4.9 Given that I 4 amps when Vs 40 volts and Is 4 amps and I 1 amp when Vs 20 volts and Is 0, use superposition and linearity to
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Answer:
Your question is lacking some information attached is the missing part and the solution
A) AB = AD = BD = 0, BC = LC
AC =
B) AB = AD = BC = BD = 0
AC =
Explanation:
A) Forces in all members due to the load L in position A
assuming that BD goes slack from an inspection of Joint B
AB = 0 and BC = LC from Joint D, AD = 0 and CD = 4L/3 C
B) steps to arrive to the answer is attached below
AB = AD = BC = BD = 0
AC =
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