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Vera_Pavlovna [14]
3 years ago
12

A sound wave moving through the air causes particles in the air to move. If air particles are moving, why doesn’t this motion cr

eate a breeze?
Physics
1 answer:
mylen [45]3 years ago
3 0

Answer:

The motion of air particle caused by sound wave is vibrational and cannot create breeze because breeze involves translational motion (permanent displacement) of the air particles.

Explanation:

This sound wave results from the back and forth vibration of the particles of air through which the sound wave is moving. The sound wave causes disturbance within the particles of air by transfering energy from one particle to the adjacent particles of air without causing permanent displacement of the particles of air. The motion between these particles of air caused by sound wave is vibrational, while breeze involves displacement of particles of air through translational motion. Hence, breeze is a moving mass of air.

Thus, the motion of air particle caused by sound wave is vibrational and cannot create breeze because breeze involves translational motion (permanent displacement) of the air particles.

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E
N76 [4]

Answer:

Weight

Explanation:

Weight is the downward pull on an object due to gravity.

For example, the moon has less gravity than Earth so we would weigh less on the moon. Our Mass and volume always stay the same but our weight could change.

4 0
4 years ago
A photon of wavelength 7.33 pm scatters at an angle of 157° from an initially stationary, unbound electron. What is the de Brogl
Ann [662]

Answer:

4.63 p.m.

Explanation:

The problem given here can be solved by the Compton effect which is expressed as

\lambda^{'}-\lambda=\frac{h}{m_e c}(1-cos\theta)

here, \lambda  is the initial photon wavelength, \lambda^{'} is the scattered photon wavelength, h is he Planck's constant, m_e is the free electron mass, c is the velocity of light, \theta  is the angle of scattering.

Given that, the scattering angle is, \theta=157^{\circ}

Putting the respective values, we get

\lambda^{'}-\lambda=\frac{6.626\times 10^{-34} }{9.11\times 10^{-31}\times 3\times 10^{8}  } (1-cos157^\circ ) m\\\lambda^{'}-\lambda=2.42\times 10^{-12} (1-cos157^\circ ) m\\\lambda^{'}-\lambda=2.42(1-cos157^\circ ) p.m.

Therfore,

\lambda^{'}-\lambda=4.64 p.m.

Here, the photon's incident wavelength is \lamda=7.33pm

So,

\lambda^{'}=7.33+4.64=11.97 p.m

From the conservation of momentum,

\vec{P_\lambda}=\vec{P_{\lambda^{'}}}+\vec{P_e}

here, \vec{P_\lambda} is the initial photon momentum, \vec{P_{\lambda^{'}}} is the final photon momentum and \vec{P_e} is the scattered electron momentum.

Expanding the vector sum, we get

P^2_{e}=P^2_{\lambda}+P^2_{\lambda^{'}}-2P_\lambda P_{\lambda^{'}}cos\theta

Now expressing the momentum in terms of De-Broglie wavelength

P=h/\lambda and putting it in the above equation we get,

\lambda_{e}=\frac{\lambda \lambda^{'}}{\sqrt{\lambda^{2}+\lambda^{2}_{'}-2\lambda \lambda^{'} cos\theta}}

Therfore,

\lambda_{e}=\frac{7.33\times 11.97}{\sqrt{7.33^{2}+11.97^{2}-2\times 7.33\times 11.97\times cos157^\circ }} p.m.\\\lambda_{e}=\frac{87.7401}{18.935} = 4.63 p.m.

This is the de Broglie wavelength of the electron after scattering.

8 0
4 years ago
What would most likely happen as a result of the generator in a wind turbine breaking? The blades would not be turned. Less stea
Aleonysh [2.5K]

Answer:

What would most likely happen as a result of the generator in a wind turbine breaking?

The What would most likely happen as a result of the generator in a wind turbine breaking?

The blades would not be turned.

Less steam would be produced.

Electricity would not be generated.

Solar energy would not be absorbed.

The blades would not be turned.

Less steam would be produced.

Electricity would not be generated.

Solar energy would not be absorbed.

Explanation:

F

6 0
3 years ago
Read 2 more answers
How do you test hydrogen gas ​
neonofarm [45]

Explanation:

The test for hydrogen consists of holding a burning splint held at the open end of a test tube of gas. If the gas is hydrogen it burns with a loud “squeaky pop” which is the result of the rapid combustion of hydrogen with oxygen to produce water.

4 0
3 years ago
To measure the wavelength of any longitudnal wave, one should measure distance
makkiz [27]
I think it’s either A or B
6 0
3 years ago
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