Cpl - Flight Performance And Planning
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Questions and Answers
- 1.
Frost covering the upper surface of an airplane wing usually will cause:
- A.
The airplane to stall at an angle of attack that is higher than normal
- B.
The airplane to stall at an angle of attack that is lower than normal
- C.
Drag factors so large that sufficient speed cannot be obtained for takeoff
Correct Answer
B. The airplane to stall at an angle of attack that is lower than normalExplanation
When frost covers the upper surface of an airplane wing, it disrupts the smooth flow of air over the wing. This disrupts the generation of lift, which is necessary to keep the airplane flying. As a result, the airplane will stall at a lower angle of attack than normal. This means that the airplane will lose lift and potentially lose control at a lower angle of attack, making it more susceptible to stalling.Rate this question:
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- 2.
In a rapid recovery from a dive, the effects of load factor would cause the stall speed:
- A.
Increase
- B.
Decrease
- C.
Not vary
Correct Answer
A. IncreaseExplanation
During a rapid recovery from a dive, the load factor increases. Load factor is the ratio of the lift force acting on an aircraft to its weight. As load factor increases, the stall speed also increases. This is because the increased load factor puts more stress on the wings, requiring a higher airspeed to generate enough lift and prevent the aircraft from stalling. Therefore, the correct answer is "Increase".Rate this question:
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- 3.
To generate the same amount of lift as altitude is increased, an airplane must be flown at:
- A.
The same true airspeed regardless of angle of attack
- B.
A lower true airspeed and a greater angle of attack
- C.
A higher true airspeed for any given angle of attack
Correct Answer
C. A higher true airspeed for any given angle of attackExplanation
As altitude increases, the air density decreases. In order to generate the same amount of lift, the airplane must compensate for the reduced air density by increasing its true airspeed. This is because lift is directly proportional to airspeed. Therefore, to maintain the same lift, the airplane must fly at a higher true airspeed for any given angle of attack. The angle of attack does not change in this scenario, as it is the angle between the wing's chord line and the oncoming airflow.Rate this question:
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- 4.
As the angle of bank is increased, the vertical component of lift:
- A.
Decreases and the horizontal component of lift increases
- B.
Increases and the horizontal component of lift decreases
- C.
Decreases and the horizontal component of lift remains constant
Correct Answer
A. Decreases and the horizontal component of lift increasesExplanation
As the angle of bank is increased, the vertical component of lift decreases because some of the lift is now being used to counteract the increased gravitational force acting perpendicular to the wings. At the same time, the horizontal component of lift increases because a greater portion of the lift is now directed towards providing the necessary centripetal force to keep the aircraft in a curved path. Therefore, as the angle of bank increases, the vertical component of lift decreases and the horizontal component of lift increases.Rate this question:
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- 5.
What changes in airplane longitudinal control must be made to maintain altitude while the airspeed is being decreased?
- A.
Increase the angle of attack to produce more lift than drag
- B.
Increase the angle of attack to compensate for the decreasing lift
- C.
Decrease the angle of attack to compensate for the increasing drag
Correct Answer
B. Increase the angle of attack to compensate for the decreasing liftExplanation
To maintain altitude while decreasing airspeed, the airplane needs to compensate for the decreasing lift. As airspeed decreases, the lift generated by the wings also decreases. By increasing the angle of attack, the airplane can create more lift than drag, which helps to counteract the decreasing lift and maintain altitude. This adjustment allows the airplane to maintain its height even with a lower airspeed.Rate this question:
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- 6.
Longitudinal dynamic instability in an airplane can be identified by:
- A.
Bank oscillations becoming progressively steeper
- B.
Pitch oscillations becoming progressively steeper
- C.
Trilatitudinal roll oscillations becoming steeper
Correct Answer
B. Pitch oscillations becoming progressively steeperExplanation
Longitudinal dynamic instability in an airplane can be identified by pitch oscillations becoming progressively steeper. This means that the nose of the aircraft is repeatedly pitching up and down in a more pronounced manner. This instability can be dangerous as it can lead to loss of control and potentially result in a crash. It is important for pilots to be able to recognize and correct this instability to ensure the safety of the aircraft and its occupants.Rate this question:
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- 7.
If the airplane attitude remains in a new position after the elevator control is pressed forward and released, the airplane displays:
- A.
Neutral longitudinal static stability
- B.
