Answer:
The appropriate solution is "1481.76 N".
Explanation:
According to the question,
Mass,
m = 540 kg
Coefficient of static friction,
= 0.28
Now,
The applied force will be:
⇒ 
By substituting the values, we get
Answer:
It is a non profit organization that dedicates to licensing professional engineers and surveyors
Explanation:
Answer:
Machine 2 has a higher process capability index, it would be best considered for purchase.
Explanation:
Process capability index: Cpk= Min [(mean-L spec)/3sd; (U spec-mean)/3sd]
For machine 1, mean= 48mm and L spec= 46 and U spec= 50, Standard deviation sd= 0.7
Cpk= [0.952;0.952]= 0.952
For machine 2, mean= 47 and L spec= 46 and U spec= 50, Standard deviation sd= 0.3
Cpk= [1.111;3.333]= 1.111
It is clearly observed from the calculations above that the Cpk value of machine 2 is higher than that of machine 1.
Since machine 2 has a higher process capability index, it would be best considered for purchase.
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.
