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Question 184-1 : The great circle distance between position a 59°34.1'n 008°08.4'e and b 30°25.9'n 171°51.6'w is ? [ Question security ]
5 400 nm.
.the shortest route is right over the north pole. distance = differences in the latitudes.90° 60°n exactly 59°34.1'n = 30°.90° 30°n exactly 30°25.9'n = 60°..total 90°..90° x 60 nm = 5400 nm.
Question 184-2 : Given.position a 45°n, °e, position b 45°n, 45°15'e, distance a b = 280 nm, b is to the east of a..required longitude of position a ?
38°39'e.
Question 184-3 : 265 us gal equals specific gravity 0.80 ?
803 kg.
.1 usg = 3.79 litres.265 x 3.785 = 1004 litres...1004 x 0.8 = 803 kg.
Question 184-4 : On the 27th of february, at 52°s and 040°e, the sunrise is at 0243 utc. on the same day, at 52°s and 035°w, the sunrise is at ?
0743 utc.
.the earth moves around the sun at a rate of 4 minutes per degree of longitude 15°/h. latitude remains the same, only the longitude changes 040°e to 035°w = 75°...75° and 4 minutes per degree = 5h..the sunrise is at 0743 utc 02h43 + 5h at longitude 35°w... dalton.utc is utc everywhere. sorry but the good answer is 0243 utc... .so the sun rises everywhere on earth at the same utc hour at 0243 utc.. caestudent109.the utc is the same everywhere around the earth. the good answer is 0243 utc.. .example.it is 8 o'clock utc and the sun is rising at london..does the sun rise at 08h00 utc at new york.. cesarmachado. are right, as usual utc it's used to give a common time but it is only the same as local time at prime meridian, anywhere else will give different utc for the same local time. over greenwich at 1000 utc will be 1000 lt but over new york, the same 1000 utc will be utc 4 hours so, 0600 lt. if the sun rises everywhere at 0600, will be in ny at 1000 utc and in london at 0600 utc.
Question 184-5 : The rhumb line distance between points a 60°00'n 002°30'e and b 60°00'n 007°30'w is ?
300 nm.
.between 002°30 east and 007°30 west , there is 10° difference of longitude...distance = difference of longitude x 60 nm x cos latitude..10° x 60 nm x cos 60 = 300 nm.
Question 184-6 : An aircraft is flying a great circle orthodromic route between two points.wpt 1 60°n 030°w, wpt 2 60°n 020°w..what will be the approximate latitude at longitude 025°w ?
60°05.7'n.
.the great circle track is on the pole side of the rhumb line 60°n , then at 025°w your northerly latitude will be more than 60°n...f = &.d / 230..f= greatest vertical distance in nm, between ortho & loxo route this is what we are looking for...& = givry correction = 1/2 g sin lm ==>1/2.10.sin 60 = 4,33°.d = 10 x 60 x cos60 = 300 nm..f = 4.33 x 300 / 230 = 5.64'.. 1403.the aeroplane is flying a great circle orthodromic route = shortest distance..to memorise/visualise the relative position of orthodrome route vs loxodrome route loxodrome route= following a parallel or flying a constant latitude , remember the acronym 'p o l e'..p = north or south pole.o= orthodrome.l= loxodrome.e= equator.
Question 184-7 : An aircraft travels from point a to point b, flying a great circle route..the coordinates of a is 45°s 010°w and b is 45°s 030°w..the true course of the aircraft on its arrival at b, to the nearest degree, is ?
277°.
.flying a great circle route = orthodromic route = shortest distance.loxodrome route= following a parallel or flying a constant latitude. 1405.conversion angle = 1/2 g sin lm.g change of longitude 20°.lm = mean latitude 45°.conversion angle = 1/2 x 20° x sin45°.conversion angle =7°..the true course of the aircraft on its arrival at b = 270° + 7° = 277°.
Question 184-8 : Which is the highest latitude listed below at which the sun will rise above the horizon and set every day ?
62°.
.the times for sunrise and sunset are based on the ideal situation, where no hills or mountains obscure the view and the flat horizon is at the same altitude as the observer. sunrise is the time when the upper part of the sun is visible, and sunset is when the last part of the sun is about to disappear below the horizon in clear weather conditions...for locations north of 66°34' n or south of 66°34' s latitude, the sun will be above the horizon all day in the summer and below the horizon all day in the winter...the sun will be seen to rise and set every day up to 62°.