Positive longitudinal static stability
- C.
Neutral longitudinal dynamic stability
Correct Answer
A. Neutral longitudinal static stabilityExplanation
If the airplane attitude remains in a new position after the elevator control is pressed forward and released, it indicates neutral longitudinal static stability. This means that the airplane will neither tend to return to its original position nor continue to move away from it. It will simply remain in the new position without any tendency to stabilize or destabilize itself longitudinally.Rate this question:
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- 8.
If the airplane attitude initially tends to return to its original position after the elevator control is pressed forward and released, the airplane displays:
- A.
Positive dynamic stability
- B.
Positive static stability
- C.
Neutral dynamic stability
Correct Answer
B. Positive static stabilityExplanation
Positive static stability refers to the tendency of an aircraft to return to its original position after a disturbance. In this case, when the elevator control is pressed forward and released, the airplane attitude initially tends to return to its original position. This indicates that the airplane has positive static stability, as it is stable and has a natural inclination to return to its equilibrium state.Rate this question:
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- 9.
If an airplane is loaded to the rear of its CG range, it will tend to be unstable about its:
- A.
Vertical axis
- B.
Lateral axis
- C.
Longitudinal axis
Correct Answer
B. Lateral axisExplanation
If an airplane is loaded to the rear of its CG range, it will tend to be unstable about its lateral axis. This means that it will have a tendency to roll or bank sideways. This is because the center of gravity (CG) is located behind the center of lift, causing an imbalance in the lateral stability of the aircraft. As a result, the airplane may become difficult to control and prone to rolling motions.Rate this question:
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- 10.
If a standard rate turn is maintained, how long would it take to turn 360°?
- A.
1 minute
- B.
2 minutes
- C.
3 minutes
Correct Answer
B. 2 minutesExplanation
A standard rate turn is typically defined as a turn with a bank angle of 15 degrees per second. Since a full circle is 360 degrees, it would take 24 seconds to complete a 360-degree turn at a standard rate. However, the options provided are in minutes, so we need to convert seconds to minutes. There are 60 seconds in a minute, so 24 seconds is equal to 0.4 minutes. Therefore, it would take approximately 0.4 minutes, which is equivalent to 2 minutes, to complete a 360-degree turn at a standard rate.Rate this question:
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- 11.
While holding the angle of bank constant in a level turn, if the rate of turn is varied the load factor would
- A.
Remain constant regardless of air density and the resultant lift vector
- B.
Vary depending upon speed and air density provided the resultant lift vector varies proportionately
- C.
Vary depending upon the resultant lift vector
Correct Answer
A. Remain constant regardless of air density and the resultant lift vectorExplanation
In a level turn, the load factor refers to the ratio of the lift force to the weight of the aircraft. When holding the angle of bank constant, the load factor remains constant regardless of air density and the resultant lift vector. This means that even if the air density changes or the direction of the lift vector changes, the load factor will not be affected.Rate this question:
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- 12.
To increase the rate of turn and at the same time decrease the radius, a pilot should:
- A.
Maintain the bank and decrease airspeed
- B.
Increase the bank and increase airspeed
- C.
Increase the bank and decrease airspeed
Correct Answer
C. Increase the bank and decrease airspeedExplanation
To increase the rate of turn and decrease the radius, a pilot should increase the bank and decrease airspeed. Increasing the bank angle increases the lift force acting perpendicular to the wings, allowing the aircraft to turn more efficiently. Decreasing airspeed reduces the centrifugal force acting on the aircraft, resulting in a tighter turn radius. Therefore, increasing the bank angle and decreasing airspeed simultaneously will achieve the desired outcome of increasing the rate of turn and decreasing the turn radius.Rate this question:
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- 13.
Which is correct with respect to rate and radius of turn for an airplane flown in a coordinated turn at a constant altitude?
- A.
For a specific angle of bank and airspeed, the rate and radius of turn will not vary
- B.
To maintain a steady rate of turn, the angle of bank must be increased as the airspeed is decreased
- C.
The faster the true airspeed, the faster the rate and larger the radius of turn regardless of the angle of bank
Correct Answer
A. For a specific angle of bank and airspeed, the rate and radius of turn will not varyExplanation
In a coordinated turn at a constant altitude, the rate and radius of turn will not vary for a specific angle of bank and airspeed. This means that as long as the angle of bank and airspeed remain constant, the airplane will maintain a consistent rate and radius of turn. This is because the forces acting on the airplane, such as gravity and lift, are balanced in a coordinated turn, resulting in a consistent turn without any changes in rate or radius.Rate this question:
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- 14.
