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.
Explanation:
Note: Refer the diagram below
Obtaining data from property tables
State 1:

State 2:

State 3:

State 4:
Throttling process 
(a)
Magnitude of compressor power input


(b)
Refrigerator capacity



(c)
Cop:


Answer:
a, c
Explanation:
As the problem statement tells you, the phasor technique cannot be used when the frequencies are different. The frequencies are different when the coefficients of t are different. The different ones are highlighted.
a. 45 sin(2500t – 50°) + 20 cos(1500t +20°)
b. 25 cos(50t + 160°) + 15 cos(50t +70°)
c. 100 cos(500t +40°) + 50 sin(500t – 120°) – 120 cos(500t + 60°) -100 sin(10,000t +90°) + 40 sin(10, 100t – 80°) + 80 cos(10,000t)
d. 75 cos(8t+40°) + 75 sin(8t+10°) – 75 cos(8t + 160°)