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Question 93-1 : Induced drag is caused by ? [ Deepen EASA ]

Movement of the aerofoil through the air

Ecqb04 feedback november 2017 exemple 193 Movement of the aerofoil through the air.

Question 93-2 : How does an increase in aircraft mass affect the gliding range ?

Has no effect on gliding range

exemple 197 Has no effect on gliding range.

Question 93-3 : Under what condition does pressure altitude has the same value as density altitude ?

At standard temperature

Ecqb04 revised november 2017 exemple 201 At standard temperature.

Question 93-4 : What is true about a clearway ?

A clearway can be water or land and must be under the control of the aerodrome authority

Ecqb04 november 2017 exemple 205 A clearway can be water or land and must be under the control of the aerodrome authority.

Question 93-5 : The effect an increase of weight has on the value of stalling speed las is that vs ?

Increases

Admin .stall speed of an aircraft changes in proportion to the square root of the change in mass .for example if an aeroplane has a stall speed of 78 kt at its mass of 5000 kg the stall speed when the mass is 6850 kg will be .change in mass ratio = new mass / old mass = 6850/5000 = 1 37.square root of 1 37 = 1 17.original stall speed = 78 kt.stall speed for a weight of 6850 kg ==> 78 x 1 17 = 91 kt .vs increases for an increase of weight stall will occur earlier exemple 209 Increases.

Question 93-6 : An aircraft has a stall speed of 83 kt on short final its speed will be ?

108 kt

Admin .on short final you are supposed to be at 1 3 x vs exemple 213 108 kt.

Question 93-7 : The parameters taken into account for take off performance calculations are ?

Outside temperature pressure altitude wind mass

exemple 217 Outside temperature, pressure altitude, wind, mass.

Question 93-8 : For a piston engine aeroplane the speed for maximum range is ?

That which gives the maximum lift to drag ratio

Admin .if you want to stay in flight the longest time possible maximum endurance you need to fly at the minimum power required speed vmp velocity for minimum power .if you want to travel the maximum distance possible maximum range you need to fly at the speed that wich gives maximum lift to drag ratio vmd velocity for minimum drag . 1066 exemple 221 That which gives the maximum lift to drag ratio.

Question 93-9 : Maximum endurance for a piston engine aeroplane is achieved at ?

The speed that approximately corresponds to the maximum rate of climb speed

Admin .maximum endurance in a piston engine aeroplane will occur at vmp minimum power speed you want to use as little fuel as possible in order to stay airborne as longer as possible you are not concerned with distance ..the best rate of climb will occur where you have excess power available that is vy best rate of climb speed and in a piston engine aeroplane that is definitely slower than speed for minimum drag vmd thus maximum endurance for a piston engine aeroplane is achieved at a speed that approximates the best rate of climb speed vmp .a common mistake is done between the trhust required curve showing drags and power required curve showing required power . 1135.for the propeller driven aircraft curve the lowest point of the curve is the tas at wich the least power is needed as opposed to producing the least drag and is therefore the best for endurance in level flight it is also the maximum rate of climb speed because the gap between power required and power available is greatest more power is needed above and below the minimum power speed exemple 225 The speed that approximately corresponds to the maximum rate of climb speed.

Question 93-10 : The maximum indicated air speed of a piston engine aeroplane without turbo charger in level flight is reached ?

At the lowest possible altitude

Admin .maximum ias will always be achieved at sea level because this is where density is highest the dynamic pressure is 1/2 rho v² which is sensed by the pitot system exemple 229 At the lowest possible altitude.

Question 93-11 : At reference or see performance manual sep1 1 figure 2 4 .with regard to the graph for landing performance what is the minimum headwind component required in order to land at helgoland airport .given .runway length 1300 ft.runway elevation msl.weather assume isa conditions.mass 3200 lbs.obstacle ?

10 kt

Admin .please find hereafter the full completed graph . 1072.be aware that we must assume a factored runway a piston engine airplane has to land in the 70% of the runway length .1300 ft x 0 7 = 910 ft with this distance none of the answers are correct exemple 233 10 kt.

Question 93-12 : How does the thrust of a propeller vary during take off run assuming unstalled flow conditions at the propeller blades .the thrust ?

