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harkovskaia [24]
2 years ago
8

Discuss the effect of PRE and CLR in JK flip-flop.​

Engineering
1 answer:
HACTEHA [7]2 years ago
6 0

Answer:

Explanation:

Asynchronous inputs on a flip-flop have control over the outputs (Q and not-Q) regardless of clock input status. These inputs are called the preset (PRE) and clear (CLR). The preset input drives the flip-flop to a set state while the clear input drives it to a reset state.

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A well-insulated tank in a vapor power plant operates at steady state. Saturated liquid water enters at inlet 1 at a rate of 125
kompoz [17]

Answer:

a. The mass flow rate (in lbm/s) is 135lbm/s

b. The temperature (in o F) is 200.8°F

Explanation:

We assume that potential energy and kinetic energy are negligible and the control volume operates at a steady state.

Given

a. The mass flow rate (in lbm/s) is 135lbm/s

b.

m1 = Rate at inlet 1 = 125lbm/s

m2 = Rate at inlet 2 = 10lbm/s

The mass flow rate (in lbm/s) is calculated as m1 + m2

Mass flow rate = 125lbm/s + 10lbm/s

Mass flow rate = 135lbm/s

Hence, the mass flow rate (in lbm/s) is 135lbm/s

b. To calculate the temperature.

First we need to determine the enthalpy h1 at 14.7psia

Using table A-3E (thermodynamics)

h1 = 180.15 Btu/Ibm

h2 at 14.7psia and 60°F = 28.08 Btu/Ibm

Calculating h3 using the following formula

h3 = (h1m1 + h2m2) / M3

h3 = (180.15 * 125 + 28.08 * 10)/135

h3 = 168.8855555555555

h3 = 168.89 Btu/Ibm

To get the final temperature; we make use of table A-2E of thermodynamics.

Because h3 < h1, it means the liquid is at a compressed state.

The corresponding temperature at h3 = 168.89 is 200.8°F

The temperature (in o F) is 200.8°F

6 0
2 years ago
Write a program with total change amount as an integer input, and output the change using the fewest coins, one coin type per li
pshichka [43]

Answer:

The C code is given below with appropriate comments

Explanation:

#include<stdio.h>

//defining constants

#define DOLLAR 100

#define QUARTER 25

#define DIME 10

#define NICKEL 5

#define PENNIES 1

//converting method

void ExactChange(int userTotal,int coinVals[])

{

//checking dollars

if (userTotal >=100)

{

coinVals[0]=userTotal/DOLLAR;

userTotal=userTotal-(100*coinVals[0]);

}

//checking quarters

if (userTotal >=25)

{

coinVals[1]=userTotal/QUARTER;

userTotal=userTotal-(25*coinVals[1] );

}

//checking dimes

if (userTotal >=10)

{

coinVals[2]=userTotal/DIME;

userTotal=userTotal-(10*coinVals[2]);

}

//checking nickels

if (userTotal >=5)

{

coinVals[3]=userTotal/NICKEL;

userTotal=userTotal-(5*coinVals[3]);

}

//checking pennies

if (userTotal >=1)

{

coinVals[4]=userTotal/PENNIES;

userTotal=userTotal-coinVals[4];

}

}

//main method

int main() {

//defining the variables

int amount;

//asking for input

printf("Enter the amount in cents :");

//reading the input

scanf("%d",&amount);

//validating the input

if(amount<1)

{

//printing the message

printf("No change..!");

}

//when the input is >0

else

{

int coinVals[5]={0,0,0,0,0};

ExactChange(amount,coinVals);

//checking dollars

if (coinVals[0]>0)

{

//printing dollars

printf("%d Dollar",coinVals[0]);

if(coinVals[0]>1) printf("s");

}

//checking quarters

if (coinVals[1]>0)

{

//printing quarters

printf(" %d Quarter",coinVals[1]);

if(coinVals[1]>1) printf("s");

}

//checking dimes

if (coinVals[2]>0)

{

//printing dimes

printf(" %d Dime",coinVals[2]);

if(coinVals[2]>1) printf("s");

}

//checking nickels

if (coinVals[3]>0)

{

//prinitng nickels

printf(" %d Nickel",coinVals[3]);

if(coinVals[3]>1) printf("s");

}

//checking pennies

if (coinVals[4]>0)

{

//printing pennies

printf(" %d Penn",coinVals[4]);

if(coinVals[4]>1) printf("ies");

else printf("y");

}

}

//end of main method

}

6 0
2 years ago
A ________ is only achieved through control and involves a specific change in one event (dependent variable) that can reliably b
prohojiy [21]

Answer:

Functional Relationship

Explanation:

A functional relationship is only achieved through control and involves a specific change in one event (dependent variable) that can reliably be produced by specific manipulations of another event (independent variable), and the change in the dependent variable is unlikely to be the result of other extraneous factors (confounding variables

6 0
2 years ago
Refrigerant-134a enters the coils of the evaporator of a refrigeration system as a saturated liquid–vapor mixture at a pressure
dezoksy [38]

Answer:

(a). Entropy change of refrigerant  is  = 0.7077 \frac{KJ}{K}

(b). Entropy change of cooled space dS_{space} = - 0.6844 \frac{KJ}{K}

(c). Total entropy change is dS = 0.0232 \frac{KJ}{K}

Explanation:

Given data

Saturation pressure = 140 K pa

Saturation temperature from property table

T_{sat} = - 18.77 °c =  - 18.77 + 273 = 254.23 K

(a). Entropy change of refrigerant  is given by

dS_{ref} = \frac{Q}{T_{sat}}

Since heat absorbed by refrigerant Q = 180 KJ

dS = \frac{180}{254.23}

dS = 0.7077 \frac{KJ}{K}

(b). Entropy change of cooled space

dS_{space} = - \frac{Q}{T_{space}}

T_{space} = - 10 °c = 263 K

dS_{space} = - \frac{180}{263}

dS_{space} = - 0.6844 \frac{KJ}{K}

(c). Total entropy change is given by

dS = dS_{ref} + dS_{space}

dS = 0.7077 - 0.6844

dS = 0.0232 \frac{KJ}{K}

This is the value of total entropy change.

4 0
2 years ago
Technician A says that TSBs are typically updates to the owner's manual. Technician B says that TSBs are generally
vichka [17]

Answer:

Technician A says that TSBs are typically updates to the owner's manual. Technician B says that TSBs are generally

updated information on model changes that do not affect the technician. Who is correct? the answer is c

4 0
2 years ago
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