TEST 6 Flashcards

1
Q

What was a design flaw of the De Havilland Comet?
Metal fatigue was causing them to break up in flight.
Expensive.
Camber Fatigue.
Under powered.

A

Metal fatigue was causing them to break up in flight.

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2
Q

You are 30NM east of Miami, when do you slow down to 250KIAS?
10000ft within 14nm of the coast.
12000ft within 12nm of the coast.
10000ft within 12nm of the coast.

A

10000ft within 12nm of the coast.
Speed restrictions of 250 knots do not apply to aircraft operating beyond 12 NM from the coastline within the U.S. Flight Information Region, in offshore Class E airspace below 10,000 feet MSL. However, in airspace underlying a Class B airspace area designated for an airport, or in a VFR corridor designated through such as a Class B airspace area, pilots are expected to comply with the 200 knot speed limit specified in 14 CFR Section 91.117(c). (See 14 CFR Sections 91.117(c) and 91.703.)

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3
Q
In standard holding on the 020R 40DME fix at 6000 feet with 20nm legs, where will the tail of your bearing pointer be on the outbound turn?
  18 degrees.
  12 degrees.
  014 degrees.
  16 degrees.
A

014 degrees.
You are flying inbound on the 020R so the head of your bearing pointer is pointed at the VOR (200 deg). You begin your standard rate turn at 200KIAS (holding speed at 6000ft). To find the radius (90 deg) of your turn use the 1% of your GS (no wind so its 200Kts) which is 2nm. The diameter (180 deg) of the turn is twice that…so 4NM. To find how many radials you have gone in your outbound turn use the 60-to-1 rule. At 40DME 1nm = 1.5radials. So in your outbound right hand turn you ended up 6 radials to the north or on the 014R (4nm turn x 1.5 radials / NM @ 40NM) = 6 radials. So assuming you are now heading 020 deg on the outbound leg with no wind, the head is still pointing at the VOR so the tail of your bearing pointer will be pointing to 014 degrees (the radial you are on at the completion of the outbound turn). Remember the tail will fall slightly in the outbound leg and the head will rise.

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4
Q

If you are at FL370 and descending to FL200 at a VSI of 2500fpm, how long will it take to descend?

  1. 8min.
  2. 6min.
  3. 4min.
  4. 8min.
A

6.8min.Incorrect

(37000-20000) ft / 2500 ft / min = 6.8min

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5
Q

What does L/DMAX give you for a jet aircraft?
Max endurance, Max angle of climb, Min power off glide range (or best glide speed).
Max range, Max angle of climb, Max power off glide range (or best glide speed).
Max endurance, Max angle of climb, Max power off glide range (or best glide speed).

A

Max endurance, Max angle of climb, Max power off glide range (or best glide speed).
Ratio determines the airspeed at which the most lift is produced for the least amount of drag.

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6
Q
What part of the wing do you want to stall first?
  Trailing edge.
  Center of the wing.
  Wing Tips.
  The wing root.
A

The wing root.
Correct
The wing root, which allows control surfaces (ailerons) to remain functional. A stalled wing root moves the CP aft providing a nose down moment helping in stall recovery.

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7
Q
Fuel burn in holding is 3000lb/hr per engine. You have 5000lbs of fuel, how long can you hold?
  35min
  45min
  40min
  50min
A

50min
Correct
5000lbs / 6000lbs/hr = 5/6 hr or 50min

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8
Q
The bearing pointer moves from 5 degrees in front of the wing to 5 degrees behind the wing in 8 minutes, you are doing 360kts. How far from the station are you?
  268nm
  285nm
  290nm
  288nm
A

288nm
Correct
360 NM/hr / 60 min/hr = 6 NM/min. 6NM / min x 8 min = 48 NM. 48NM / 10 degrees = 4.8 degrees / NM x 60 = 288nm. At 60nm, 1 degree = 1nm. At 120nm 1 degree = 2nm. At 180nm 1 degree = 3nm, at 240nm 1 degree = 4nm. So the answer makes sense. At 288nm 1 degree = 4.8nm

