Answer:
a. true
Explanation:
Firstly, we need to understand what takes places during the compression process in a quasi-equilibrium process. A quasi-equilibrium process is a process in during which the system remains very close to a state of equilibrium at all times. When a compression process is quasi-equilibrium, the work done during the compression is returned to the surroundings during expansion, no exchange of heat, and then the system and the surroundings return to their initial states. Thus a reversible process.
While for a non-quasi equilibrium process, it takes more work to move the piston against this high-pressure region.
Answer with Explanation:
The modulus of elasticity has an profound effect on the mechanical design of any machine part as explained below:
1) Effect on the stiffness of the member: The ability of any member of a machine to resist any force depends on the stiffness of the member. For a member with large modulus of elasticity the stiffness is more and hence in cases when the member has to resist a direct load the member with more modulus of elasticity resists the force better.
2)Effect on the deflection of the member: The deflection caused by a force in a member is inversely proportional to the modulus of elasticity of the member thus in machine parts in which we need to resist the deflections caused by the load we can use materials with greater modulus of elasticity.
3) Effect to resistance of shear and torque: Modulus of rigidity of a material is found to be larger if the modulus of elasticity of the material is more hence for a material with larger modulus of elasticity the resistance it offer's to shear forces and the torques is more.
While designing a machine element since the above factors are important to consider thus we conclude that modulus of elasticity has a profound impact on machine design.
Answer:
insert (array[] , value , currentsize , maxsize )
{
if maxsize <=currentsize
{
return -1
}
index = currentsize-1
while (i>=0 && array[index] > value)
{
array[index+1]=array[index]
i=i-1
}
array[i+1]=value
return 0
}
Explanation:
1: Check if array is already full, if it's full then no component may be inserted.
2: if array isn't full:
- Check parts of the array ranging from last position of range towards initial range and determine position of that initial range that is smaller than the worth to be inserted.
- Right shift every component of the array once ranging from last position up to the position larger than the position at that smaller range was known.
- assign new worth to the position that is next to the known position of initial smaller component.
The answer is true because if the effect is neglected, the saturation id region is considered true
Answer:
minimum factor of safety for fatigue is = 1.5432
Explanation:
given data
AISI 1018 steel cold drawn as table
ultimate strength Sut = 63.800 kpsi
yield strength Syt = 53.700 kpsi
modulus of elasticity E = 29.700 kpsi
we get here
= 
...........1
here kb and kt = 1 combined bending and torsion fatigue factor
put here value and we get
= 
= 12 kpsi
and
= 
...........2
put here value and we get
= 
= 17.34 kpsi
now we apply here goodman line equation here that is
...................3
here Se = 0.5 × Sut
Se = 0.5 × 63.800 = 31.9 kspi
put value in equation 3 we get
solve it we get
FOS = 1.5432
