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11111nata11111 [884]
3 years ago
13

How much energy will an electron gain if it moves through a potential difference of 1.0 V?

Physics
1 answer:
Setler [38]3 years ago
6 0

Answer:

a. 1.0 eV

Explanation:

Given that

Voltage difference ,ΔV = 1 V

From work power energy

Work =Change in the kinetic energy

We know that work on the charge W= q ΔV

For electron ,e= 1.6 x 10⁻¹⁹ C

q=e= 1 x 1.6 x 10⁻¹⁹ C

Change in the kinetic energy

ΔKE= q ΔV

Now by putting the values

ΔKE=  1 x 1.6 x 10⁻¹⁹   x 1   C.V

We can also say that

ΔKE=  1 e.V

Therefore the answer will be a.

a). 1.0 eV

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Answer:

(C) The frequency decrease and intensity decrease

Explanation:

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if the observer and the source move away from each other as is the case for this problem, the wavelength heard by the observer is bigger.

The frequency is the inverse from the wavelength, so the frequency heard will increase.

The sound intensity depends inversely on the area in which the sound propagates. When the buzzer is close, the area is from a small sphere, but as the buzzer moves further away, the wave area will be from a larger sphere and therefore the intensity will decrease.

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A charge of 2.00 μC flows onto the plates of a capacitor when it is connected to a 12.0-V potential source. What is the minimum
uysha [10]

To develop this problem we will apply the concept of energy conservation. For which the work carried out must be equivalent to the potential energy stored on the capacitor. We will start by finding the capacitance to later be able to calculate the energy and therefore the work in the capacitor

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Here,

C = Capacitance

V = Potential difference between the plates

Q = Charge between the capacitor plates

At the same time the energy stored in the capacitor can be defined as,

U = \frac{1}{2} CV^2

We will start by finding the value of the capacitance, so we will have to,

C = \frac{2\mu C}{12.0V}

C = 0.166\mu F

Finally using the expression for the energy we have that,

U = \frac{1}{2} CV^2

U = \frac{1}{2} (0.166\muF)(12.0V)^2

U = 11.0*10^{-6} J

Therefore the minimum amount of work that must be done in charging this capacitor is 11.0*10^{-6} J

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(a) 1.15\cdot 10^9 N

For an electromagnetic wave incident on a surface, the radiation pressure is given by (assuming all the radiation is absorbed)

p=\frac{I}{c}

where

I is the intensity

c is the speed of light

In this problem, I=1360 W/m^2; substituting this value, we find the radiation pressure:

p=\frac{1360 W/m^2}{3\cdot 10^8 m/s}=4.5\cdot 10^{-6} Pa

the force exerted on the Earth depends on the surface considered. Assuming that the sunlight hits half of the Earth's surface (the half illuminated by the Sun), we have to consider the area of a hemisphere, which is

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\frac{1.15\cdot 10^9 N}{3.58\cdot 10^{22}N}=3.2\cdot 10^{-14}

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The other 4 kg may have left the scene in the form of
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