Answer:
......................
Explanation:
vehicles are abreast of each other, the police car starts to pursue the automobile at
a constant acceleration of 1.96 m/s². The motorist noticed the police car in his rear
view mirror 12 s after the police car started the pursuit and applied his brakes and decelerates at 3.05 m/s². (Hint: The police will not go against the law.)
a) Find the total time required for the police car to overtake the automobile.
b) Find the total distance travelled by the police car while overtaking the
automobile.
c) Find the speed of the police car at the time it overtakes the automobile.
d) Find the speed of the automobile at the time it was overtaken by the police car.
Answer:
a.) Time = 17.13 seconds
b.) 31.88 m
c.) V = 11.18 m/s
d.) V = 7.1 m/s
Explanation:
The initial velocity U of the automobile is 15.65 m/s.
At the instant that the two vehicles are abreast of each other, the police car starts to pursue the automobile with initial velocity U = 0 at a constant acceleration of 1.96 m/s². Because the police is starting from rest.
For the automobile, let us use first equation of motion
V = U - at.
Acceleration a is negative since it is decelerating with a = 3.05 m/s² . And
V = 0.
Substitute U and a into the formula
0 = 15.65 - 3.05t
15.65 = 3.05t
t = 15.65/3.05
t = 5.13 seconds
But the motorist noticed the police car in his rear view mirror 12 s after the police car started the pursuit and applied his brakes and decelerates at 3.05 m/s².
The total time required for the police car to overtake the automobile will be
12 + 5.13 = 17.13 seconds.
b.) Using the third equation of motion formula for the police car at V = 11.18 m/s and a = 1.96 m/s²
V^2 = U^2 + 2aS
Where S = distance travelled.
Substitute V and a into the formula
11.18^2 = 0 + 2 × 1.96 ×S
124.99 = 3.92S
S = 124.99/3.92
S = 31.88 m
c.) The speed of the police car at the time it overtakes the automobile will be in line with the speed zone which is 11.18 m/s
d.) That will be the final velocity V of the automobile car.
We will use third equation of motion to solve that.
V^2 = U^2 + 2as
V^2 = 15.65^2 - 2 × 3.05 × 31.88
V^2 = 244.9225 - 194.468
V = sqrt( 50.4545)
V = 7.1 m/s
What type of search do you think this would be?
This type of search would be a linear search. A linear search involves looking at each item in the array one by one until the desired item is found.
In this case, you would look at each square in the array until you find the one that matches the object pictured on the left. This is a common search method for small arrays or when the array is unsorted.
The algorithm continues until the target element is found, or until it reaches the end of the array and fails to find the target element. Linear searches are useful in scenarios where the array is small and unsorted, as the algorithm does not need to compare every element in the array to find the target element.
Learn more about linear search:
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(A) contouring
(B) roughing out
(C) forming
(D) tapping out
Answer:
tapping out I think sorry if I'm wrong
commutative property of addition
identity property of multiplication
associative property of addition
commutative property of multiplication
The property of realnumbers is shown below is associative property of addition. The correct option is C.
According to the associativeproperty of addition, you can arrange the addends in several ways without changing the result.
According to the commutative property of addition, you can rearrange the addends without altering the result.
When more than two numbers are added together or multiplied, the outcome is always the same, regardless of how the numbers are arranged.
This is known as the associativeproperty. As an illustration, 2 (7 6) = (2 7) 6. 2 + (7 + 6) = (2 + 7) + 6.
Thus, the correct option is C.
For more details regarding associative property, visit:
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Answer:
C
Explanation:
Answer:
//This Program is written in C++
// Comments are used for explanatory purpose
#include <iostream>
using namespace std;
enum mailbox{open, close};
int box[149];
void closeAllBoxes();
void OpenClose();
void printAll();
int main()
{
closeAllBoxes();
OpenClose();
printAll();
return 0;
}
void closeAllBoxes()
{
for (int i = 0; i < 150; i++) //Iterate through from 0 to 149 which literarily means 1 to 150
{
box[i] = close; //Close all boxes
}
}
void OpenClose()
{
for(int i = 2; i < 150; i++) {
for(int j = i; j < 150; j += i) {
if (box[j] == close) //Open box if box is closed
box[j] = open;
else
box[j] = close; // Close box if box is opened
}
}
// At the end of this test, all boxes would be closed
}
void printAll()
{
for (int x = 0; x < 150; x++) //use this to test
{
if (box[x] = 1)
{
cout << "Mailbox #" << x+1 << " is closed" << endl;
// Print all close boxes
}
}
}
Explanation:
Answer:
Volume of aeration tank = 1.29 x 10^4 m³
Explanation:
Food/Micro- organism Ratio = 0.2/day
Feed Rate (Q) = 0.438 m³/s
Influent BOD = 150 mg/L
MLVSS = 2200 mg/L
The above mentioned parameters are related by the equation
F/M = QS₀/VX
where S₀ is the influent BOD and X is the cell concentration, with V being the Volume of the tank. Re- arranging the above equation for V and putting the values in, we get
V = 0.4380 x 150/0.2 x 2200
V = 0.1493 (m³/s) x day
V = 0.1493 x 24 x 60 x 60
V = 1.29 x 10^4 m³