Correct: C)
Explanation: The correct answer is C because to convert pounds to kilograms, divide by 2.2: 1235 / 2.2 = 561.4 kg, which rounds to approximately 560 kg. A (620 kg) would correspond to about 1364 lbs. B (2720 kg) results from multiplying instead of dividing. D (2470 kg) is also the result of a multiplication error. The key formula is: mass in kg = weight in lbs / 2.2.
Correct: C)
Explanation: The correct answer is C because on an upsloping field with a tailwind, the competing effects partially cancel each other: the upslope shortens the ground roll while the tailwind lengthens it. The normal approach speed (yellow triangle on the ASI) provides the correct balance of energy management. A is wrong because a faster approach would result in excessive float on the upslope. B is wrong because flaring higher risks ballooning on the slope. D is wrong because full airbrakes may cause an excessively steep descent on short final.
Correct: A)
Explanation: The correct answer is A because at 2000 m AMSL above Langenthal, you are in Class E airspace. VFR flight in Class E requires 5 km horizontal visibility, 1500 m horizontal cloud clearance, and 300 m vertical cloud clearance. B is wrong because Class G with its reduced minima applies only at very low altitudes. C is wrong because there is no Class D TMA at this location and altitude. D is wrong because Class C begins at FL 130 in this region, far above 2000 m AMSL.
Correct: C)
Explanation: The correct answer is C because when the C.G. is too far aft, the glider loses longitudinal static stability — the nose tends to pitch up without returning to equilibrium, potentially leading to uncontrollable divergent oscillations or a stall/spin. A (too far forward) is less dangerous because the aircraft remains stable, though elevator authority may be insufficient for landing. B and D are wrong because vertical C.G. displacement is not the primary concern in standard glider mass-and-balance analysis.
Correct: C)
Explanation: The correct answer is C because the airspeed indicator measures dynamic pressure, which inherently accounts for air density. The V_NE marking on the ASI (red line) represents a fixed IAS value that corresponds to the structural limit. However, note that the allowable maximum IAS must actually be reduced at high altitude per the flight manual's speed-altitude table — the ASI marking itself does not change, but the pilot must observe a lower limit. A and B/D are wrong because the physical mark on the instrument does not move. The subtlety is that while the ASI reading mechanism inherently accounts for density, glider pilots must consult the altitude-correction table for the actual limit at high altitude.
Correct: C)
Explanation: The correct answer is C because ground speed = distance / time = 150 km / 1.25 hours = 120 km/h. The key step is converting 1 hour 15 minutes to decimal hours: 15 minutes = 0.25 hours, so total time = 1.25 hours. A (125 km/h) results from dividing by 1.2 hours. B (115 km/h) and D (110 km/h) do not correspond to any correct calculation with these inputs.
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- A) The extended CTR/TMA Payerne and restricted zone LS-R4 must be strictly avoided every day from 02 to 06 September 2013, between sunrise and sunset.
- B) An airshow is taking place in the Payerne area from 02 to 06 September 2013. The TMA Payerne and restricted zone LS-R4 are active each day during this period between 0600 UTC and 1500 UTC as holding areas and airshow demonstration sectors.
- C) Due to an airshow from 02 to 06 September 2013, the extended CTR/TMA Payerne is active each day between 0600 UTC and 1500 UTC. The TMA is used as a holding area, the restricted zone LS-R4 as a demonstration and holding area. The area must be strictly avoided.
- D) Due to an airshow, a transit clearance for the extended CTR/TMA Payerne and restricted zone LS-R4 must be requested on frequency 135.475 (Payerne TWR) from 02 to 06 September 2013.
Correct: C)
Explanation: The correct answer is C because the NOTAM establishes that from 2 to 6 September 2013, between 0600 and 1500 UTC, the extended CTR/TMA Payerne is activated as a holding area, while LS-R4 serves as both a demonstration and holding area for an airshow. These areas must be strictly avoided during the active period. A is wrong because the times are 0600-1500 UTC, not sunrise to sunset. B incorrectly states both areas serve as holding and demonstration areas. D is wrong because transit is not permitted — the area must be avoided entirely, not transited with clearance.
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- A) 95km/h
- B) 75km/h
- C) 55km/h
- D) 135km/h
Correct: B)
Explanation: The correct answer is B (75 km/h) because the best glide speed is found by drawing a tangent from the origin to the speed polar curve for 450 kg. The point where this tangent touches the curve gives the speed for maximum lift-to-drag ratio (best glide). A (95 km/h) is too fast and would correspond to a heavier mass or a different polar. C (55 km/h) is near the stall speed. D (135 km/h) is deep in the high-speed range where the glide ratio is significantly reduced.
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- A) The DABS can be ignored as it solely applies to military aircraft.
