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antoniya [11.8K]
2 years ago
11

(a) Suppose you charge a 2.5 F capacitor with two 1.5 volt batteries. How much charge was on each plate? 7.5 Correct: Your answe

r is correct. C (b) How many excess electrons were on the negative plate? electrons
Physics
1 answer:
natali 33 [55]2 years ago
7 0

Answer:

C = Q / V      

Q = C * V = 2.5 F * 3 V = 7.5 Coulombs (assuming the batteries are in series)

N e = Q       where N is the number of electrons and e the electronic charge

N = 7.5 / (1.60E-19) = 4.7E19 electrons

or 4.7 * 10^19 electrons

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Answer:

The total distance is 381.5 [m]

Explanation:

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v_{f} =v_{o} +(a*t)

where:

Vf = final velocity [m/s]

Vo = initial velocity = 0

a = acceleration = 2 [m/s²]

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With this velocity, we can calculate the displacement using the following expression.

v_{f} ^{2} =v_{o} ^{2} +2*a*x

where

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The other important clue to solve this problem in the second part is that the final velocity is now the initial velocity.

We must calculate the final velocity.

v_{f}= v_{i} +(a*t)

Vf = final velocity [m/s]

Vi = initial velocity = 14 [m/s]

a = desacceleration = 4 [m/s²]

t = time = 8 [s]

Vf = 24 + (4*8)

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With this velocity, we can calculate the displacement using the following expression.

v_{f} ^{2} =v_{o} ^{2} +2*a*x

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Now Potential energy is defined by this formula,

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