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m_a_m_a [10]
3 years ago
15

A closed, rigid tank is lled with a gas modeled as an ideal gas, initially at 27°C and a gage pressure of 300 kPa. The gas is he

ated, and the gage pressure at the nal state is 367 kPa. Determine the nal temperature, in °C. The local atmospheric pressure is 1 atm.
Engineering
1 answer:
sergejj [24]3 years ago
8 0

Answer:

T₂ =93.77  °C

Explanation:

Initial temperature ,T₁ =27°C= 273 +27 = 300 K

We know that

Absolute pressure = Gauge pressure + Atmospheric pressure

Initial pressure ,P₁ = 300+1=301 kPa

Final pressure  ,P₂= 367+1 = 368  kPa

Lets take  temperature=T₂

We know that ,If the volume of the gas is constant ,then we can say that

\dfrac{P_2}{P_1}=\dfrac{T_2}{T_1}

{T_2}=T_1\times \dfrac{P_2}{P_1}

Now by putting the values in the above equation we get

{T_2}=300\times \dfrac{368}{301}\ K

The temperature in  °C

T₂ = 366.77 - 273  °C

T₂ =93.77  °C

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Modify the Rainfall Statistics program you wrote for Programming Challenge 2 of Chapter 7 . The program should display a list of
rjkz [21]

Answer:

#include<iostream>

#include <iomanip>

using namespace std;

const int NUM_MONTHS = 12;

double getTotal(double [], int);

double getAverage(double [], int);

double getLargest(double [], int, int &);

double getSmallest(double [], int, int &);

double getTotal(int rainFall,double NUM_MONTHS[])

{

double total = 0;

for (int count = 0; count < NUM_MONTH; count++)

total += NUM_MONTH[count];

return total;

}

double getAverage(int rainFall,double NUM_MONTH[])

{getTotal(rainFall,NUM_MONTH)

average= total/NUM_MONTHS;

return average;

}

double getHighest(int rainFall, double NUM_MONTHS[]) //I left out the subScript peice as I was not sure how to procede with that;

{

double largest;

largest = NUM_MONTHS[0];

for ( int month = 1; month <= NUM_MONTHS; month++ ){

                     if ( values[month] > largest ){

                 largest = values[month];

return largest;

          }

double getSmallest(int rainFall, double NUM_MONTHS[])

{

double smallest;

smallest = NUM_MONTHS[0];

for ( int month = 1; month <= NUM_MONTHS; month){

                     if ( values[month] < smallest ){

                 smallest = values[month];

return smallest;

          }

 

int main()

{

double rainFall[NUM_MONTHS];

 for (int month = 0; month < NUM_MONTHS; month++)

  {

     cout << "Enter the rainfall (in inches) for month #";

     cout << (month + 1) << ": ";

     cin >> rainFall[month];

 

     while (rainFall[month] < 0)

     {

      cout << "Rainfall must be 0 or more.\n"

             << "Please re-enter: ";

      cin >> rainFall[month];

     }

  }

  cout << fixed << showpoint << setprecision(2) << endl;

  cout << "The total rainfall for the year is ";

  cout << getTotal(rainFall, NUM_MONTHS)

      << " inches." << endl;

   cout << "The average rainfall for the year is ";

  cout << getAverage(rainFall, NUM_MONTHS)

      << " inches." << endl;

   int subScript;

cout << "The largest amount of rainfall was ";

  cout << getLargest(rainFall, NUM_MONTHS, subScript)

      << " inches in month ";

  cout << (subScript + 1) << "." << endl;

  cout << "The smallest amount of rainfall was ";

  cout << getSmallest(rainFall, NUM_MONTHS, subScript)

      << " inches in month ";

  cout << (subScript + 1) << "." << endl << endl;

  return 0;

}

8 0
3 years ago
Build a 32-bit accumulator circuit. The circuit features a control signal inc and enable input en. If en is 1 and inc is 1, the
Alex_Xolod [135]

Sorry need.points I'm new

7 0
3 years ago
For the SR-latch below high levels of Set and Reset result in Q= 1 and 0, respectively. The next state is unknown when both inpu
dusya [7]

Answer:

hello your question lacks the required image attached to this answer is the image required

answer :  NOR1(q_) wave is complementary to NOR2(q)

Explanation:

Note ; NOR 2 will be addressed as q in the course of this solution while NOR 1 will be addressed as q_

Initial state is unknown i.e q = 0 and q_= 1

from the diagram the waveform reset and set

= from 0ns to 20ns reset=1 and set=0.from the truth table considering this given condition q=0 and q_bar=1 while  

from 30ns to 50ns reset=0 and set=1.from the truth table considering this condition q=1 and q_bar=1.so from 35ns also note there is a delay of 5 ns for the NOR gate hence the NOR 2 will be higher ( 1 )

From 50ns to 65ns both set and reset is 0.so NOR2(q)=0.

From 65 to 75 set=1 and reset=0,so our NOR 2(q)=1 checking from the truth table

also  from 75 to 90 set=1 and reset=1 , NOR2(q) is undefined "?" and is mentioned up to 95ns.

since q_ is a complement of q, then NOR1(q_) wave is complementary to NOR2(q)

3 0
3 years ago
Compute the solution to x + 2x + 2x = 0 for Xo = 0 mm, vo = 1 mm/s and write down the closed-form expression for the response.
Nutka1998 [239]

Answer:

β = \frac{c}{\sqrt{km} } =  0.7071 ≈ 1 ( damping condition )

closed-form expression for the response is attached below

Explanation:

Given :  x + 2x + 2x = 0   for Xo = 0 mm and Vo = 1 mm/s

computing a solution :

M = 1,

c = 2,

k = 2,

Wn = \sqrt{\frac{k}{m} }  = \sqrt{2}  

next we determine the damping condition using the damping formula

β = \frac{c}{\sqrt{km} } =  0.7071 ≈ 1

from the condition above it can be said that the damping condition indicates underdamping

attached below is the closed form expression for the response

6 0
2 years ago
we wish to send at a rate of 10Mbits/s over a passband channel. Assuming that an excess bandwidth of 50% is used, how much bandw
gayaneshka [121]

Answer:

QPSK: 7.5 MHz

64-QAM:2.5 MHz

64-Walsh-Hadamard: 160 MHz

Explanation:

See attached picture.

6 0
3 years ago
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