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Amiraneli [1.4K]
2 years ago
6

Describe the pattern you see in the chart PLEASE HURRY IS URGENT

Physics
1 answer:
klasskru [66]2 years ago
6 0

Answer:

0-0

Explanation:

Forgive me..... I'm getting points 0-0

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For copper, ρ = 8.93 g/cm3 and M = 63.5 g/mol. Assuming one free electron per copper atom, what is the drift velocity of electro
viktelen [127]

Answer:

V_d = 1.75 × 10⁻⁴ m/s

Explanation:

Given:

Density of copper, ρ = 8.93 g/cm³

mass, M = 63.5 g/mol

Radius of wire = 0.625 mm

Current, I = 3A

Area of the wire, A = \frac{\pi d^2}{4} = A = \frac{\pi 0.625^2}{4}

Now,

The current density, J is given as

J=\frac{I}{A}=\frac{3}{ \frac{\pi 0.625^2}{4}}= 2444619.925 A/mm²

now, the electron density, n = \frac{\rho}{M}N_A

where,

N_A=Avogadro's Number

n = \frac{8.93}{63.5}(6.2\times 10^{23})=8.719\times 10^{28}\ electrons/m^3

Now,

the drift velocity, V_d

V_d=\frac{J}{ne}

where,

e = charge on electron = 1.6 × 10⁻¹⁹ C

thus,

V_d=\frac{2444619.925}{8.719\times 10^{28}\times (1.6\times 10^{-19})e} = 1.75 × 10⁻⁴ m/s

4 0
3 years ago
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In a particular metal, the mobility of the mobile electrons is 0.0033 (m/s)/(N/C). At a particular moment, the electric field ev
Lapatulllka [165]

Answer:

the average drift speed of the mobile electrons in the metal is 1.089 x 10⁻⁴ m/s.

Explanation:

Given;

mobility of the mobile electrons in the metal, μ = 0.0033 (m/s)/(N/C)

the electric field strength inside the cube of the metal, E = 0.033 N/C

The average drift speed of the mobile electrons in the metal is calculated as;

v = μE

v =  0.0033 (m/s)/(N/C) x 0.033 N/C

v = 1.089 x 10⁻⁴ m/s.

Therefore, the average drift speed of the mobile electrons in the metal is 1.089 x 10⁻⁴ m/s.

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