Decreases while the aeroplane speed builds up

Admin .the angle of attack as in the case of a aircraft wing is defined as the angle between the chord line and relative airflow with a propeller however the relative airflow is the resultant of the airflow due to rotation and forward speed change either of these values and the angle of attack will change . 1079.the angle of attack decreases similar to a decreases of angle of attack for an aircraft wing less lift will be produce for a propeller it results in less thrust exemple 237 Decreases while the aeroplane speed builds up.

Question 93-13 : In a given configuration the endurance of a piston engine aeroplane only depends on ?

Altitude speed mass and fuel on board

Oszklarska .the fuel on board weight is contained in airplane mass is it not .therefore the 'altitude speed and mass' answer is equivalent to 'altitude speed mass and fuel' . .no you need to know the quantity of fuel on board to be able to calculate endurance to stay airborne as longer as possible

Question 93-14 : At a higher gross mass on a piston engined aeroplane in order to maintain a given angle of attack configuration and altitude ?

The airspeed must be increased and the drag will also increase

Admin .at a higher gross mass we need to increase lift if we consider to maintain same altitude .lift = cl x 1/2 rho v² x s.cl = lift coefficient.rho = density.v = tas in m/s .s = surface.since density surface and lift coefficient e g change in angle of attack do not change only speed can changed .therefore the airspeed must be increased and the drag will also increase exemple 245 The airspeed must be increased and the drag will also increase.

Question 93-15 : On a reciprocating engine aeroplane to maintain a given angle of attack configuration and altitude at higher gross mass ?

An increase in airspeed and power is required

Admin .at a higher gross mass we need to increase lift if we consider to maintain same altitude .lift = cl x 1/2 rho v² x s.cl = lift coefficient.rho = density.v = tas in m/s .s = surface.since density surface and lift coefficient e g change in angle of attack do not change only speed can changed .therefore power required increases airspeed will increased and the drag will also increase .notice whatever the propulsion system prop or jet it doesn't play a role as seen by its absence in the lift formula

Question 93-16 : An aeroplane with reciprocating engines is flying at a constant angle of attack mass and configuration .with increasing altitude the drag ?

Remains unchanged but the tas increases

Admin .with increasing altitude density rho decreases angle of attack mass and configuration remain constant to maintain lift tas has to increase .drag = cd x 1/2 x rho x v² velocity = tas .v increases while rho decreases drag remains constant exemple 253 Remains unchanged but the tas increases.

Question 93-17 : On a reciprocating engine aeroplane with increasing altitude at constant gross mass angle of attack and configuration the power required ?

Increases and the tas increases by the same percentage

Admin .this question compares an aircraft at a certain weight and angle of attack for a first altitude and then in the same configuration at a second altitude .you have to understand that the aircraft is in straight and level flight it is not climbing .our aircraft has to produce the same lift .lift l = 1/2 rho cl s v².for a higher altitude density reduces cl and s remain unchanged thus only tas v can and must increase .for that reason the power required must increase exemple 257 Increases and the tas increases by the same percentage.

Question 93-18 : At reference or see performance manual sep 1 figure 2 4 .with regard to the landing chart for the single engine aeroplane determine the landing distance from a height of 50 ft .given .o a t 27 °c.pressure altitude 3000 ft.aeroplane mass 2900 lbs.tailwind component 5 kt.flaps landing position ?

Approximately 1850 feet

Admin . 1091 exemple 261 Approximately : 1850 feet.

Question 93-19 : Performance manual sep 1 figure 2 4 .with regard to the landing chart for the single engine aeroplane determine the landing distance from a height of 50 ft.given .o a t isa +15°c.pressure altitude 0 ft.aeroplane mass 2940 lb.tailwind component 10 kt.flaps landing position down .runway tarred and ?

Approximately 1930 feet

Admin . 1128 exemple 265 Approximately : 1930 feet.

Question 93-20 : For this question use annex 032 005 or performance manual sep 1 figure 2 4 .with regard to the landing chart for the single engine aeroplane determine the landing distance from a height of 50 ft .given .oat isa +15°c.pressure altitude 0 ft.aeroplane mass 2940 lb.headwind component 10 kt.flaps ?

Approximately 1800 feet

Img /com en/com032 295 jpg. oat isa +15°c so oat is +30°c at 0 ft pressure altitude .runway wet grass .landing factor for short grass is 1 15 and landing factor for wet runway is 1 15.1275 ft x 1 15 = 1466 ft.1466 ft x 1 15 = 1686 ft.approximately 1800 feet is our answer exemple 269 Approximately: 1800 feet.

