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denis-greek [22]
1 year ago
13

A light bulb is rated at 25 w when operated at 110 v. how much charge enters (and leaves) the light bulb in 1.0 hour?

Physics
1 answer:
DedPeter [7]1 year ago
4 0

According to the current, the charge is 818.18 coulombs.

We need to know about current to solve this problem. Current can be defined as the flow rate of charge through the medium. It can be determined as

I = Q / t

where I is current, Q is charge and t is time.

From the question above, we know that

P = 25 W

V = 110 V

h = 1 hour = 3600 second

Calculate the current

P = V . I

25 = 110 . I

I = 0.23 A

By substituting the given parameters, we can calculate the charge

I = Q / t

0.23 = Q / 3600

Q = 818.18 coulombs

Hence, the charge is 818.18 coulombs.

Find more on current at: brainly.com/question/24858512

#SPJ4

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Answer:

For proton

a=8.8\times 10^{17}\ m/s^2

For electron

a=1.5\times 10^{21}\ m/s^2

Explanation:

We know that

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m=9.1\times 10^{-31}\ kg

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q₁=q₂=q

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Electrostatics force

F=K\dfrac{q_1q_2}{r^2}

Now by putting the values

F=9\times 10^9\times \dfrac{1.6\times 10^{-19}\times 1.6\times 10^{-19}}{(4\times 10^{-10})^2}

F=1.44\times 10^{-9}\ N

For proton

F = m a

a =F/m

a=\dfrac{1.4\times 10^{-9}}{1.67\times 10^{-27}}\ m/s^2

a=8.8\times 10^{17}\ m/s^2

For electron

a=\dfrac{1.4\times 10^{-9}}{9.1\times 10^{-31}}\ m/s^2

a=1.5\times 10^{21}\ m/s^2

5 0
3 years ago
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A force F₁ of magnitude 6.50 units acts at the origin in a direction 49.0° above the positive x axis. A second force F₂ of magni
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A bare 4 AWG copper conductor Installed horizontally near the bottom or vertically, and within that portion of a concrete foun d
neonofarm [45]

Answer:

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Explanation:

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3 years ago
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Answer:

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Explanation:

From the question we are told that

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substituting values

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3 0
3 years ago
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