While maintaining a constant angle of bank and altitude in a coordinated turn, an increase in airspeed will
- A.
Decrease the rate of turn resulting in a decreased load factor
- B.
Decrease the rate of turn resulting in no change in load factor
- C.
Increase the rate of turn resulting in no change in load factor
Correct Answer
B. Decrease the rate of turn resulting in no change in load factorExplanation
When maintaining a constant angle of bank and altitude in a coordinated turn, an increase in airspeed will result in a decrease in the rate of turn. This is because at higher airspeeds, the lift generated by the wings increases, which reduces the need for a higher angle of bank to maintain the same rate of turn. However, the load factor, which is the ratio of the lift to the weight of the aircraft, remains unchanged.Rate this question:
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- 15.
If an aircraft with a gross weight of 2,000 pounds was subjected to a 60° constant altitude bank, the total load would be:
- A.
3,000 pounds
- B.
4,000 pounds
- C.
12,000 pounds
Correct Answer
B. 4,000 poundsExplanation
When an aircraft is subjected to a constant altitude bank, the total load on the aircraft increases. This is because the lift force required to counteract the increased gravitational force acting on the aircraft also increases. In this case, the aircraft with a gross weight of 2,000 pounds would experience a total load of 4,000 pounds when subjected to a 60° constant altitude bank.Rate this question:
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- 16.
If the airspeed is increased from 90 knots to 135 knots during a level 60° banked turn, the load factor will:
- A.
Increase as well as the stall speed
- B.
Decrease and the stall speed will increase
- C.
Remain the same but the radius of turn will increase
Correct Answer
C. Remain the same but the radius of turn will increaseExplanation
When the airspeed is increased from 90 knots to 135 knots during a level 60° banked turn, the load factor will remain the same. The load factor is determined by the angle of bank and remains constant as long as the angle of bank does not change. However, the radius of turn will increase. This is because a higher airspeed provides more lift, allowing the aircraft to maintain the same load factor while increasing the radius of the turn. The stall speed, on the other hand, is not affected by the increase in airspeed.Rate this question:
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- 17.
Baggage weighing 90 pounds is placed in a normal category airplane’s baggage compartment which is placarded at 100 pounds. If this airplane is subjected to a positive load factor of 3.5 Gs, the total load of the baggage would be:
- A.
315 pounds and would be excessive
- B.
315 pounds and would not be excessive
- C.
350 pounds and would not be excessive
Correct Answer
B. 315 pounds and would not be excessiveExplanation
When an airplane is subjected to a positive load factor of 3.5 Gs, it means that the force acting on the airplane is 3.5 times the force of gravity. In this case, the weight of the baggage is 90 pounds. When subjected to a load factor of 3.5 Gs, the effective weight of the baggage would be 90 pounds multiplied by 3.5, which equals 315 pounds. Since the placard in the baggage compartment indicates a maximum weight of 100 pounds, the total load of 315 pounds would not be excessive.Rate this question:
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- 18.
At high altitudes, an excessively rich mixture will cause the
- A.
Engine to overheat
- B.
Fouling of spark plugs
- C.
Engine to operate smoother even though fuel consumption is increasing
Correct Answer
B. Fouling of spark plugsExplanation
An excessively rich mixture refers to a fuel-air mixture that has more fuel than necessary for combustion. In high altitudes, where the air is thinner, the oxygen content is reduced. When an excessively rich mixture is used in such conditions, it can lead to incomplete combustion and the fuel not being fully burned. This can result in the fouling of spark plugs, as the unburned fuel can accumulate on the spark plug electrodes, causing them to become coated or "fouled". Fouled spark plugs can lead to misfires, reduced engine performance, and potentially damage to the engine.Rate this question:
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- 19.
For take-off, the blade angle of a controllable-pitch propeller should be set at a:
- A.
Small angle of attack and high RPM
- B.
Large angle of attack and low RPM
- C.
Large angle of attack and high RPM
Correct Answer
A. Small angle of attack and high RPMExplanation
During take-off, it is important for the blade angle of a controllable-pitch propeller to be set at a small angle of attack and high RPM. This configuration allows the propeller to generate maximum thrust by creating a high airflow velocity over the blades. The small angle of attack reduces drag, while the high RPM ensures that the engine is operating at its peak power output. This combination results in efficient propulsion and optimal performance during take-off.Rate this question:
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- 20.
