Answer:
These drive fittings come in four common sizes: 1⁄4 inch, 3⁄8 inch, 1⁄2 inch, and 3⁄4 inch (referred to as "drives", as in "3⁄8 drive").
The answer is number 2) Increase the resistance of the concrete to freeze-thaw damage.
Answer:
The solution code is written in Python:
- def convertCSV(number_list):
- str_list = []
- for num in number_list:
- str_list.append(str(num))
-
- return ",".join(str_list)
- result = convertCSV([22,33,44])
- print(result)
Explanation:
Firstly, create a function "convertCSV" with one parameter "number_list". (Line 1)
Next, create an empty list and assign it to a new variable <em>str_list</em>. (Line 2)
Use for-loop to iterate through all the number in the <em>number_list</em>.(Line 4). Within the loop, each number is converted to a string using the Python built-in function <em>str() </em>and then use the list append method to add the string version of the number to <em>str_list</em>.
Use Python string<em> join() </em>method to join all the elements in the str_list as a single string. The "," is used as a separator between the elements (Line 7) . At the end return the string as an output.
We can test the function by calling the function and passing [22,33,34] as an argument and we shall see "22,33,44" is printed as an output. (Line 9 - 10)
Answer:
Explanation:
First we calculate the mass of the aire inside the rigid tank in the initial and end moments.
(i could be 1 for initial and 2 for the end)
State1
State2
So, the total mass of the aire entered is
At this point we need to obtain the properties through the tables, so
For Specific Internal energy,
For Specific enthalpy
For the second state the Specific internal Energy (6bar, 350K)
At the end we make a Energy balance, so
No work done there is here, so clearing the equation for Q
The sign indicates that the tank transferred heat<em> to</em> the surroundings.
Answer:
M2 = 0.06404
P2 = 2.273
T2 = 5806.45°R
Explanation:
Given that p1 = 10atm, T1 = 1000R, M1 = 0.2.
Therefore from Steam Table, Po1 = (1.028)*(10) = 10.28 atm,
To1 = (1.008)*(1000) = 1008 ºR
R = 1716 ft-lb/slug-ºR cp= 6006 ft-lb/slug-ºR fuel-air ratio (by mass)
F/A =???? = FA slugf/slugaq = 4.5 x 108ft-lb/slugfx FA slugf/sluga = (4.5 x 108)FA ft-lb/sluga
For the air q = cp(To2– To1)
(Exit flow – inlet flow) – choked flow is assumed For M1= 0.2
Table A.3 of steam table gives P/P* = 2.273,
T/T* = 0.2066,
To/To* = 0.1736 To* = To2= To/0.1736 = 1008/0.1736 = 5806.45 ºR Gives q = cp(To* - To) = (6006 ft-lb/sluga-ºR)*(5806.45 – 1008)ºR = 28819500 ft-lb/slugaSetting equal to equation 1 above gives 28819500 ft-lb/sluga= FA*(4.5 x 108) ft-lb/slugaFA =
F/A = 0.06404 slugf/slugaor less to prevent choked flow at the exit
