Correct: A)
Explanation: The correct answer is A because the airspeed indicator measures dynamic pressure, which directly relates to aerodynamic forces regardless of altitude. At 1800 m AMSL, air density is lower, so the TAS will be higher for the same IAS — but the aerodynamic forces (lift, stall characteristics) depend on IAS, not TAS. Therefore, the same indicated approach speed provides the same safety margins as at sea level. B is wrong because flying at a lower IAS would reduce the stall margin. D is wrong because a higher IAS is unnecessary and would result in excessive float. C is wrong because the minimum sink speed is not the correct approach speed.
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- A) 130km/h
- B) 90km/h
- C) 70km/h
- D) 110km/h
Correct: B)
Explanation: The correct answer is B (90 km/h) because the best glide ratio speed is found where the tangent from the origin touches the speed polar curve for 450 kg. For this glider type at 450 kg, this occurs at approximately 90 km/h. A (130 km/h) is too fast — at this speed the glide ratio is significantly reduced. C (70 km/h) is closer to the minimum sink speed, which maximises endurance but not distance. D (110 km/h) would give a reduced glide ratio compared to the optimum.
Correct: C)
Explanation: The correct answer is C because when the aft C.G. limit is exceeded, the useful load must be redistributed to move mass forward — for example, adding nose ballast, repositioning equipment, or adjusting the pilot's seating position. This physically moves the C.G. within approved limits. A is wrong because trimming aft would worsen the situation aerodynamically. B is wrong because being within mass limits does not compensate for a C.G. out of limits — both must be satisfied independently. D is wrong because trim adjusts aerodynamic forces but does not change the actual C.G. position.
Correct: D)
Explanation: The correct answer is D because high temperature reduces air density, decreasing the lift generated at any given groundspeed, requiring a longer acceleration to reach flying speed. A tailwind reduces the headwind component, meaning the aircraft needs a higher groundspeed to achieve the same airspeed, further lengthening the takeoff run. A is wrong because low temperature increases air density (more lift) and headwind shortens the run. B is wrong because a strong headwind shortens the takeoff distance. C is wrong because high atmospheric pressure increases density, which helps rather than hinders takeoff performance.
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- A) On 18 November, a military night flying exercise will take place in the ZUGERSEE, SUSTEN and TICINO areas. Lower limit: Class E airspace, upper limit: max. FL150.
- B) On 18 November from 1800 LT to 2100 LT, a military night flying exercise will take place in the ZUGERSEE, SUSTEN and TICINO areas.
- C) On 18 November from 1800 UTC to 2100 UTC, a military night flying exercise with helicopters will take place.
- D) On 18 November from 1800 UTC to 2100 UTC, a military night flying exercise will take place in the ZUGERSEE, SUSTEN and TICINO areas. Lower limit: GND, upper limit: max. 15,000 ft AMSL.
Correct: D)
Explanation: The correct answer is D because the NOTAM specifies a military night flying exercise on 18 November from 1800 to 2100 UTC in the ZUGERSEE, SUSTEN, and TICINO areas, with vertical limits from GND to 15,000 ft AMSL. A is wrong because the lower limit is GND, not Class E airspace, and the upper limit is 15,000 ft AMSL, not FL150. B is wrong because the times are in UTC, not local time. C is wrong because it incorrectly specifies helicopter-only operations and omits the geographic areas.
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- A) 5500 ft GND.
- B) 4500 ft AMSL.
- C) 5000 ft AMSL
- D) 3000 ft AMSL.
Correct: D)
Explanation: The correct answer is D because the CTR (Control Zone) of Bern-Belp airport has an upper limit of 3000 ft AMSL. Above this altitude, you exit the CTR and enter different airspace. A (5500 ft GND) does not match the published limit. B (4500 ft AMSL) is too high. C (5000 ft AMSL) is also too high. VFR flight within the CTR requires a clearance from Bern Tower and must remain below the published upper limit.
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- A) Class G airspace, horizontal visibility 1.5 km, clear of cloud with continuous sight of the ground.
- B) Class C airspace, horizontal visibility 8 km, cloud clearance 1.5 km horizontally, 300 m vertically.
- C) Class C airspace, horizontal visibility 5 km, cloud clearance 1.5 km horizontally, 300 m vertically.
- D) Class E airspace, horizontal visibility 5 km, cloud clearance 1.5 km horizontally, 300 m vertically.
Correct: D)
Explanation: The correct answer is D because at 1700 m AMSL above Bex aerodrome, you are in Class E airspace. VFR minima in Class E require 5 km horizontal visibility, 1500 m horizontal cloud clearance, and 300 m vertical cloud clearance. A is wrong because Class G applies at lower altitudes with reduced requirements. B is wrong because Class C has the right visibility minimum (5 km in Switzerland, not 8 km) but starts at a much higher altitude. C is wrong for the same airspace classification reason — Class C begins at FL 130, well above 1700 m.
