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anyanavicka [17]

3 years ago

8

The voltage across a resistor is found to be 1.5 V. It is also found that there is a charge of 2 Coulombs passing through the re

sistor within 10 seconds. What is the value of the resistor in ohms

1 answer:

djverab [1.8K]3 years ago

8 0

Answer:

R = 7.5 Ω

Explanation:

If the resistor is in the linear zone of operation, the resistance must obey Ohm's Law:

$V = I*R (1)$

By definition, the current flowing through the resistor, is equal to the charge passing through the resistor, per unit time.
So, we can write the following expression for the current I:

$I =\frac{\Delta Q}{\Delta t} = \frac{2C}{10s} = 0.2 A (2)$

From (1) and (2) we can solve for R, as follows:

$R =\frac{V}{I} = \frac{1.5V}{0.2A} = 7.5 \Omega$

The value of the resistor is 7.5 Ω.

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1 year ago

You hang a tv on your wall. What kind of energy does it have?

lara31 [8.8K]

Answer:

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Explanation:

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3 years ago

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

A 2kg bowling ball rolls at a speed of 5 m/s on a roof of the building that is 40 meters tall. What is the kinetic energy

MrRa [10]

The kinetic energy is $\frac 1 2 m v^2$ and the height of the building doesn't matter at all.

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8 0

3 years ago

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amm1812

Answer:

C

Explanation:

8 0

3 years ago