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:
When the uneven burning of the fuel takes place due to the incorrect air/fuel mixture inside the engine cylinder, a knocking sound is observed. This is called as the engine knocking.
Explanation:
When the uneven burning of the fuel takes place due to the incorrect air/fuel mixture inside the engine cylinder, a knocking sound is observed. This is called as the engine knocking.
The engine knock problem can be caused due to the following reason
a) When the octane rating of the fuel used is low.
b) The deposition of the carbon around the cylinder walls takes place.
c) The spark plug used in the vehicle is not correct.
Answer:
The most damaged people are the wisest is a fact
Explanation:
Answer:
a)W=12.62 kJ/mol
b)W=12.59 kJ/mol
Explanation:
At T = 100 °C the second and third virial coefficients are
B = -242.5 cm^3 mol^-1
C = 25200 cm^6 mo1^-2
Now according isothermal work of one mole methyl gas is
W=-
a=
b=
from virial equation
And
a=
b=
Now calculate V1 and V2 at given condition
Substitute given values 
= 1 x 10^5 , T = 373.15 and given values of coefficients we get
Solve for V1 by iterative or alternative cubic equation solver we get
Similarly solve for state 2 at P2 = 50 bar we get
Now
a=241.33
b=30780
After performing integration we get work done on the system is
W=12.62 kJ/mol
(b) for Z = 1 + B' P +C' P^2 = PV/RT by performing differential we get
dV=RT(-1/p^2+0+C')dP
Hence work done on the system is
a=
b=
by substituting given limit and P = 1 bar , P2 = 50 bar and T = 373 K we get work
W=12.59 kJ/mol
The work by differ between a and b because the conversion of constant of virial coefficients are valid only for infinite series
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
