Answer:
Shearing stresses are the stresses generated in any material when a force acts in such a way that it tends to tear off the material.
Generally the above definition is valid at an armature level, in more technical terms shearing stresses are the component of the stresses that act parallel to any plane in a material that is under stress. Shearing stresses are present in a body even if normal forces act on it along the centroidal axis.
Mathematically in a plane AB the shearing stresses are given by
Yes the shearing force which generates the shearing stresses is similar to frictional force that acts between the 2 surfaces in contact with each other.
Answer:
- def median(l):
- if(len(l) == 0):
- return 0
- else:
- l.sort()
- if(len(l)%2 == 0):
- index = int(len(l)/2)
- mid = (l[index-1] + l[index]) / 2
- else:
- mid = l[len(l)//2]
- return mid
-
- def mode(l):
- if(len(l)==0):
- return 0
-
- mode = max(set(l), key=l.count)
- return mode
-
- def mean(l):
- if(len(l)==0):
- return 0
- sum = 0
- for x in l:
- sum += x
- mean = sum / len(l)
- return mean
-
- lst = [5, 7, 10, 11, 12, 12, 13, 15, 25, 30, 45, 61]
- print(mean(lst))
- print(median(lst))
- print(mode(lst))
Explanation:
Firstly, we create a median function (Line 1). This function will check if the the length of list is zero and also if it is an even number. If the length is zero (empty list), it return zero (Line 2-3). If it is an even number, it will calculate the median by summing up two middle index values and divide them by two (Line 6-8). Or if the length is an odd, it will simply take the middle index value and return it as output (Line 9-10).
In mode function, after checking the length of list, we use the max function to estimate the maximum count of the item in list (Line 17) and use it as mode.
In mean function, after checking the length of list, we create a sum variable and then use a loop to add the item of list to sum (Line 23-25). After the loop, divide sum by the length of list to get the mean (Line 26).
In the main program, we test the three functions using a sample list and we shall get
20.5
12.5
12
Answer:
ω=314.15 rad/s.
0.02 s.
Explanation:
Given that
Motor speed ,N= 3000 revolutions per minute
N= 3000 RPM
The speed of the motor in rad/s given as
Now by putting the values in the above equation
ω=314.15 rad/s
Therefore the speed in rad/s will be 314.15 rad/s.
The speed in rev/sec given as
ω= 50 rev/s
It take 1 sec to cover 50 revolutions
That is why to cover 1 revolution it take
