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20問 • 1年前
  • ユーザ名非公開
    • 通報

    問題一覧

  • 1

    An observer is located north of the plane of the ecliptic Which statement is correct

    The Earth moves on a circular orbit around the sun, and the rotational direction is clockwise

  • 2

    Which of the following ax does not have earth center as its center?

    Small cirle

  • 3

    Point A on the earth surface lies exactly on the parallel of longitude of 47° 50 27 N which point is exactly 240 NM north of A

    51 50 27

  • 4

    The sun moves 10° of longitude. What is the difference in time?

    0.66 H

  • 5

    The term Sunset is defined

    The point in time when the upper edge of the sun appears below the horizon

  • 6

    The term magnetic cause MC is defined as

    The angle between magnetic north and and arbitrary direction

  • 7

    Given TC 168° CA +012° MH 178° CH 176° what are VAR and DVE

    Var 002 e dev +002

  • 8

    3500m equal

    1.89 nm

  • 9

    Which are the properties of a conformal projection chart?

    The scale on an arbitrary point must be independent of direction in the angle between longitude and latitudes on the chart is always 90°

  • 10

    10. Given the following information, what is the aircraft position at the cross bearing? VOR Hamburg (HAM) (53°412N, 010°12?E): Radial 119°; VOR Brünkendorf (BKD) (53°02?N, 011°33?E): Radial 318°; See annex (NAV-031)

    5320n 1110e

  • 11

    11.Given: True course from A to B: 250°. Ground distance: 210 NM. TAS: 130 kt. Headwind component: 15 kt. Estimated time of departure (ETD): 0915 UTC.The estimated time of arrival (ETA) is

    1105 utc

  • 12

    An aircraft is flying with a true airspeed (TAS) of 120 kt and experiences 35 kt tailwind. How much time is needed for a distance of 185 NM?

    1h 12min

  • 13

    What is the distance from Neustadt (EDAN) (53°22'N, 011°37'E) to Uelzen (EDVU) (52°59?N, 10°28?E)

    46nm

  • 14

    Given true course 255° TAS 100 KT wind 200° 10 KT true heading equals

    250

  • 15

    Given true course 270° TAS 100 KT wind 90° 25 KT distance 100 NM Flight time equals

    48 min

  • 16

    A well-known ground feature along the flight track is passed 5 minutes ahead of the planned flight schedule. The expected ground speed was 120 kts and the distance of the previously flown leg was 30 NM. The wind component (WC) equals...

    60KT tailwind

  • 17

    The approximate propagation speed of electromagnetic waves is

    300000kms

  • 18

    A VHF direction finder VDF can determine

    Magnetic bearing

  • 19

    Given QDR 152 VAR 5 W DEV 5 E the QUJ equals

    327

  • 20

    The pilot receives a QDR of 225° from the VDF ground station. Where is the aircraft located i relation to the ground station?

    southwest

    • operational procedures 1

    operational procedures 1

    ユーザ名非公開 · 18問 · 1年前

    operational procedures 1

    operational procedures 1

    18問 • 1年前
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    air law 1

    air law 1

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    aerodynamics 1

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    aerodynamics 1

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    flight planning 1

    flight planning 1

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    flight planning 1

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    flight planning calculations

