Answer:
a) 28 stations
b) Rp = 21.43
E = 0.5
Explanation:
Given:
Average downtime per occurrence = 5.0 min
Probability that leads to downtime, d= 0.01
Total work time, Tc = 39.2 min
a) For the optimum number of stations on the line that will maximize production rate.
Maximizing Rp =minimizing Tp
Tp = Tc + Ftd
At minimum pt. = 0, we have:
dTp/dn = 0
Solving for n²:
The optimum number of stations on the line that will maximize production rate is 28 stations.
b) 
Tp = 1.4 +1.4 = 2.8
The production rate, Rp =
The proportion uptime,
Answer:
Taking responsibility for your own learning makes it easier to identify your strengths and weaknesses. Once these have been identified you can work on a learning plan that focuses on the areas that you need most help with, increasing the speed of your learning, and build the skills you have been trying to perfect.
Explanation:
Answer:
The code is given below in Python with appropriate comments
Explanation:
# convert list to set
male_names = set(['Oliver','Declan','Henry'])
# get remove and add name from user
remove_name = input("Enter remove name: ")
add_name = input("Enter add name: ")
# remove name from set
male_names.remove(remove_name)
# add new name ij set
male_names.add(add_name)
# sort the set
a = sorted(male_names)
# print the set
print(a)
Answer:
Fossil Fuels 67% (Non-Renewable Source): Coal 41%, Natural Gas 21% & Oil 5.1%
Renewable Energy 16%
Mainly Hydroelectric 92%: Wind 6%, Geothermal 1%, Solar 1%
Nuclear Power 13%
Explanation:
Answer:
Explanation:
(a) Given that 620g moisture and 330g decomposable organic matter in yard trimming is represented by C₁₂.₇₆H₂₁.₂₈O₉.₂₆N₀.₅₄
Given the atomic mass of Carbon C = 12, Hydrogen H = 1, Oxygen O = 16 and Nitrogen N = 14
1 mole of trimming = 12*12.76 + 1*21.28 + 16*9.26 + 14*0.54
= 153.12 + 21.28 + 148.16 + 7.56
= 330.12 g/mol
which means 1 kg of as received trimming has 330 g of decomposable that produce 1 mole of decomposable
The moles of methane produced will be given as
m = (4a + b -2c - 3d)/8
= (4*12.76 + 21.28 - 2*9.26 - 3*0.54)/8
= (51.04 + 21.28 - 18.52 - 1.62)/8
= 52.18/8
= 6.5225
(b) Volume of methane V is given as
V = (0.0224 m³ CH₄mol/CH₄) × (6.5225 mol CH₄/ kg)
= 0.1461 m³ CH₄/kg lawn trimmings
(c) Energy will be given as
CH₄Energy = 6.5225 mol of CH₄/kg × 890 kJ/mol
= 5805.025
≈ 5805 kJ/kg
