Answer:
Time delay increases
Explanation:
Time delay is the delay between occurance of signal. If sampling time that is time between two samples is increased, the delay in the occurance of regenerated samples is also increased.
Answer:
//Convert any decimal number to binary number
//Program is written in C++ Programming Language
// Comments are used for explanatory purpose
// Program starts here
#include <iostream>
using namespace std;
// Main Method declared here
int main()
{
int x;
cout<<"Enter any integer number: ";
cin>>x;
DecBin(x);
return 0;
}
// Here a function named DecBin is declared along with an integer variable, x
void DecBin(int x)
{
// Declare an array to store the resulting binary digits
int bindigit[32];
// counter for binary array
int kount = 0;
while (x > 0) {
// Store the remainder of each division in the declared array
bindigit[kount] = x % 2;
x = x / 2;
kount++;
}
// Loop to print the binary digits in reverse order
for (int j = i - 1; j >= 0; j--)
{
cout << bindigit[j];
}
}
// End of Program
Answer:
The pressure in tank will become 541.3 KPa.
Explanation:
From general gas equation we know that:
PV = nRT
PV = (m/M)RT
P/Tm = R/VM
Now, for initial state:
P1/T1 m1 = R/VM ______ eqn (1)
For Final State, as the gas constant (R), volume (V) and molecular mass of air (M) remains constant;
P2/T2 m2 = R/VM ______ eqn (2)
Comparing eqn (1) and eqn (2):
P1/T1 m1 = P2/T2 m2
Now, we have:
P1 = 400 KPa
P2 = ?
m1 = 2 kg
m2 = 0.6 kg + 2 kg = 2.6 kg
T1 = 20°C = 293 K
T2 = 32°C = 305 K
Therefore,
(400 KPa)/(293 K)(2 kg) = P2/(305 K)(2.6 kg)
<u>P2 = 541.3 KPa</u>
Answer:
Explanation:
Given data:
Ammonia Nitrogen 30 mg/L
pH = 8.5
-log[H +] = 8.5
[H +] = 10^{-8.5}
Rate constant is given as
...........1
Total ammonia as NItrogen is given as 30 mg/l
=
.....2
from equation 1 we have
{10^{8.5}}
plug this value in equation 2 we get
Total ammonia as N = 30 mg/lt