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:
R = 31.9 x 10^(6) At/Wb
So option A is correct
Explanation:
Reluctance is obtained by dividing the length of the magnetic path L by the permeability times the cross-sectional area A
Thus; R = L/μA,
Now from the question,
L = 4m
r_1 = 1.75cm = 0.0175m
r_2 = 2.2cm = 0.022m
So Area will be A_2 - A_1
Thus = π(r_2)² - π(r_1)²
A = π(0.0225)² - π(0.0175)²
A = π[0.0002]
A = 6.28 x 10^(-4) m²
We are given that;
L = 4m
μ_steel = 2 x 10^(-4) Wb/At - m
Thus, reluctance is calculated as;
R = 4/(2 x 10^(-4) x 6.28x 10^(-4))
R = 0.319 x 10^(8) At/Wb
R = 31.9 x 10^(6) At/Wb
Answer:
The final velocity of the rocket is 450 m/s.
Explanation:
Given;
initial velocity of the rocket, u = 0
constant upward acceleration of the rocket, a = 18 m/s²
time of motion of the rocket, t = 25 s
The final velocity of the rocket is calculated with the following kinematic equation;
v = u + at
where;
v is the final velocity of the rocket after 25 s
Substitute the given values in the equation above;
v = 0 + 18 x 25
v = 450 m/s
Therefore, the final velocity of the rocket is 450 m/s.
Answer:
The average thickness of the blubber is<u> 0.077 m</u>
Explanation:
Here, we want to calculate the average thickness of the Walrus blubber.
We employ a mathematical formula to calculate this;
The rate of heat transfer(H) through the Walrus blubber = dQ/dT = KA(T2-T1)/L
Where dQ is the change in amount of heat transferred
dT is the temperature gradient(change in temperature) i.e T2-T1
dQ/dT = 220 W
K is the conductivity of fatty tissue without blood = 0.20 (J/s · m · °C)
A is the surface area which is 2.23 m^2
T2 = 37.0 °C
T1 = -1.0 °C
L is ?
We can rewrite the equation in terms of L as follows;
L × dQ/dT = KA(T2-T1)
L = KA(T2-T1) ÷ dQ/dT
Imputing the values listed above;
L = (0.2 * 2.23)(37-(-1))/220
L = (0.2 * 2.23 * 38)/220 = 16.948/220 = 0.077 m
