Answer: hello your question is incomplete below is the complete question
answer:
N010 GO2 X7.0 Y2.0 15.0 J2.0 ( option 1 )
Explanation:
Given that the NC machining has to be moved from point ( 5,4 ) to point ( 7,2 ) along a circular path
GO2 = circular interpolation in a clockwise path
G91 = incremental dimension
<em>hence the correct option is </em>:
N010 GO2 X7.0 Y2.0 15.0 J2.0
Answer:
// Program is written in C++
// Comments are used to explain some lines
// Only the required function is written. The main method is excluded.
#include<bits/stdc++.h>
#include<iostream>
using namespace std;
int divSum(int num)
{
// The next line declares the final result of summation of divisors. The variable declared is also
//initialised to 0
int result = 0;
// find all numbers which divide 'num'
for (int i=2; i<=(num/2); i++)
{
// if 'i' is divisor of 'num'
if (num%i==0)
{
if (i==(num/i))
result += i; //add divisor to result
else
result += (i + num/i); //add divisor to result
}
}
cout<<result+1;
}
Answer:

Explanation:
From the question we are told that:
Thickness 
Internal Pressure
Shear stress 
Elastic modulus 
Generally the equation for shear stress is mathematically given by

Where
r_i=internal Radius
Therefore


Generally



Generally the equation for outer diameter is mathematically given by


Therefore
Assuming that the thin cylinder is subjected to integral Pressure
Outer Diameter is

Answer:
hello your question is incomplete attached below is the missing equation related to the question
answer : 40.389° , 38.987° , 38° , 39.869° , 40.265°
Explanation:
<u>Determine the friction angle at each depth</u>
attached below is the detailed solution
To calculate the vertical stress = depth * unit weight of sand
also inverse of Tan = Tan^-1
also qc is in Mpa while σ0 is in kPa
Friction angle at each depth
2 meters = 40.389°
3.5 meters = 38.987°
5 meters = 38.022°
6.5 meters = 39.869°
8 meters = 40.265°
Answer:
38 kJ
Explanation:
The solution is obtained using the energy balance:
ΔE=E_in-E_out
U_2-U_1=Q_in+W_in-Q_out
U_2=U_1+Q_in+W_in-Q_out
=38 kJ