Question 184-9 : The utc of sunrise on 6 december at winnipeg canada 49°50'n 097°30'w is. 2471 ?
14 13
On the almanac, at position 50°n close to 49°50'n we can read that sunrise is at 07 40 december 4th, and at 07 43 on december 7th..for december 6th, sunrise will be at 07 42 local time at winnipeg...now, we must convert local time to utc.the conversion factor between lmt and utc is arc change of longitude converted to time...change of longitude = 097°30'..097°30' x 4 minutes per degree = 6 30..7 43 + 6 30 = 14 13.
Question 184-10 : When it is 1000 standard time in kuwait, the standard time in algeria is. 2472 ?
08 00
.on the almanach, you can read on top.'the time given should be > substracted from standard time to give utc '.10 00 3 = 07 00 utc in kuwait..'the time given should be > added to utc to give standard time '.07 00 utc +1 = 08 00 st in algeria.
Question 184-11 : The value of magnetic variation ?
Has a maximum of 180°.
.the minimum magnetic variation declination is 0°. the maximum magnetic variation declination is 180°and this will be along a line between the magnetic north pole and the geographic north pole. 1417
Question 184-12 : The horizontal component of the earth's magnetic field ?
Is stronger closer to the magnetic equator.
Question 184-13 : When it is 0600 standard time in queensland australia the standard time in hawaii usa is. 2473 ?
10 00
.it is 0600 standard time in queensland first page list i places fast on utc austrialia+queensland = 10 h.they state the times given below should be substractd from standard time to give ut. 06h00 10h = 20h00 utc the previous day...standard time in hawaii.20h00 utc refer to list iii places slow on ut united sates of america+hawaii 10 h..they state the times given below should be substraced from ut to give standard time.20h00 utc 10 h = 10h00.
Question 184-14 : The circumference of the earth is approximately ?
21600 nm.
.360° x 60 nm = 21600 nm.
Question 184-15 : Isogonic lines connect positions that have ?
The same variation.
. 1425.isogonic lines connect positions that have the same variation..agonic line a line which joins all points where the value of magnetic variation is zero.
Question 184-16 : The local mean time at longitude 095°20'w, at 0000 utc, is ?
17h38 40 previous day.
.the greenwich meridian is selected as standard meridian, lmt local mean time at the greenwich meridian is equal to greenwich mean time gmt...at 00h00 utc, it's midnight at the greenwich meridian...on earth we have 24 timezones, so 360°/24h = 15°/h..each timezone equals 15° on earth's surface...if we move west ot the greenwich meridian, we 'come back'..as we move of 95.3°, thus 953°/15 = 6.3533 h 6h41min 20 sec...6h41 20 sec before midnight, it is 17h 38 40 the previous day.
Question 184-17 : 5h 20min 20sec corresponds to a longitude difference of ?
80°05'.
.5h 20min 20sec = 5 x 60 + 20.3 = 320.3 min...each degree of longitude is equivalent to four minutes in time or 1h = 15°...320.3 min = 320.2/4° = 80.075' 80°05'.
Question 184-18 : What is the value of the magnetic dip at the magnetic south pole ?
90°.
.magnetic dip is the angle between the horizontal and vertical forces acting on a compass needle toward the nearer pole.. 1430.at the magnetic poles, the earth's magnetic field is perpendicular to the earth's surface. magnetic dip is 90° at the magnetic poles a dip needle stands vertical.
Question 184-19 : What is the meaning of the term standard time ?
It is the time set by the legal authorities for a country or part of a country.
Question 184-20 : What is the local mean time, position 65°25'n 123°45'w at 2200 utc ?
1345.
.123.75' x 4 minutes per degree = 495 minutes..495 minutes = 8h35..west of greenwich, we substract 8h35 to 22h00 = 13h45.
Question 184-21 : The main reason that day and night, throughout the year, have different duration, is due to the ?
Inclination of the ecliptic to the equator.
Question 184-22 : The lines on the earth's surface that join points of equal magnetic variation are called ?
Isogonals.
. 1425.isogonic lines connect positions that have the same variation..agonic line a line which joins all points where the value of magnetic variation is zero.
Question 184-23 : An aircraft departing a n40°00' e080°00' flies a constant true track of 270° at a ground speed of 120 kt. what are the coordinates of the position reached in 6 hours ?