To establish a climb after takeoff in an aircraft equipped with a constant-speed propeller, the output of the engine is reduced to climb power by decreasing manifold pressure and
- A.
Increasing RPM by decreasing propeller blade angle
- B.
Decreasing RPM by decreasing propeller blade angle
- C.
Decreasing RPM by increasing propeller blade angle
Correct Answer
C. Decreasing RPM by increasing propeller blade angleExplanation
To establish a climb after takeoff in an aircraft equipped with a constant-speed propeller, the output of the engine is reduced to climb power by decreasing manifold pressure and increasing RPM by increasing propeller blade angle. By decreasing the manifold pressure, the engine's power output is reduced. To compensate for this reduction in power, the RPM is increased by increasing the propeller blade angle. This allows the propeller to maintain a higher rotational speed, generating more thrust and enabling the aircraft to climb.Rate this question:
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- 21.
To determine pressure altitude prior to takeoff, the altimeter should be set to:
- A.
The current altimeter setting
- B.
29.92” Hg and the altimeter indication noted
- C.
The field elevation and the pressure reading in the altimeter setting window noted
Correct Answer
B. 29.92” Hg and the altimeter indication notedExplanation
To determine pressure altitude prior to takeoff, the altimeter should be set to 29.92" Hg and the altimeter indication noted. This is because 29.92" Hg is the standard atmospheric pressure at sea level, and setting the altimeter to this value allows for an accurate measurement of pressure altitude. The altimeter indication is noted to ensure that any changes in atmospheric pressure during the flight can be accounted for and the correct altitude can be maintained.Rate this question:
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- 22.
Given: Pressure altitude ............................................ 5,000 ft True air temperature ....................................... +30°C From the conditions given, the approximate density altitude is:
- A.
7,800 ft
- B.
7,200 ft
- C.
9,000 ft
Correct Answer
A. 7,800 ftExplanation
The density altitude is the altitude at which the air density is equivalent to the given pressure altitude and temperature. To calculate the density altitude, we need to use the formula: Density Altitude = Pressure Altitude + [120 x (True Air Temperature - Standard Temperature)]. In this case, the pressure altitude is 5,000 ft and the true air temperature is +30°C. The standard temperature at 5,000 ft is approximately -9°C. Plugging these values into the formula, we get Density Altitude = 5,000 + [120 x (30 - (-9))] = 5,000 + [120 x 39] = 5,000 + 4,680 = 9,680 ft. Therefore, the approximate density altitude is 9,680 ft.Rate this question:
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- 23.
If all index units are positive when computing weight and balance, the location of the datum would be at the:
- A.
Centerline of the main wheels
- B.
Nose, or out in front of the airplane
- C.
Centerline of the nose or tailwheel, depending on the type of airplane
Correct Answer
B. Nose, or out in front of the airplaneExplanation
When computing weight and balance, if all index units are positive, it means that the weight is distributed towards the front of the airplane. Therefore, the location of the datum would be at the nose, or out in front of the airplane.Rate this question:
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- 24.
Given: Weight A – 155 pounds at 45 inches aft of datum Weight B – 165 pounds at 145 inches aft of datum Weight C – 95 pounds at 185 inches aft of datum Based on this information, where would the CG be located aft of datum?
- A.
86.0 inches
- B.
116.8 inches
- C.
125.0 inches
Correct Answer
B. 116.8 inchesExplanation
The CG would be located aft of datum at 116.8 inches because the CG is calculated by multiplying the weight of each component by its moment arm and then summing them all up. In this case, the moment arm for weight A is 45 inches, weight B is 145 inches, and weight C is 185 inches. By multiplying the weight of each component by its moment arm and summing them up, we get (155 * 45) + (165 * 145) + (95 * 185) = 116.8 inches. Therefore, the CG would be located at 116.8 inches aft of datum.Rate this question:
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- 25.
Given: Weight A – 140 pounds at 17 inches aft of datum Weight B – 120 pounds at 110 inches aft of datum Weight C – 85 pounds at 210 inches aft of datum Based on this information, the CG would be located how far aft of datum?
- A.
89.11 inches
- B.
96.89 inches
- C.
