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nignag [31]
4 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]4 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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Which of the following ranges depicts the 2% tolerance range to the full 9 digits provided?
Lyrx [107]

Answer:

the only one that meets the requirements is option C .

Explanation:

The tolerance of a quantity is the maximum limit of variation allowed for that quantity.

To find it we must have the value of the magnitude, its closest value is the average value, this value can be given or if it is not known it is calculated with the formula

         x_average = ∑ x_{i} / n

The tolerance or error is the current value over the mean value per 100

         Δx₁ = x₁ / x_average

         tolerance = | 100 -Δx₁  100 |

bars indicate absolute value

let's look for these values ​​for each case

a)

    x_average = (2.1700000+ 2.258571429) / 2

    x_average = 2.2142857145

fluctuation for x₁

        Δx₁ = 2.17000 / 2.2142857145

        Tolerance = 100 - 97.999999991

        Tolerance = 2.000000001%

fluctuation x₂

        Δx₂ = 2.258571429 / 2.2142857145

        Δx2 = 1.02

        tolerance = 100 - 102.000000009

        tolerance 2.000000001%

b)

    x_average = (2.2 + 2.29) / 2

    x_average = 2,245

fluctuation x₁

         Δx₁ = 2.2 / 2.245

         Δx₁ = 0.9799554

         tolerance = 100 - 97,999

         Tolerance = 2.00446%

fluctuation x₂

          Δx₂ = 2.29 / 2.245

          Δx₂ = 1.0200445

          Tolerance = 2.00445%

c)

   x_average = (2.211445 +2.3) / 2

   x_average = 2.2557225

       Δx₁ = 2.211445 / 2.2557225 = 0.9803710

       tolerance = 100 - 98.0371

       tolerance = 1.96%

       Δx₂ = 2.3 / 2.2557225 = 1.024624

       tolerance = 100 -101.962896

       tolerance = 1.96%

d)

   x_average = (2.20144927 + 2.29130435) / 2

   x_average = 2.24637681

       Δx₁ = 2.20144927 / 2.24637681 = 0.98000043

       tolerance = 100 - 98.000043

       tolerance = 2.000002%

       Δx₂ = 2.29130435 / 2.24637681 = 1.0200000017

       tolerance = 2.0000002%

e)

   x_average = (2 +2,3) / 2

   x_average = 2.15

   Δx₁ = 2 / 2.15 = 0.93023

   tolerance = 100 -93.023

   tolerance = 6.98%

   Δx₂ = 2.3 / 2.15 = 1.0698

   tolerance = 6.97%

Let's analyze these results, the result E is clearly not in the requested tolerance range, the other values ​​may be within the desired tolerance range depending on the required precision, for the high precision of this exercise the only one that meets the requirements is option C .

4 0
3 years ago
A cylinder 8 in. in diameter and 3 ft long is concentric with a pipe of 8.25 in internal diameter. Between the cylinder and the
yarga [219]

Answer:

What force is requiere to move the cylinder along

Explanation:

The kinematic viscosity of the ol is 0.006 fr2

5 0
3 years ago
An insulated piston-cylinder device initially contains 0.16 m2 of CO2 at 150 kPa and 41 °C. Electric resistance heater supplied
lyudmila [28]

Answer:

I=0.3636

Explanation:

See the attached picture for explanation.

4 0
3 years ago
QUESTIONS
tigry1 [53]

the answer is C cause drivers dont always follow the rules

8 0
4 years ago
A levee will be constructed to provide some flood protection for a residential area. The residences are willing to accept a one-
777dan777 [17]

Answer:

1709.07 ft^3/s

Explanation:

Annual peak streamflow = Log10(Q [ft^3/s] )

mean = 1.835

standard deviation = 0.65

Probability of levee been overtopped in the next 15 years = 1/5

<u>Determine the design flow ins ft^3/s </u>

P₁₅ = 1 - ( q )^15 = 1 - ( 1 - 1/T )^15 = 0.2

                         ∴  T = 67.72 years

Q₁₅ = 1 - 0.2 = 0.8

Applying Lognormal distribution : Zt = mean + ( K₂ * std ) --- ( 1 )

K₂ = 2.054 + ( 67.72 - 50 ) / ( 100 - 50 ) * ( 2.326 - 2.054 )

    = 2.1504

back to equation 1

Zt = 1.835 + ( 2.1504 * 0.65 )  = 3.23276

hence:

Log₁₀ ( Qt(ft^3/s) ) = Zt  = 3.23276

hence ; Qt = 10^3.23276

                  = 1709.07 ft^3/s

4 0
3 years ago
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