Answer:
b). Occurs at the outer surface of the shaft
Explanation:
We know from shear stress and torque relationship, we know that
where, T = torque
J = polar moment of inertia of shaft
τ = torsional shear stress
r = raduis of the shaft
Therefore from the above relation we see that
Thus torsional shear stress, τ is directly proportional to the radius,r of the shaft.
When r= 0, then τ = 0
and when r = R , τ is maximum
Thus, torsional shear stress is maximum at the outer surface of the shaft.
Answer:
1.44 mm
Explanation:
Compute the maximum allowable surface crack length using
where E is the modulus of elasticity, 
is surface energy and 
is tensile stress
Substituting the given values
The maximum allowable surface crack is 1.44 mm
Answer:
The time required to elute the two species is 53.3727 min
Explanation:
Given data:
tA = retention time of A=16.63 min
tB=retention time of B=17.63 min
WA=peak of A=1.11 min
WB=peak of B=1.21 min
The mathematical expression for the resolution is:
The mathematical expression for the time to elute the two species is:
Here
ReB = 1.5
1(A)
2(B)
3(E)
4(C)
5(D)
6(B)
7(C)
Answer:
b
Explanation:
the NEC has expanded the requirements for ground-fault circuit interrupters (GFCI) to protect anyone who plugs into an electrical system. Initially, it was only required for temporary wiring at construction sites and in dwelling unit bathrooms, but in recent years the Code requirements for GFCI protection have expanded to include many other areas, including commercial occupancies, fountains and swimming pools, and temporary installations, to name a few. (For a complete list of 2002 NEC references, see the sidebar below)