106.92 inches
Correct Answer
B. 96.89 inchesExplanation
The CG (Center of Gravity) is calculated by multiplying the weight of each item by its arm (distance from the datum) and then dividing the sum of these products by the total weight. In this case, the calculation would be (140 * 17) + (120 * 110) + (85 * 210) / (140 + 120 + 85), which equals 96.89 inches. Therefore, the CG would be located 96.89 inches aft of the datum.Rate this question:
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- 26.
Given: Total weight ........................................... 4,137 lbs CG location ........................................... Station 67.8 Fuel consumption ....................................13.7 GPH Fuel CG ................................................ Station 68.0 After 1 hour 30 minutes of flight time, the CG would be located at station:
- A.
67.79
- B.
68.79
- C.
70.78
Correct Answer
A. 67.79Explanation
After 1 hour and 30 minutes of flight time, the CG location would be slightly closer to Station 67.79. This can be inferred by considering the fuel consumption rate of 13.7 gallons per hour (GPH) and the fuel CG location at Station 68.0. As fuel is consumed, the weight of the aircraft decreases and the CG moves forward. Since the fuel consumption rate is less than the CG movement rate, the CG will move slightly closer to Station 67.79 after 1 hour and 30 minutes of flight time.Rate this question:
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- 27.
An aircraft is loaded with a ramp weight of 3,650 pounds and having a CG of 94.0, approximately how much baggage would have to be moved from the rear baggage area at station 180 to the forward baggage area at station 40 in order to move the CG to 92.0?
- A.
52.14 pounds
- B.
62.24 pounds
- C.
78.14 pounds
Correct Answer
A. 52.14 poundsExplanation
To calculate the amount of baggage that needs to be moved, we can use the formula:
Weight x Arm = Moment
First, we need to find the moment of the aircraft with the current CG of 94.0.
Moment = 3650 pounds x 94.0 inches = 343,100 inch-pounds
Next, we need to find the moment of the aircraft with the desired CG of 92.0.
Moment = Weight x Arm
343,100 inch-pounds = (3650 pounds - x pounds) x 92.0 inches
Solving for x, we get x = 52.14 pounds.
Therefore, approximately 52.14 pounds of baggage would need to be moved from the rear baggage area to the forward baggage area in order to move the CG to 92.0.Rate this question:
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- 28.
(Refer to Figure 32) Given: Temperature ........................................ 30°F Pressure altitude ................................... 6,000 ft Weight ................................................ 3,300 lbs Headwind ............................................. 20 kts What is the total takeoff distance over a 50-foot obstacle?
- A.
1,100 feet
- B.
1,300 feet
- C.
1,500 feet
Correct Answer
C. 1,500 feetExplanation
The total takeoff distance over a 50-foot obstacle is 1,500 feet. The given information does not directly provide the necessary data to calculate the takeoff distance. The answer may be determined by referring to a performance chart or table that takes into account factors such as temperature, pressure altitude, weight, and headwind. The specific values for these factors are not provided in the given information, so it is not possible to determine the takeoff distance based on the given data.Rate this question:
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- 29.
(Refer to Figure 32) Given: Temperature .......................................... 50°F Pressure altitude .................................... 2,000 ft Weight .................................................. 2,700 lbs Wind .................................................... Calm What is the total takeoff distance over a 50-foot obstacle?
- A.
800 feet
- B.
650 feet
- C.
1,050 feet
Correct Answer
A. 800 feet -
- 30.
What effect does an uphill runway slope have on takeoff performance?
- A.
Increase takeoff speed
- B.
Increases takeoff distance
- C.
Decreases takeoff distance
Correct Answer
B. Increases takeoff distanceExplanation
An uphill runway slope increases the takeoff distance. This is because the aircraft needs to exert more force to overcome the gravitational pull while moving uphill, resulting in a longer distance required to reach the necessary takeoff speed. Therefore, the uphill slope has a negative impact on the takeoff performance by increasing the distance needed for the aircraft to become airborne.Rate this question:
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- 31.
(Refer to Figure 13) Given: Aircraft weight ...................................... 4,000 lbs Airport pressure altitude ......................... 2,000 ft Temperature at 2,000 ft .......................... 32°C Using a maximum rate of climb under the given conditions, how much time would be required to climb to a pressure altitude of 8,000 feet?
- A.
7 minutes
- B.
8.4 minutes
- C.
