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labwork [276]
3 years ago
14

Which of these forces can increase the effect of electric force?

Physics
1 answer:
Sati [7]3 years ago
8 0

Answer:

Electrostatic force. can increase the effect of electric force .

hope it is helpful to you

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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
3 years ago
98 POINTS, 5 simple questions!! HELP
lozanna [386]

25,000 Feet = 7620m

PE = mgh where m is mass, g is gravity accel: 9.8 n h is height

= 90 x 9.8 x 7620

= 6720840J

= 6.72MJ

F = ma where m is mass, a is accel = gravity = 9.8

= 90 x 9.8

= 882N

Accel = gravity = 9.8m/s^2

KE = 1/2mv^2 where m is mass n v is vel

if no wind resistance, PE leaving airplane = KE at net

6720840 = 1/2 x 90 x v^2

v^2 = 149352

v = 386.5m/s


3 0
3 years ago
Read 2 more answers
How can accuracy be limited?
lozanna [386]

Answer:

accuracy refers to the deviation of a measurement from a standard or true value of the quantity being measured

5 0
3 years ago
What effect does an unbalanced force have on an object?
Mars2501 [29]
<span>it either changes the direction of the object or it stops the object in motion.</span>
5 0
3 years ago
Read 2 more answers
In one scene in the movie The Godfather II, a solid gold phone is passed around a large table for everyone to see. Suppose the v
dangina [55]
Volume of gold in the phone = 10 cm^3
                                              = 0.<span>00001 m^3 </span>
Density of gold = 19300 kg/m^3
1 kg mass = 2.2 pounds
Mass of 10 cm^3 of gold = 0<span>.00001 m^3 * (19300 kg/m^3)
                                        = 0.193 kg 
So
0.193 kg = 0.193 * 2.2 pounds
               = 0.43 pounds
I think there is something wrong with the options given in the question.</span>
7 0
3 years ago
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