Answer:
V = 125.7m/min
Explanation:
Given:
L = 400 mm ≈ 0.4m
D = 150 mm ≈ 0.15m
T = 5 minutes
F = 0.30mm ≈ 0.0003m
To calculate the cutting speed, let's use the formula :

We are to find the speed, V. Let's make it the subject.

Substituting values we have:

V = 125.68 m/min ≈ 125.7 m/min
Therefore, V = 125.7m/min
Explanation:
First of all get the input from the user, number of rows and number of columns where rows represents seat digit number and column represents the seat letter
rows is initialized to 1 to ensure that row starts at 1 or you can remove it then seat number will start from 0.
The first loop is used for digits starting from 1 to number of rows
The second loop is used for letters starting from 1 to number of columns
since rows and cols are not of the same type that's why we are converting the int type to string type
print(str(rows)+cols) counter will keep updating the columns A, B, C.....
rows= rows + 1 counter will keep updating the rows 1, 2, 3....
Code:
Please refer to the attached image.
Output:
Please enter the number of rows: 2
Please enter the number of columns: 3
1A
1B
1C
2A
2B
2C
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:
You need a 120V to 24V commercial transformer (transformer 1:5), a 100 ohms resistance, a 1.5 K ohms resistance and a diode with a minimum forward current of 20 mA (could be 1N4148)
Step by step design:
- Because you have a 120V AC voltage supply you need an efficient way to reduce that voltage as much as possible before passing to the rectifier, for that I recommend a standard 120V to 24V transformer. 120 Vrms = 85 V and 24 Vrms = 17V = Vin
- Because 17V is not 15V you still need a voltage divider to step down that voltage, for that we use R1 = 100Ω and R2 = 1.3KΩ. You need to remember that more than 1 V is going to be in the diode, so for our calculation we need to consider it. Vf = (V*R2)/(R1+R2), V = Vin - 1 = 17-1 = 16V and Vf = 15, Choosing a fix resistance R1 = 100Ω and solving the equation we find R2 = 1.5KΩ
- Finally to select the diode you need to calculate two times the maximum current and that would be the forward current (If) of your diode. Imax = Vf/R2 = 10mA and If = 2*Imax = 20mA
Our circuit meet the average voltage (Va) specification:
Va = (15)/(pi) = 4.77V considering the diode voltage or 3.77V without considering it