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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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2. A pebble is dropped down a well and hits the water 1.5 s later. Using the equations for motion with constant acceleration, de
kow [346]

Answer:

S = 11.025 m

Explanation:

Given,

The time taken by the pebble to hit the water surface is, t = 1.5 s

Acceleration due to gravity, g = 9.8 m/s²

Using the II equations of motion

                          S = ut + 1/2 gt²

Here u is the initial velocity of the pebble. Since it is free-fall, the initial velocity

                                u = 0

Therefore, the equation becomes

                            S = 1/2 gt²

Substituting the given values in the above equation

                              S = 0.5 x 9.8 x 1.5²

                                 = 11.025 m

Hence, the distance from the edge of the well to the water's surface is, S = 11.025 m

3 0
3 years ago
HELP PLS ILL GIVE U BRAINLIST
jeyben [28]

B. insulator

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6 0
3 years ago
Read 2 more answers
PLEASE WILL MARK BRAINLIEST
devlian [24]
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3 years ago
Read 2 more answers
What is the mass of an object if a force of 17 N causes it to accelerate at 1.5 m/s/s?
DochEvi [55]
The answer is m=11.4g
4 0
2 years ago
10. A hockey puck with mass 0.3 kg is sliding along ice that can be considered frictionless. The puck’s velocity is 20 m/s when
SVEN [57.7K]

Answer:

d = \frac{v^2_i}{2a}= \frac{(20m/s)^2}{2* 3.43 m/s^2}=58.309m

Explanation:

For this case  we can use the second law of Newton given by:

\sum F = ma

The friction force on this case is defined as :

F_f = \mu_k N = \mu_k mg

Where N represent the normal force, \mu_k the kinetic friction coeffient and a the acceleration.

For this case we can assume that the only force is the friction force and we have:

F_f = ma

Replacing the friction force we got:

\mu_k mg = ma

We can cancel the mass and we have:

a = \mu_k g = 0.35 *9.8 \frac{m}{s^2}= 3.43 \frac{m}{s^2}

And now we can use the following kinematic formula in order to find the distance travelled:

v^2_f = v^2_i - 2ad

Assuming the final velocity is 0 we can find the distance like this:

d = \frac{v^2_i}{2a}= \frac{(20m/s)^2}{2* 3.43 m/s^2}=58.309m

5 0
3 years ago
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