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2 years ago

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1) Consider an electric power transmission line that carries a constant electric current of i = 500 A. The cylindrical copper ca

ble used to transmit this current has a diameter o = 2.00 cm and a length L = 150 km. If there are 8.43x10^28 free electrons per cubic meter (m^3 ) in the cable, calculate how long it would take for an electron to cross the entire length of the transmitter line.

1 answer:

JulsSmile [24]2 years ago

7 0

Answer:

t = 1.27 x 10⁹ s

Explanation:

First, we will find the volume of the wire:

Volume = V = AL

where,

A = Cross-sectional area of wire = πr² = π(1 cm)² = π(0.01 m)² = 3.14 x 10⁻⁴ m²

L = Length of wire = 150 km = 150000 m

Therefore,

V = 47.12 m³

Now, we will find the number of electrons in the wire:

No. of electrons = n = (Electrons per unit Volume)(V)

n = (8.43 x 10²⁸ electrons/m³)(47.12 m³)

n = 3.97 x 10³⁰ electrons

Now, we will use the formula of current to find out the time taken by each electron to cross the wire:

$I =\frac{q}{t}$

where,

t = time = ?

I = current = 500 A

q = total charge = (n)(chareg on one electron)

q = (3.97 x 10³⁰ electrons)(1.6 x 10⁻¹⁹ C/electron)

q = 6.36 x 10¹¹ C

$500\ A = \frac{6.36\ x\ 10^{11}\ C}{t}\\\\t = \frac{6.36\ x\ 10^{11}\ C}{500\ A}$

Therefore,

<u>t = 1.27 x 10⁹ s</u>

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Three children are riding on the edge of a merry-go-round that is a solid disk with a mass of 102 kg and a radius of 1.53 m. The

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Three children of masses and their position on the merry go round

M1 = 22kg

M2 = 28kg

M3 = 33kg

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Final moment of inertial when R2 =0

I(final)=½MR²+M1•R²+M2•R2²+M3•R²

Since R2 = 0

I(final) = ½MR²+ M1•R² + M3•R²

I(final) = ½MR² + (M1 + M3)• R²

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I(final) = 151.725 + 158.95

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L(initial) = L(final)

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