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shutvik [7]
3 years ago
12

A debugging process where you, the programmer, pretend you are a computer and step through each statement while recording the va

lue of each variable at each step is known as
Engineering
1 answer:
Ipatiy [6.2K]3 years ago
5 0

Answer:

hand tracing

Explanation:

as a programmer when we pretend  computer in the  debugging process by the step of each statement in recording    

then there value of variable is hand tracing because as The hand tracking feature is the use of hands as an input method      

so while recording value of each variable each step is hand tracing

You might be interested in
The drag force, Fd, imposed by the surrounding air on a
Ad libitum [116K]

Answer:

a)  23.551 hp

b)  516.89 hp

Explanation:

<u>given:</u>

F_{d} =\frac{1}{2} C_{d} A_{p} V^{2} \\V_{a}=25 m/hr-->25*\frac{5280}{3600} =36.67ft/s\\V_{b}=70 m/hr-->70*\frac{5280}{3600} =102.67ft/s\\\\C_{d}=.28\\A=25 ft^2\\p=.075lb/ft^2

<u>required:</u>

the power in hp

<u>solution:</u>

(F_{d})_{a}  =\frac{1}{2} C_{d} A_{p} V_{a} ^{2}.............(1)

by substituting in the equation (1)

         =353.27 lbf

(F_{d})_{b}  =\frac{1}{2} C_{d} A_{p} V_{b} ^{2}..........(2)

by substituting in the equation (2)

         = 2769.29 lbf

power is defined by

             P=F.V

     P_{a}=353.27*36.67

           =12954.411 lbf.ft/s

           =12954.411*.001818

           =23.551 hp

      P_{a}=2769.29*102.67

           = 284323 lbf.ft/s

           = 284323*.001818

           = 516.89 hp

3 0
3 years ago
Select the creative imaging fields that require knowledge of programming.
denis23 [38]
Video game designer for sure
8 0
3 years ago
For this problem, you may not look at any other code or pseudo-code (even if it is on the internet), other than what is on our w
sergiy2304 [10]

Answer:

(a)

(i) pseudo code :-

current = i

// assuming parent of root is -1

while A[parent] < A[current] && parent != -1 do,

if A[parent] < A[current] // if current element is bigger than parent then shift it up

swap(A[current],A[parent])

current = parent

(ii) In heap we create a complete binary tree which has height of log(n). In shift up we will take maximum steps equal to the height of tree so number of comparison will be in term of O(log(n))

(b)

(i) There are two cases while comparing with grandparent. If grandparent is less than current node then surely parent node also will be less than current node so swap current node with parent and then swap parent node with grandparent.

If above condition is not true then we will check for parent node and if it is less than current node then swap these.

pseudo code :-

current = i

// assuming parent of root is -1

parent is parent node of current node

while A[parent] < A[current] && parent != -1 do,

if A[grandparent] < A[current] // if current element is bigger than parent then shift it up

swap(A[current],A[parent])

swap(A[grandparent],A[parent])

current = grandparent

else if A[parent] < A[current]

swap(A[parent],A[current])

current = parent

(ii) Here we are skipping the one level so max we can make our comparison half from last approach, that would be (height/2)

so order would be log(n)/2

(iii) C++ code :-

#include<bits/stdc++.h>

using namespace std;

// function to return index of parent node

int parent(int i)

{

if(i == 0)

return -1;

return (i-1)/2;

}

// function to return index of grandparent node

int grandparent(int i)

{

int p = parent(i);

if(p == -1)

return -1;

else

return parent(p);

}

void shift_up(int A[], int n, int ind)

{

int curr = ind-1; // because array is 0-indexed

while(parent(curr) != -1 && A[parent(curr)] < A[curr])

{

int g = grandparent(curr);

int p = parent(curr);

if(g != -1 && A[g] < A[curr])

{

swap(A[curr],A[p]);

swap(A[p],A[g]);

curr = g;

}

else if(A[p] < A[curr])

{

swap(A[p],A[curr]);

curr = p;

}

}

}

int main()

{

int n;

cout<<"enter the number of elements :-\n";

cin>>n;

int A[n];

cout<<"enter the elements of array :-\n";

for(int i=0;i<n;i++)

cin>>A[i];

int ind;

cout<<"enter the index value :- \n";

cin>>ind;

shift_up(A,n,ind);

cout<<"array after shift up :-\n";

for(int i=0;i<n;i++)

cout<<A[i]<<" ";

cout<<endl;

}

Explanation:

8 0
3 years ago
What is your understanding and opinion on the validity of the stand-by equipment maintenance requirement?
Fed [463]

Answer:

  The stand-by equipment is technically required in case where the we need some urgent equipment for the purpose of maintenance in emergency and if the other equipment system get fails. the term stand by means backup equipment and component.

In the reliability engineering, we always need to provide an extra equipment or component in case of emergency. It is basically used so that it does not affect any type of productivity in the organization. It is also increase the redundancy of the equipment.

6 0
3 years ago
The less surface area of gasoline exposed to the air, the faster a given amount will burn. True of False
romanna [79]

Answer:

true

Explanation:

it depends on how fast the car goes because the gas burns faster or slower depending on the speed you go or drive. if your drive 50 MPH, the gas will burn slowly. if your drive 45 MPH, the gas will burn faster if you go slow too much. if you press the glass button and hold it for 5 minutes, the more you hold the gas for too long, the more the gas will burn inside the gas engine. go look at your car or your parents car and see how it goes and that will help.

4 0
3 years ago
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