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3 years ago

In a 5V system if you were asked to take one input HIGH and another LOW what would you do (i.e. where would you connect them)?

Engineering

1 answer:

dangina [55]3 years ago

4 0

Answer:

HIGH from the supply voltage

LOW from ground

Explanation:

The answer depends on the kind of system and the purpose of the signal. But for practical reasons, in a DIGITAL system where 5V is HIGH and 0 V is LOW, 5 volts can be taken from the supply voltage (usually the same as high, BUT must be verified), and the LOW signal from ground.

If the user has a multimeter, it must be set to continuous voltage on 0 to 20 V range. Then place the probe in the ground of the circuit (must be a big copper area). Finally leave one probe in the circuit ground and place the other probe in some test points to identify 5 v.

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Consider the thermocouple and convection conditions of Example 1, but now allow for radiation exchange with the walls of a duct

alex41 [277]

Answer:

hello your question has some missing part attached below is the complete question

answer : steady state temperature = 419.713k ≈ 218.7⁰c

Time required to reach a junction ≈ 5 secs

Explanation:

The detailed solution of the given problem is attached below but the solution to the subsequent problem from which the question you asked is referenced to( problem 1 ), is not attached because it was not part of the question you asked

6 0

3 years ago

An air-standard Carnot cycle is executed in a closed system between the temperature limits of 350 and 1200 K. The pressures befo

SVEN [57.7K]

This question is incomplete, the complete question is;

An air-standard Carnot cycle is executed in a closed system between the temperature limits of 350 and 1200 K. The pressures before and after the isothermal compression are 150 and 300 kPa, respectively. If the net work output per cycle is 0.5 kJ, determine (a) the maximum pressure in the cycle, (b) the heat transfer to air, and (c) the mass of air. Assume variable specific heats for air.

Answer:

a) the maximum pressure in the cycle is 30.01 Mpa

b) the heat transfer to air is 0.7058 KJ

c) mass of Air is 0.002957 kg

Explanation:

Given the data in the question;

We find the relative pressure of air at 1200 K (T1) and 350 K ( T4)

so from the "ideal gas properties of air table"

Pr1 = 238

Pr4 = 2.379

we know that Pressure P1 is only maximum at the beginning of the expansion process,

now we express the relative pressure and pressure relation for the process 4-1

P1 = (Pr2/Pr4)P4

so we substitute

P1 = (238/2.379)300 kPa

P1 = 30012.6 kPa = 30.01 Mpa

Therefore the maximum pressure in the cycle is 30.01 Mpa

the Thermal heat efficiency of the Carnot cycle is expressed as;

ηth = 1 - (TL/TH)

we substitute

ηth = 1 - (350K/1200K)

ηth = 1 - 0.2916

ηth = 0.7084

now we find the heat transferred

Qin = W_net.out / ηth

given that the net work output per cycle is 0.5 kJ

we substitute

Qin = 0.5 / 0.7084

Qin = 0.7058 KJ

Therefore, the heat transfer to air is 0.7058 KJ

first lets express the change in entropy for process 3 - 4

S4 - S3 = (S°4 - S°3) - R.In(P4/P3)

S4 - S3 = - (0.287 kJ/Kg.K) In(300/150)kPa

= -0.1989 Kj/Kg.K = S1 - S2

so that; S2 - S1 = 0.1989 Kj/Kg.K

Next we find the net work output per unit mass for the Carnot cycle

W"_netout = (S2 - S1)(TH - TL)

we substitute

W"_netout = ( 0.1989 Kj/Kg.K )( 1200 - 350)K

= 169.065 kJ/kg

Finally we find the mass

mass m = W_ net.out / W"_netout

we substitute

m = 0.5 / 169.065

m = 0.002957 kg

Therefore, mass of Air is 0.002957 kg

5 0

3 years ago

A particle moves along a straight line such that its position is defined by s = (t2 - 6t + 5) m. Determine the average velocity,

Dominik [7]

Answer:

0 m/s , 3 m/s , 2 m/s^2

Explanation:

Given : s(t) = ( t^2 - 6t + 5)

v(t) = ds / dt = 2t - 6

s(0) = 5 m

s(6) = (6)^2 - 6*6 + 5 = 5 m

Vavg = ( s(6) - s(0) ) / 2 = 0 m\s

Find the turning point of particle:

ds/dt = 0 = 2t - 6

t = 3 sec

s(3) = 3^2 -6*3 + 5 = - 4

Total distance = 5 - (-4) + (5 - (-4)) = 18 m

Total time = 6s

Average speed = Total distance / Total time = 18 / 6 = 3 m/s

Taking derivative of v(t) to obtain a(t)

a (t) = dv(t) / dt = 2 m/s^2

7 0

3 years ago

This assignment will give you more experience on the use of loops In this project, we are going to compute the number of times a

musickatia [10]

Answer:

import java.io.BufferedReader;

import java.io.File;

import java.io.FileReader;

import java.io.IOException;

import java.util.Scanner;

import java.util.StringTokenizer;

public class Tester {

public static void main(String[] args) throws IOException {

Scanner in=new Scanner(System.in);

System.out.println("Enter a Number =======>");

long N ;

while (!in.hasNextLong()) {

System.out.println("That's not a number!");

in.next();

}

N=in.nextLong();

System.out.println("Number Entered is =======>"+N);

System.out.println("Enter a Digit =======>");

int D;

while (!in.hasNextLong()) {

System.out.println("That's not a number!");

in.next();

}

D=in.nextInt();

System.out.println("Digit Entered is =======> "+ D);

long que=N;

int rem;

int count=0;

while(true){

long temp;

temp = (long)que/10;

rem = (int) (que % 10);

System.out.println(temp+" "+ que+" "+rem);

if(rem==D) count++;

que=temp;

if(que==0) break;

}

System.out.println("The number of "+ D+"'s in "+ N + " is "+ count );;

}

Explanation:

Divide the que variable by 10 and assign its result to the temp variable.
Calculate the remainder.
Increment the count if the remainder is equal to the value of D variable and assign the value of que to temp variable.

3 0

3 years ago

There are four spheres of earth. These include __________, _________, ________ and _________. They are important because when th

ololo11 [35]

Answer:

"lithosphere" , "hydrosphere" , "biosphere" , "atmosphere"

4 0

3 years ago