Answer:
/* C Program to rotate matrix by 90 degrees */
#include<stdio.h>
int main()
{
int matrix[100][100];
int m,n,i,j;
printf("Enter row and columns of matrix: ");
scanf("%d%d",&m,&n);
/* Enter m*n array elements */
printf("Enter matrix elements: \n");
for(i=0;i<m;i++)
{
for(j=0;j<n;j++)
{
scanf("%d",&matrix[i][j]);
}
}
/* matrix after the 90 degrees rotation */
printf("Matrix after 90 degrees roration \n");
for(i=0;i<n;i++)
{
for(j=m-1;j>=0;j--)
{
printf("%d ",matrix[j][i]);
}
printf("\n");
}
return 0;
}
Answer:
The radius of a wind turbine is 691.1 ft
The power generation potential (PGP) scales with speed at the rate of 7.73 kW.s/m
Explanation:
Given;
power generation potential (PGP) = 1000 kW
Wind speed = 5 mph = 2.2352 m/s
Density of air = 0.0796 lbm/ft³ = 1.275 kg/m³
Radius of the wind turbine r = ?
Wind energy per unit mass of air, e = E/m = 0.5 v² = (0.5)(2.2352)²
Wind energy per unit mass of air = 2.517 J/kg
PGP = mass flow rate * energy per unit mass
PGP = ρ*A*V*e

r = 210.64 m = 691.1 ft
Thus, the radius of a wind turbine is 691.1 ft
PGP = CVᵃ
For best design of wind turbine Betz limit (c) is taken between (0.35 - 0.45)
Let C = 0.4
PGP = Cvᵃ
take log of both sides
ln(PGP) = a*ln(CV)
a = ln(PGP)/ln(CV)
a = ln(1000)/ln(0.4 *2.2352) = 7.73
The power generation potential (PGP) scales with speed at the rate of 7.73 kW.s/m
Answer:
The correct option is;
D. Market...Command
Explanation:
In a mixed economy, socialism and capitalism are combined in the economic policy, such the economy allows ownership of and offers protection for private property as well as provision of certain degree of allowance of capital utilization by the private sector. The government of a country with a mixed economic system can disrupt, obstruct, or restrict the activities taking place in the economy in order to meet the social objectives of the country's.
Answer:
293 kg
Explanation:
Let's say the tension in each cable is Tb, Tc, and Td.
First, find the length of cable AD:
r = √(2² + 2² + 1²)
r = 3
Using similar triangles:
Tdx = 2/3 Td
Tdy = 2/3 Td
Tdz = 1/3 Td
Sum of the forces in the x direction:
∑F = ma
Tb − 2/3 Td = 0
Td = 3/2 Tb
Sum of the forces in the y direction:
∑F = ma
2/3 Td − Tc = 0
Td = 3/2 Tc
Sum of the forces in the z direction:
∑F = ma
1/3 Td − mg = 0
Td = 3mg
From the first two equations, we know Td is greater than Tb or Tc. So we need to set Td to 8.6 kN, or 8600 N.
8600 N = 3mg
m = 8600 N / (3 × 9.8 m/s²)
m ≈ 292.5 kg
Rounded to three significant figures, the maximum mass of the crate is 293 kg.