Answer:
See attachment and explanation.
Explanation:
- The following question can be solved better with the help of a MATLAB program as follows. The code is given in the attachment.
- The plot of the graph is given in attachment.
- The code covers the entire spectrum of the poly-tropic range ( 1.2 - 1.6 ) and 20 steps ( cases ) have been plotted and compared in the attached plot.
Question:
The question is not complete. See the complete question and the answer below.
A well that pumps at a constant rate of 0.5m3/s fully penetrates a confined aquifer of 34 m thickness. After a long period of pumping, near steady state conditions, the measured drawdowns at two observation wells 50m and 100m from the pumping well are 0.9 and 0.4 m respectively. (a) Calculate the hydraulic conductivity and transmissivity of the aquifer (b) estimate the radius of influence of the pumping well, and (c) calculate the expected drawdown in the pumping well if the radius of the well is 0.4m.
Answer:
T = 0.11029m²/sec
Radius of influence = 93.304m
expected drawdown = 3.9336m
Explanation:
See the attached file for the explanation.
Answer:
The code is attached.
Explanation:
I created a string s including 6 colors with spaces in between. Then I converted the string into a list x by using split() method. I used three different methods for removing elements from the list. These methods are remove(), pop() and del.
Then I used methods append(), insert() and extend() for adding elements to the list.
Finally I converted list into a string using join() and adding space in between the elements of the list.
Answer:
maximum speed for safe vehicle operation = 55mph
Explanation:
Given data :
radius ( R ) = 678 ft
old building located ( m )= 30 ft
super elevation = 0.06
<u>Determine the maximum speed for safe vehicle operation </u>
firstly calculate the stopping sight distance
m = R ( 1 - cos ) ---- ( 1 )
R = 678
m ( horizontal sightline ) = 30 ft
back to equation 1
30 = 678 ( 1 - cos (28.655 *s / 678 ) )
( 1 - cos (28.655 *s / 678 ) ) = 30 / 678 = 0.044
cos = 1.044
hence ; 28.65 * s = 678 * 0.2956
s = 6.99 ≈ 7 ft
next we will calculate the design speed ( u ) using the formula below
S = 1.47 ut + ---- ( 2 )
t = reaction time, a = vehicle acceleration, G1 = grade percentage
assuming ; t = 2.5 sec , a = 11.2 ft/sec^2, G1 = 0
back to equation 2
6.99 = 1.47 * u * 2.5 +
3.675 u + 0.0958 u^2 - 6.99 = 0
u ( 3.675 + 0.0958 u ) = 6.99