- B) You may pass through all relevant danger and restricted zones below 1000 ft AGL or above 10,000 ft AMSL.
- C) The route can be flown without coordination between 1200 and 1300 LT.
- D) It is not possible to fly the planned route that day.
Correct: C)
Explanation: The correct answer is C because checking the DABS for 19 March 2013 (winter time, CET = UTC+1), the planned time of 1205-1255 LT converts to 1105-1155 UTC. During this period, the relevant danger and restricted zones along the route are not active, allowing the route to be flown without coordination. A is wrong because the DABS applies to all airspace users, including gliders. B is wrong because altitude-based exemptions do not automatically apply to all restricted areas. D is wrong because the route is flyable during the specified time window.
Correct: A)
Explanation: The correct answer is A because stall speed (and therefore minimum speed) is proportional to the square root of wing loading. If wing loading increases by 40% (factor 1.4), the new minimum speed is the original multiplied by the square root of 1.4, which equals approximately 1.183 — an increase of about 18.3%. B is wrong because the speed does not increase linearly with wing loading. C is wrong because a 100% increase would mean doubling the speed. D is wrong because any mass increase raises the minimum speed.
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- A) the sink rate of the ASK21 is independent of its mass
- B) the ASK21 has a worse glide ratio at lower flying mass
- C) the ASK21 has a higher sink rate at higher flying mass
- D) the ASK21 has a better glide ratio at lower flying mass
Correct: A)
Explanation: The correct answer is A because at 150 km/h, the two polar curves for different masses of the ASK21 intersect, meaning both configurations have the same sink rate at this particular speed. This is an aerodynamic property of the polar: the curves cross at one speed where mass has no effect on sink rate. B is wrong because at 150 km/h the glide ratio is equal for both masses. C is wrong because the sink rates are identical at this intersection point. D is also wrong because neither mass has a better glide ratio at this specific speed.
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- A) 700 ft.
- B) 780m.
- C) 780 ft
- D) 700m.
Correct: B)
Explanation: The correct answer is B (780 m) because the AIP chart for Amlikon aerodrome shows a maximum landing distance available of 780 metres in the eastward direction. A and C are wrong because landing distances in Switzerland are given in metres, not feet. D (700 m) does not match the published data for the eastward heading. Always verify the unit and the specific runway direction when reading aerodrome charts.
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- A) 2400 ft AMSL.
- B) 3500 ft AMSL.
- C) 2000ft GND.
- D) 5000 ft AMSL.
Correct: B)
Explanation: The correct answer is B because the EMMEN TMA lower boundary between Cham and Hitzkirch is at 3500 ft AMSL. Below this altitude, you remain in uncontrolled airspace and no clearance is needed. Above 3500 ft AMSL, you enter the TMA and must obtain an ATC clearance. A (2400 ft) is too low and does not correspond to the published limit. C (2000 ft GND) references above ground level, which is not how this TMA boundary is expressed. D (5000 ft) is too high.
Correct: D)
Explanation: The correct answer is D because when the maximum permitted payload is exceeded, the only correct action is to reduce the payload until it complies with the limit. The maximum payload is a certification limit based on structural strength and C.G. envelope. A and C are wrong because trimming adjusts aerodynamic forces on the tail but does not change the aircraft's mass or C.G. — it cannot make an overloaded aircraft safe. B is wrong because increasing takeoff speed does not solve an overweight condition and may actually overstress the structure further.
Correct: D)
Explanation: The correct answer is D because a headwind reduces groundspeed while the sink rate in the airmass remains unchanged. Since the glider covers less horizontal ground distance per unit of altitude lost, the descent angle relative to the ground steepens (increases). A is wrong because a tailwind decreases (flattens) the glide angle over the ground by increasing groundspeed. B is wrong because a headwind increases, not decreases, the ground glide angle. C is wrong because wind significantly affects the ground track glide angle, even though it does not affect the airmass glide angle.
Correct: B)
Explanation: The correct answer is B because as altitude increases, air density decreases. For the same true airspeed, the pitot tube measures less dynamic pressure, so the IAS reading is lower than TAS. Conversely, to maintain the same IAS at altitude, the aircraft must fly at a higher TAS. The relationship is approximately TAS = IAS x square root of (sea-level density / actual density). A is wrong because IAS does not rise relative to TAS with altitude. C is wrong because IAS can always be measured. D is wrong because IAS and TAS diverge increasingly with altitude.
Correct: A)
Explanation: The correct answer is A because heavy rain on the wing surface increases roughness and can degrade the boundary layer, potentially raising the stall speed and reducing maximum lift coefficient. A higher approach speed provides a safety margin against these effects. B is wrong because deliberately increasing wing loading in rain would require adding ballast, which is impractical and counterproductive. C is wrong because a shallower approach reduces obstacle clearance in poor visibility. D is wrong because a lower approach speed reduces the safety margin when aerodynamic degradation is already a risk.
