Answer:
All will be Explained below.
Explanation:
1) Under which Condition can a soil be chemically Stabilize.
Answer
a). Plasticity Index :A soil with a high value of plasticity Index is not good for various engineering projects. The introduction of line helps in reducing plasticity due cation exchange reaction.Pozzolanic reaction over time reduces plasticity and increase index strength due to the formation of calcium - silicate hydrate.
Answer: a)True
Explanation: Takt time is defined as the average time difference between the production of the two consecutive unit of goods by the manufacturer and this rate is matched with the demand of the customer. This is the time which is calculated to find the acceptable time for which the goods unit must be produced by the factory to meet the needs of the customer. Therefore , the statement is true that takt time is the rate at which a factory must produce to satisfy the customer's demand.
Answer:
true
Explanation:
Creep is known as the time dependent deformation of structure due to constant load acting on the body.
Creep is generally seen at high temperature.
Due to creep the length of the structure increases which is not fit for serviceability purpose.
When time passes structure gain strength as the structure strength increases with time so creep tends to decrease.
When we talk about Creep rate for new structure the creep will be more than the old structure i.e. the creep rate decreases with time.
Answer:
a) 

b)

Explanation:
Given that:
diameter d = 12 in
thickness t = 0.25 in
the radius = d/2 = 12 / 2 = 6 in
r/t = 6/0.25 = 24
24 > 10
Using the thin wall cylinder formula;
The valve A is opened and the flowing water has a pressure P of 200 psi.
So;




b)The valve A is closed and the water pressure P is 250 psi.
where P = 250 psi






The free flow body diagram showing the state of stress on a volume element located on the wall at point B is attached in the diagram below
Answer:
The work of the cycle.
Explanation:
The area enclosed by the cycle of the Pressure-Volume diagram of a Carnot engine represents the net work performed by the cycle.
The expansions yield work, and this is represented by the area under the curve all the way to the p=0 line. But the compressions consume work (or add negative work) and this is substracted fro the total work. Therefore the areas under the compressions are eliminated and you are left with only the enclosed area.