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9
Q

What is ground effect?
A reduction in the downwash and wingtip vortices caused within one wingspan of the ground, reduces induced drag and gives the feeling of “floating.”
An increase in the downwash and wingtip vortices caused within one wingspan of the ground, reduces induced drag and gives the feeling of “floating.”
A reduction in the upwash and wingtip vortices caused within one wingspan of the ground, reduces induced drag and gives the feeling of “floating.”

A

A reduction in the downwash and wingtip vortices caused within one wingspan of the ground, reduces induced drag and gives the feeling of “floating.”

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10
Q
10. Question
You’re at 1500’ AGL, and you are flying at 230KIAS. Your approach speed is 130KIAS, and your aircraft slows at a rate of 10kts/NM. How far out from the runway do you need to slow down in order to shoot the ILS?
  22nm
  15nm
  18nm
  24nm
A

15nm

Correct
100kt speed reduction / 10kts/NM = 10nm. The outer marker of a typical ILS is at 5nm, where you would want to be at approach speed. So,15nm from the airport.

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11
Q
You are flying Mach 0.7 with a 50kt tailwind. What is your ground speed?
  425kts GS
  470kts GS
  450kts GS
  415kts GS
A

470kts GS
Correct
0.7IMN = 7NM/min x 60min/hr = 420KTAS + 50kts wind = 470kts GS

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12
Q
You are flying at 20000 feet with a thunderstorm 20nm in front of you. The radar look up angle is 5 degrees. How high is the thunderstorm?
  24000 ft
  30000 ft
  28000 ft
  22000 ft
A

30000 ft
Correct
1 degree of radar elevation is equal to 100ft at 1nm.
5deg x 100feet x 20nm = 10000 feet. (Add to 20000ft to get 30000ft)

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13
Q
You are at 25000 ft, there is a Level 5 thunderstorm 80nm in front of you. You tilt your radar up 1.5 degrees, and the cell disappears. How high is the top of the thunderstorm?
  35000 ft
  31000 ft
  32000 ft
  37000 ft
A

37000 ft
Correct
1 degree of radar elevation is equal to 100ft at 1nm.
1.5deg x 100feet x 80nm = 150 x 80 = 12000 feet. (Add to 25000ft to get 37000)

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14
Q
You are at 90DME at FL300 inbound to the VOR. You are requested to cross the station at 3000ft. If you are at .6IMN, when do you start down and what is descent gradient?
  4 degrees / 1600fpm/ 90nm.
  6 degrees / 1800fpm / 90nm.
  3 degrees / 1800fpm / 90nm.
  3 degrees / 1800fpm / 80nm.
A

3 degrees / 1800fpm / 90nm.
27000/90NM = 300ft/NM 0r 3 deg gradient
300ft/NM x 6NM/min = 1800fpm
** One sheet had 81NM (27K x 3 = 81NM) and a descent gradient of 3.3 deg (27000ft/81NM = 330ft/NM = 3.3 deg gradient)

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15
Q
You are on a northerly heading and the VOR bearing pointer is showing a bearing of 260. Eight minutes later the bearing is showing 280. You’re traveling 420KTAS and the wind is 360 @ 60KTS. How far away is the VOR?
  145nm
  144nm
  124nm
  141nm
A

144nm
Correct
420KTAS – 60KTS headwind = 360KTS GS. 360KTS/60min = 6NM/min
6NM/min x 8 min = 48NM / 20 degrees = 2.4 NM per degree
Apply the 60-to-1 rule so 60 x 2.4 = 144nm

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16
Q
17. Question
You are established on the 20DME arc to the north and you are going to intercept the 120R inbound to the VOR. Your flying 300KTAS, what is your lead point?
  129 radial.
  125 radial.
  128 radial.
  139 radial.
A

129 radial.
Incorrect
Turn radius = TAS/60min/hr – 2, so 300 / 60 – 2 = 3NM
Lead point = 60/DME x Turn radius (NM) = 60 / 20NM x 3NM = 9 degrees
Lead Radial = Radial +/- Lead radial. In this case we are approaching R120 from the south (north heading) so we would start a left hand turn at R129 to intercept the R120 inbound.