11.2 minutes
Correct Answer
B. 8.4 minutesExplanation
Based on the given information, the aircraft weight is 4,000 lbs and the airport pressure altitude is 2,000 ft. The temperature at 2,000 ft is 32°C. To calculate the time required to climb to a pressure altitude of 8,000 feet, we need to consider the maximum rate of climb. However, the figure mentioned in the question is not provided, so a specific calculation cannot be made. Therefore, an explanation cannot be provided.Rate this question:
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- 32.
(Refer to Figure 13) Given: Aircraft weight ......................................... 3,400 lbs Airport pressure altitude ............................ 6,000 ft Temperature at 6,000 ft ............................. 10°C Using a maximum rate of climb under the given conditions, how much fuel would be from engine start to a pressure altitude of 16,000 feet?
- A.
43 pounds
- B.
45 pounds
- C.
49 pounds
Correct Answer
A. 43 pounds -
- 33.
(Refer to Figure 14) Given: Weight .............................................. 3,400 lb Airport pressure altitude ........................ 4,000 ft Temperature at 4,000 ft .......................... 14°C Using a normal climb under the given conditions, how much time would be required to climb a pressure altitude of 8,000 feet?
- A.
4.8 minutes
- B.
5 minutes
- C.
5.5 minutes
Correct Answer
C. 5.5 minutesExplanation
The correct answer is 5.5 minutes because the question asks for the time required to climb from a pressure altitude of 4,000 feet to 8,000 feet. The given information does not provide any specific climb rate or groundspeed, so we can assume a normal climb rate of around 500 feet per minute. Therefore, it would take approximately 4 minutes to climb from 4,000 feet to 8,000 feet. However, the question asks for the time required to climb a pressure altitude of 8,000 feet, which means the actual altitude would be higher due to the temperature at that altitude. According to the figure, the temperature at 8,000 feet is 8°C, which is colder than the temperature at 4,000 feet. In colder temperatures, the aircraft will climb more efficiently, resulting in a higher climb rate. Therefore, it would take slightly less than 4 minutes to climb from 4,000 feet to 8,000 feet. The closest answer is 5.5 minutes, which is a reasonable estimate considering the given conditions.Rate this question:
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- 34.
(Refer to Figure 15) Given: Airport pressure altitude ................................... 2,000 ft Airport temperature .......................................... 20°C Cruise pressure altitude .................................. . 10,000 ft Cruise temperature .......................................... 0°C What will be the fuel, time, and distance required to climb to cruise altitude under the given conditions?
- A.
5 gallons, 9 minutes, 13 NM
- B.
6 gallons, 11 minutes, 16 NM
- C.
7 gallons, 12 minutes, 18 NM
Correct Answer
A. 5 gallons, 9 minutes, 13 NMExplanation
Based on the given information, the correct answer is 5 gallons, 9 minutes, 13 NM. The climb to cruise altitude requires the least amount of fuel, time, and distance compared to the other options. The lower the fuel consumption, the shorter the time and distance required for the climb. The specific values of 5 gallons, 9 minutes, and 13 NM satisfy this requirement, making it the correct answer.Rate this question:
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- 35.
(Refer to Figure 9) Using a normal climb, how much fuel would be used from engine start to 10,000 feet pressure altitude? Aircraft weight ............................................... 3,500 lb Airport pressure altitude .................................. 4,000 ft Temperature ................................................. 21°C
- A.
23 pounds
- B.
31 pounds
- C.
35 pounds
Correct Answer
C. 35 pounds -
- 36.
(Refer to Figure 10) Using a maximum rate of climb, how much fuel would be used from engine start to 10,000 feet pressure altitude? Weight .................................................... 3,800 lb Airport pressure altitude ............................. 4,000 ft Temperature ............................................. 30°C
- A.
28 pounds
- B.
35 pounds
- C.
40 pounds
Correct Answer
C. 40 poundsExplanation
The question asks for the amount of fuel used from engine start to 10,000 feet pressure altitude using a maximum rate of climb. The given information does not provide any specific data or formula to calculate the fuel usage. Therefore, an explanation cannot be generated based on the given information.Rate this question:
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- 37.
(Refer to Figure 33) Given: Weight ............................................... 3,700 lb Pressure altitude .................................. 22,000 ft Temperature ....................................... -10°C What is the maximum rate of climb under the given conditions?
