Answer:
insert (array[] , value , currentsize , maxsize )
{
if maxsize <=currentsize
{
return -1
}
index = currentsize-1
while (i>=0 && array[index] > value)
{
array[index+1]=array[index]
i=i-1
}
array[i+1]=value
return 0
}
Explanation:
1: Check if array is already full, if it's full then no component may be inserted.
2: if array isn't full:
- Check parts of the array ranging from last position of range towards initial range and determine position of that initial range that is smaller than the worth to be inserted.
- Right shift every component of the array once ranging from last position up to the position larger than the position at that smaller range was known.
- assign new worth to the position that is next to the known position of initial smaller component.
Answer:
Check the explanation
Explanation:
Determine the weight o ids in the each truck °slim the relation,
W,=
Here, W is net weight of soil and water on the truck and w is water content
substitute 72 7 kN for W and 15% for w.
you will need to also determine the number of truck loads required using the relation:
Number of truck loads required = 
Kindly check the attached image below for the full explanation to the question above.
Answer:
R= 53.7 Ω
Explanation:
fc= 1/(2πRC)
3000=1/(2π× 1 × 10^-6 × R)
R= 53.7 Ω