17
Q
Your DME is inoperative. You turn 90 degrees to the station and time for 1min. The needle falls from 5 degrees above the wingline to 5 degrees below the wingline. You are doing 480KTAS. What is your distance from the station?
  42nm
  40nm
  45nm
  48NM
A
48NM
Correct
480KTAS (assuming no wind) = 8NM/min in 1 minute, you have traveled 8NM.
8NM / 10 degrees = 0.8 NM per degree
60 x 0.8 NM per degree = 48NM
18
Q
On the 10DME arc you need a 2NM lead radial, how many radials is that?
  12 radials.
  6 radials.
  3 radials.
  4 radials.
A

12 radials.

60-to-1 Rule, at 60NM, 1 Radial = 1NM, so at 10NM from the VOR there are 6 radials/NM, multiplied by 2NM and you get 12 radials.

19
Q

If you set take off EPR, and you only get 80% RPM what do you do?
Continue takeoff.
Max power continue takeoff.
Abort take off.

A

Abort take off.

20
Q
At Mach 0.6, what is the VSI in a 1degree descent?
  500 fpm
  600fpm
  400 fpm
  300 fpm
A
600fpm
Correct
0.6IMN = 6NM/min
1 degree = 100ft/NM
VSI = 6NM/min x 100ft/NM = 600ft/min
21
Q

MDA for the LOC 27 is 450’AGL, Time from FAF to MAP is 3:00. How long after the FAF will your VDP occur?

2: 45
2: 15
3: 00
2: 05

A

2:15
Correct
Take 10% of the MDA (AGL) and subtract it from the total time. That is your time on the approach until reaching the VDP 450 x 10% = 45 sec. 3:00 – 0:45 = 2:15

22
Q
Inbound to the VOR heading 180 degrees at 300KTAS. You want to intercept the 20 DME arc, what is your lead point?
  22nm.
  23nm.
  21nm.
  18nm.
A

23nm.
Correct
Turn radius = 300 / 60 – 2 = 3NM
Lead point = Arc DME + turn radius

23
Q
Which of the following is not normally connected to the accessory gear?
  AC Packs.
  Oil pump.
  Hydraulic pump.
  Fuel pumps.
A

AC Packs.

24
Q
You are at FL270, flying at .75IMN, and are 25 DME from the VOR. What descent gradient and VSI is required to be at 12000 feet over the VOR?
  1600 ft/min 4 degrees.
  4500ft/min and 6 degrees.
  1800 ft/min 3 degrees.
  4200 ft/min and 3 degrees.
A

4500ft/min and 6 degrees.
15000ft/25NM = 600ft/NM (remember 100ft/NM = 1 deg gradient), so 600ft/nm is 6 deg gradient.
.75IMN = 7.5NM/min x 600ft/NM = 4500fpm descent rate.

25
Q
You are at FL350 going Mach .84, 112nm from XYZ VOR. You are cleared to cross 12nm prior to XYZ VOR at 10000 feet and 250KIAS. Indicated airspeed in the descent is 300KIAS. When do you start your descent?
  92nm
  88nm
  95nm
  85nm
A

92nm
Correct
25000ft of altitude to lose / 1000 x 3 =75nm. Reduce speed by 50kts at a rate of 10kts / 1 NM = 5nm. So far we have 80nm. We must be at 10000 feet by 12nm prior to the VOR, so we’ll add the 12 to the 80nm for a total of 92nm from the VOR.