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A
46980 Kg
B
46680 KG
C
51500 KG
D
49350 KG
Explanation
Refer to image.
Find the obstacle limited TOM as 51,400kg.
Since PMC OFF, find the table at the bottom left and adjust for temperature.
51,400 - 4420 = 46,980kg
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Question ID: 626994
CAP 698 Figure 3.8
Find the balked landing rate of climb if the temperature is 20 °C and the pressure altitude of the airfield is 4 000 ft:
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A
1200ft /min
B
800ft /min
C
500ft /min
D
950ft /min
Explanation
Refer to image
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Question ID: 626993
Performance Manual MEP 1 Figure 3.7 Given the following conditions
OAT: - 20° C
Pressure altitude: 14000 ft
Gross mass: 4000 lb
Mixture: full rich
Other conditions as associated in the header of the graph.
What is the 2 engine rate of climb?
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A
1300 ft/min
B
850 ft/min
C
170 ft/min
D
1550 ft/min
Explanation
Refer to image
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Question ID: 626972
The centre of gravity is located aft. How does this affect take-off performance and climb gradient?
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A
Take-off ground roll distance is increased and climb gradient is higher.
B
Take-off ground roll distance is reduced and climb gradient is lower.
C
Take-off ground roll distance is reduced and climb gradient is higher.
D
Take-off ground roll distance is increased and climb gradient is lower.
Explanation
A more aft CG will reduce the Tailplane-force required to compensate so less lift will need to be generated. This causes and increase in performance with a shorter ground roll and an increased climb gradient.
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Question ID: 626373
Departing SID requires 6.1% gradient with T/O flaps, speed 130 kts to 1800ft and thereafter a 5% gradient to FL80 clean at 230 kts. What is the ROC required to 1800 ft?
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A
1250 fpm with standard speed change altitude.
B
1150 fpm with standard speed change altitude.
C
800 fpm to 1800 ft with non-standard speed change altitude.
D
600 fpm to 1800 ft with non-standard speed change altitude.
Explanation
Gradient = (ROC / TAS) x (6000 / 6080)
Make ROC the subject
ROC = (Gradient X TAS x 6080) / 6000)
= (6.1 x 130 x 6080) / 6000
= 803 fpm
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Question ID: 625205
Climb power: Full throttle 2 500,RPM
Altitude: 6 500 ft Temperature
ISA +15 °C QNH
1030 hPa Weight
1 250 kgFind the rate of climb:
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A
1 450 ft/min
B
1 620 ft/min
C
1 420 ft/min
D
1 560 ft/min
Explanation
Outside air temperature is ISA+15°
Isa at 6,500 feet = 15-13 = +2°C
2°C+ 15 = OAT 17°
Refer t
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Question ID: 625166
Required net climb gradients in the 1st segment for 3-engine aircraft with one engine out are:
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A
0.1 %
B
2.7%
C
0.3%
D
5.5%
Explanation
(a) Takeoff; landing gear extended. In the critical takeoff configuration existing along the flight path (between the points at which the aircraft reaches VLOF and at which the landing gear is fully retracted) and in the configuration used in Sec. 25.111 but without ground effect, the steady gradient of climb must be:
- Not less than 0.3 percent for three-engine aircraft
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Question ID: 625136
The One-Engine-Inoperative Level-Off Altitude is affected by the aircraft mass and which of the following parameters?
1. Temperature
2. Headwind component
3. Flight Director mode
4. Power setting of the remaining engine
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A
1 and 4
B
2 and 3
C
1 only
D
2 and 4
Explanation
Engine Failure and Drift Down
In the case of an engine failure, the remaining thrust is not sufficient to balance the drag force and the cruise speed cannot be maintained. The only solution is to descend to a lower flight altitude, where the remaining engine can provide enough thrust to balance the drag and allow level flight. As the airplane descends into the lower atmosphere where density is greater, the remaining engine can develop more thrust which will equal the drag force, this is the GROSS level-off altitude, but would give no performance margin. So the DRIFT DOWN PROCEDURE is continued to a lower altitude, the NET level-off altitude.
The main purpose of the drift down procedure is to bring the aircraft to an altitude where the aircraft can generate enough trust with the remaining to balance the drag. The level-off altitude is determined by the actual air density.
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Question ID: 625105
Given that the characteristics of a three engined turbojet aeroplane are as follows:
Thrust = 50000 Newton / Engine
g = 10 m/s
Drag = 72 569 N
Minimum steady gradient of climb 2nd segment = 2.7% SIN
Angle of climb = Thrust Drag / Weight
The Maximum Take-Off Mass under the 2nd segment conditions is?
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A
209 064 kg.
B
101 596 kg.
C
286 781 kg.
D
74 064 kg.
Explanation
Since the question stated we are in 2nd segment conditions, we can assume we are in the class A climb requirements with the critical engine failed. This leaves us 2 engines to deal with. We know that (T - D) ÷ W = Climb Gradient. If we now rewrite this formula to find the weight we get: W = (T - D) ÷ Climb Gradient. This gives us (100.000 N - 72569 N) ÷ 0.027 = 1.015.963 N. Since g = 10 m/s², the MTOM to meet the minimum climb gradient is 101.596 kg.
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Question ID: 625082
All other things being equal, if the temperature decreases, the climb performance
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A
deteriorates at low altitude, and improves at high altitude.
B
deteriorates
C
remains unchanged.
D
improves.
Explanation
Colder temperatures are better for our performance.
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Question ID: 127660
Given Oat 25°C
Pressure altitude 3000ft
Rwy 26L
Wind 310 / 20 kt
TOM 4400 lb
Heavy duty brakes installed
Other conditions as associated in the header of graph.
What is the accelerate and stop distance under the conditions given?
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A
4300FT
B
3350FT
C
3500FT
D
3800FT
Explanation
See attachment. Graphical solution is approx 3750, but you must apply the correction for heavy duty brakes.
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