The ground temperature a few meters below the surface is fairly constant throughout the year and is near the average value of th
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Answer:
i) S–N plot is attached
ii) fatigue strength = 100 MPa
iii) fatigue life = 5.62 x 10^(5) cycles
Explanation:
i) I have attached the S–N plot (stress amplitude versus logarithm of cycles to failure)
ii) The question says we should find the fatigue strength at 4 × 10^(6) cycles.
So let's find the log of this and trace it on the graph attached.
Log(4 × 10^(6)) = 6.6
From the graph attached, at log of cycle value of 6.6, the fatigue strength is approximately 100 MPa
iii) The question says we should find the fatigue life for 120 MPa.
Thus, from the graph, at stress amplitude of 120 MPa, the log of cycles is approximately 5.75.
Thus,the fatigue life will be the inverse log of 5.75.
Thus, fatigue life = 10^(5.75)
Fatigue life = 5.62 x 10^(5)
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