    flight planning calculations

    ユーザ名非公開 · 34問 · 1年前

    flight planning calculations

    flight planning calculations

    34問 • 1年前
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    AGK 1

    AGK 1

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    AGK 1

    AGK 1

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    AGK 2

    AGK 2

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    AGK 2

    AGK 2

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    AGK 3

    AGK 3

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    AGK 3

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    AGK 4

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    AGK 4

    AGK 4

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    meteo 1

    meteo 1

    ユーザ名非公開 · 19問 · 1年前

    meteo 1

    meteo 1

    19問 • 1年前
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    meteo 2

    meteo 2

    ユーザ名非公開 · 19問 · 1年前

    meteo 2

    meteo 2

    19問 • 1年前
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    meteo 3

    meteo 3

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    meteo 3

    meteo 3

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    meteo 4

    meteo 4

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    meteo 4

    meteo 4

    19問 • 1年前
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    meteo 5

    meteo 5

    ユーザ名非公開 · 19問 · 1年前

    meteo 5

    meteo 5

    19問 • 1年前
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    meteo 6

    meteo 6

    ユーザ名非公開 · 19問 · 1年前

    meteo 6

    meteo 6

    19問 • 1年前
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    meteo 7

    meteo 7

    ユーザ名非公開 · 20問 · 1年前

    meteo 7

    meteo 7

    20問 • 1年前
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    nav 2

    nav 2

    ユーザ名非公開 · 20問 · 1年前

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    nav 2

    20問 • 1年前
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    nav 3

    ユーザ名非公開 · 20問 · 1年前

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    nav 3

    20問 • 1年前
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    nav 4

    ユーザ名非公開 · 20問 · 1年前

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    nav 4

    20問 • 1年前
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    nav 5

    nav 5

    ユーザ名非公開 · 20問 · 1年前

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    nav 5

    20問 • 1年前
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    nav 6

    ユーザ名非公開 · 20問 · 1年前

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    nav 7

    ユーザ名非公開 · 20問 · 1年前

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    20問 • 1年前
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    nav 8

    ユーザ名非公開 · 20問 · 1年前

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    nav 8

    20問 • 1年前
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    nav 9

    ユーザ名非公開 · 20問 · 1年前

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    nav 9

    20問 • 1年前
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    nav 10

    ユーザ名非公開 · 19問 · 1年前

    nav 10

    nav 10

    19問 • 1年前
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    nav test

    nav test

    ユーザ名非公開 · 77問 · 1年前

    nav test

    nav test

    77問 • 1年前
    ユーザ名非公開

    問題一覧

  • 1

    An observer is located north of the plane of the ecliptic Which statement is correct

    The Earth moves on a circular orbit around the sun, and the rotational direction is clockwise

  • 2

    Which of the following ax does not have earth center as its center?

    Small cirle

  • 3

    Point A on the earth surface lies exactly on the parallel of longitude of 47° 50 27 N which point is exactly 240 NM north of A

    51 50 27

  • 4

    The sun moves 10° of longitude. What is the difference in time?

    0.66 H

  • 5

    The term Sunset is defined

    The point in time when the upper edge of the sun appears below the horizon

  • 6

    The term magnetic cause MC is defined as

    The angle between magnetic north and and arbitrary direction

  • 7

    Given TC 168° CA +012° MH 178° CH 176° what are VAR and DVE

    Var 002 e dev +002

  • 8

    3500m equal

    1.89 nm

  • 9

    Which are the properties of a conformal projection chart?

    The scale on an arbitrary point must be independent of direction in the angle between longitude and latitudes on the chart is always 90°

  • 10

    10. Given the following information, what is the aircraft position at the cross bearing? VOR Hamburg (HAM) (53°412N, 010°12?E): Radial 119°; VOR Brünkendorf (BKD) (53°02?N, 011°33?E): Radial 318°; See annex (NAV-031)

    5320n 1110e

  • 11

    11.Given: True course from A to B: 250°. Ground distance: 210 NM. TAS: 130 kt. Headwind component: 15 kt. Estimated time of departure (ETD): 0915 UTC.The estimated time of arrival (ETA) is

    1105 utc

  • 12

    An aircraft is flying with a true airspeed (TAS) of 120 kt and experiences 35 kt tailwind. How much time is needed for a distance of 185 NM?

    1h 12min

  • 13

    What is the distance from Neustadt (EDAN) (53°22'N, 011°37'E) to Uelzen (EDVU) (52°59?N, 10°28?E)

    46nm

  • 14

    Given true course 255° TAS 100 KT wind 200° 10 KT true heading equals

    250

  • 15

    Given true course 270° TAS 100 KT wind 90° 25 KT distance 100 NM Flight time equals

    48 min

  • 16

    A well-known ground feature along the flight track is passed 5 minutes ahead of the planned flight schedule. The expected ground speed was 120 kts and the distance of the previously flown leg was 30 NM. The wind component (WC) equals...

    60KT tailwind

  • 17

    The approximate propagation speed of electromagnetic waves is

    300000kms

  • 18

    A VHF direction finder VDF can determine

    Magnetic bearing

  • 19

    Given QDR 152 VAR 5 W DEV 5 E the QUJ equals

    327

  • 20

    The pilot receives a QDR of 225° from the VDF ground station. Where is the aircraft located i relation to the ground station?

    southwest