Answer:
(a) Precipitation hardening - 1, 2, 4
(b) Dispersion strengthening - 1, 3, 5
Explanation:
The correct options for each are shown as follows:
Precipitation hardening
From the first statement; Dislocation movement is limited by precipitated particles. This resulted in an expansion in hardness and rigidity. Precipitates particles are separated out from the framework after heat treatment.
The aging process occurs in the second statement; because it speaks volumes on how heated solutions are treated with alloys above raised elevated temperature. As such when aging increases, there exists a decrease in the hardness of the alloy.
Also, for the third option for precipitation hardening; This cycle includes the application of heat the alloy (amalgam) to a raised temperature, maintaining such temperature for an extended period of time. This temperature relies upon alloying components. e.g. Heating of steel underneath eutectic temperature. Subsequent to heating, the alloy is extinguished and immersed in water.
Dispersion strengthening
Here: The effect of hearting is not significant to the hardness of alloys hardening by the method in statement 3.
In statement 5: The process only involves the dispersion of particles and not the application of heat.
Answer: Due to the way that spur gears work, starters that use them require an offset armature, which is achieved by placing the starter drive in separate gear housing. In starters that use planetary gears, the gears can be contained in an in line drive-end housing.
Explanation: true
Answer:
A) 
B) 
Explanation:
Given data:
P-1 = 100 lbf/in^2
degree f
effeciency = 80%
from steady flow enerfy equation
where h1 and h2 are inlet and exit enthalpy
for P1 = 100 lbf/in^2 and T1 = 500 degree F
for P1 = 40 lbf/in^2
exit enthalapy h_2
from above equation
[1 Btu/lbm = 25037 ft^2/s^2]
b) amount of entropy
at ![h_2 = 1197.77 Btu/lbm [\tex] and [tex]P_2 = 40 lbf/in^2](https://tex.z-dn.net/?f=h_2%20%3D%201197.77%20Btu%2Flbm%20%5B%5Ctex%5D%20%20and%20%5Btex%5DP_2%20%3D%2040%20lbf%2Fin%5E2)
Answer:
A) 209.12 GPa
B) 105.41 GPa
Explanation:
We are given;
Modulus of elasticity of the metal; E_m = 67 GPa
Modulus of elasticity of the oxide; E_f = 390 GPa
Composition of oxide particles; V_f = 44% = 0.44
A) Formula for upper bound modulus of elasticity is given as;
E = E_m(1 - V_f) + (E_f × V_f)
Plugging in the relevant values gives;
E = (67(1 - 0.44)) + (390 × 0.44)
E = 209.12 GPa
B) Formula for upper bound modulus of elasticity is given as;
E = 1/[(V_f/E_f) + (1 - V_f)/E_m]
Plugging in the relevant values;
E = 1/((0.44/390) + ((1 - 0.44)/67))
E = 105.41 GPa
Answer:
As P is continually increased, the block will now slip, with the friction force acting on the block being: f = muK*N, where muK is the coefficient of kinetic friction, with f remaining constant thereafter as P is increased.