N40°00' e064°20'.
.6h x 120 kt = 720 nm...720 cos mean latitude = 720 / cos40° = 940..940 / 60 = 15.66 15°39'..from longitude e080°00', flying west, e080°00' 15°39' = e64°21'.
Question 184-24 : A nautical mile is equivalent to ?
1852 m.
Question 184-25 : An aircraft flies the following rhumb line tracks and distances from position 04°00'n 030°00'w 600 nm south, then 600 nm east, then 600 nm north, then 600 nm west. the final position of the aircraft is ?
04°00'n 029°58'w
.the aircraft flies 600 nm 600/60 = 10° south from position 04°00'n 030°00'w.it is now at 06°00's 030°00'w...the aircraft flies 600 nm east from position 06°00's 030°00'w.it is now at 10°/cos6 around 10°03' east, at 06°00's 019°57'w...the aircraft flies 600 nm north from position 06°00's 019°57'w.it is now 10° north, at 04°00'n 019°57'w...the aircraft flies 600 nm west from position 04°00'n 019°57'w.it is now at 10°/cos4 around 10°01' west, at 04°00'n 029°58'w.
Question 184-26 : What is the final position after the following rhumb line tracks and distances have been followed from position 60°00'n 030°00'w south for 3600 nm, east for 3600 nm, north for 3600 nm, west for 3600 nm. the final position of the aircraft is ?
60°00'n 090°00'w.
.the aircraft flies 3600 nm 3600/60 = 60° south from position 60°00'n 030°00'w..it is now at 00°00'n/s 030°00'w...the aircraft flies 3600 nm east from position 00°00'n/s 030°00'w.it is now at 60°/cos0 around 60° east, at 00°00'n/s 030°00'e...the aircraft flies 3600 nm north from position 00°00'n/s 030°00'e.it is now 60° north, at 60°00'n 030°00'e...the aircraft flies 3600 nm west from position 60°00'n 030°00'e.it is now at 60°/cos 60 around 120° west, at 60°00'n 090°00'w.
Question 184-27 : Complete the following statement regarding magnetic variation. the charted values of magnetic variation on earth normally change annually due to ?
Magnetic pole movement causing numerical values at all locations to increase or decrease.
Question 184-28 : In which two months of the year is the difference between the transit of the apparent sun and mean sun across the greenwich meridian the greatest ?
February and november.
.apparent solar time or true solar time is given by the daily apparent motion of the true, or observed, sun. it is based on the apparent solar day, which is the interval between two successive returns of the sun to the local meridian..mean solar time conceptually is the hour angle of the fictitious mean sun. the duration of daylight varies during the year the length of a mean solar day is nearly constant, unlike that of an apparent solar day... true solar time duration is between 23 h 59 min 39 s et 24 h 0 min 30 s..because many of these long or short days occur in succession, the difference builds up so that mean time is greater than apparent time by about 16 minutes in february and mean time is less than apparent time by about 16 minutes in november.. 2474.the difference between the transit of the apparent sun and mean sun across the greenwich meridian is the greatest in february and november.
Question 184-29 : What is the highest latitude listed below at which the sun will reach an altitude of 90° above the horizon at some time during the year ?
23°.
.the axis of the earth is tilted 23.5° degrees from 'vertical'. this causes the northern hemisphere to be tilted toward the sun during half the year, and tilted away from the sun the other half of the year.. 2475.the sun is vertical perpendicular to the surface or as states an altitude of 90° above the horizon , at 23.5°s and 23.5°n for solstices.
Question 184-30 : Assuming mid latitudes 40° to 50°n/s. at which time of year is the relationship between the length of day and night, as well as the rate of change of declination of the sun, changing at the greatest rate ?
Spring equinox and autumn equinox.
. 2476.the length of daylight/night at a given latitude varies with the declination of the sun and the greatest rate of change of declination is when the sun is crossing the equator at the spring and autumn equinox. the rate of change of the length of daylight will therefore be greatest when the rate of change of declination is greatest.
Question 184-31 : At what approximate date is the earth closest to the sun perihelion ?
Beginning of january.
. 2477
Question 184-32 : At what approximate date is the earth furthest from the sun aphelion ?
Beginning of july.
. 2477.the aphelion is the point in the orbit of a planet or comet where it is farthest from the sun.
