Answer: LED have the lowest cost of operation.
Explanation:
If we ignore the initial procurement cost of the items the operational cost of any device consuming electricity is given by
Among the three item's LED consumes the lowest power to give the same level of brightness as compared to the other 2 item's thus LED's shall have the lowest operational cost.
Technician A is correct. Technician B is wrong because a gear's transmission is used to increase or decrease torque.
The relation torque is relying on multiplying the circumferential detail with the resource of the usage of the radius; massive gears experience a greater amount of torque, at the same time as smaller gears experience a great deal much less torque. Similarly, the torque ratio is equal to the ratio of the gears' radii. A gear's transmission torque modifications as it will boom or decreases speed. Commonly, with the resource of the usage of lowering the speed, a small torque on the doorway issue is transferred as a massive torque at the output issue. The calculation of torque is quantified with the resource of the usage of an extensive form of teeth.
Learn more about the torque at brainly.com/question/28220969
#SPJ4
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:
hshdhriwjajaldh skshdjdywuusg
Explanation:
null
Answer: At time 18.33 seconds it will have moved 500 meters.
Explanation:
Since the acceleration of the car is a linear function of time it can be written as a function of time as
Integrating both sides we get
Now since car starts from rest thus at time t = 0 ; v=0 thus c=0
again integrating with respect to time we get
Now let us assume that car starts from origin thus D=0
thus in the first 15 seconds it covers a distance of
Thus the remaining 125 meters will be covered with a constant speed of
in time equalling 
Thus the total time it requires equals 15+3.33 seconds
t=18.33 seconds
