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:
(3) the compressor power, in kW. = 4.17KW
(4) the refrigeration capacity, in tons.= 4.905tons
(5) the coefficient of performance. = 4.1336
check attached files for other answers.
Explanation:
Answer: Fracture will not occur since Kc (32.2 MPa√m) ∠ KIc (35 MPa√m).
Explanation:
in this question we are asked to determine if an aircraft will fracture for a given fracture toughness.
let us begin,
from the question we have that;
stress = 325 MPa
fracture toughness (KIc) = 35 MPa√m
the max internal crack length = 1.0 m
using the formula;
Y = KIc/σ√(πα) ---------------(1)
solving for Y we have;
Y = 35 (MPa√m) / 250 (MPa) √(π × 2×10⁻3/2m)
Y = 2.50
so to calculate the fracture roughness;
Kc = Y × σ√(πα) = 2.5 × 3.25√(π × 1×10⁻³/2) = 32.2 MPa√m
Kc = 32.2 MPa√m
From our results we can say that fracture will not occur since Kc (32.2 MPa√m) is less than KIc (35 MPa√m) of the material.
cheers i hope this helps!!!!
Answer:
company should focus on their goals
Answer:
(c) 5.71 V
Explanation:
The circuit can be redrawn to a Thevenin equivalent that is 6V through a 5-ohm resistor into a 100-ohm load. Then the voltage at the load is ...
(6 V)(100/(100 +5) ≈ 5.71 V