Given that the skin depth of graphite at 100 (MHz) is 0.16 (mm), determine (a) the conductivity of graphite, and (b) the distance that a 1 (GHz) wave travels in graphite such that its field intensity is reduced by 30 (dB).

Answers

Answer 1

Answer:

the answer is below

Explanation:

a) The conductivity of graphite (σ) is calculated using the formula:

[tex]\delta=\frac{1}{\sqrt{\pi f \mu \sigma} }\\\\\sigma =\frac{1}{\pi f \mu \delta^2}[/tex]

where f = frequency = 100 MHz, δ = skin depth = 0.16 mm = 0.00016 m, μ = 0.0000012

Substituting:

[tex]\sigma =\frac{1}{\pi *10^6* 0.0000012*0.00016^2}=0.99*10^4\ S/m[/tex]

b) f = 1 GHz = 10⁹ Hz.

[tex]\alpha=\sqrt{\pi f \mu \sigma} = \sqrt{0.0000012*10^9*\pi*0.99*10^5}=1.98*10^4\ Np/m\\\\20log_{10} e^{-\alpha z}=-30\ dB\\\\(-\alpha z)log_{10} e=-1.5 \\\\z=\frac{-1.5}{log_{10} e*-\alpha} =1.75*10^{-4}\ m=0.175\ mm[/tex]


Related Questions

Which metal has the ability to rust

A gold

B silver

C iron

D aluminum

Answers

Answer:

I got iron

Explanation:

on my plato test

It’s iron :)
I’ve took a test with this question

Why is the reversed Carnot cycle executed within the saturation dome not a realistic model for refrigeration cycles?

Answers

Explanation:

The paths of warmth and job interactions are inverted by repeating the Carnot process. A Carnot fridge or a Carnot heating system is considered a fridge or heat engine that acts on the reverse Carnot cycle.

The Carnot process comprises liquid phase mixture compression and expanding, which limits mechanical components.

A blown fuse or tripped circuit breaker shows

Answers

Answer:

Your house is in need of a service upgrade, or it may indicate that your house has too few circuits.

Explanation:

It's a sign that you are making excessive demands on the circuit and need to move some appliances and devices to other circuits.

What is the primary reason that heating, ventilating, and air conditioning (HVAC) is critical to a data center

Answers

Answer:

It prevents hardware from overheating.

Explanation:

HVAC stands for Heating, ventilation and air conditioning. It may be defined as technology which provides both vehicular as well as indoor environmental comfort. It provides thermal comfort and also used in places where overheating of the equipment or object is not desired.

A data center is a dedicated space which is used to store computer systems and the associated components like telecommunication as well as storage system.

Now proper HVAC system is necessary in data center so as to ensure that the components or the computer system does not get over heated and gets damaged. Overheating of the storage system may lead to loss of valuable information and other important data.

who is the strongest avenger i say hulk but who knows at this point

Answers

Answer:

or is the strongest evenger she hulk

Explanation:

?????????

Answer:

Thor!

Explanation:

In Thor: Ragnarok he beat the Hulk in order for Hulk to win thor had to be electrocuted and in Avengers: Endgame Thor is seen holding open the  "Floodgates" and withstanding the radiation from a dying star, also the fact that Thor is a god means that he is all powerful and the rightful heir to the throne to Asgard, plus the fact that he has defeated Loki multiple times a feat that not even the Hulk has done.

According to the Bureau of Labor Statistics, which occupation employed 4.2 million people – more than any other occupation – in 2010? Customer Service Representatives Retail Salespersons Cashiers Marketing Specialists

Answers

Answer:

Retail Salespersons

Explanation:

The Bureau of Labor Statistics or the BLS in short is the federal unit or agency of the United States Labor Department. It is the main stats finding agency of the United States federal government in the field of statistics and labor economics.

According to a publication published by the Labor Statistics Bureau, the retail salesperson sector of occupation employed the most people in the U.S. in the year 2010. It provided employment to about 4.2 million people.

Answer:

b. retail salespersons

Explanation:

A brine solution is 26% salt with 70.0 kg of water evaporated per hour. To produce 195 kg of pure salt (0% moisture) per day, how long should the process operate each day and how much brine must be fed to the evaporator per hour?
The process should operate each day for ___ hours.
The amount of brine that must be fed to the evaporator is ___ kg/h.

Answers

Answer:

- the process should operate each day for 7.9286 hours

- the amount of brine that must be fed to the evaporator is 94.594 kg/hr

Explanation:

Given that;

concentration of brine = 26%

so water concentration will be 100% - 26% = 74%

for evaporation of 70kg water per hour, residual is pure salt( 0% moisture)

so mass flow rate of brine = 70/(74%) = 70/0.74 = 94.5945 kg/hr

Amount of pure dry salt produced(0%) = 94.5945 - 70 = 24.5945 kg/hr

Now for production of 195kg of pure dry salt, number of hours required will be

T = 195 / 24.5945 = 7.9286 hrs

Therefore the process should operate each day for 7.9286 hours.

Total brine solution required ( 26% salt conc.) = 195/0.26 = 750 kg

Feed rate of brine solution ( 26% salt conc.) = 750 / 7.9286 = 94.594 kg/hr

Therefore the amount of brine that must be fed to the evaporator is 94.594 kg/hr

why carbon is not used as a semiconductor material​

Answers

Carbon is not used as semiconductor it has 4 valence electrons in it valence shell but the energy gap is very small it will conduct electricity even at room temperature ,the size of carbon is very small .

If the roller at A and the pin at B can support a load upto 4kN and 8kN, repsectively, determine the maximum intensity of the distributed load w, measured in kN/m, so that failure of the supports does not occur.

Answers

Solution :

Taken moment about point B;

[tex]$\sum M_B = 0$[/tex]

[tex]$\left[(w)(4)(2)-N_A \sin 30^\circ (3)(\sin 30^\circ)-N_A \cos 30^\circ(3 \cos 30^\circ +4) \right] =0$[/tex]

[tex]$N_A = 1.2376w$[/tex]

[tex]$\sum F_x = 0$[/tex]

[tex]$N_A \sin 30^\circ - B_x = 0$[/tex]

[tex]$B_x = 1.2376w \times \sin 30^\circ$[/tex]

[tex]$B_x = 0.6188w$[/tex]

[tex]$\sum F_y = 0$[/tex]

[tex]$B_y+N_A \cos^\circ - (w \times 4) = 0$[/tex]

[tex]$B_y+(1.2376w) \cos 30^\circ - (w \times 4)=0 $[/tex]

[tex]$B_y = 2.9282w$[/tex]

Now calculating the resultant force at B,

[tex]$F_B= \sqrt{B_x^2+B_y^2}$[/tex]

[tex]$F_B= \sqrt{(0.6188w)^2+(2.9282w)^2}$[/tex]

     = 2.9929w

For no failure,

[tex]$N_A<4 \ kN $[/tex]

2.9929 w < 4 kN

[tex]$w < 3.232 \ kN/m$[/tex]

And,

[tex]$F_B < 8 \ kN$[/tex]

2.9929 w < 8 kN

w < 2.673 kN/m

For the safe operation, w = 2.673 kN/m  

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