Answer:
Explanation: All real solids are impure. A very high purity material, say 99.9999% pure. Impurities are often added to materials to improve the properties. For instance, carbon added in small amounts to iron makes steel, which is stronger than iron. Boron impurities added to silicon drastically change its electrical properties.
The properties of materials are profoundly influenced by the presence of imperfections.
TYPES OF IMPURITIES
1. Point defects
• Vacancy atoms (vacant atomic sites in a structure causing a distortion of planes). They are common, especially at high temperatures when atoms are frequently and randomly change their positions leaving behind empty lattice sites. In most cases diffusion (mass transport by atomic motion) can only occur because of vacancies.
• Self-Interstitial atoms ("extra" atoms positioned between atomic sites causing a distortion of planes) A self interstitial atom is an extra atom that has crowded its way into an interstitial void in the crystal structure. Self interstitial atoms occur only in low concentrations in metals because they distort and highly stress the tightly packed lattice structure.
2. Impurities:
- Substitutional (mpurity atom in lattice). Substitutional impurity atoms are usually close in size (within approximately 15%) to the bulk atom. An example of substitutional impurity atoms is the zinc atoms in brass. In brass, zinc atoms with a radius of 0.133 nm have replaced some of the copper atoms, which have a radius of 0.128 nm.
- Interstitial atoms (impurity atom not in regular lattice sit) . An example of interstitial impurity atoms is the carbon atoms that are added to iron to make steel. Carbon atoms, with a radius of 0.071 nm, fit nicely in the open spaces between the larger (0.124 nm) iron atoms.
3.
Line defects
• Dislocations
4. Area defects
• Grain Boundaries
Answer:
Gage pressure at the bottom of the tank = 46.1kP
Explanation:
Note that it is required to expressed the gage pressure ,pressure at the free surface=0 gage
Pressure due to the oil layer= (3m)*((800kg/m^3)(9.81m/s^2))/1000=25.9kpa
Pressure due to the brine layer=(2m)*((1030kg/m^3)(9.81m/s^2))/1000=20.2ka
Therefore, the pressure at the bottom=25.9+20.2=46.1kP
Answer:
a) 280MPa
b) -100MPa
c) -0.35
d) 380 MPa
Explanation:
GIVEN DATA:
mean stress
stress amplitude
a)
--------------1
-----------2
solving 1 and 2 equation we get
b)
c)
stress ratio
d)magnitude of stress range
= 280 -(-100) = 380 MPa
Answer:
R = 4.75 lb (↑)
Explanation:
Number of books = n = 19
Weight of each book = W = 1 lb
Length of the bookshelf = L = 40 inches
We can get the value of the distributed load as follows
q = n*W/L = 19*1 lb/ 40 inches = 0.475 lb/in
then the reactions at 4 ends (supports) of the bookshelf are
R = (q/2)/2 = 4.75 lb
We can see the bookshelf in the pic.
Answer:
The voltage input to the inverting terminal is 60μV
Explanation:
Given;
open-loop gain, A = 150,000
output voltage, V₀ = 15 V
voltage at the inverting input, = −40 μV = 40 x 10⁻⁶ V
The relationship between output voltage and voltage at the inverting input is given as;
Therefore, the voltage input to the inverting terminal is 60μV