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3 years ago

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Question 1: Final Results = What are the values of the resistances such that the gain = -100, Rin = 1 MI2. Don't use resistances

greater than 1 M2.
is this question wrong??? me and mh freinds are trying to figure oyt the answer for 3 days but we can't

1 answer:

lidiya [134]3 years ago

4 0

Answer:

Explanation:

In a study of algebra, you will encounter many families of equations, or groups of

equations that share common characteristics. Of interest to us here is the family of

linear equations in one variable, a study that lays the foundation for understanding

more advanced families. In addition to solving linear equations, we’ll use the skills we

develop to solve for a specified variable in a formula, a practice widely used in science,

business, industry, and research.

A. Solving Linear Equations Using Properties of Equality

An equation is a statement that two expressions are

equal. From the expressions and

we can form the equation

which is a linear equation in one variable. To solve

an equation, we attempt to find a specific input or xvalue that will make the equation true, meaning the

left-hand expression will be equal to the right. Using

Table 1.1, we find that is a

true equation when x is replaced by 2, and is a false

equation otherwise. Replacement values that make

the equation true are called solutions or roots of the equation.

You might be interested in

An Ideal gas is being heated in a circular duct as while flowing over an electric heater of 130 kW. The diameter of duct is 500

max2010maxim [7]

Answer: The exit temperature of the gas in deg C is $32^{o}C$ .

Explanation:

The given data is as follows.

$C_{p}$ = 1000 J/kg K, R = 500 J/kg K = 0.5 kJ/kg K (as 1 kJ = 1000 J)

$P_{1}$ = 100 kPa, $V_{1} = 15 m^{3}/s$

$T_{1} = 27^{o}C = (27 + 273) K = 300 K$

We know that for an ideal gas the mass flow rate will be calculated as follows.

$P_{1}V_{1} = mRT_{1}$

or, m = $\frac{P_{1}V_{1}}{RT_{1}}$

= $\frac{100 \times 15}{0.5 \times 300}$

= 10 kg/s

Now, according to the steady flow energy equation:

$mh_{1} + Q = mh_{2} + W$

$h_{1} + \frac{Q}{m} = h_{2} + \frac{W}{m}$

$C_{p}T_{1} - \frac{80}{10} = C_{p}T_{2} - \frac{130}{10}$

$(T_{2} - T_{1})C_{p} = \frac{130 - 80}{10}$

$(T_{2} - T_{1})$ = 5 K

$T_{2}$ = 5 K + 300 K

$T_{2}$ = 305 K

= (305 K - 273 K)

= $32^{o}C$

Therefore, we can conclude that the exit temperature of the gas in deg C is $32^{o}C$ .

7 0

3 years ago

Your program will be a line editor. A line editor is an editor where all operations are performed by entering commands at the co

soldier1979 [14.2K]

Answer:

Java program given below

Explanation:

import java.util.*;

import java.io.*;

public class Lineeditor

{

private static Node head;

class Node

{

int data;

Node next;

public Node()

{data = 0; next = null;}

public Node(int x, Node n)

{data = x; next =n;}

}

public void Displaylist(Node q)

{if (q != null)

{

System.out.println(q.data);

Displaylist(q.next);

}

}

public void Buildlist()

{Node q = new Node(0,null);

head = q;

String oneLine;

try{BufferedReader indata = new

BufferedReader(new InputStreamReader(System.in)); // read data from terminals

System.out.println("Please enter a command or a line of text: ");

oneLine = indata.readLine(); // always need the following two lines to read data

head.data = Integer.parseInt(oneLine);

for (int i=1; i<=head.data; i++)

{System.out.println("Please enter another command or a new line of text:");

oneLine = indata.readLine();

int num = Integer.parseInt(oneLine);

Node p = new Node(num,null);

q.next = p;

q = p;}

}catch(Exception e)

{ System.out.println("Error --" + e.toString());}

}

public static void main(String[] args)

{Lineeditor mylist = new Lineeditor();

mylist.Buildlist();

mylist.Displaylist(head);

}

}

7 0

3 years ago

Consider a refrigerator that consumes 320 W of electric power when it is running. If the refrigerator runs only one quarter of t

Semenov [28]

Answer:

B. $5.18

Explanation:

Cost of electricity per kWh = $0.09

Power consumption of refrigerator = 320W = 320/1000 = 0.32kW

In a month (30 days) the refrigerator works 1/4 × 30 days = 7.5 days = 7.5 × 24 hours = 180 hours

Energy consumed in 180 hours = 0.32kW × 180h = 57.6kWh

Cost of electricity of 57.6kWh energy consumed by the refrigerator = 57.6 × $0.09 = $5.18

3 0

4 years ago

Read 2 more answers

An ideal fluid flows through a pipe made of two sections with diameters of 1.0 and 3.0 inches, respectively. The speed of the fl

geniusboy [140]

Answer:

$(\frac{r_1}{r_2})^2=\frac{1}{9}$

Explanation:

From the question we are told that:

Diameter 1 $d_1=1.0$

Diameter 2 $d_2=3.0$

Generally the equation for Radius is mathematically given by

At Diameter 1

$r_{1}=\frac{1}{2} inch$

At Diameter 2

$r_{2}=\frac{3}{2} inch$

Generally the equation for continuity is mathematically given by

$A_1V_1=A_2V_2$

Therefore

$(\frac{r_1}{r_2})^2=(\frac{1/2}{3/2})^2$

$(\frac{r_1}{r_2})^2=\frac{1}{9}$

5 0

3 years ago

Write the Different types of Trainings and explain Anyone in short

Eddi Din [679]

Answer:

Coaching or mentoring

Coaching or mentoring can share similar qualities to hands-on training, but in this type of employee training, the focus is on the relationship between an employee and a more experienced professional, such as their supervisor, a coach, or a veteran employee.

The one-on-one mentoring style creates a relationship between employees that carries far beyond training. Lecture-style training

Important for getting big chunks of information to a large employee population, lecture-style training can be an invaluable resource for communicating required information quickly.

However, use this type of employee training sparingly.

3 0

3 years ago