They are positive and remain inside the nucleus.
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:
So the minimum force is
32.2Newton
Explanation:
To solve for the minimum force, let us assume it to be F (N)
So
F=mgsinA
But
=>>>> coefficient of static friction x (F + mgcosA
=>3 x 9.8 x sin35 = 0.3 x (F + 3 x 9.8 x cos35)
So making F subject of formula
F + 24.0 = 56.2
F = 32.2N
Answer:
A. The target nucleus split into two nuclei, each with fewer nucleons than the original.
Explanation:
I think it’s C b/c it works for me