Question 93-21 : With regard to the take off performance chart for the single engine aeroplane determine the maximum allowable take off mass .given .o a t isa.pressure altitude 4000 ft.headwind component 5 kt.flaps up.runway tarred and dry.factored runway length 2000 ft.obstacle height 50 ft . 2143 ?

3200 lbs

Admin .complete the graph with the data . 1093.you will find 3200 lbs exemple 273 3200 lbs.

Question 93-22 : At reference or see performance manual sep 1 figure 2 2 .with regard to the take off performance chart for the single engine aeroplane determine the take off distance to a height of 50 ft .given .oat 7°c.pressure altitude 7000 ft.aeroplane mass 2950 lbs.headwind component 5 kt.flaps approach ?

Approximately 2050 ft

Complete the graph with the data . /com en/com032 297 jpg.you find 1900 ft the closest answer is 2050 ft exemple 277 Approximately: 2050 ft.

Question 93-23 : At reference or see performance manual sep 1 figure 2 1 .with regard to the take off performance chart for the single engine aeroplane determine the take off speed for 1 rotation and 2 at a height of 50 ft .given .o a t isa+10°c.pressure altitude 5000 ft.aeroplane mass 3400 lbs.headwind component ?

71 and 82 kias

Img /com en/com032 298 jpg. exemple 281 71 and 82 kias.

Question 93-24 : At reference or see performance manual sep 1 figure 2 2 .with regard to the take off performance chart for the single engine aeroplane determine the take off distance to a height of 50 ft .given .o a t 38°c.pressure altitude 4000 ft.aeroplane mass 3400 lbs.tailwind component 5 kt.flaps approach ?

Approximately 3960 ft

Img /com en/com032 299 jpg.do not forget to apply 'grass.correction factor' .3300 ft x 1 2 = 3960 ft exemple 285 Approximately: 3960 ft.

Question 93-25 : With regard to the climb performance chart for the single engine aeroplane determine the climb speed ft/min .o a t isa + 15°c.pressure altitude 0 ft.aeroplane mass 3400 lbs.flaps up.speed 100 kias . 2139 ?

1290 ft/min

Admin . 1129.youssef92 .for me i find that the answer is 1370ft/min . .if you start at 15°c instead of 30°c you will find around 1350 1370 ft but the question states o a t isa + 15°c .it means an outside temperature of 30°c at sea level exemple 289 1290 ft/min.

Question 93-26 : At reference or see performance manual sep 1 figure 2 2 .with regard to the take off performance chart for the single engine aeroplane determine the take off distance over a 50 ft obstacle height .given .o a t 30°c.pressure altitude 1000 ft.aeroplane mass 2950 lbs.tailwind component 5 kt.flaps ?

2275 ft

Img /com en/com032 3587 jpg..1750 ft x surface factor 1 3 = 2275 ft exemple 293 2275 ft.

Question 93-27 : Using the landing diagram for single engine aeroplane determine the landing distance from a screen height of 50 ft in the following conditions .given .pressure altitude 4000 ft.o a t 5°c.aeroplane mass 3530 lbs.headwind component 15 kt.flaps down.runway tarred and dry.landing gear down . 2135 ?

1350 ft

Admin . 1097.1320 ft close to 1350 ft exemple 297 1350 ft.

Question 93-28 : The pilot of a single engine aircraft has established the climb performance .the carriage of an additional passenger will cause the climb performance to be ?

Degraded

exemple 301 Degraded.

Question 93-29 : An extract of the flight manual of a single engine propeller aircraft is reproduced in annex .airport characteristics hard dry and zero slope runway.actual conditions are .pressure altitude 1 500 ft.outside temperature +18°c.wind component 4 knots tailwind.for a take off mass of 1 270 kg the take ?

465 m

Admin . 1099.1270 kg = 2800 lbs .convert 1520 ft to meters .1520 x 0 3048 = 463 m exemple 305 465 m.

Question 93-30 : With regard to the landing chart for the single engine aeroplane determine the landing distance from a height of 50 ft .given .o a t isa.pressure altitude 1000 ft.aeroplane mass 3500 lbs.tailwind component 5 kt.flaps landing position down .runway tarred and dry . 2136 ?

Approximately 1700 feet

Admin . 1102.you will find 1760 ft close enough to 1700 feet exemple 309 Approximately : 1700 feet.