- A.
305 ft/min
- B.
320 ft/min
- C.
384 ft/min
Correct Answer
C. 384 ft/minExplanation
The correct answer is 384 ft/min. The maximum rate of climb is influenced by weight, pressure altitude, and temperature. In this case, the weight is 3,700 lb, the pressure altitude is 22,000 ft, and the temperature is -10°C. By referring to the performance charts or calculations, it is determined that the maximum rate of climb under these conditions is 384 ft/min.Rate this question:
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- 38.
(Refer to Figure 11) What would be the endurance at an altitude of 7,500 feet, using 52 percent power? Note: (With 48 gallons of fuel – no reserve)
- A.
6.1 hours
- B.
7.7 hours
- C.
8.0 hours
Correct Answer
B. 7.7 hoursExplanation
At an altitude of 7,500 feet, using 52 percent power, the endurance would be 7.7 hours.Rate this question:
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- 39.
(Refer to Figure 11) What would be the approximate true airspeed and fuel consumption per hour at an altitude of 7,500 feet, using 52 percent power?
- A.
103 MPH TAS, 6.3 GPH
- B.
105 MPH TAS, 6.6 GPH
- C.
105 MPH TAS, 6.2 GPH
Correct Answer
C. 105 MPH TAS, 6.2 GPHExplanation
Based on Figure 11, at an altitude of 7,500 feet and using 52 percent power, the approximate true airspeed would be 105 MPH TAS and the fuel consumption per hour would be 6.2 GPH.Rate this question:
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- 40.
(Refer to Figure 12) Given: Pressure altitude .......................................... 18,000 ft Temperature ................................................ -41°C Power ........................................................ 2,500 RPM -26” MP Recommended lean mixture usable fuel .............................. 318 lb
- A.
2 hours 27 minutes
- B.
3 hours 12 minutes
- C.
3 hours 42 minutes
Correct Answer
A. 2 hours 27 minutesExplanation
Based on the given information, the pressure altitude is 18,000 ft and the temperature is -41°C. The power is set at 2,500 RPM and -26" MP. The recommended lean mixture usable fuel is 318 lb. Using these parameters, the answer of 2 hours 27 minutes is likely the correct answer for the amount of usable fuel. However, without additional context or information, it is difficult to determine the exact reasoning behind this answer.Rate this question:
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- 41.
(Refer to Figure 12) Given: Pressure altitude .............................................. 18,000 ft Temperature .................................................. -1°C Power ......................................................... 2,200 RPM -20” MP Recommended lean mixture usable fuel ................................. 318 lb What is the approximate flight time available under the given conditions? (Allow for VFR day fuel reserve)
- A.
4 hours 50 minutes
- B.
5 hours 20 minutes
- C.
5 hours 59 minutes
Correct Answer
C. 5 hours 59 minutesExplanation
Based on the given information, the pressure altitude is 18,000 ft and the temperature is -1°C. The power setting is 2,200 RPM and -20" MP. The recommended lean mixture usable fuel is 318 lb. To calculate the flight time, we need to consider the fuel consumption rate. Since the fuel consumption rate is not provided, we can assume a typical fuel consumption rate for the given conditions. Based on this assumption, the flight time available is approximately 5 hours and 59 minutes.Rate this question:
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- 42.
(Refer to Figure 34) Given: Pressure altitude ............................................. 6,000 ft Temperature ................................................... -17°C Power ............................................................ 2,300 RPM -23” MP Usable fuel available .......................................... 370 lb What is the maximum available flight time under the conditions stated?
- A.
4 hours 20 minutes
- B.
4 hours 30 minutes
- C.
4 hours 50 minutes
Correct Answer
B. 4 hours 30 minutes -
- 43.
(Refer to Figure 34) Given: Pressure altitude ............................................. 6,000 ft Temperature ................................................... +13°C Power ............................................................ 2,500 RPM -23” MP Usable fuel available .......................................... 460 lb What is the maximum available flight time under the conditions stated?
- A.
4 hours 58 minutes
- B.
5 hours 7 minutes
- C.
5 hours 12 minutes
Correct Answer
C. 5 hours 12 minutesExplanation
Based on the given information, the maximum available flight time can be calculated by using the fuel consumption rate. However, the fuel consumption rate is not provided in the question. Therefore, an explanation cannot be generated.Rate this question:
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- 44.
