Question:
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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.
Algorithm of the Nios II assembly program.
- Attain data for simulation from the SW11-0, on the DE2-115 Simulator
- The data will be read from the switches in loop.
- The decimal output is displayed using the seven-segment displays and done using the loop.
- The program is ended by the user operating the SW1 switch
and
The decimal equivalent on the seven-segment displays HEX3-0 is
- DE2-115
- DE2-115_SW11
- DE2-115_HEX3
- DE2-115_HEX4
- DE2-115_HEX5
- DE2-115_HEX6
- DE2-115_HEX7
<h3>The Algorithm and
decimal equivalent on the
seven-segment displays HEX3-0</h3>
Generally, the program will be written using a cpulator simulator in order to attain best result.
We are to
- Attain data for simulation from the SW11-0, on the DE2-115 Simulator
- The data will be read from the switches in loop.
- The decimal output is displayed using the seven-segment displays and done using the loop.
- The program is ended by the user operating the SW1 switch
This will be the Algorithm of the Nios II assembly program .
Hence, the decimal equivalent on the seven-segment displays HEX3-0 is
- DE2-115
- DE2-115_SW11
- DE2-115_HEX3
- DE2-115_HEX4
- DE2-115_HEX5
- DE2-115_HEX6
- DE2-115_HEX7
For more information on Algorithm
brainly.com/question/11623795
Answer:
Angle grinders are used mostly for copper, iron, steel, lead, and other metals.
Explanation:
I hope it helps! Have a great day!
Lilac~
Answer:
R = 31.9 x 10^(6) At/Wb
So option A is correct
Explanation:
Reluctance is obtained by dividing the length of the magnetic path L by the permeability times the cross-sectional area A
Thus; R = L/μA,
Now from the question,
L = 4m
r_1 = 1.75cm = 0.0175m
r_2 = 2.2cm = 0.022m
So Area will be A_2 - A_1
Thus = π(r_2)² - π(r_1)²
A = π(0.0225)² - π(0.0175)²
A = π[0.0002]
A = 6.28 x 10^(-4) m²
We are given that;
L = 4m
μ_steel = 2 x 10^(-4) Wb/At - m
Thus, reluctance is calculated as;
R = 4/(2 x 10^(-4) x 6.28x 10^(-4))
R = 0.319 x 10^(8) At/Wb
R = 31.9 x 10^(6) At/Wb
E. Parts they don’t resemble
