(E. Call the hospital to take them away
Answer:
hello below is missing piece of the complete question
minimum size = 0.3 cm
answer : 0.247 N/mm2
Explanation:
Given data :
section span : 10.9 and 13.4 cm
minimum load applied evenly to the top of span : 13 N
maximum load for each member ; 4.5 N
lets take each member to be 4.2 cm
Determine the max value of P before truss fails
Taking average value of section span ≈ 12 cm
Given minimum load distributed evenly on top of section span = 13 N
we will calculate the value of by applying this formula
=
= 1.56 * 10^-5
next we will consider section ; 4.2 cm * 0.3 cm
hence Z (section modulus ) = BD^2 / 6
= ( 0.042 * 0.003^2 ) / 6 = 6.3*10^-8
Finally the max value of P( stress ) before the truss fails
= M/Z = ( 1.56 * 10^-5 ) / ( 6.3*10^-8 )
= 0.247 N/mm2
Answer:
n=2.32
w= -213.9 KW
Explanation:
![V_1=0.3m^3,T_1=298 K](https://tex.z-dn.net/?f=V_1%3D0.3m%5E3%2CT_1%3D298%20K)
![V_2=0.1m^3,T_1=1273 K](https://tex.z-dn.net/?f=V_2%3D0.1m%5E3%2CT_1%3D1273%20K)
Mass of air=1 kg
For polytropic process
,n is the polytropic constant.
![Tv^{n-1}=C](https://tex.z-dn.net/?f=Tv%5E%7Bn-1%7D%3DC)
![T_1v^{n-1}_1=T_2v^{n-1}_2](https://tex.z-dn.net/?f=T_1v%5E%7Bn-1%7D_1%3DT_2v%5E%7Bn-1%7D_2)
![298\times .3^{n-1}_1=1273\times .1^{n-1}_2](https://tex.z-dn.net/?f=298%5Ctimes%20.3%5E%7Bn-1%7D_1%3D1273%5Ctimes%20.1%5E%7Bn-1%7D_2)
n=2.32
Work in polytropic process given as
w=![\dfrac{P_1V_1-P_2V_2}{n-1}](https://tex.z-dn.net/?f=%5Cdfrac%7BP_1V_1-P_2V_2%7D%7Bn-1%7D)
w=![mR\dfrac{T_1-T_2}{n-1}](https://tex.z-dn.net/?f=mR%5Cdfrac%7BT_1-T_2%7D%7Bn-1%7D)
Now by putting the values
w=![1\times 0.287\dfrac{289-1273}{2.32-1}](https://tex.z-dn.net/?f=1%5Ctimes%200.287%5Cdfrac%7B289-1273%7D%7B2.32-1%7D)
w= -213.9 KW
Negative sign indicates that work is given to the system or work is done on the system.
For T_V diagram
We can easily observe that when piston cylinder reach on new position then volume reduces and temperature increases,so we can say that this is compression process.
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:A rectangular region ABCD is to be built inside a semicircle of radius 10 m with points A and B on the line for the diameter and points C and D on the semicircle with CD parallel to AB. The objective is to find the height h * that maximizes the area of ABCD.
Formulate the optimization problem.
Explanation:A rectangular region ABCD is to be built inside a semicircle of radius 10 m with points A and B on the line for the diameter and points C and D on the semicircle with CD parallel to AB. The objective is to find the height h * that maximizes the area of ABCD.
Formulate the optimization problem.