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Sophie [7]
2 years ago
6

What source of energy is responsible for the water cycle

Physics
1 answer:
Eduardwww [97]2 years ago
7 0
Water? The sun. I DUNNO I FEEL BAD :(
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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
?during the first few days of a fast, what energy source provides about 90% of the glucose needed to fuel the body?
Kobotan [32]
<span>The energy source provides about 90% of the glucose needed to fuel the body in the first few days of the fast is protein. protein is a string of amino acids that peforms a variety of functions to the body. These are found in foods like eggs, cheese, chicken breasts, meat and many more/.</span>
8 0
3 years ago
Una prenda de 320gramos de ropa gira en el interior de una lavadora si dicha lavadora tiene 40 cm y gira con una frecuencia de 4
Nitella [24]

Answer:

Período del tambor: T = 0.25\,s, fuerza sobre la prenda: F \approx 80.852\,N, velocidad lineal del tambor: v \approx 10.053\,\frac{m}{s}, velocidad angular del tambor: \omega \approx 25.133\,\frac{rad}{s}.

Explanation:

La expresión tiene un error por omisión, su forma correcta queda descrita a continuación:

<em>"Una prenda de 320 gramos de ropa gira en el interior de una lavadora si dicha lavadora tiene un radio de 40 centímetros y gira con una frecuencia de 4 hertz. Halle </em><em>a)</em><em> el período, </em><em>b) </em><em>la velocidad angular, </em><em>c) </em><em>la fuerza con la que gira la prenda y </em><em>d) </em><em>la velocidad lineal de la lavadora."</em>

El tambor gira a velocidad angular constante (\omega), en radianes por segundo, lo cual significa que la prenda experimenta una aceleración centrífuga (a), en metros por segundo al cuadrado. En primer lugar, calculamos el período de rotación del tambor (T), en segundos:

T = \frac{1}{f} (1)

Donde f es la frecuencia, en hertz.

(f = 4\,hz)

T = \frac{1}{4\,hz}

T = 0.25\,s

Ahora determinamos la fuerza aplicada sobre la prenda (F), en newtons:

F = m\cdot a (2)

F = \frac{4\pi^{2}\cdot m \cdot r}{T^{2}} (2b)

Donde:

m - Masa de la prenda, en kilogramos.

r - Radio interior del tambor, en metros.

(m = 0.32\,kg, r = 0.4\,m, T = 0.25\,s)

F = \frac{4\pi^{2}\cdot (0.32\,kg)\cdot (0.4\,m)}{(0.25\,s)^{2}}

F \approx 80.852\,N

La velocidad lineal de la lavadora es:

v = \frac{2\pi\cdot r}{T} (3)

(r = 0.4\,m, T = 0.25\,s)

v = \frac{2\pi\cdot (0.4\,m)}{0.25\,s}

v \approx 10.053\,\frac{m}{s}

Y la velocidad angular del tambor de la lavadora:

\omega = \frac{2\pi}{T}

(T = 0.25\,s)

\omega = \frac{2\pi}{0.25\,s}

\omega \approx 25.133\,\frac{rad}{s}

7 0
2 years ago
A person's center of mass is easily found by having the person lie on a reaction board. A horizontal, 2.1-m-long, 6.6 kg reactio
MrRissso [65]
Answer is x hopefully this helps your stupid head
5 0
3 years ago
How are the oscillating magnetic and electric fields of an electromagnetic wave positioned relative to each other?​
Pepsi [2]

Answer:

Electromagnetic waves consist of both electric and magnetic field waves. These waves oscillate in perpendicular planes with respect to each other, and are in phase. The creation of all electromagnetic waves begins with an oscillating charged particle, which creates oscillating electric and magnetic fields.

Explanation:

7 0
2 years ago
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