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zaharov [31]
3 years ago
11

While discussing the testing of a Hall-effect switch:

Engineering
1 answer:
sergejj [24]3 years ago
6 0

Answer:

your answer is technician A

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When all conductors of the same alternating-current circuit are in the same raceway, the magnetic flux around each conductor flu
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B. Cancels out

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Which of the following is true about how the universe is expanding?
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B. The space between the galaxy is getting better
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There are little to no benefits to supplementary field identification programs
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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.

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4 years ago
g Consider a thin opaque, horizontal plate with an electrical heater on its backside. The front end is exposed to ambient air th
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Answer:

The electrical power is 96.5 W/m^2

Explanation:

The energy balance is:

Ein-Eout=0

qe+\alpha sGs+\alpha skyGsky-EEb(Ts)-qc=0

if:

Gsky=oTsky^4

Eb=oTs^4

qc=h(Ts-Tα)

\alpha s=\frac{\int\limits^\alpha _0 {\alpha l Gl} \, dl }{\int\limits^\alpha _0 {Gl} \, dl }

\alpha s=\frac{\int\limits^\alpha _0 {\alpha lEl(l,5800 } \, dl }{\int\limits^\alpha _0 {El(l,5800)} \, dl }

if Gl≈El(l,5800)

\alpha s=(1-0.2)F(0-2)+(1-0.7)(1-F(0-2))

lt= 2*5800=11600 um-K, at this value, F=0.941

\alpha s=(0.8*0.941)+0.3(1-0.941)=0.77

The hemispherical emissivity is equal to:

E=(1-0.2)F(0-2)+(1-0.7)(1-F(0-2))

lt=2*333=666 K, at this value, F=0

E=0+(1-0.7)(1)=0.3

The hemispherical absorptivity is equal to:

qe=EoTs^{4}+h(Ts-T\alpha  )-\alpha sGs-\alpha oTsky^{4}=(0.3*5.67x10^{-8}*333^{4})+10(60-20)-(0.77-600)-(0.3*5.67x10^{-8}*233^{4})=96.5 W/m^{2}

3 0
3 years ago
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