Answer:
The answer is "
".
Explanation:
Please find the correct question in the attachment file.
using formula:



Answer:
The fluid level difference in the manometer arm = 22.56 ft.
Explanation:
Assumption: The fluid in the manometer is incompressible, that is, its density is constant.
The fluid level difference between the two arms of the manometer gives the gage pressure of the air in the tank.
And P(gage) = ρgh
ρ = density of the manometer fluid = 60 lbm/ft³
g = acceleration due to gravity = 32.2 ft/s²
ρg = 60 × 32.2 = 1932 lbm/ft²s²
ρg = 1932 lbm/ft²s² × 1lbf.s²/32.2lbm.ft = 60 lbf/ft³
h = fluid level difference between the two arms of the manometer = ?
P(gage) = 9.4 psig = 9.4 × 144 = 1353.6 lbf/ft²
1353.6 = ρg × h = 60 lbf/ft³ × h
h = 1353.6/60 = 22.56 ft
A diagrammatic representation of this setup is presented in the attached image.
Hope this helps!
Answer:
- using System;
- public class Program
- {
- public static void Main()
- {
- Console.WriteLine("Enter number of students: ");
- int num = Convert.ToInt32(Console.ReadLine());
- string [] firstName = new string[num];
- string [] lastName = new string[num];
-
- for(int i=0 ; i < num; i++){
- Console.WriteLine("Enter first name: ");
- firstName[i] = Console.ReadLine();
-
- Console.WriteLine("Enter last name: ");
- lastName[i] = Console.ReadLine();
- }
-
- for(int j=0; j < num; j++){
- Console.WriteLine(lastName[j] + "," + firstName[j]);
- }
- }
- }
Explanation:
Firstly, prompt user to enter number of student to be stored (Line 6- 7). Next, create two array, firstName and lastName with num size (Line 8-9).
Create a for-loop to repeat for num times and prompt user to enter first name and last name and then store them in the firstName and lastName array, respectively (Line 11 - 17).
Create another for loop to traverse through the lastName and firstName array and display the last name and first name by following the format given in the question (Line 19 - 21).
Answer:
q = 1.73 W
Explanation:
given data
small end = 5 cm
large end = 10 cm
high = 15 cm
small end is held = 600 K
large end at = 300 K
thermal conductivity of asbestos = 0.173 W/mK
solution
first we will get here side of cross section that is express as
...............1
here x is distance from small end and S1 is side of square at small end
and S2 is side of square of large end and L is length
put here value and we get
S = 5 +
S =
m
and
now we get here Area of section at distance x is
area A = S² ...............2
area A =
m²
and
now we take here small length dx and temperature difference is dt
so as per fourier law
heat conduction is express as
heat conduction q =
...............3
put here value and we get
heat conduction q =
it will be express as
now we intergrate it with limit 0 to 0.15 and take temp 600 to 300 K
solve it and we get
q (30) = (0.173) × (600 - 300)
q = 1.73 W
Answer:
Departure rate = 7.65 vehicle/min
Explanation:
See the attached file for the calculation.