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nignag [31]
3 years ago
15

Match each situation with the type of material (conductor or inductor) you would want to use in it. You need to connect a recent

ly landed plane to the Earth in order to ground it and remove the built-up precipitation static. You would want to use this kind of material: You need to move a live power line out of a puddle of water. There is a lot of charge moving through this line, and if any makes it to your hands it's going to hurt. You would want to use this kind of material: You need a smooth sphere to put a sensitive piece of equipment inside that will minimize any sparks between the sphere and pieces of equipment outside the sphere. You would want to use this kind of material:
Engineering
1 answer:
Marina CMI [18]3 years ago
5 0

Answer: for the following process, I will explain each process and where the material is best to be used.

1. You need to connect a recently landed plane to the Earth in order to ground it and remove the built-up precipitation static. You would want to use this kind of material:

Answer: Conductor

Explanation: for you to ground the plane, you need a conductor that can be able to direct the current down to the earth. Because electron can only flow freely in a conductor.

2. You need to move a live power line out of a puddle of water. There is a lot of charge moving through this line, and if any makes it to your hands it's going to hurt. You would want to use this kind of material:

Answer: Inductor

Explanation: an Inductor resist the flow of electric current through it. You have to use an inductor, to avoid been electrocuted by the live wire. If a conductor is used current will flow through it, which may lead to electrocutions, as the water is also a conductor of electricity.

3. You need a smooth sphere to put a sensitive piece of equipment inside that will minimize any sparks between the sphere and pieces of equipment outside the sphere. You would want to use this kind of material:

Answer: Inductor

Explanation: to avoid spark, an inductor should be used, because when they is a friction between a conductors and an electric current, they will be a spark. So an inductor should be used to avoid spark. Inductors does not give a spark when in friction with an electric current

AN INDUCTOR IS ANY MATERIAL THAT RESIST THE FLOW OF ELECTRIC CURRENT THROUGH IT.

A CONDUCTOR IS ANY MATERIAL THAT ALLOWS THE FLOW ELECTRIC CURRENT THROUGH IT.

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Air at a pressure of 6000 N/m^2 and a temperature of 300C flows with a velocity of 10 m/sec over a flat plate of length 0.5 m. E
White raven [17]

Answer:

Q=hA(T_{w}-T_{inf})=16.97*0.5(27-300)=-2316.4J

Explanation:

To solve this problem we use the expression for the temperature film

T_{f}=\frac{T_{\inf}+T_{w}}{2}=\frac{300+27}{2}=163.5

Then, we have to compute the Reynolds number

Re=\frac{uL}{v}=\frac{10\frac{m}{s}*0.5m}{16.96*10^{-6}\rfac{m^{2}}{s}}=2.94*10^{5}

Re<5*10^{5}, hence, this case if about a laminar flow.

Then, we compute the Nusselt number

Nu_{x}=0.332(Re)^{\frac{1}{2}}(Pr)^{\frac{1}{3}}=0.332(2.94*10^{5})^{\frac{1}{2}}(0.699)^{\frac{1}{3}}=159.77

but we also now that

Nu_{x}=\frac{h_{x}L}{k}\\h_{x}=\frac{Nu_{x}k}{L}=\frac{159.77*26.56*10^{-3}}{0.5}=8.48\\

but the average heat transfer coefficient is h=2hx

h=2(8.48)=16.97W/m^{2}K

Finally we have that the heat transfer is

Q=hA(T_{w}-T_{inf})=16.97*0.5(27-300)=-2316.4J

In this solution we took values for water properties of

v=16.96*10^{-6}m^{2}s

Pr=0.699

k=26.56*10^{-3}W/mK

A=1*0.5m^{2}

I hope this is useful for you

regards

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3 years ago
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7 0
2 years ago
214Bi83 --&gt; 214Po84 + eBismuth-214 undergoes first-order radioactive decay to polonium-214 by the release of a beta particle,
Zolol [24]

Answer:

(C) ln [Bi]

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3 0
3 years ago
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provides steady-state operating data for a solar power plant that operates on a Rankine cycle with Refrigerant 134a as its worki
Vaselesa [24]

Answer:

hello some parts of your question is missing attached below is the missing part ( the required fig and table )

answer : The solar collector surface area = 7133 m^2

Explanation:

Given data :

Rate of energy input to the collectors from solar radiation = 0.3 kW/m^2

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Determine the solar collector surface area

The solar collector surface area = 7133 m^2

attached below is a detailed solution of the problem

8 0
3 years ago
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