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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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When a spacecraft travels away from Earth, the gravitational force acting on it

S_A_V [24]

The correct answer is decreases

The further away you are the weaker it would be. That's why at one point you stop being in the field and ti doesn't pull you towards it anymore. Proportionally, if you move towards the Earth then it increases.

6 0

3 years ago

Read 2 more answers

A long, current-carrying solenoid with an air core has 1800 turns per meter of length and a radius of 0.0165 m. A coil of 210 tu

Hoochie [10]

Answer:

The mutual inductance is $M = 0.000406 \ H$

Explanation:

From the question we are told that

The number of turns per unit length is $N = 1800$

The radius is $r = 0.0165 \ m$

The number of turns of the solenoid is $N_s = 210 \ turns$

Generally the mutual inductance of the system is mathematically represented as

$M = \mu_o * N * N_s * A$

Where A is the cross-sectional area of the system which is mathematically represented as

$A = \pi * r^2$

substituting values

$A = 3.142 * (0.0165)^2$

$A = 0.0008554 \ m^2$

also $\mu_o$ is the permeability of free space with the value $\mu_o = 4\pi * 10^{-7} N/A^2$

$M = 4\pi * 10^{-7} *1800 * 210 * 0.0008554$

$M = 0.000406 \ H$

3 0

3 years ago

If the frequency of the motion of a simple harmonic oscillator is doubled, by what factor does the maximum speed of the oscillat

Mamont248 [21]

<h2>The option ( c ) is correct </h2>

Explanation:

As the frequency of oscillation of any oscillator is doubled

The velocity of sound v = νλ

here ν is the frequency and λ is the wavelength

Now if ν becomes double , the wavelength λ becomes one half . The velocity of sound remains the same in the same medium .

Thus option ( c ) is correct

7 0

3 years ago

A capacitance C and an inductance L are operated at the same angular frequency.

Levart [38]

A) $\omega = \frac{1}{\sqrt{LC}}$

The magnitude of the capacitive reactance is given by

$X_C = \frac{1}{\omega C}$

where

$\omega$ is the angular frequency

C is the capacitance

While the magnitude of the inductive capacitance is given by

$X_L = \omega L$

where L is the inductance.

Since we want the two reactances to be equal, we have

$X_C = X_L$

So we find

$\frac{1}{\omega C}= \omega L\\\omega^2 = \frac{1}{LC}\\\omega = \frac{1}{\sqrt{LC}}$

B) 7449 rad/s

In this case, we have

$L=5.30 mH = 5.3\cdot 10^{-3}H$ is the inductance

$C= 3.40 \mu F= 3.40 \cdot 10^{-6}F$ is the capacitance

Therefore, substituting in the formula for the angular frequency, we find

$\omega=\frac{1}{\sqrt{LC}}=\frac{1}{\sqrt{(5.30\cdot 10^{-3}H)(3.40\cdot 10^{-6} F)}}=7449 rad/s$

C) $39.5 \Omega$

Now we can us the formulas of the reactances written in part A). We have:

- Capacitive reactance:

$X_C = \frac{1}{\omega C}=\frac{1}{(7449 rad/s)(3.40\cdot 10^{-6}F)}=39.5 \Omega$

- Inductive reactance:

$X_L = \omega L=(7449 rad/s)(5.30\cdot 10^{-3}H)=39.5 \Omega$

7 0

3 years ago

If the resulting trajectory of the charged particle is a circle,what is , the angular frequency of the circular motion?Express i

Maru [420]

Answer:

ω = B₀q/m

Explanation:

The centripetal force on the charge equal the magnetic force on the charge. (Since the trajectory is a circle). So, mrω² = B₀qv.

v = rω The speed of the charge and r = radius of path

mrω² = B₀qrω

ω = B₀q/m

The angular frequency ω = B₀q/m

8 0

3 years ago