The work done in moving an electron through the 3 ohm resistor is

Physics

1 answer:

Orlov [11]3 years ago

8 0

If the 3 ohm resistor is in parallel to the 12 ohm (I can't see it in the image) then the current through it is 12V/3ohm = 4A Amps are Coulombs/s The charge on the electron is 1.6e-19C So 4 amps means 4/1.6e-19 = 2.5e19 electrons/s.
The reciprocal of this gives the number of electrons per second:
1/2.5e19 = 4e-20 s/(elect. charge)
The power in the 3 ohm resistor is 12*4=48W. This is Joules/sec. We know it takes on average 4e-20 s to get one electron charge through the resistor so we multiply: 48W * 4e-20s = 1.9e-18J
Note: In reality there are many electrons moving through the resistor together, and a single electron does not move this fast, but the total energy to do it would be equivalent to this.

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ASAP

nikitadnepr [17]

Answer:

A. 59.4

Explanation:

The refractive index of the glass, n₁ = 1.50

The angle of incidence of the light, θ₁ = 35°

The refractive index of air, n₂ = 1.0

Snell's law states that n₁·sin(θ₁) = n₂·sin(θ₂)

Where;

θ₂ = The angle of refraction of the light, which is the angle the light will have when it passes from the glass into the air

Therefore;

θ₂ = arcsin(n₁·sin(θ₁)/n₂)

Plugging in the values of n₁, n₂ and θ₁ gives;

θ₂ = arcsin(1.50 × sin(35°)/1.0) ≈ 59.357551° ≈ 59.4°

The angle the light will have when it passes from the glass into the air, θ₂ ≈ 59.4°.

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2 years ago

Question 25

Kruka [31]

Answer:

The one with equal forces on both sides, D is your answer

Explanation:

A, 5 and 10 No

B 13 and 8 no

C 5 and 8 no

D 6 and 6 YES

3 0

2 years ago

A negative ion of charge -2e is located at the origin and a second negative ion of charge -3e is located nearby at x = 3.8 nm ,

Rus_ich [418]

Answer:

$\vec{F}_{21}=-5.63\times 10^{-11}N\\\\\vec{F}_{21}=\\$

Explanation:

Given that

$Q_1 = -2e\, C\\\\Q_2=-3e\,C\\\\x= 3.8 \times 10^{-9}\,m\\\\y= 3.2 \times 10^{-9}\,m\\\\r=\sqrt{x^2+y^2}\\\\r= 4.96\times 10^{-9} m\\$

As both charges are negative so there exist force of repulsion in direction as shown in figure.

$F_{12}=\frac{kQ_1Q_2}{r^2}\\\\F_{12}= \frac{(9\times 10^9)(6)(1.602\times 10^{-19})^2}{(4.96\times 10^{-9})^2}\\\\F_{12}=5.63\times 10^{-11}N$

Angle at which force F12 is acting is

$\theta=tan^{-1}\frac{3.2}{3.8}\\\\\theta=tan^{-1}\frac{y}{x}\\\\\theta= 40.1^o$

$F_{x}=F_{12}cos\theta\\\\F_{x}=(5.63\times 10^{-11})cos(40.1)\\\\F_{x}=4.306\times 10^{-11}N\\\\F_{y}=F_{12}sin\theta\\\\F_{y}=(5.63\times 10^{-11})sin(40.1)\\\\F_{y}=3.62\times 10^{-11}N\\\\$