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.
According to the question of the pulsating brake pedal, both A and B are correct.
What causes brake pulsation?
Brake pulsation is mainly caused by warped rotors/brake discs. Excessive hard braking or quick stops, which can significantly overheat the discs, are the primary causes of deformed rotors. When the discs overheat, the composition of the metal disc material changes, resulting in imperfections in the metal's surface. Hotspots are noticeable irregularities. They appear as discoloured areas of the disc material, which are often bluish or blackish in appearance. The brake pedal is the pedal which you press with your foot to slow or stop a vehicle. When the driver presses the brake pedal, the system automatically delivers the appropriate pressure required to prevent colliding with the vehicle in front.
To learn more about brake pulsation
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Answer:
Impossible.
Explanation:
The ideal Coefficient of Performance is:


The real Coefficient of Performance is:


Which leads to an absurds, since the real Coefficient of Performance must be equal to or lesser than ideal Coefficient of Performance. Then, the cycle is impossible, since it violates the Second Law of Thermodynamics.
Answer:
minimum factor of safety for fatigue is = 1.5432
Explanation:
given data
AISI 1018 steel cold drawn as table
ultimate strength Sut = 63.800 kpsi
yield strength Syt = 53.700 kpsi
modulus of elasticity E = 29.700 kpsi
we get here
=
...........1
here kb and kt = 1 combined bending and torsion fatigue factor
put here value and we get
=
= 12 kpsi
and
=
...........2
put here value and we get
=
= 17.34 kpsi
now we apply here goodman line equation here that is
...................3
here Se = 0.5 × Sut
Se = 0.5 × 63.800 = 31.9 kspi
put value in equation 3 we get
solve it we get
FOS = 1.5432
Answer: D. All of the choice A, B and C are correct.