A ball is thrown straight up into the air with an initial speed of 8.0 m/s. How long does it take for the rocket to reach its hi
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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.
Learn more here: brainly.com/question/22406248
1) Analyze the equation
Vₓ = α + β t²
Vₓ = 3.00 + 0.100 t²
That is a quadratic equation, so the graph is a parabola.
2) Therefore, take into account that the main points to draw a parabola are:
i) vertex
ii) concavity
iii) y-intercepts
iv) x - intercepts
Also, for all graphs you need the domain and the range.
3) Find the y-intercept (t = 0)
t = 0 ⇒ Vₓ = 3.00 + 0.100 (0)² = 3.00
4) Find the x-intercepts (Vₓ = 0)
Vₓ = 0 ⇒ 0 = 3.00 + 0.100 t²
⇒ t² = - 3.00 / 0.100 = -30.0. Since t² cannot be negative, there is not x-intercepts.
5) Concavity
Since, the coefficient of t² is positive, the parabola open upwards.
6) Vertex
It is the local minimum of the equation. You can find it by the first derivative
Vₓ' = 2×0.100 t = 0 ⇒ t = 0
Vₓ = 3.00 + 0.100 (0)² = 3.00 m/s
⇒ vertex = (0,3.00)
7) The domain is given t ∈ [0,5.00]
8) You can also build a table with several points in the domain
t =0; Vₓ = 3.00 + 0.100 (0)² = 0
t = 1; Vₓ = 3.00 + 0.100 (1)² = 3.10
t = 2; Vₓ = 3.00 + 0.100 (2)² = 3.40
t = 3; Vₓ = 3.00 + 0.100 (3)² = 3.90
t = 4; Vₓ = 3.00 + 0.100 (4)² = 4.60
t = 5; Vₓ = 3.00 + 0.100 (5)² = 5.50
9) Range: Vₓ ∈ [ 3.00, 5.50]
10) All that information permits you to put several points in a coordinate sysment and sketch the same graph as the shown in the figure attached.
