S Four identical particles, each having charge q and mass m, are released from rest at the vertices of a square of side L. How f
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Answer:
a) 75.5 degree relative to the North in north-west direction
b) 309.84 km/h
Explanation:
a)If the pilot wants to fly due west while there's wind of 80km/h due south. The north-component of the airplane velocity relative to the air must be equal to the wind speed to the south, 80km/h in order to counter balance it
So the pilot should head to the West-North direction at an angle of
relative to the North-bound.
b) As the North component of the airplane velocity cancel out the wind south-bound speed. The speed of the plane over the ground would be the West component of the airplane velocity, which is
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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