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storchak [24]
3 years ago
10

A semiconductor wafer with n-type impurities has been stripped from all its free electrons. The wafer is in the form of a flat d

isk and has positive charges uniformly distributed on its top and bottom surface. The charge density is 10 C/cm^2. Determine the magnitude of the electric flux density D above the wafer in C/m^2.a. 0.5 x10+5 c/m2 b. 0.5x10-5 C/m2 c. 0.5x10 C/m2 d. 10+5 C/m2
Engineering
1 answer:
skelet666 [1.2K]3 years ago
4 0

Answer:

a. 0.5x10⁵

Explanation:

The charge density = 10C/cm²

We are asked to find the magnitude of the electric flux density

The electric field of infinite sheet

E = σ/2Eo

We have electric density as follows:

D = EEo = EoE

Since charge density σ = 10C/cm²

σ = 10x10²C/m²

D = σ/2

10x10⁴/2

= 5x10⁴C/m²

= 0.5x10⁵C/m²

Therefore the magnitude is equal to 0.5x10⁵C/m². Which gives us option a as the answer

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A train starts from rest at station A and accelerates at 0.4 m/s^2 for 60 s. Afterwards it travels with a constant velocity for
mash [69]
<h3><u>The distance between the two stations is</u><u> </u><u>3</u><u>7</u><u>.</u><u>0</u><u>8</u><u> km</u></h3>

\\

Explanation:

<h2>Given:</h2>

a_1 \:=\:0.4\:m/s²

t_1 \:=\:60\:s

v_{i1} \:=\:0\:m/s

a_2 \:=\:0\:m/s²

t_2 \:=\:25\:min\:=\:1500\:s

a_3 \:=\:-0.8\:m/s²

v_{f3} \:=\:0\:m/s

\\

<h2>Required:</h2>

Distance from Station A to Station B

\\

<h2>Equation:</h2>

a\:=\:\frac{v_f\:-\:v_i}{t}

v_{ave}\:=\:\frac{v_i\:+\:v_f}{2}

v\:=\:\frac{d}{t}

\\

<h2>Solution:</h2><h3>Distance when a = 0.4 m/s²</h3>

Solve for v_{f1}

a\:=\:\frac{v_f\:-\:v_i}{t}

0.4\:m/s²\:=\:\frac{v_f\:-\:0\:m/s}{60\:s}

24\:m/s\:=\:v_f\:-\:0\:m/s

v_f\:=\:24\:m/s

\\

Solve for v_{ave1}

v_{ave}\:=\:\frac{v_i\:+\:v_f}{2}

v_{ave}\:=\:\frac{0\:m/s\:+\:24\:m/s}{2}

v_{ave}\:=\:12\:m/s

\\

Solve for d_1

v\:=\:\frac{d}{t}

12\:m/s\:=\:\frac{d}{60\:s}

720\:m\:=\:d

d_1\:=\:720\:m

\\

<h3>Distance when a = 0 m/s²</h3>

v_{f1}\:=\:v_{i2}

v_{i2}\:=\:24\:m/s

\\

Solve for v_{f2}

a\:=\:\frac{v_f\:-\:v_i}{t}

0\:m/s²\:=\:\frac{v_f\:-\:24\:m/s}{1500\:s}

0\:=\:v_f\:-\:24\:m/s

v_f\:=\:24\:m/s

\\

Solve for v_{ave2}

v_{ave}\:=\:\frac{v_i\:+\:v_f}{2}

v_{ave}\:=\:\frac{24\:m/s\:+\:24\:m/s}{2}

v_{ave}\:=\:24\:m/s

\\

Solve for d_2

v\:=\:\frac{d}{t}

24\:m/s\:=\:\frac{d}{1500\:s}

36,000\:m\:=\:d

d_2\:=\:36,000\:m

\\

<h3>Distance when a = -0.8 m/s²</h3>

v_{f2}\:=\:v_{i3}

v_{i3}\:=\:24\:m/s

\\

Solve for v_{f3}

a\:=\:\frac{v_f\:-\:v_i}{t}

-0.8\:m/s²\:=\:\frac{0\:-\:24\:m/s}{t}

(t)(-0.8\:m/s²)\:=\:-24\:m/s

t\:=\:\frac{-24\:m/s}{-0.8\:m/s²}

t\:=\:30\:s

\\

Solve for v_{ave3}

v_{ave}\:=\:\frac{v_i\:+\:v_f}{2}

v_{ave}\:=\:\frac{24\:m/s\:+\:0\:m/s}{2}

v_{ave}\:=\:12\:m/s

\\

Solve for d_3

v\:=\:\frac{d}{t}

12\:m/s\:=\:\frac{d}{30\:s}

360\:m\:=\:d

d_3\:=\:360\:m

\\

<h3>Total Distance from Station A to Station B</h3>

d\:= \:d_1\:+\:d_2\:+\:d_3

d\:= \:720\:m\:+\:36,000\:m\:+\:360\:m

d\:= \:37,080\:m

d\:= \:37.08\:km

\\

<h2>Final Answer:</h2><h3><u>The distance between the two stations is </u><u>3</u><u>7</u><u>.</u><u>0</u><u>8</u><u> km</u></h3>
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Consider a thermal energy reservoir at 1500 K that can supply heat at a rate of 150,000 kJ/h. Determine the exergy of this suppl
anzhelika [568]

Answer:

33.4KW

Explanation:

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n = 1 - T1/T2

T1 = atmospheric temperature 298k

T2 = reservoir temperature 1500k

n = 1 - 298/1500

n = 0.8013

output energy = n × energy input

0.8013×150000 = 120195KJ/hr

Power output = 120195KJ/hr/3600 = 33.4KW

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Determine the adiabatic flame temperature of carbon monoxide (CO) burning in air at an equivalence ratio of unity. The reactants
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Answer:

Explanation:

The detailed analysis and step by step calculation is as shown in the attachment.

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Exercise 19
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Assume we have already defined a variable of type String called password with the following line of code: password' can have any
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Answer:

The Java code is given below with appropriate comments for better understanding

Explanation:

import java.util.Scanner;

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          String password = input.nextLine();

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  }

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      int count = 0;

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          count++;

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          count++;

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          count++;

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          count++;

      return count;

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      return pwd.length() >= 7;

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          if (Character.isLetter(pwd.charAt(i)) && Character.isUpperCase(pwd.charAt(i)))

              return true;

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  }

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      for (int i = 0; i < pwd.length(); i++)

          if (Character.isLetter(pwd.charAt(i)) && Character.isLowerCase(pwd.charAt(i)))

              return true;

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  public static boolean checkForDigit(String pwd) {

      for (int i = 0; i < pwd.length(); i++)

          if (Character.isDigit(pwd.charAt(i)))

              return true;

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