Answer:
= 1.75 × 10⁻⁴ m/s
Explanation:
Given:
Density of copper, ρ = 8.93 g/cm³
mass, M = 63.5 g/mol
Radius of wire = 0.625 mm
Current, I = 3A
Area of the wire, =
Now,
The current density, J is given as
= 2444619.925 A/mm²
now, the electron density,
where,
=Avogadro's Number
Now,
the drift velocity,
where,
e = charge on electron = 1.6 × 10⁻¹⁹ C
thus,
= 1.75 × 10⁻⁴ m/s
Answer:
The maximum acceleration of the system is 359.970 centimeters per square second.
Explanation:
The motion of the mass-spring system is represented by the following formula:
Where:
- Position of the mass with respect to the equilibrium position, measured in centimeters.
- Amplitude of the mass-spring system, measured in centimeters.
- Angular frequency, measured in radians per second.
- Time, measured in seconds.
- Phase, measured in radians.
The acceleration experimented by the mass is obtained by deriving the position equation twice:
Where the maximum acceleration of the system is represented by .
The natural frequency of the mass-spring system is:
Where:
- Spring constant, measured in newtons per meter.
- Mass, measured in kilograms.
If and
, the natural frequency is:
Lastly, the maximum acceleration of the system is:
The maximum acceleration of the system is 359.970 centimeters per square second.