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An aircraft is following a true course (TC) of 040° at a constant true airspeed (TAS) of 180 kt. The wind vector is 350°/30 kt. The groundspeed (GS) equals...4
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An aircraft is following a true course (TC) of 040° at a constant true airspeed (TAS) of 180 kt. The wind vector is 350°/30 kt. The wind correction angle (WCA) equals...4
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Given: True course: 270°. TAS: 100 kt. Wind: 090°/25 kt. Distance: 100 NM. The ground speed (GS) equals...4
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Given: True course: 270°. TAS: 100 kt. Wind: 090°/25 kt. Distance: 100 NM. The flight time equals...4
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An aircraft is following a true course (TC) of 040° at a constant true airspeed (TAS) of 180 kt. The wind vector is 350°/30 kt. The wind correction angle (WCA) equals...4
Answers
Given: True course: 120°. TAS: 120 kt. Wind: 150°/12 kt. The WCA equals...4
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The distance from 'A' to 'B' measures 120 NM. At a distance of 55 NM from 'A' the pilot realizes a deviation of 7 NM to the right. What approximate course change must be made to reach 'B' directly?4
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An aeroplane has a heading of 090°. The distance which has to be flown is 90 NM. After 45 NM the aeroplane is 4.5 NM north of the planned flight path. What is the corrected heading to reach the arrival aerodrome directly?4
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What is the meaning of the 1:60 rule?4
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An aircraft is flying from 'A' to 'B' (distance 220 NM) at an average ground speed (GS) of 120 kt. It departs 'A' at 1200 UTC. After 70 NM along the course from 'A', the aircraft is 5 min ahead of the planned schedule. Using the actual GS, what is the revised estimated time of arrival (ETA) at B?4
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