3.5-m-diameter merry-go-round is rotating freely with an angular velocity of 0.78 rad/s . Its total moment of inertia is 1710 kg
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Answer:
0.07756 m
Explanation:
Given mass of object =0.20 kg
spring constant = 120 n/m
maximum speed = 1.9 m/sec
We have to find the amplitude of the motion
We know that maximum speed of the object when it is in harmonic motion is given by where A is amplitude and
is angular velocity
Angular velocity is given by where k is spring constant and m is mass
So
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