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Ilia_Sergeevich [38]
3 years ago
15

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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What is a electron orbit?
Advocard [28]

the path of an electron around the nucleus of an atom

5 0
3 years ago
Read 2 more answers
A point charge is placed at each corner of square with side leanth a. The charges all have same magnitude q. My question, What i
nexus9112 [7]

Answer:

 E = k q / a²   (1.3535) (- i ^ + j ^)

  E = k q / a²  1.914  ,      θ’= 135

Explanation:

For this exercise we will use Newton's second law where we must add as vectors

        E_total = E₁₂ i ^ + E₁₄ j ^ + E₁₃

Let's look for the value of each term

On the x axis

       E₁₂ = k q / a²

On the y axis

       E₁₄ = k q / a²

For the charge in the opposite corner we look for the distance

        d = √ (a² + a²) = a √2

let's look for the field

      E₁₃ = k q / d²

      E₁₃ = k q / 2a²

let's use trigonometry to find the two components of this field

       cos 45 = E₁₃ₓ / E₁₃

       E₁₃ₓ = E₁₃ cos 45

       

       sin 45 = E_{13y} / E₁₃

       E_{13y} = E₁₃ sin 45

       E₁₃ₓ = k q / 2a²  cos 45

       E_{13y} = k q / 2a²  sin 45

let's find each component of the electric field

X axis

      Eₓ = -E₁₂ - E₁₃ₓ

      Eₓ = - k q / a² - k q / 2a² cos 45

      Eₓ = - k q / a² (1 + cos 45/2)

      cos 45 = sin 45 = 0.707

      Eₓ = - k q / a²   (1 + 0.707 / 2)

      Eₓ = - k q / a²    (1.3535)

Y axis  

      E_y = E₁₄ + E_{13y}

       E_y = k q / a² + k q / 2a²     sin 45

       E_y = k q / a² (1 + sin 45/2)

       E_y = k q / a²       (1.3535)

we can give the results in two ways

       E = k q / a²   (1.3535) (- i ^ + j ^)

In modulus and angle form, let's use Pythagoras' theorem for the angle

       E = √ (Eₓ² + E_y²)

        E = k q / a²    1.3535 √2

        E = k q / a²     1.914

we use trigonometry for the angle

        tan θ = E_y / Eₓ

         θ = tan⁻¹  (E_y / Eₓ)

         θ = tan⁻¹ (1 / -1)

         θ = 45

in the third quadrant, if we measure the angle of the positive side of the x-axis

           θ‘= 90 + 45

           θ’= 135

4 0
3 years ago
What event will produce the greatest increase in the gravitational force between the two masses?
TiliK225 [7]

Answer:

Doubling the large mass

Explanation:

Doubling the destance bewteen the masses will simply make the gravitational force weaker

same with every answer exce

6 0
2 years ago
If you are six feet tall how far back from a 3 foot mirror do you have to stand in order to see yourself completely?
OverLord2011 [107]

Answer:

you would have to stand 6 ft back

Explanation:

7 0
3 years ago
20 cubic inches of a gas with an absolute pressure of 5 psi is compressed until its pressure reaches 10 psi. What's the new volu
Anna71 [15]

Answer:

B. V_{f}= 10\,cubic\,inches

Explanation:

Assuming we are dealing with a perfect gas, we should use the perfect gas equation:

PV=nRT

With T the temperature, V the volume, P the pressure, R the perfect gas constant and n the number of mol, we are going to use the subscripts i for the initial state when the gas has 20 cubic inches of volume and absolute pressure of 5 psi, and final state when the gas reaches 10 psi, so we have two equations:

P_{i}V_{i}=n_{i}RT_{i} (1)

P_{f}V_{f}=n_{f}RT_{f} (2)

Assuming the temperature and the number of moles remain constant (number of moles remain constant if we don't have a leak of gas) we should equate equations (1) and (2) because T_{i}=T_{f}, n_{i}=n_{f} and R is an universal constant:

P_{i}V_{i}= P_{f}V_{f}, solving for V_{f}

V_{f} =\frac{P_{i}V_{i}}{P_{f}} =\frac{(5)(20)}{10}

V_{f}= 10 cubic\,inches

6 0
3 years ago
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