Answer:
Following are the solution to this question:
Explanation:
Whenever a chemical reaction occurs between water and cement the heat is released, and a 
(C-S-H gel) gel constructs gel is also recognized as "tobermorite gel."
This one gel acts like a pack of gum and also has a cement quality, that holds its particles intact and therefore contributes to the overall compression mix. An increase in supply explicitly causes the movement in the outcome of power. C3S and C2S are both the compounds of Bouge that produce hydration C-S-H gel.
It mixture must be balanced as 
with C-S-H gel also is given as a byproduct. It , that cause sudden with sulphate and form 
, is an unacceptable substance. Sulfate attack or later deterioration of its cement is caused by this .
All C3S and C2S generate various amounts of C-S-H gel so, the required strength can be maintained without compromising on real term durability.
Answer: False
Explanation:
The given statement are false as, the gauss quadrature predicted same value only when their is a minute error at one point and two point rule. It basically given the more precise answer and the answer are only changed by the decimal place. The gauss quadrature method are basically used for calculating the certain integral.
Answer:
The hardenability increases with increasing austenite grain size, because the grain boundary area is decreasing. This means that the sites for the nucleation of ferrite and pearlite are being reduced in number, with the result that these transformations are slowed down, and the hardenability is therefore increased.
Answer:
(i) 12 V in series with 18 Ω.
(ii) 0.4 A; 1.92 W
(iii) 1,152 J
(iv) 18Ω — maximum power transfer theorem
Explanation:
<h3>(i)</h3>
As seen by the load, the equivalent source impedance is ...
10 Ω + (24 Ω || 12 Ω) = (10 +(24·12)/(24+12)) Ω = 18 Ω
The open-circuit voltage seen by the load is ...
(36 V)(12/(24 +12)) = 12 V
The Thevenin's equivalent source seen by the load is 12 V in series with 18 Ω.
__
<h3>(ii)</h3>
The load current is ...
(12 V)/(18 Ω +12 Ω) = 12/30 A = 0.4 A . . . . load current
The load power is ...
P = I^2·R = (0.4 A)^2·(12 Ω) = 1.92 W . . . . load power
__
<h3>(iii)</h3>
10 minutes is 600 seconds. At the rate of 1.92 J/s, the electrical energy delivered is ...
(600 s)(1.92 J/s) = 1,152 J
__
<h3>(iv)</h3>
The load resistance that will draw maximum power is equal to the source resistance: 18 Ω. This is the conclusion of the Maximum Power Transfer theorem.
The power transferred to 18 Ω is ...
((12 V)/(18 Ω +18 Ω))^2·(18 Ω) = 144/72 W = 2 W
Answer:
Hook's law holds good up to. A elastic limit. B. plastic limit. C.yield point. D.Breaking point
