Answer and Explanation:
Core self-evaluations (CSEs) stands for a wide personality trait that comprises of 4 positive individual traits, namely:
(1) self-efficacy
(2) self esteem
(3) locus of control
(4) emotional stability.
Baiscally, when people have a positive evaluation about themselves, or quality core self-evaluation, they believe that they are worthy and fit for a task. They trust their capability and effectiveness. This leads to some implications in their managers duties and careers, which could either be positive or negative.
A group leader can use CSEs to create a more effective unit by implementing the ten items points of Generalized Self Efficacy Tool to test the self efficiency of individual personnels in that unit.
Answer:
a) 0.697*10³ lb.in
b) 6.352 ksi
Explanation:
a)
For cylinder AB:
Let Length of AB = 12 in


For cylinder BC:
Let Length of BC = 18 in




b) Maximum shear stress in BC

Maximum shear stress in AB

Answer:
the saturated density should be
Explanation:
Answer:
1.3cm
Explanation:
the arrow is 3 lines past the 1 so it is 1.3cm
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