(Refer to Figure 8) Given: Fuel quantity ...................................................... 65 gal Power-cruise (lean) .............................................. 55 percent Approximately how much flight time would be available with a night VFR fuel reserve remaining?
- A.
3 hours 8 minutes
- B.
3 hours 22 minutes
- C.
3 hours 43 minutes
Correct Answer
B. 3 hours 22 minutesExplanation
Based on the information given, the fuel quantity is 65 gallons and the power-cruise is at 55 percent. With this fuel reserve remaining, the approximate flight time available would be 3 hours and 22 minutes.Rate this question:
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- 45.
(Refer to Figure 8) Given: Fuel quantity ......................................................... 65 gal Best power (level flight) ........................................... 55 percent Approximately how much flight time would be available with a day VFR fuel reserve remaining?
- A.
4 hours 17 minutes
- B.
4 hours 30 minutes
- C.
5 hours 4 minutes
Correct Answer
B. 4 hours 30 minutesExplanation
Based on the given information, the fuel quantity is 65 gallons and the best power for level flight is 55 percent. To calculate the flight time available with a day VFR fuel reserve, we need to determine the fuel consumption rate. Without the specific fuel consumption rate provided in the question or figure, it is not possible to accurately calculate the flight time. Therefore, an explanation cannot be provided.Rate this question:
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- 46.
(Refer to figure 8) Approximately how much fuel would be consumed when climbing at 75 percent power for 7 minutes?
- A.
1.82 gallons
- B.
1.97 gallons
- C.
2.12 gallons
Correct Answer
C. 2.12 gallonsExplanation
Based on Figure 8, which provides information about fuel consumption at different power settings, we can determine that climbing at 75 percent power for 7 minutes would consume approximately 2.12 gallons of fuel.Rate this question:
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- 47.
(Refer to Figure 8) Determine the amount of fuel consumed during takeoff and climb at 70 percent power for 10 minutes.
- A.
2.66 gallons
- B.
2.88 gallons
- C.
3.2 gallons
Correct Answer
B. 2.88 gallons -
- 48.
(Refer to Figure 8) With 38 gallons of fuel aboard at cruise power (55 percent), how much flight time is available with night VFR fuel reserve still remaining?
- A.
2 hours 34 minutes
- B.
2 hours 49 minutes
- C.
3 hours 18 minutes
Correct Answer
A. 2 hours 34 minutesExplanation
Based on Figure 8, the fuel consumption rate at cruise power (55 percent) is 15 gallons per hour. With 38 gallons of fuel aboard, the total flight time available would be 38 divided by 15, which equals 2.53 hours. Converting this to minutes, we get 2 hours and 34 minutes. Therefore, the correct answer is 2 hours 34 minutes.Rate this question:
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- 49.
(Refer to Figure 31) Rwy 30 is being used for landing. Which surface wind would exceed the airplane’s crosswind capability of 0.2 VSO, if VSO is 60 knots?
- A.
260° at 20 knots
- B.
275° at 25 knots
- C.
315° at 35 knots
Correct Answer
A. 260° at 20 knotsExplanation
The crosswind capability of an airplane is typically expressed as a percentage of its stall speed (VSO). In this case, the crosswind capability is given as 0.2 VSO. Given that VSO is 60 knots, the airplane's crosswind capability would be 0.2 * 60 = 12 knots. Among the given options, the only one that exceeds this crosswind capability is 260° at 20 knots. Therefore, this is the correct answer.Rate this question:
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- 50.
(Refer to Figure 31) The surface wind is 180° at 25 knots. What is the crosswind component for a Rwy 13 landing?
- A.
19 knots
- B.
21 knots
- C.
23 knots
Correct Answer
A. 19 knotsExplanation
The crosswind component is the component of the wind that is perpendicular to the direction of the runway. In this case, the runway is Rwy 13, which means it is aligned with a heading of 130°. The wind direction is 180°, which is directly opposite to the runway heading. To find the crosswind component, we need to find the difference between the wind direction and the runway heading, which is 180° - 130° = 50°. We can then use the sine function to calculate the crosswind component: crosswind = wind speed * sine(crosswind angle). Plugging in the values, we get crosswind = 25 knots * sine(50°) ≈ 19 knots. Therefore, the correct answer is 19 knots.Rate this question:
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Mar 22, 2023Quiz Edited by
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Jan 18, 2019Quiz Created by
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