Answer:
True, supplementary field identification programs tend to limit the use of routine programs that target service delivery using routine systems.
Explanation:
When supplementary field identification programs are applied in a study, they have damaging effects to other systems and programs already in progress targeting certain/similar variables in a study group.Such programs are initiated to boost the already existing systems of programs that are in continuous application( routine basis). As a supplement , we expect more positive results in the rates per the variables included in a study.However, results has proved the opposite.For example, supplementary immunization activities applied in programs targeting demographic and health systems services reveled that such programs reduce the probability of receiving the services provided by other routine health systems conducting continuous vaccination programs to the target groups.
Answer:
<em>Object-oriented</em>
Explanation:
<em>Object Oriented programming</em> <em>(OOP)</em> is a specific way of programming, where the code is organized in units called classes, from which objects are created that are related to each other to achieve the objectives of the applications. Object-oriented programming took over as the dominant programming style in the mid-1980s, largely due to the influence of C ++. Its dominance was consolidated thanks to the rise of graphical user interfaces, for which object-oriented programming is particularly well suited. Its most important characteristics are the following:
Answer:
79 kW.
Explanation:
The equation for enthalpy is:
H2 = H1 + Q - L
Enthalpy is defined as:
H = G*(Cv*T + p*v)
This is specific volume.
The gas state equation is:
p*v = R*T (with specific volume)
The specific gas constant for air is:
287 K/(kg*K)
Then:
T1 = 60 + 273 = 333 K
T2 = 200 + 273 = 473 K
p1*v1 = 287 * 333 = 95.6 kJ/kg
p2*v2 = 287 * 473 = 135.7 kJ/kg
The Cv for air is:
Cv = 720 J/(kg*K)
So the enthalpies are:
H1 = 0.8*(0.72 * 333 + 95.6) = 268 kW
H2 = 0.8*(0.72 * 473 + 135.7) = 381 kW
Ang the heat is:
Q = 34 kW
Then:
H2 = H1 + Q - L
381 = 268 + 34 - L
L = 268 + 34 - 381 = -79 kW
This is the work from the point of view of the air, that's why it is negative.
From the point of view of the machine it is positive.
Answer: 35.3 °
Explanation:
Body-centered cubic lattice (bcc or cubic-I), just like all lattices, has lattice points at the eight corners of the unit cell with an additional points at the center of the cell. It has unit cell vectors a = b = c and interaxial angles α=β=γ=90°.
The simplest crystal structures are those that have present only a single atom at each lattice point.
body-centered cubic unit cell has atoms at each of the eight corners of a cube (like the cubic unit cell) plus one atom in the center of the cube. Each of the corner atoms is the corner of another cube so the corner atoms are shared between eight unit cells. It is said to have a coordination number of 8. The bcc unit cell consists of a net total of two atoms; one in the center and eight eighths from corners atoms
With the use of BCC unit cell, if a applied stress is in [110] direction, but slip applies in [111] direction, the angle between applied direction and slip direction is given as:
[1 1 0] [1 1 1]
λ = Cos^-1 ( 1×1 + 1×1 + 0×1 ÷ (1^2 + 1^2 +0^2) (1^2 + 1^2+ 1^2))
Cos^-1 2/ sqrt 6
= 35.386°
Answer
1056
Explanation:
for example
A soil is to be excavated from a borrow pit which has a density of 1.75g/cc and water content of 12%. The G is 2.7 the soil is compacted to that water content of 18% and dry density of 1.65g/cc. for 1000 m3 of soil used in fill estimate
Quantity of soil to be excavated from pit in m3