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- A) The traffic pattern for runway 33 is clockwise.
- B) The traffic pattern for runway 15 is clockwise.
- C) The traffic pattern for runway 33 is counter-clockwise.
- D) Depending on wind, the traffic pattern for runway 33 may be either clockwise or counter-clockwise.
Correct: D)
Explanation: The correct answer is D because at Bex aerodrome, terrain constraints (the Rhone valley and surrounding mountains) mean the traffic pattern direction for runway 33 depends on the prevailing wind conditions. The chart shows that either a left or right circuit may be used. A is wrong because it limits the pattern to clockwise only. B relates to runway 15, not 33. C is wrong because it limits the pattern to counter-clockwise only. Pilots must check the local procedures and wind conditions before joining the circuit.
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- A) 3500 ft AGL.
- B) FL 100.
- C) FL 35.
- D) 3500 ft AMSL.
Correct: D)
Explanation: The correct answer is D because the lower limit of TMA BERN 1 over Biel Kappelen is 3500 ft AMSL. By staying below this altitude, you remain in uncontrolled airspace and do not need a transit clearance. A (3500 ft AGL) is wrong because TMA boundaries are referenced to MSL, not AGL. B (FL 100) is far above the relevant boundary. C (FL 35) converts to approximately 3500 ft in standard atmosphere, but flight levels use the standard pressure setting (1013.25 hPa), not QNH, so this is not the correct way to express the limit.
Correct: A)
Explanation: The correct answer is A because applying the mass-and-balance calculation with the data provided (from the attached sheet), the new C.G. position computes to 76.7, which falls within the approved forward and aft C.G. limits. B (78.5) is an incorrect calculation result. C (82.0) is too far aft and would be outside limits. D (75.5) is incorrectly calculated and would also fall outside the forward limit. Always verify your calculation by checking whether the result is between the published forward and aft limits.
Correct: A)
Explanation: The correct answer is A because a waterlogged grass surface creates greater friction and drag on the landing gear during the ground roll, causing the glider to decelerate faster and stop in a shorter distance. The water acts as a braking medium. B is wrong because wet grass increases, not decreases, rolling resistance for a glider. C is wrong because while directional control may be slightly affected, the primary effect is shortened stopping distance. D is wrong because surface conditions always affect landing distance.
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- A) 520 m.
- B) 470m.
- C) 520 ft.
- D) 470 ft.
Correct: B)
Explanation: The correct answer is B (470 m) because the AIP chart for Schanis aerodrome shows a maximum landing distance available of 470 metres in the NNW direction. A (520 m) does not match the published data for this heading. C and D are wrong because Swiss aerodrome distances are given in metres, not feet. Always read the correct runway direction and corresponding distance from the aerodrome chart.
Correct: D)
Explanation: The correct answer is D (45.71 lbs). The calculation uses the shift formula: when mass x is moved from the current C.G. position (80) to arm 150, the C.G. shifts aft. The new C.G. must not exceed 80.5. Using the formula: delta CG = (x × delta arm) / total mass, we get: 0.5 = (x × 70) / 6400, therefore x = (0.5 × 6400) / 70 = 45.71 lbs. A (27.82), B (56.63), and C (39.45) result from incorrect calculations using wrong distances or mass values.
Correct: C)
Explanation: The correct answer is C because correct loading requires satisfying two independent conditions simultaneously: the total mass must not exceed the maximum allowable mass (MTOM), and the payload must be distributed so that the C.G. remains within the approved envelope. A is wrong because respecting the mass limit alone does not guarantee the C.G. is within limits. B is wrong because correct distribution alone does not ensure the total mass is within limits. D is wrong because it addresses only one specific baggage compartment rather than the complete loading requirements.
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- A) in the speed range up to 100 km/h, an increase in flying mass reduces the sink rate.
- B) minimum speed is independent of flying mass.
- C) both glide ratio and minimum speed are independent of flying mass.
- D) only the maximum glide ratio is independent of flying mass, apart from a minor Reynolds number effect.
Correct: D)
Explanation: The correct answer is D because when comparing polar curves for different masses, the tangent from the origin touches each curve at the same angle, meaning the maximum lift-to-drag ratio (best glide ratio) is essentially unchanged by mass, apart from minor Reynolds number effects. However, the speed at which this best glide ratio occurs increases with mass. A is wrong because increasing mass always increases the sink rate at any given speed. B is wrong because minimum speed increases with mass (proportional to the square root of mass ratio). C is wrong because while glide ratio is mass-independent, minimum speed is not.
