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miss Akunina [59]
3 years ago
3

The electric field between two parallel plates has a magnitude of 180 N/C. The two plates are 2.5 cm apart. Recall that the magn

itude of the charge on an electron is 1.602 × 10-19 C.
How much work is done to move an electron from the positive plate to the negative plate?

4.0 × 10-21 J
7.2 × 10-19 J
4.5 J
7200 J
Physics
2 answers:
Sunny_sXe [5.5K]3 years ago
7 0

Answer:

b

Explanation:

PilotLPTM [1.2K]3 years ago
3 0

The amount of work done to move an electron from the positive plate to the negative plate is :  7.2 * 10^{-19} J

<u>Step 1: Given data</u>

magnitude of two parallel plate ( E ) = 180 N/C

Distance between plates ( d ) = 2.5 cm

magnitude of charge on electron  ( e ) = 1.602 * 10^{-19}  C

<u> Step 2 :  determine the work done</u>

Work done = qV = еEd

                           = 1.602 * 10^{-19}  *  180 * 2.5

                           = 720 * 10^{-22} J  ≈   7.2 * 10^{-19} J

Hence the amount of work done to move an electron from the positive plate to the negative plate  =  7.2 * 10^{-19} J

learn more about electric field : brainly.com/question/11509296

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The relationship between gravitational force and distance of separation is an "inverse relationship".

<u>Explanation:</u>

Gravitational force has been explained by "Newton's Universal Law of Gravitation" which says "In this universe every single particle attract every other particle with a force which directly proportional to the product of their masses while inversely proportional to the square of the distance in between them".

The gravitational force is inversely proportional to square of distance because as the distance between the objects with masses increases the force decreases. The gravitational force between any two objects with masses can be understood by given formula:

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Here the F is gravitational force,

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4 years ago
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nata0808 [166]

force acting on the object is given as

F = 2.00\hat i + 3.00 \hat j

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now the displacement is given as

d = r_f - r_i

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now the work done is given as

W = F.d

W = (2.00\hat i + 3.00\hat j).(3\hat i - 2\hat j)

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Explanation:

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u is the initial velocity

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For the bullet in this problem:

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v = 0 is the final velocity (the bullet comes to a stop)

s = 0.125 m is the stopping distance of the bullet

Therefore, by solving the equation for a, we find its acceleration:

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