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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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51.4 Ohms

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A motorcycle starts from rest with an initial acceleration of 3 m/s^2, and the acceleration then changes with the distance s as
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Follows are the solution to this question:

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   =\frac{1}{2}(3 +6 \frac{m}{s^2})(100 \ m)+ \frac{1}{2}(6+4 \frac{m}{s^2})(100 m) \\\\=\frac{1}{2}(9 \frac{m}{s^2})(100 \ m)+ \frac{1}{2}(10\frac{m}{s^2})(100 m) \\\\=\frac{1}{2}(900 \frac{m^2}{s^2})+ \frac{1}{2}(1,000\frac{m^2}{s^2}) \\\\=(450 \frac{m^2}{s^2})+ (500\frac{m^2}{s^2}) \\\\= 950 \ \frac{m^2}{s^2}

Calculating the kinematics equation:

\to v^2 = v^2_{o} + 2as\\\\

        =0+ \sqrt{2as}\\\\ = \sqrt{2(A)}\\\\= \sqrt{2(950 \frac{m^2}{s^2})}\\\\= 43.59 \frac{m}{s}

Calculating the value of acceleration:  

\to a= \frac{dv}{dt}

=\frac{dv}{ds}(\frac{ds}{dt}) \\\\=v\frac{dv}{ds}\\\\\to \frac{dv}{ds}=\frac{a}{v}

\to \frac{dv}{ds} =\frac{4 \frac{m}{s^2}}{43.59 \frac{m}{s}} \\\\

         =\frac{0.092}{s}

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