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.
Answer:
by getting a house permit or build permit.
Explanation:
Answer: Regardless of feature size (RFS)
Explanation:
It is a pre-selected condition for every geometric tolerance which is the second rule of GD&T and it requires no repairs, it is adopted when size feature has no effect on defined tolerance. RFS implies that GD&T callout is not dependent on size dimension. Least Material Condition (LMC) cancels out this rule.
•Reasons for using RFS.
1. It can not be easily determined.
2. It can be used anytime but can be ignored if it is specified.
Complex ratings because if they don’t the appliances might not be safe or good to use sorry if it’s incorrect I’m not very smart
Answer:
the final temperature would have been 2.81 °C if the pressure drop was 2 psi
Explanation:
if we assume that is no change in volume, there are not leaks present and also that the gas inside the football behaves as an ideal gas, we have:
initial state) P1 V = n R T1
final state) P2 V = n R T2
where P = absolute pressure , V = volume occupied by the gas, n = number of moles of gas, R = ideal gas constant T= absolute temperature
therefore since V= constant (constant volume) and n= constant ( no leaks), if we divide both equations
P2/P1 = T2/T1
therefore
T2 = T1 *(P2/P1)
since P1 absolute = P1 relative + P atmospheric (14.7 psi) = 12.5 psi + 14.7 psi = 27.2 psia
also P2 = P1 - 2 psi = 25.2 psia
T1 = 24.7°C + 273 °C = 297.7 K
therefore
T2 = T1 * (P2/P1) = 297.7 K ( 25.2 psia/27.2 psia) = 275.81 K
thus T2 = 275.81 K = 2.81 °C