The change in mean drift velocity for electrons as they pass from one end of the wire to the other is 3.506 x 10⁻⁷ m/s and average acceleration of the electrons is 4.38 x 10⁻¹⁵ m/s².
The given parameters;
- <em>Current flowing in the wire, I = 4.00 mA</em>
- <em>Initial diameter of the wire, d₁ = 4 mm = 0.004 m</em>
- <em>Final diameter of the wire, d₂ = 1 mm = 0.001 m</em>
- <em>Length of wire, L = 2.00 m</em>
- <em>Density of electron in the copper, n = 8.5 x 10²⁸ /m³</em>
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The initial area of the copper wire;

The final area of the copper wire;

The initial drift velocity of the electrons is calculated as;

The final drift velocity of the electrons is calculated as;

The change in the mean drift velocity is calculated as;

The time of motion of electrons for the initial wire diameter is calculated as;

The time of motion of electrons for the final wire diameter is calculated as;

The average acceleration of the electrons is calculated as;

Thus, the change in mean drift velocity for electrons as they pass from one end of the wire to the other is 3.506 x 10⁻⁷ m/s and average acceleration of the electrons is 4.38 x 10⁻¹⁵ m/s².
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Answer:
20 degrees.
Explanation:
From Snell’s law of refraction:
sinθ1•n1 = sinθ2•n2
where θ1 is the incidence angle, θ2 is the refraction angle, n1 is the refraction index of light in medium1, and n2 is the refraction index for virgin olive oil. The incidence angle of the red light is θ1 = 30 degrees.
The red light is in air as medium1, so n1 (air) = 1.00029
So, to find θ2, the refracted angle:
sinθ1•1.00029 = sinθ2•1.464
sin(30)•1.00029 / 1.464 = sinθ2
0.5•1.00029 / 1.464 = sinθ2
sinθ2 = 0.3416291
θ2 = arcsin(0.3416291)
θ2 = 19.976 degrees
To the nearest degree,
θ2 = 20 degrees.
Answer:
option C
Explanation:
given,
Force on the object = 10 N
distance of push = 5 m
Work done = ?
we know,
work done is equal to Force into displacement.
W = F . s
W = 10 x 5
W = 50 J
Work done by the object when 10 N force is applied is equal to 50 J
Hence, the correct answer is option C
Answer:
An object is called a horizontal projectile if it is launched from a certain height with some initial horizontal velocity only. The initial vertical velocity of such an object is zero. But as the object falls through the atmosphere the horizontal component of velocity remains constant but vertical component increases due to gravitational acceleration.
Explanation: