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Question ID: 627520

Before take-off, the setting of the angle of incidence of a horizontal trimmable stabilizer depends on :

Select 3 options from the below

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The temperature

The position of the CG

The actual take-off mass

The flaps setting

Explanation

When taking off you need to have enough elevator authority to enable sufficient rotation during take-off. A forward CG, heavier mass, and lower flaps all require more elevator authority. More aft CG, lighter, and more flaps mean you don't need quite as much elevator control to rotate.The effect of temperature is negligible.

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Question ID: 627252

Select the correct characteristics of a swept wing:

Select 3 options from the below

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More effective leading edge high lift devices

Increase lift compares to a straight wing

Delay onset of Mcrit and shockwaves

More effective trailing edge flaps

Increase in stall speed

Increase in take-off and landing distance required

Explanation

A swept wing reduces the component of the relative airflow impacting the wing in the direction of the wing. If the sweepback is 10°, the component of the velocity V that the aircraft 'feels' is Vcos(10). This has the positive effect of increasing the critical mach number. However, it reduces lift. This means the stall speed is increased, and it will take longer to accelerate to a speed sufficient to take off. It also means you must fly faster to maintain lift on approach, increasing your landing distance.

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Question ID: 627250

What is the VA of a transport aircraft in the clean configuration with a maximum take-off weight is 344 Tonnes who's VS0 speed = 145 knots and VS1 speed = 162 Knots

Enter the answer in KTS

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Answer:

Correct answer : 256

Explanation

Square root of 2.5 (limit load factor utility) x 162 = 256

VS0 is not used as question states clean config - which is VS1

if you got 229 you used the wrong speed to start with

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Question ID: 627249

With regard to Propeller Icing:

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A

ice only accumulates on the middle third of the prop

B

25% decrease in efficiency but thrust is unaffected in a power-on state

C

rotational velocity is too high and ice doesn’t stick to the propeller

D

thrust and propeller efficiency are reduced

Explanation

With ice on the propeller, the airflow over the blade will be disturbed and not be as efficient, there will be a decrease in efficiency and a decrease in thurst

ice also accumulates on the 1st third of a prop - where the rotation speeds are slower

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Question ID: 627248

Approximately what is the radius of an aircraft who flies a complete circle during a horizontal steady co-ordinated turn with a bank angle of 60° and a TAS of 120kt?

Use Gravity = 10 for this question

Answer in meters

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Answer:

Correct answer : 219

Explanation

Convert the TAS into SI unit
120 x 0.514 = 61.68 m/sec

Then calculate the radius

Radius of turn (m) = Square of TAS / g x tan angle
Radius of turn = 61.68 square / 10 x tan 60
Radius of turn = 219 Meters

If you used a gravity of 9.81 you will get 221 , the exam reports it gives you gravity to use as 10

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Question ID: 627090

Approximately how long does it take to fly a complete circle during a horizontal steady co-ordinated turn with a bank angle of 45° and a TAS of 380 kt?

Answer in Seconds

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Answer:

Correct answer : 125

Explanation

Rate of turn = G×tan(bank angle)/TAS

Rate of turn = 9.8 × tan(45)/(380×0.514)

Rate of turn = 0.05 radians per second

Time to turn = 2×Pi/0.05

Time to turn = 125.7 seconds

Units must be kept the same i.e TAS in knots needs to be converted to m/s by multiplying TAS by 0.514.

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Question ID: 627084

In straight, horizontal steady flight, the function of the tailplane is to maintain equilibrium by supplying..

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A

an up force when the Centre of Pressure is behind the CG.

B

an up force when the CG approaches its permissible values.

C

a down force when the Centre of Pressure is behind the CG.

D

a down force when the Centre of Pressure coincides with the CG.

Explanation

If the aircraft has a forward CG, then the tailplane needs to provide a balancing force to maintain equilibrium. Nose down = tail up, so tail DOWN force is required.

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Question ID: 627077

An aircraft is flying in steady, straight and level flight with a Coefficient of Lift of 0.47. Calculate the final load factor if a vertical gust increases the angle of attack by 3°.Consider that a change in 1° of Angle of Attack produces a change in CL of 0.1.

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A

0.61

B

1.47

C

1.78

D

1.64

Explanation

Increase aoa by 3° = increase CL by 0.3

0.47 + 0.3 = 0.77

0.77 / 0.47 = 1.638

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Question ID: 627070

What happens to the airflow’s velocity (V), static pressure (Ps), and density when passing through an oblique shock wave?

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A

V increases, Ps decreases, density increases.

B

V decreases, Ps increases, density increases.

