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Eduardwww [97]
3 years ago
13

A current of I = 3.8 A is passing through a conductor with cross sectional area A = 2.5 x 10^-4 m^2. The charge carriers in the

conductor, electrons, have a number density n = 1.3 x 10^27 m^-3.
(a) Express the drift velocity of electrons through 1.A. n, and the fundamental charge e.
(b) Calculate the numerical value of v, in m/s.
Physics
1 answer:
bulgar [2K]3 years ago
8 0

Answer:

Explanation:

current, i = 3.8 A

Area of crossection, A = 2.5 x 10^-4 m²

number density, n = 1.3 x 10^27 m^-3

(a) the relation between the current and the drift velocity is given by

i = n e A v

where, e is the electronic charge, A be the area of crossection area of the wire and v be the drift velocity

v = i / neA

(b)

3.8 = 1.3 x 10^27 x 1.6 x 10^-19 x 2.5 x 10^-4 x v

3.8 = 52000 v

v = 7.3 x 10^-5 m/s

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What is the magnitude of electrical force of attraction between an copper nucleus (29 protons) and its innermost electron if the
Agata [3.3K]
The charge of the copper nucleus is 29 times the charge of one proton:
Q=29 q= 29 \cdot 1.6 \cdot 10^{-19}C=4.64 \cdot 10^{-18}C
the charge of the electron is
e=-1.6 \cdot 10^{-19}C
and their separation is
r=1.0 \cdot 10^{-12} m

The magnitude of the electrostatic force between them is given by:
F=k_e  \frac{Qe}{r^2}
where k_e is the Coulomb's constant. If we substitute the numbers, we find (we can ignore the negative sign of the electron charge, since we are interested only in the magnitude of the force)
F=(8.99 \cdot 10^9 Nm^2C^{-2}) \frac{(4.64 \cdot 10^{-18}C)(1.6 \cdot 10^{-19}C)}{(1.0 \cdot 10^{-12} m)^2}=6.68 \cdot 10^{-3} N
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3 years ago
A cable is 100-m long and has a cross-sectional area of 1.0 mm2. A 1000-N force is applied to stretch the cable. Young's modulus
Blizzard [7]

Answer:

1 m

Explanation:

L = 100 m

A = 1 mm^2 = 1 x 10^-6 m^2

Y = 1 x 10^11 N/m^2

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Let the cable stretch be ΔL.

By the formula of Young's modulus

Y=\frac{F\times L}{A\times\Delta L}

\Delta L=\frac{F\times L}{A\times\Y}

\Delta L=\frac{1000\times 100}{10^{-6}\times10^{11}}

ΔL = 1 m

Thus, the cable stretches by 1 m.

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3 years ago
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