Answer:
void bubble_sort( int A[ ], int n ) {
int temp;
for(int k = 0; k< n-1; k++) {
// (n-k-1) to ignore comparisons of already compared iterations
for(int i = 0; i < n-k-1; i++) {
if(A[ i ] > A[ i+1] ) {
// swapping occurs here
temp = A[ i ];
A[ i ] = A[ i+1 ];
A[ i + 1] = temp ;
}
}
}
}
Answer
Assuming
At 10000 m height temperature T = -55 C = 218 K
At 1000 m height temperature T = 0 C = 273 K
R = 287 J/kg K
V₂ = V₁ ×1.1222
V₁ = 0.5 × C₁ = 0.5 × 295 = 147.5 m/s
V₂ = 1.1222 × 147.5 = 165.49 m/s
so, the jetliner need to increase speed by ( V₂ -V₁ )
= 165.49 - 147.5
= 17.5 m/s
Answer:
Enthalpy, hsteam = 2663.7 kJ/kg
Volume, Vsteam = 0.3598613 m^3 / kg
Density = 2.67 kg/ m^3
Explanation:
Mass of steam, m = 1 kg
Pressure of the steam, P = 0.5 MN/m^2
Dryness fraction, x = 0.96
At P = 0.5 MPa:
Tsat = 151.831°C
Vf = 0.00109255 m^3 / kg
Vg = 0.37481 m^3 / kg
hf = 640.09 kJ/kg
hg = 2748.1 kJ/kg
hfg = 2108 kJ/kg
The enthalpy can be given by the formula:
hsteam = hf + x * hfg
hsteam = 640.09 + ( 0.96 * 2108)
hsteam = 2663.7 kJ/kg
The volume of the steam can be given as:
Vsteam = Vf + x(Vg - Vf)
Vsteam = 0.00109255 + 0.96(0.37481 - 640.09)
Vsteam = 0.3598613 m^3 / kg
From the steam table, the density of the steam at a pressure of 0.5 MPa is 2.67 kg/ m^3
Answer:
time management concept
Explanation:
it says time management concept.
Answer: 
Explanation:
Given
Again differentiate for the second derivative