Question 93-31 : With regard to the take off performance chart for the single engine aeroplane determine the take off distance to a height of 50 ft .given .o a t 30°c.pressure altitude 1000 ft.aeroplane mass 3450 lbs.tailwind component 2 5 kt.flaps up.runway tarred and dry . 2134 ?

Approximately 2470 feet

Admin . 1131 exemple 313 Approximately : 2470 feet.

Question 93-32 : For this question use reference or performance manual sep 1 figure 2 1.with regard to the take off performance chart for the single engine aeroplane determine the maximum allowable take off mass .given .oat isa.pressure altitude 4000 ft.headwind component 5 kt.flaps up.runway tarred and ?

3240 lbs

Complete the graph with the data . /com en/com032 359 jpg.you will find 3240 lbs exemple 317 3240 lbs.

Question 93-33 : Consider the graphic representation of the power required versus true air speed tas for a piston engined aeroplane with a given mass .when drawing the tangent from the origin the point of contact a determines the speed of . 2133 ?

Maximum specific range

Admin . 1085.maximum specific range is reached at minimum drag speed vmd for a piston engined aeroplane . 1135.vmd is the speed for maximum range in a prop aircraft .vmd is the speed for maximum endurance in a jet aircraft exemple 321 Maximum specific range.

Question 93-34 : For this question use reference sep 1 figure 2 3 .using the climb performance chart for the single engine aeroplane determine the ground distance to reach a height of 1500 ft above the reference zero in the following conditions .given .o a t at take off isa.airport pressure altitude 5000 ?

15640 ft

At 5000 ft isa t° is 5°c ..1500/1080= 1 39 minute.ground speed 100 + 5 = 105 kt.. 105/60 x 1 39 = 2 43 nm..2 43 = 4 5 km = 14722 ft . /com en/com032 362 jpg.closest answer is 15640 ft normally we should use the climb gradient since the question gives a speed in ias but rate of climb can also lead closely to the correct answer. ebbr1000 .tas= 108 kt.height difference x 100/ climb gradient= 1500 x 100 / 9 9 = 15151 ft. 15151 x gs / tas= 15151 x 113 / 108 = 15852 ft exemple 325 15640 ft.

Question 93-35 : For this question use reference .using the climb performance chart for the single engine aeroplane determine the rate of climb and the gradient of climb in the following conditions .given .o a t at take off isa.airport pressure altitude 3000 ft.aeroplane mass 3450 lbs.speed 100 kias . err a 032 363 ?

1140 ft/min and 10 6%

Isa at 3000 ft .15° 2° x 3 = +9°c.for a given ias the true airspeed is about 2% higher than ias per 1000ft of altitude above sea level . /com en/com032 363 jpg.rate of climb 1140 ft/min .gradient of climb 10 6% . atplea .figure cap698 figure 2 3 exemple 329 1140 ft/min and 10.6%.

Question 93-36 : At reference or use performance manual sep 1 figure 2 1.airport characteristics hard dry and runway slope zero.actual conditions are .pressure altitude 1500 ft.outside temperature +18°c.wind component 4 kt tailwind.for a take off mass of 2800 lbs the take off distance will be . err a 032 405 ?

1500 ft

Img /com en/com032 405 png. exemple 333 1500 ft.

Question 93-37 : Unless otherwise specified in the afm for a performance class b aeroplane landing on a downhill runway what factor must be applied for each 1% of downslope ?

1 05

Ecqb04 october 2017..the factor to be applied for each 1% of downslope is 5% or 1 05 exemple 337 1.05

Question 93-38 : Unless otherwise specified in the afm for a performance class b aeroplane taking off on a uphill runway what factor must be applied for each 1% of uphill slope ?

1 05

Ecqb04 october 2017..the factor to be applied for each 1% of uphill slope is 5% or 1 05 exemple 341 1.05

Question 93-39 : The following conditions are observed at an airport runway 13 wind 140° for 30 kt .a pilot can determine a crosswind component of ?

5 kt

Admin .vt = sin 10° x 30 = 5 kt exemple 345 5 kt.

Question 93-40 : Maximum crosswind demonstrated is equal to 0 2 vs0 and the following conditions are observed at an airport .vs0 70kt landing runway 35 wind 300° for 20 kt ?

Admin .350 300=50°.crosswind sin50x20=15 3kt.demonstrated crosswind 70x0 2=14kt exemple 349


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