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4 years ago

What is the potential energy at the origin due to an electric field of 5 x 10^6 N/C located at x=43cm,y=28cm?

Physics

1 answer:

Fudgin [204]4 years ago

3 0

Answer:

potential energy at origin is $2.57*10^{6} volt$

Explanation:

given data:

electric field E = 5*10^{6} N/C

at x = 43 cm, y = 28 cm

distance btween E and origin

$\Delta r = \sqrt{43^2 +28^2}$

$\Delta r = 51.313 cm$

potential energy per unit charge $\Delta V = - Edr$

$\Delta V = 5*10^6*51.313*10^{-2} J/C$

$\Delta V = 2.57*10^{6} volt$

potential energy at origin is 2.57*10^{6} volt

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Write a 150 word paragraph or two that describes at least three everyday things that exist or occur because of science. Make sur

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4 years ago

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A copper rod of length 27.5 m has its temperature increases by 35.9 degrees celsius. how much does its length increase?(unit=m)

gavmur [86]

<h2>The increase in length = 1.87 x 10⁻²</h2>

Explanation:

When copper rod is heated , its length increases

The increase in length can be found by the relation

L = L₀ ( 1 + α ΔT )

here L is the increased length and L₀ is the original length

α is the coefficient of linear expansion and ΔT is the increase in temperature .

The increase in length = L - L₀ = L₀ x α ΔT

Substituting all these value

Increase in length = 27.5 x 1.7 x 10⁻⁵ x 35.9

= 1.87 x 10⁻² m

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4 years ago

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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.

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4 years ago

Therese plays lacrosse, but she has injured her ankle and has to take a few weeks off. She is finding that she has a lot of trou

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Answer: No endorphins

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3 years ago

A string is wrapped around a pulley with a radius of 2.0 cm. The pulley is initially at rest. A constant force of 50 N is applie

Ray Of Light [21]

Answer:

0.20kg-m^2

Explanation:

Let the linear velocity of the rope(=of pulley) is v m/s

Using kinematic equation

=> v = u + at

=>v = 0 + 4.9a

=>v = 4.9a ------------ eq1

By v^2 = u^2 + 2as

=>v^2 = 0 + 2 x v/4.9 x 1.2

=>4.9v^2 - 2.4v = 0

=>v(4.9v - 2.4) = 0

=>v = 2.4/4.9 = 0.49 m/s

Thus by v = r x omega

=>omega = v/r = 0.49/0.02 = 24.49 rad/sec

BY W = F x s = 50 x 1.2 = 60 J

=>KE(rotational) = W = 1/2 x I x omega^2

=>60 = 1/2 x I x (24.49)^2

=>I = 0.20 kg-m^2

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3 years ago