Material engineering studies the physical behavior of metallic elements.
Answer: Option C
<u>Explanation:
</u>
Material Engineering is the creation and learning about the materials at an atomic level. An engineers from this branch focus on material and model its characteristics using the computer.
Also, they combine the knowledge of solid-states, metallurgy, chemistry and ceramics to the application level. It also has a great role in building the future with the advancing study in nanotechnology, biotechnology, etc. Simply, these are meant to have vivid applications in future life.
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:
the difference in pressure between the inside and outside of the droplets is 538 Pa
Explanation:
given data
temperature = 68 °F
average diameter = 200 µm
to find out
what is the difference in pressure between the inside and outside of the droplets
solution
we know here surface tension of carbon tetra chloride at 68 °F is get from table 1.6 physical properties of liquid that is
σ = 2.69 × 
N/m
so average radius = 
= 100 µm = 100 ×
m
now here we know relation between pressure difference and surface tension
so we can derive difference pressure as
2π×σ×r = Δp×π×r² .....................1
here r is radius and Δp pressure difference and σ surface tension
Δp = 
put here value
Δp = 
Δp = 538
so the difference in pressure between the inside and outside of the droplets is 538 Pa
Answer:
The flexural strength of a specimen is = 78.3 M pa
Explanation:
Given data
Height = depth = 5 mm
Width = 10 mm
Length L = 45 mm
Load = 290 N
The flexural strength of a specimen is given by
78.3 M pa
Therefore the flexural strength of a specimen is = 78.3 M pa