Question 184-33 : An aircraft at position 60°n 005°w tracks 090° t for 315 km. on completion of the flight the longitude will be ?
000°40'e.
.315 km = 170 nm.170 / cos 60 = 340'.340' = 5° 40'..60°n 005°w 5° 40' = 60°n 000°40'e.
Question 184-34 : The 'departure' between positions 60°n 160°e and 60°n 'x' is 900 nm. what is the longitude of 'x' ?
170°w.
.distance = change in longitude x cos latitude.900 nm = change in longitude x cos 60.therefore.change in longitude = 900 / 0.5.change in longitude = 1800'.1800' / 60' = 30°..if position 'x' is west 160°e 30° = 130°..if position 'x' is east 160°e + 30° = 170°w.
Question 184-35 : An aircraft at latitude 02°20'n tracks 180° t for 685 km. on completion of the flight the latitude will be ?
03°50's.
.on a meridian, 1° = 60 nm..we are heading south..685 km / 1.852 = 370 nm..370 / 60 = 6.16°..0.16° x 60 = 10' exactly 9.6 min..02°20'n 6°10' = 03°50's.. 2478
Question 184-36 : An aircraft at latitude 10° south flies north at a gs of 890 km/h. what will its latitude be after 1.5h ?
02°00'n.
.1° of longitude = 60 nm..ground speed is 890 km/h or 890 /1.852 = 480 kt..480 x 1.5h = 720 nm travelled...720/60 = 12°..you are 10° south of the equator, you are travelling north, you will reach the equator after 10°, you are now in northern hemisphere and you travel another 2° to arrive at 02°00'n.
Question 184-37 : An aircraft at latitude 10°north flies south at a groundspeed of 445 km/h. what will be its latitude after 3 h ?
02°00's.
.1° of longitude = 60 nm..ground speed is 445 km/h or 445 /1.852 = 240 kt..240 x 3h = 720 nm travelled...720/60 = 12°..you are 10° north of the equator, you are travelling south, you will reach the equator after 10°, you are now in southern hemisphere and you travel another 2° to arrive at 02°00's.
Question 184-38 : An aircraft is flying in the shortest possible way between point 1 and 2, and between point 2 and 3.point 1 60°n 30°w.point 2 60°n 20°w.point 3 60°n 10°w.the track change when passing point 2 will be approximately ?
9°.
.the aircraft follows a great circle track, which is on the polar side of the rhumb line track...givry correction = 1/2 x 10° x sin60° = 4.33..east track.depart point 1 on heading 090° 4.33 = 85.67°..at halfway between point 1 and point 2 85.67 + 4.33 = 090°.arriving point 2 on heading 90° + 4.33 = 94.33°...crossing point 2 on course to point 3, we perform the same heading change when we have left point 1..the track change when passing point 2 will be approximately 94.33° 85.67 = 8.66° minus 9° changes.. 1418
Question 184-39 : At the magnetic equator, when accelerating after take off on heading west, a direct reading pivot suspended compass ?
Indicates the correct heading.
.magnetic dip is the angle between the horizontal and vertical forces acting on a compass needle toward the nearer pole.. 1430.a direct reading compass has a pivoted magnet that is free to aligne itself with the horizontal component of the earth's magnetic field. 1945.the compass is pendulous and symmetrical and will sit level if there is no magnetic field or if the field is horizontal with no vertical component of the earth's magnetic force, as on the magnetic equator...in fact, the pendulosity helps to counteract the effect of dip, trying to keep the compass card level...but because the cg is under the pivot point, accelerating makes the bulk of the compass lag behind the machine and displace the cg aft of the pivot point. if you were just going north or south, all you would get is extra dip, but if you are going east or west, the north bit of the compass is pointing to the side of the aircraft, and the displaced cg, not being vertically in line with the pivot point, ges towards north to create a couple that makes the compass turn clockwise to read less than 90° during the turn. a deceleration has the opposite effect... acceleration errors are maximum on eat/west headings and near the magnetic poles, and nil on north/south headings, and at the equator...a mnemonic for the effect of acceleration error is the word ' ands ' acceleration north, deceleration south..acceleration causes an indication toward north deceleration causes an indication toward south.
Question 184-40 : The circumference of the parallel of latitude at 60°n is approximately ?
10 800 nm.
.360° x 60 nm x cos 60 = 10 800 nm.
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