Answer:
Fatigue lifetimes will be ranked as B>A>C
Explanation:
Start by calculating the mean stress for all samples
σa= (σamax + σamin)/2 = (450 - 350)/2 = 50MPa
σb= (σbmax + σbmin)/2 = (400 - 300)/2 = 50MPa
σa= (σcmax + σcmin)/2 = (340 - 340)/2 = 0MPa
Now calculate the stress amplitudes of all three samples
σa= (σamax - σamin)/2 = (450 + 350)/2 = 400MPa
σb= (σbmax - σbmin)/2 = (400 + 300)/2 = 350MPa
σc= (σcmax - σcmin)/2 = (340 + 340)/2 = 340MPa
The mean stress of samples A and B is the same which is higher than sample C. Hence samples A and B will have a higher fatigue life than sample C. However, the higher stress amplitude means lower fatigue life, hence, sample B will have a higher fatigue life than sample A. So the order will be B> A> C.
Attached picture shows the justification using and S- N plot.
I think it might be DNA profiling correct me if I’m wrong
Answer:
a) ∝ and β
The phase compositions are :
C = 5wt% Sn - 95 wt% Pb
C = 98 wt% Sn - 2wt% Pb
b)
The phase is; ∝
The phase compositions is; 82 wt% Sn - 91.8 wt% Pb
Explanation:
a) 15 wt% Sn - 85 wt% Pb at 100⁰C.
The phases are ; ∝ and β
The phase compositions are :
C = 5wt% Sn - 95 wt% Pb
C = 98 wt% Sn - 2wt% Pb
b) 1.25 kg of Sn and 14 kg Pb at 200⁰C
The phase is ; ∝
The phase compositions is; 82 wt% Sn - 91.8 wt% Pb
Csn = 1.25 * 100 / 1.25 + 14 = 8.2 wt%
Cpb = 14 * 100 / 1.25 + 14 = 91.8 wt%
Answer:
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advantages: </h3>
<em>lower power consumption, modulation system is simple</em>
<h3>disadvantages<em>:</em></h3>
<em>complex detection</em>
<h3><em>applications:</em></h3>
analog TV systems: to transmit color information
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Explanation: