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dalvyx [7]
3 years ago
14

Suppose we want to determine how many of the bits in a twelve-bit unsigned number are equal to zero. Implement the simplest circ

uit that can accomplish this task.

Engineering
1 answer:
timurjin [86]3 years ago
8 0

Answer:

See the picture for circuit figure.

Explanation:

First load 12 bit unsigned number on shift register and apply clock tick to increment counter from LSB bit.

Logic 1 is shifted into register to ensure counter doesn't count more than 12 bits.

You might be interested in
You are using a Geiger counter to measure the activity of a radioactive substance over the course of several minutes. If the rea
Reika [66]

Answer: 33.35 minutes

Explanation:

A(t) = A(o) *(.5)^[t/(t1/2)]....equ1

Where

A(t) = geiger count after time t = 100

A(o) = initial geiger count = 400

(t1/2) = the half life of decay

t = time between geiger count = 66.7 minutes

Sub into equ 1

100=400(.5)^[66.7/(t1/2)

Equ becomes

.25= (.5)^[66.7/(t1/2)]

Take log of both sides

Log 0.25 = [66.7/(t1/2)] * log 0.5

66.7/(t1/2) = 2

(t1/2) = (66.7/2 ) = 33.35 minutes

4 0
3 years ago
A square loop of wire surrounds a solenoid. The side of the square is 0.1 m, while the radius of the solenoid is 0.025 m. The sq
Semmy [17]

Answer:

I=9.6×e^{-8}  A

Explanation:

The magnetic field inside the solenoid.

B=I*500*muy0/0.3=2.1×e ^-3×I.

so the total flux go through the square loop.

B×π×r^2=I×2.1×e^-3π×0.025^2

=4.11×e^-6×I

we have that

(flux)'= -U

so differentiating flux we get

so the inducted emf in the loop.

U=4.11×e^{-6}×dI/dt=4.11×e^-6×0.7=2.9×e^-6 (V)

so, I=2.9×e^{-6}÷30

I=9.6×e^{-8}  A

6 0
3 years ago
The section should span between 10.9 and 13.4 cm (4.30 and 5.30 inches) as measured from the end supports and should be able to
Sergeeva-Olga [200]

Answer:

hello below is missing piece of the complete question

minimum size = 0.3 cm

answer : 0.247 N/mm2

Explanation:

Given data :

section span : 10.9 and 13.4 cm

minimum load applied evenly to the top of span  : 13 N

maximum load for each member ; 4.5 N

lets take each member to be 4.2 cm

Determine the max value of P before truss fails

Taking average value of section span ≈ 12 cm

Given minimum load distributed evenly on top of section span = 13 N  

we will calculate the value of   by applying this formula

= \frac{Wl^2}{12}  =  (0.013 * 0.0144 )/ 12  =  1.56 * 10^-5

next we will consider section ; 4.2 cm * 0.3 cm

hence Z (section modulus ) = BD^2 / 6  

                                             = ( 0.042 * 0.003^2 ) / 6  = 6.3*10^-8

Finally the max value of P( stress ) before the truss fails

= M/Z = ( 1.56 * 10^-5 ) / ( 6.3*10^-8 )  

          = 0.247 N/mm2

5 0
3 years ago
Your Java program will be reading input from a file name strInput.txt. Each record contains String firstname String lastName Str
stiks02 [169]

Answer:

The program requires that you have the specified input files and it reads from each file at a time and processes salary in digits, states the city, state and bonus with respective first and last name as requested in the question. Note that you must have access to the mentioned output files for the program to work properly. Below is the java version of the program.

import java.io.File;

import java.io.FileNotFoundException;

import java.io.PrintWriter;

import java.util.Scanner;

class Driver

{

public static void main(String[] args) throws FileNotFoundException

{

Scanner sc = new Scanner(new File("strInput.txt"));

PrintWriter pd = new PrintWriter(new File("strOutputD"));

PrintWriter prf = new PrintWriter(new File("strOutputRF"));

String firstname = "", lastname = "", strSalary = "", status = "", cityState = "", city = "", state = "";

double salary = 0, bonus = 0;

int incorrectRecords = 0;

int dRecords = 0;

int fRecords = 0;

while(sc.hasNextLine())

{

firstname = sc.next();

lastname = sc.next();

strSalary = sc.next();

status = sc.next();

cityState = sc.next();

if(!status.equals("D") && !status.equals("F"))

{

System.out.println("Records is neither D nor F. Skipping this...");

incorrectRecords++;

continue;

}

else if(status.equals("D") || status.equals("F"))

{

char c = ' ';

int i = 0;

for(i=0; i<strSalary.length() && c != '.'; i++)

{

c = strSalary.charAt(i);

if(!Character.isDigit(c))

{

System.out.println("Char at position " + (i+1) + " in salary is not a digit");

incorrectRecords++;

continue;

}

}

if(c == '.')

{

if(i+1 == strSalary.length()-1)

{

if(!Character.isDigit(strSalary.charAt(i)))

{

System.out.println("Char at position " + (i+1) + " in salary is not a digit");

incorrectRecords++;

continue;

}

if(!Character.isDigit(strSalary.charAt(i+1)))

{

System.out.println("Char at position " + (i+1+1) + " in salary is not a digit");

incorrectRecords++;

continue;

}

}

else

{

System.out.println("Period is in the wrong position. Expected at " + (strSalary.length()-3) + " but found at " + (i+1));

continue;

}

}

city = cityState.split(",")[0];

state = cityState.split(",")[1];

salary = Double.parseDouble(strSalary);

if(status.equals("D"))

{

bonus = salary * 0.125;

dRecords++;

pd.write(firstname + " " + lastname + " " + status + " " + salary + " " + bonus + " " + city + " " + state);

}

else

{

bonus = salary * 0.18;

fRecords++;

prf.write(firstname + " " + lastname + " " + status + " " + salary + " " + bonus + " " + city + " " + state);

}

}

}

System.out.println("No of D records : " + dRecords);

System.out.println("No of F records : " + fRecords);

System.out.println("No of incorrect records : " + incorrectRecords);

}

}

6 0
3 years ago
Let be a real-valued signal for which when . Amplitude modulation is preformed to produce the signal . A proposed demodulation t
densk [106]

Answer:

hello your question is incomplete attached below is the complete question

answer : attached below

Explanation:

let ; x(t)  be a real value signal for x ( jw ) = 0 , |w| > 200\pi

g(t) = x ( t ) sin ( 2000 \pi t )

x_{1} (t) = \frac{1}{2}  x(t)  sin ( 4000\pi t )

next we apply Fourier transform

attached below is the remaining part of the solution

6 0
3 years ago
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