Consider a very long rectangular fin attached to a flat surface such that the temperature at the end of the fin is essentially t
hat of the surrounding air, i.e. 20°C. Its width is 5.0 cm; thickness is 1.0 mm; thermal conductivity is 200 W/m·K; and base temperature is 40°C. The heat transfer coefficient is 20 W/m2 ·K. Estimate the fin temperature at a distance of 5.0 cm from the base and the rate of heat loss from the entire fin.
1 answer:
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Answer:
The impedance of the circuit depends on the angular frequency of the voltage source.
Explanation:
- In a electric circuit, the magnitude of the impedance, is given by the following expression:
where R = Resistance
Xl = Inductive reactance = ω*L
Xc = Capacitive Reactance = 1/ωC
and ω = angular frequency of the voltage source.
- So, it can be seen that the impedance depends on the value of the constants R,L and C, and on the angular frequency ω.
Answer: Inherent width in the emission line: 9.20 × 10⁻¹⁵ m or 9.20 fm
length of the photon emitted: 6.0 m
Explanation:
The emitted wavelength is 589 nm and the transition time is ∆t = 20 ns.
Recall the Heisenberg's uncertainty principle:-
∆t∆E ≈ h ( Planck's Constant)
The transition time ∆t corresponds to the energy that is ∆E
.
The corresponding uncertainty in the emitted frequency ∆v is:
∆v= ∆E/h = (5.273*10^-27 J)/(6.626*10^ J.s)= 7.958 × 10^6 s^-1
To find the corresponding spread in wavelength and hence the line width ∆λ, we can differentiate
λ = c/v
dλ/dv = -c/v² = -λ²/c
Therefore,
∆λ = (λ²/c)*(∆v) = {(589*10⁻⁹ m)²/(3.0*10⁸ m/s)} * (7.958*10⁶ s⁻¹)
= 9.20 × 10⁻¹⁵ m or 9.20 fm
The length of the photon (<em>l)</em> is
l = (light velocity) × (emission duration)
= (3.0 × 10⁸ m/s)(20 × 10⁻⁹ s) = 6.0 m