Answer:
h_f = 15 ft, so option A is correct
Explanation:
The formula for head loss is given by;
h_f = [10.44•L•Q^(1.85)]/(C^(1.85))•D^(4.8655))
Where;
h_f is head loss due to friction in ft
L is length of pipe in ft
Q is flow rate of water in gpm
C is hazen Williams constant
D is diameter of pipe in inches
We are given;
L = 1,800 ft
Q = 600 gpm
C = 120
D = 8 inches
So, plugging in these values into the equation, we have;
h_f = [10.44*1800*600^(1.85)]/(120^(1.85))*8^(4.8655))
h_f = 14.896 ft.
So, h_f is approximately 15 ft
flip flops are shoes and they look like shoes
Answer:
e.Fire resistance,Inexpensive,Non-toxic.
Explanation:
Desirable hydraulic property of fluid as follows
1. Good chemical and environment stability
2. Low density
3. Ideal viscosity
4. Fire resistance
5. Better heat dissipation
6. Low flammability
7. Good lubrication capability
8. Low volatility
9. Foam resistance
10. Non-toxic
11. Inexpensive
12. Demulsibility
13. Incompressibility
So our option e is right.
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
Answer:
a.) -147V
b.) -120V
c.) 51V
Explanation:
a.) Equation for potential difference is the integral of the electrical field from a to b for the voltage V_ba = V(b)-V(a).
b.) The problem becomes easier to solve if you draw out the circuit. Since potential at Q is 0, then Q is at ground. So voltage across V_MQ is the same as potential at V_M.
c.) Same process as part b. Draw out the circuit and you'll see that the potential a point V_N is the same as the voltage across V_NP added with the 2V from the other box.
Honestly, these things take practice to get used to. It's really hard to explain this.