Question 9.1 from the textbook. Consider the following workload: Process Burst Time Priority Arrival Time P1 50 4 0 P2 20 1 20 P
3 100 3 40 P4 40 2 60 Show the schedule using shortest remaining time, non-preemptive priority (a smaller number implies higher priority) and round robin with quantum 30, Make each entry in your answer equal to 10 time units. For example if your algorithm was FCFS then your answer would be: P1 P1 P1 P1 P1 P2 P2 P3 P3 P3 P3 P3 P3 P3 P3 P3 P3 P4 P4 P4 P4
1 answer:
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Answer:
attached below
Explanation:
Answer:
The break force that must be applied to hold the plane stationary is 12597.4 N
Explanation:
p₁ = p₂, T₁ = T₂
The heat supplied = × Heating value of jet fuel
The heat supplied = 0.5 kg/s × 42,700 kJ/kg = 21,350 kJ/s
The heat supplied = ·
= 20 kg/s
The heat supplied = 20* = 21,350 kJ/s
= 1.15 kJ/kg
T₃ = 21,350/(1.15*20) + 485.03 = 1413.3 K
p₂ = p₁ × p₂/p₁ = 95×9 = 855 kPa
p₃ = p₂ = 855 kPa
T₃ - T₄ = T₂ - T₁ = 485.03 - 280.15 = 204.88 K
T₄ = 1413.3 - 204.88 = 1208.42 K
T₅ = 1208.42*(2/2.333) = 1035.94 K
= √(1.333*287.3*1035.94) = 629.87 m/s
The total thrust = ×
= 20*629.87 = 12597.4 N
Therefore;
The break force that must be applied to hold the plane stationary = 12597.4 N.
Answer:
The speed at point B is 5.33 m/s
The normal force at point B is 694 N
Explanation:
The length of the spring when the collar is in point A is equal to:
The length in point B is:
lB=0.2+0.2=0.4 m
The equation of conservation of energy is:
(eq. 1)
Where in point A: Tc = 1/2 mcVA^2, Ts=0, Vc=mcghA, Vs=1/2k(lA-lul)^2
in point B: Ts=0, Vc=0, Tc = 1/2 mcVB^2, Vs=1/2k(lB-lul)^2
Replacing in eq. 1:
Replacing values and clearing vB:
vB = 5.33 m/s
The balance forces acting in point B is:
Fc-NB-Fs=0
Replacing values and clearing NB:
NB = 694 N