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- A) 1,6m/s
- B) 0,8m/s
- C) 2,0m/s
- D) 1,2m/s
Correct: C)
Explanation: The correct answer is C (2.0 m/s) because reading the speed polar curve for a flying mass of 580 kg at 160 km/h, the sink rate is approximately 2.0 m/s. A (1.6 m/s) would correspond to a lighter mass or lower speed. B (0.8 m/s) is near the minimum sink rate at much lower speed. D (1.2 m/s) is also too low for this speed and mass combination. When reading a speed polar, always identify the correct curve for the given mass before reading the value at the specified speed.
Correct: B)
Explanation: The correct answer is B because to convert kilograms to pounds, multiply by 2.2: 550 x 2.2 = 1,210 lbs. A (12,100 lbs) results from multiplying by 22 instead of 2.2. C (2,500 lbs) does not correspond to any correct calculation. D (250 lbs) results from dividing instead of multiplying. The key formula is: weight in lbs = mass in kg x 2.2.
Correct: D)
Explanation: The correct answer is D because the best glide ratio speed (also called best L/D speed) maximises the horizontal distance covered per unit of altitude lost in still air. This speed is found on the polar curve where the tangent from the origin touches the curve. A is wrong because minimum sink speed maximises endurance (time aloft), not distance. B is wrong because maximum speed produces the worst glide ratio due to high parasite drag. C is wrong because minimum flying speed is near the stall and gives a poor glide ratio due to high induced drag.
Correct: A)
Explanation: The correct answer is A because the maximum glide ratio (best L/D) is essentially independent of mass — both the lift coefficient and drag coefficient at the optimal angle of attack remain the same, so their ratio is unchanged. Only a minor Reynolds number effect exists. B is wrong because wing loading = mass / wing area, which directly increases with mass. C is wrong because sink rate increases with mass at any given speed. D is wrong because the speeds corresponding to best glide and minimum sink both increase with mass.
Correct: D)
Explanation: The correct answer is D because time = distance / speed = 150 km / 100 km/h = 1.5 hours = 1 hour 30 minutes. A (1 hour 50 minutes) would correspond to a distance of about 183 km. B (1 hour 40 minutes = 1.667 hours) would correspond to about 167 km. C (2 hours) would correspond to 200 km. The calculation is straightforward: 150 / 100 = 1.5 hours. Convert the decimal 0.5 hours to 30 minutes.
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- A) The route can be flown without restriction after contacting 128.375 MHz.
- B) Restricted zones LS-R8 and LS-R8A may be transited below 28,000 ft AMSL.
- C) It is not possible to fly this route while the restricted zones are active.
- D) Restricted zones LS-R8 and LS-R8A may be overflown at 9200 ft AMSL or above.
Correct: C)
Explanation: The correct answer is C because when restricted zones LS-R8 and LS-R8A are active, they cover the planned alpine route between Munster and Amsteg, making it impossible to fly through them. Restricted zones with "entry prohibited" status cannot be transited, regardless of altitude or radio contact. A is wrong because radio contact does not grant transit rights through active restricted zones. B is wrong because a 28,000 ft ceiling does not help a glider. D is wrong because overflying at 9,200 ft may still be within the zone's vertical limits.
Correct: A)
Explanation: The correct answer is A because to transit the Zurich TMA, the pilot must make first radio contact on frequency 124.7 MHz (Zurich Information) at least 10 minutes before entering the controlled airspace. This provides ATC sufficient time to assess traffic, issue a clearance or alternative instructions, and ensure separation. B is wrong because 5 minutes is insufficient lead time. C is wrong because 118.975 is not the correct frequency for Zurich TMA transit requests. D is wrong on both the frequency and the lead time.
Correct: A)
Explanation: The correct answer is A because in a turn, the stall speed increases by the square root of the load factor: Vsturn = Vsstraight x sqrt(n). With n = 2.0: Vs_turn = 60 x sqrt(2) = 60 x 1.414 = 84.85 kts. The increase is (84.85 - 60) / 60 x 100 = 41.4%, which rounds to approximately 40%. B is wrong because the stall speed always increases in a turn. C (5%) and D (20%) significantly underestimate the effect. This relationship between bank angle, load factor, and stall speed is fundamental to safe manoeuvring flight.
