Answer:
attached below
Explanation:
Consider a drainage basin having 60% soil group A and 40% soil group B. Five years ago the land use pattern in the basin was ½ w
1 answer:
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Answer:
The time required is 10.078 hours or 605 min
Explanation:
The formula to apply here is ;
K=(d²-d²₀ )/t
where t is time in hours
d is grain diameter to be achieved after heating in mm
d₀ is the grain diameter before heating in mm
Given
d=5.5 × 10^-2 mm
d₀=2.4 × 10^-2 mm
t₁= 500 min = 500/60 =25/3 hrs
t₂=?
n=2.2
First find K
K=(d²-d²₀ )/t₁
K={ (5.1 × 10^-2 mm)²-(2.4 × 10−2 mm)² }/ 25/3
K=(0.051²-0.024²) ÷25/2
K=0.000243 mm²/h
Re-arrange equation for K ,to get the equation for d as;
d=√(d₀²+ Kt) where now t=t₂
Answer:
Check the explanation
Explanation:
Points to consider:
We need to take the input from the user
We need to find the manhatan distance and euclidian using the formula
(x1, y1) and (x2, y2) are the two points
Manhattan:
Euclidian Distance:
Code
#include<stdio.h>
#include<math.h>
struct Point{
int x, y;
};
int manhattan(Point A, Point B){
return abs(A.x - B.x) + abs(A.y- B.y);
}
float euclidean(Point A, Point B){
return sqrt(pow(A.x - B.x, 2) + pow(A.y - B.y, 2));
}
int main(){
struct Point A, B;
printf("Enter x and Y for first point: ");
int x, y;
scanf("%d%d", &x, &y);
A.x = x;
A.y = y;
printf("Enter x and Y for second point: ");
scanf("%d%d", &x, &y);
B.x = x;
B.y = y;
printf("Manhattan Distance: %d\n", manhattan(A, B));
printf("Euclidian Distance: %f\n", euclidean(A, B));
}
Sample output
