Answer:
105.70 mm
Explanation:
Poisson’s ratio, v is the ratio of lateral strain to axial strain.
E=2G(1+v) where E is Young’s modulus, v is poisson’s ratio and G is shear modulus
Since G is given as 25.4GPa, E is 65.5GPa, we substitute into our equation to obtain poisson’s ratio
Original length 
Where 
is final diameter, 
is original diameter, 
is final length and 
is original length.
Therefore, the original length is 105.70 mm
Answer:
yes, it is
Explanation:
The sequence: (+4)
23,27,31,35,39,43,47,51,55,59,63,67,71,75,79,83
Hope this helps! :)
Answer:
all of the above hidtsybkhitonp
Explanation:
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Answer:
M2 = 0.06404
P2 = 2.273
T2 = 5806.45°R
Explanation:
Given that p1 = 10atm, T1 = 1000R, M1 = 0.2.
Therefore from Steam Table, Po1 = (1.028)*(10) = 10.28 atm,
To1 = (1.008)*(1000) = 1008 ºR
R = 1716 ft-lb/slug-ºR cp= 6006 ft-lb/slug-ºR fuel-air ratio (by mass)
F/A =???? = FA slugf/slugaq = 4.5 x 108ft-lb/slugfx FA slugf/sluga = (4.5 x 108)FA ft-lb/sluga
For the air q = cp(To2– To1)
(Exit flow – inlet flow) – choked flow is assumed For M1= 0.2
Table A.3 of steam table gives P/P* = 2.273,
T/T* = 0.2066,
To/To* = 0.1736 To* = To2= To/0.1736 = 1008/0.1736 = 5806.45 ºR Gives q = cp(To* - To) = (6006 ft-lb/sluga-ºR)*(5806.45 – 1008)ºR = 28819500 ft-lb/slugaSetting equal to equation 1 above gives 28819500 ft-lb/sluga= FA*(4.5 x 108) ft-lb/slugaFA =
F/A = 0.06404 slugf/slugaor less to prevent choked flow at the exit
