Answer:
a. Fraction of Atom = 2.41E-5 when T = 600K
b. Fraction of Atom = 5.03E-10 when T = 298K
Explanation:
a.
Given
T = Temperature = 600K
Qv = Energy for formation = 0.55eV/atom
To calculate the fraction of atom sites, we make use of the following formula
Nv/N = exp(-Qv/kT)
Where k = Boltzmann Constant = 8.62E-5eV/K
Nv/N = exp(-0.55/(8.62E-5 * 600))
Nv/N = 0.000024078672493307
Nv/N = 2.41E-5
b. When T = 298K
Nv/N = exp(-0.55/(8.62E-5 * 298))
Nv/N = 5.026591237904E−10
Nv/N = 5.03E-10 ----- Approximated
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:
Iron
Explanation:
Once a star starts fusing iron in its core it gives off very large amounts of energy. Also helium, hydrogen, carbon, oxygen, and silicon still exist in the star in different shells while fusin is taking place at different parts of the star. Right at the surface, hydrogen continues to fuse to helium, as e go further down helium is fusing to carbon and oxygen; right inside the core we have silicon that is fusing with iron. At this point, Iron being of stable atomic structure and quite heavier will no longer be fused into anything because of the quite large amounts of energy and force required to fuse iron atoms. The process comes to a standstill at this point.
Answer :
<h3>Flow rate in pipe B is = 0.3094
</h3>
Explanation:
Given :
Length of pipe A 
m
Length of pipe B 
m
Flow rate through pipe A 
Diameter of pipe 
m
Velocity from pipe A,
Here, head loss is same because height is same.
Now rate of flow from pipe B is,
Answer:
i) 43.55 kg/s
ii) 40 m/s
iii) -199.32 KW
Explanation:
To resolve the above question we have to make some assumptions :
- mass flow through the system is constant
- The only interactions that are between the system and the surrounding are work and heat
- The fluid is uniform
i) first we have to determine the mass flow rate of the air
M = 
= 
---------- (1) hence M = 43.55 kg/s
ii) using this relationship : A1V1 = A2V2 hence V1 = (0.2/0.5) * 100 = 40m/s ( inlet velocity )
input this value into equation 1
iii) Next we will determine the power required to run compressor
attached below
power required = -199.32 KW ( this value indicates that there is power supplied )
