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yanalaym [24]
3 years ago
14

Amy counts the wave crest traveling down a stretched string five wave crests pass Amy in two seconds what is the frequency of th

e wave
Physics
1 answer:
Gwar [14]3 years ago
8 0
<span>One end of a uniform meter stick is placed against a vertical wall. The other end is held by a lightweight cord that makes an angle, theta, with the stick. The coefficient of static friction between the end of the meter stick and the wall is 0.390. A. what is the maximum value...</span>
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A photon of wavelength 2.78 pm scatters at an angle of 147° from an initially stationary, unbound electron. What is the de Brogl
Elena-2011 [213]

Answer:

2.07 pm

Explanation:

The problem given here is the very well known 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=147^{\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-cos147^\circ ) m\\\lambda^{'}-\lambda=2.42\times 10^{-12} (1-cos147^\circ ) m.\\\lambda^{'}-\lambda=2.42(1-cos147^\circ ) p.m.\\\lambda^{'}-\lambda=4.45 p.m.

Here, the photon's incident wavelength is \lamda=2.78pm

Therefore,

\lambda^{'}=2.78+4.45=7.23 pm

From the conservation of momentum,

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

where,\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}}

Therefore,

\lambda_{e}=\frac{2.78\times 7.23}{\sqrt{2.78^{2}+7.23^{2}-2\times 2.78\times 7.23\times cos147^\circ }} pm\\\lambda_{e}=\frac{20.0994}{9.68} = 2.07 pm

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

6 0
3 years ago
PLEASE HELP!!!!!!
Helga [31]
Hey there,

Density = Mass/Valume

D= 3.1/0.35

D= 8.86 g/cm3
3 0
3 years ago
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Io experiences tidal heating primarily because __________. hints io experiences tidal heating primarily because __________. io i
maxonik [38]
Lo experiences tidal heating primarily because lo’s elliptical orbit causes the tidal force on lo to vary as it orbits the Jupiter. Thus, lo’s elliptical orbit is essential to its tidal heating. This elliptical orbit, in turn, is an end result of the orbital resonance among lo, Europa and ganymade. This orbital resonance origin lo to have a more elliptical orbit than it would because lo intermittently passes Europa and ganymade in the same orbital position. We cannot perceive tidal forces of tidal heating in lo but rather we foresee that they must occur based on the orbital characteristic of the moons and active volcanoes on lo is the observational evidence that tidal heating is significant in lo.
8 0
3 years ago
A ball is traveling 24° above the horizontal at a speed of 12 m/s. What is the vertical component of its speed?
victus00 [196]

Answer:

4.88 m/s

Explanation:

Vertical component would be  12 * sin 24  =  4.88 m/s

Horizontal is   12 * cos 24

3 0
2 years ago
Un caballo parte del reposo y alcanza una velocidad de 15m/s en un tiempo de 8s. ¿ Cuál fue su aceleración y que distancia recor
Lemur [1.5K]

Answer: A 120 metros por segundo

Explanation: multiplicas la velocidad por el tiempo

7 0
2 years ago
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