Correct: C)
Explanation: The correct answer is C because restricted airspace areas (LO R) on aeronautical charts express their limits using standard altitude references. LO R 16 has an upper limit of 1,500 ft MSL (mean sea level), which is a fixed, absolute altitude. A is wrong because 1,500 m MSL would be approximately 4,900 ft — a completely different altitude that confuses feet with metres. B is wrong because FL150 (15,000 ft pressure altitude) is far too high for a typical low-level restriction. D is wrong because 1,500 ft GND (above ground level) would vary with terrain elevation and is not the published reference.
Correct: B)
Explanation: The correct answer is B because LO R 4 has its upper limit at 4,500 ft MSL, a fixed altitude above mean sea level. A is wrong because 4,500 ft AGL (above ground level) would vary with terrain, which is inappropriate for a fixed regulatory boundary. C is wrong because 1,500 ft AGL is both the wrong altitude value and the wrong reference. D is wrong because 1,500 ft MSL is too low and corresponds to a different restricted area (LO R 16).
Correct: B)
Explanation: The correct answer is B because the NOTAM prohibits overflight up to an altitude of 9,500 ft MSL, following ICAO convention where "altitude" refers to height above mean sea level. A is wrong because "height" in aviation terminology means above a local ground reference (AGL), which is not what the NOTAM specifies. C is wrong because FL 95 is a pressure altitude reference based on 1013.25 hPa, which differs from an MSL altitude depending on actual atmospheric conditions. D is wrong because 9,500 m MSL would be approximately 31,000 ft — clearly inconsistent with a typical VFR NOTAM.
Correct: B)
Explanation: The correct answer is B (symbol C in the annex) because ICAO aeronautical chart symbology (defined in ICAO Annex 4) uses specific symbols to distinguish between single and grouped obstacles, and between lighted and unlighted ones. Symbol C represents a group of unlighted obstacles. A (symbol D), C (symbol B), and D (symbol A) represent other obstacle categories such as single obstacles, lighted groups, or lighted single obstacles. Correct identification of these symbols is essential for cross-country flight planning and obstacle avoidance.
Correct: B)
Explanation: The correct answer is B (symbol A in the annex) because ICAO chart symbology uses distinct depictions for different aerodrome types — civil versus military, international versus domestic, and paved versus unpaved. Symbol A represents a civil (non-international) airport with a paved runway. A (symbol D), C (symbol C), and D (symbol B) represent other aerodrome categories such as international airports, military aerodromes, or grass-strip airfields. Glider pilots must recognise these symbols when identifying potential emergency landing options.
Correct: D)
Explanation: The correct answer is D (symbol C in the annex) because on ICAO aeronautical charts, a general spot elevation is indicated by a specific symbol showing a terrain point of known height, used for situational awareness and terrain clearance planning. A (symbol A), B (symbol B), and C (symbol D) represent other elevation-related markings such as maximum elevation figures, surveyed points, or obstruction elevations defined in ICAO Annex 4.
Correct: A)
Explanation: The correct answer is A. The center of gravity is the single point through which the resultant of all gravitational forces acts on the aircraft — it is the mass-weighted average position of all components. B is wrong because the neutral point is a distinct aerodynamic concept used for stability analysis, not another name for C.G. C duplicates the same incorrect description as A's wording, but the C.G. is defined by mass distribution, not as a geometric midpoint. D is wrong because the C.G. is not the heaviest point — it is where the total weight effectively acts.
Correct: B)
Explanation: The correct answer is B because in mass-and-balance calculations, moment is defined as the product of mass and balance arm: Moment = Mass x Arm (e.g., in kg-m or lb-in). This follows the physical definition of a torque. The total C.G. is found by summing all moments and dividing by total mass. A is wrong because adding mass and arm is dimensionally meaningless. C is wrong because dividing mass by arm does not produce a moment. D is wrong because subtracting them is equally incorrect.
Correct: C)
Explanation: The correct answer is C because the balance arm (moment arm) is the horizontal distance measured from the aircraft's datum reference point to the center of gravity of a specific mass item. A is wrong because that describes the datum itself, not the balance arm. B is wrong because balance arms are measured from the datum, not from the overall aircraft C.G. D is wrong because that is the definition of the center of gravity of a mass item, not the balance arm.
Correct: D)
Explanation: The correct answer is D because interception lines (also called catching lines or line features) are prominent linear ground features — motorways, rivers, coastlines, railways — that a pilot selects during pre-flight planning to navigate toward if orientation is lost. By flying toward a known interception line, the pilot can re-establish position and resume navigation. A is wrong because interception lines are geographic features, not airport markers. B is wrong because they are not range indicators. C is wrong because nothing authorises continuing flight below VFR minima — interception lines are a lost-procedure tool, not a visibility workaround.
