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NNADVOKAT [17]
3 years ago
11

A typical cell phone charger is rated to transfer a maximum of 1.0 Coulomb of charge per second. Calculate the maximum number of

electrons that can be transferred by this charger in 1.0 hour
Physics
1 answer:
Alexandra [31]3 years ago
3 0

Answer:

The maximum no. of electrons- 2.25\times 10^{22}

Solution:

As per the question:

Maximum rate of transfer of charge, I = 1.0 C/s

Time, t = 1.0 h = 3600 s

Rate of transfer of charge is current, I

Also,

I = \frac{Q}{t}

Q = ne

where

n = no. of electrons

Q = charge in coulomb

I = current

Thus

Q = It

Thus the charge flow in 1. 0 h:

Q = 1.0\times 3600 = 3600\ C

Maximum number of electrons, n is given by:

n = \frac{Q}{e}

where

e = charge on an electron = 1.6\times 10^{- 19}\ C

Thus

n = \frac{3600}{1.6\times 10^{- 19}} = 2.25\times 10^{22}

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Answer:

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Explanation:

1)

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  • Replacing by the givens, and solving for f, we get:

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2)

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       v =\frac{d}{t} (3)

       where v= c= speed of light in vacuum = 3*10⁸ m/s

       d= distance between mountaintops = 52 km = 52*10³ m

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       t = \frac{d}{c} = \frac{52e3m}{3e8m/s} = 17.3e-5 sec = 0.173e-3 sec = 0.173 ms (4)

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