A 15.0-μF capacitor is charged by a 130.0-V power supply, then disconnected from the power and connected in series with a 0.280-
1 answer:
The resonant frequency of a circuit is the frequency at which the equivalent impedance of a circuit is purely real (the imaginary part is null).
Mathematically this frequency is described as
Where
L = Inductance
C = Capacitance
Our values are given as
Replacing we have,
From this relationship we can also appreciate that the resonance frequency infers the maximum related transfer in the system and that therefore given an input a maximum output is obtained.
For this particular case, the smaller the capacitance and inductance values, the higher the frequency obtained is likely to be.
You might be interested in
Answer:
The amount of solid ice remaining after 6 hours is approximately 3.68664 kg
Explanation:
The given parameters are;
The side length of the cube box, s = 3(0) cm = 0.3 m
The thickness of the cube box, d = 5.0 cm = 0.05 m
The mass of ice in the box, m = 4.0 kg
The outside temperature of the cube box, T₁ = 45°C
The temperature of the melting ice inside the box, T₂ = 0°C
The latent heat of fusion of ice, = 3.35 × 10⁵ J/K/hr/kg
The surface area of the box, A = 6·s² 6 × (0.3 m)² = 0.54 m²
The coefficient of thermal conductivity, K = 0.01 J/s·m⁻¹·K⁻¹
For thermal equilibrium, we have;
The heat supplied by the surrounding = The heat gained by the ice
The heat supplied by the surrounding, Q = K·A·ΔT·t/d
Where;
ΔT = T₁ - T₂ = 45° C - 0° C = 45° C
ΔT = 45° C
Q = K·A·ΔT·t/d = 0.01 × 0.54 × 45 × 6× 60×60/0.05 = 104976
∴ The heat supplied by the surrounding, Q = 104976 J
The heat gained by the ice = ×
=3.35 × 10⁵ J/kg ×
Therefore, from Q = ×
, we have;
Q = 104976 J = ×
= 3.35 × 10⁵ J/kg ×
104976 J = 3.35 × 10⁵ J/kg ×
= 104976 J/(3.35 × 10⁵ J/kg) ≈ 0.31336 kg
The mass of melted ice, ≈ 0.31336 kg
∴ The amount of solid ice remaining after 6 hours, = m -
Which gives;
= m -
= 4.0 kg - 0.31336 kg ≈ 3.68664 kg
The amount of solid ice remaining after 6 hours, ≈ 3.68664 kg.