C

V increases, Ps decreases, density decreases.

V decreases, Ps increases, density decreases.

Explanation

Easy mnemonic... STD+
Static pressure (S), Temperature (T) and Density (D) all INCREASE through a shockwave
Velocity and Mach number DECREASE

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Question ID: 626387

The manouveuring speed (VA) of a Cessna Skyhawk at 2550 lb is 102 kt. What is the VA at mass of 1900 lb?

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A

B

118

C

102

D

Explanation

New stall speed (VA) = stall speed ÷√old mass ÷ new mass

VA = 102 ÷ √ 2550 ÷ 1900 = 88kts

because the mass is being reduced the stall speed is reduced, same applies if the mass was increased the stall speed would increase.

Explanation Provided by Tom Kinchella

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Question ID: 626078

The critical Mach number of a conventional aerofoil section decreases if:

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A

its thickness to chord ratio is reduced.

B

it is flown at lower angles of attack.

C

its leading edge radius is decreased.

D

its camber is increased.

Explanation

If the camber of the aerofoil is increased, air on the upper surface has further to travel essentially. Because the air over the upper surface is moving faster the wing will reach MCRIT as a lower speed ( E.g M0.84) than as less cambered wing which will reach MCRIT at a higher speed ( E.g M0.87)

(speeds are just examples not given values)

Explanation Provided by Tom Kinchella

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Question ID: 626060

Static lateral stability will be decreased by:

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A

the use of a high, rather than low, wing mounting.

B

increasing the size of the vertical tail.

C

reducing wing anhedral.

D

reducing wing sweepback.

Explanation

The lateral static stability is a measure of the airplane’s tendency to return to the wings-level attitude after a disturbance has caused the airplane to be disturbed in the rolling motion of the aircraft

Ways to increase the lateral static stability are:
dihedral.
low CG.
sweepback.
high-wing mounting.
increased effective dihedral.
large, high vertical fin.
Ways to decrease the lateral static stability are:
Forward-swept wings
anhedral.
ventral fin. (this is a tiny fin under the rear of the aircraft)
low-wing mounting.
extending inboard flaps.

TIP:
To remember which way round anehedral and dihedral go…
think of the phrase DI / AD

Dihedral increases stability
Anehdral Decreases stability

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Question ID: 625955

What is the value of the Mach number if the Mach angle equals 45°?

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A

1.2

B

2.0

C

1.4

D

0.7

Explanation

1÷ sin Mach angle = Mach number

in this case 1÷ sin45 = √2 ( 1.41)

closest answer 1.4

Explanation Provided by Tom Kinchella

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Question ID: 625848

The correct drag formula is:
(note: RHO = density)

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A

DRAG = CD × ½ RHO × V2 × S

B

DRAG = CD × ½ (1 / RHO) × V2 × S

C

DRAG = CD × 2 × RHO × V2 × S

D

DRAG = CD × ½ RHO × V × S

Explanation

same as lift but just with CD instead of CL
CL or CD x1/2 Rho x v² x s

Explanation Provided by Tom Kinchella

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Question ID: 625837

An aeroplane flying at 200 kts in straight and level flight is subjected to a disturbance that suddenly increases the speed by 10 kts. Assuming the angle of attack remains constant, the load factor will initially:

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A

increase to 1.21.

B

increase to 1.05.

C

increase to 1.10.

D

remain unchanged, since the angle of attack does not change.

Explanation

n=L÷W

L=Cl 1/2 d v² s
v increases by 1.05 therefore v² = 1.05² = 1.10. Therefore if v² increases by 1.10 Lift will increase by 1.10 resulting in load foactor increasing by 1.10.

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Question ID: 118680

A twin-engine turboprop aeroplane has an engine failure in flight

Select the correct statements with regard to the asymmetric effects of the propeller slipstream

Select 2 options from the below

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Slipstream-induced lift will be lost on the side with the failed engine

Total lift will be reduced, giving a tendency to descend

Use of leading edge devices can counter the effects of lost slipstream.

A rolling moment towards the live engine will be induced

Explanation

A propellor aircraft adds to the generation of lift as directly behind the propellor, a large mass of air is being accelerated backwards. This increases the lift generated directly behind the engine (Slip-stream enduced lift). The loss of an engine wil reduce this on one side, as the propellor is no longer accelerating an air mass over the wing.... A rolling moment towards the dead wing engine will be enduced due to the unbalanced force caused by slipstream enduced lift on the live engine.... Less lift as the overall lift is reduced, becuase no longer have SSE lift on one wing.

Explanation Provided by Rory shearer

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