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:
ρ=962.16kg/m^3
Explanation:
The first thing we must do to solve is to find the mass of the specimen using the weight equation
w = mg
m=w/g
m=0.45/9.81=0.04587kg
To find the volume we must make a free-body diagram on the specimen, taking into account that the weight will go down and the buoyant force up, and the result of that subtraction will be the measured weight value (0.081N).
We must bear in mind that the principle of archimedes indicates that the buoyant force is given by
F = ρgV
where V is the specimen volume and ρ is the density of alcohol = 789kg / m ^ 3
considering the above we have the following equation
0.081=0.45-(789)(9.81m/s^2)V
solving for V
V=(0.081-0.45)/(-789x9.81)
V=4.7673x10^-5m^3
finally we found the density
ρ=m/v
ρ=0.04587kg/4.7673x10^-5m^3
ρ=962.16kg/m^3
Answer:
a)Patm=135.95Kpa
b)Pabs=175.91Kpa
Explanation:
the absolute pressure is the sum of the water pressure plus the atmospheric pressure, which means that for point a we have the following equation
Pabs=Pw+Patm(1)
Where
Pabs=absolute pressure
Pw=Water pressure
Patm=
atmospheric pressure
Water pressure is calculated with the following equation
Pw=γ.h(2)
where
γ=especific weight of water=9.81KN/M^3
H=depht
A)
Solving using ecuations 1 y 2
Patm=Pabs-Pw
Patm=185-9.81*5=135.95Kpa
B)
Solving using ecuations 1 y 2, and atmospheric pressure
Pabs=0.8x5x9.81+135.95=175.91Kpa
Answer: I will ask my dad he knows
Explanation:
Answer:
The Decision Matrix
As you compare potential solutions to your design brief and the universal criteria for a good design, it may be obvious which solution is the best. ... A decision matrix is a chart with your requirements and criteria on one axis and the different solutions on the other.