Answer:
In ferrous metal iron present but on the other hand in the non ferrous material iron does not present.That is why there is a different heat treatment process for ferrous and nonferrous materials.
Ferrous materials contains iron is the main constitute.Like steel ,cast iron ,wrought iron .Steel and cast iron are the alloy element of iron ans carbon.Wrought iron is the purest from of iron.
Heat treatment process for ferrous materials :
1.Normalizing
2.Annealing
3.Quenching
4.Surface hardening
Heat treatment process for non ferrous materials :
Mostly annealing process is used for non ferrous materials.After annealing non ferrous will become soft.
When two metal plates are joined then they form a bimetallic structure.The bimetallic structure is used to find the relationship of thermal temperature and the mechanical displacement.
The use of bimetallic structure -In clock ,thermometers ,engines.
Answer:
OSHA Hazard Communication Standard is responsible.
Explanation:
Answer:
v = 1.076 m /s
Explanation:
Initial volume of balloon = 4/3 x 3.14 x (9.905/2)³
=508.56 m³
Final volume of balloon = 4/3 x 3.14 x (16.502/2)³
= 2351.73 m³
Increase in volume = 1843.17 m³
Cross sectional area of inlet A = 3.14 x( 1.458/2)²
A = 1.6687 m²
Volume rate of flow of air = cross sectional area x velocity of inflow
= 1 .6687 V [ V is velocity of inflow ]
Total time taken = Increase in volume / rate of flow of air
17.108 X 60 = 1843.17 / 1.6687 V
V = 
v = 1.076 m /s
Answer:
<u>the thickness required would be 12 inch HMA and granular base layer of 6 inches</u>
Explanation:
structural number = 4.5
stone base course material coefficient = 0.13
hma material layer coefficient = 0.40
drainage coefficient = 0.90
we will use layered analysis procedure to get thickness
D1 >= sN1/a1
when we cross multiply,
sN1 = a1D1 >=sN1
D2 >= -sN2-sN1/a2m2
sN2* + sN1* >= sN2
D3 >= sN3-(sN1*+sN2*)/a2m2
where a1,a2,a3 = layer coefficient
d1 d2 d3 = actual thickness
m2,m3 = coefficient of base
a1 = 0.4
a1 = 0.13
sN = 4.5
m2 = 0.9
D1 >= sN1/a1 = 4.5/0.4
= 11.25
thickness of surface = 12 inches
a1D1 = 0.4x12 = 4.8
we have value of sN2 = 5.5
(5.5 -4.8)/(0.13*0.9)
= 0.7/0.117
= 5.9829 inches
approximately 6 inches
so the pavement will have 12inch HMA surface and 6 inches granular base layer.
