10^{-6}(\dfrac{1}{(0.4)}-\dfrac{1}{(0.8)})$

$0.00075(400-v_{f}^2)=0.18
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$400-v_{f}^2=\dfrac{0.18}{0.00075}$

$-v_{f}^2=240-400$

$v_{f}^2=160$

$v_{f}=4\sqrt{10}\ m/s$

Hence, The speed of q₂ is $4\sqrt{10}\ m/s$

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