What drives up the cost of consumables?
1 answer:
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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: hello the complete question is attached below
answer:
A) Group symbol = SW
B) Group name = well graded sand , fine to coarse sand
C) It is not a clean sand given that ≤ 50% particles are retained on No 200
Explanation:
<u>A) Classifying the soil according to USCS system</u>
( using 2nd image attached below )
<em>description of sand</em> :
The soil is a coarse sand since ≤ 50% particles are retained on No 200 sieve, also
The soil is a sand given that more than 50% particles passed from No 4 sieve
The soil can be a clean sand given that fines ≤ 12%
The soil can be said to be a well graded sand because the percentage of particles passing through decreases gradually over time
Group symbol as per the 2nd image attached below = SW
B) Group name = well graded sand , fine to coarse sand
C) It is not a clean sand given that ≤ 50% particles are retained on No 200
Answer:
1) The three possible assumptions are
a) All processes are reversible internally
b) Air, which is the working fluid circulates continuously in a closed loop
cycle
c) The process of combustion is depicted as a heat addition process
2) The diagrams are attached
5) The net work per cycle is 845.88 kJ/kg
The power developed in horsepower ≈ 45374 hP
Explanation:
1) The three possible assumptions are
a) All processes are reversible internally
b) Air, which is the working fluid circulates continuously in a closed loop
cycle
c) The process of combustion is depicted as a heat addition process
2) The diagrams are attached
5) The dimension of the cylinder bore diameter = 3.7 in. = 0.09398 m
Stroke length = 3.4 in. = 0.08636 m.
The volume of the cylinder v₁= 0.08636 ×(0.09398²)/4 = 5.99×10⁻⁴ m³
The clearance volume = 16% of cylinder volume = 0.16×5.99×10⁻⁴ m³
The clearance volume, v₂ = 9.59 × 10⁻⁵ m³
p₁ = 14.5 lbf/in.² = 99973.981 Pa
T₁ = 60 F = 288.706 K
Otto cycle T-S diagram
T₂ = 288.706* = 592.984 K
The maximum temperature = T₃ = 5200 R = 2888.89 K
T₄ = 2888.89 / = 1406.5 K
Work done, W = ×(T₃ - T₂) -
×(T₄ - T₁)
0.718×(2888.89 - 592.984) - 0.718×(1406.5 - 288.706) = 845.88 kJ/kg
The power developed in an Otto cycle = W×Cycle per second
= 845.88 × 2400 / 60 = 33,835.377 kW = 45373.99 ≈ 45374 hP.
