2a553c CDC Volume 2 Ure
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Questions and Answers
- 1.
Which of the following is not looked for during a visual preventative maintenance inspection?
- A.
Fungus growth.
- B.
Corroded connectors.
- C.
Sand and grit deposits.
- D.
Voltages within tolerances.
Correct Answer
D. Voltages within tolerances.Explanation
During a visual preventative maintenance inspection, various factors are examined to ensure the proper functioning and condition of the equipment. Fungus growth, corroded connectors, and sand and grit deposits are all potential issues that can affect the performance and longevity of the equipment. However, voltages within tolerances are not typically assessed during a visual inspection, as they require specialized equipment and testing procedures to accurately measure. Therefore, the correct answer is "Voltages within tolerances."Rate this question:
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- 2.
What type of damage softens the protective conformal coatings on circuit boards and components?
- A.
Electrostatic discharge (ESD).
- B.
Solvent.
- C.
Abrasion.
- D.
Solvent entrapment.
Correct Answer
B. Solvent.Explanation
Solvent damage refers to the weakening or softening of the protective conformal coatings on circuit boards and components. Solvents have the ability to dissolve or break down these coatings, compromising their protective properties. This can lead to potential damage or malfunction of the circuit board or its components.Rate this question:
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- 3.
What lubricant is available in an aerosol and used for light to moderate loads?
- A.
Silicone.
- B.
General-purpose grease.
- C.
General-purpose lubricating oil.
- D.
High-temperature lubricating oil.
Correct Answer
A. Silicone.Explanation
Silicone lubricant is available in an aerosol and is commonly used for light to moderate loads. It is a versatile lubricant that can be used in various applications and environments. Silicone lubricant is known for its high resistance to heat, water, and chemicals, making it suitable for a wide range of uses. It provides smooth and long-lasting lubrication, reducing friction and wear on moving parts.Rate this question:
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- 4.
The composite tool kit (CTK) program was developed to
- A.
Reduce tool costs through the use of chits.
- B.
Make end-of-shift tool inspections more efficient.
- C.
Eliminate damage to aircraft and to control aircraft costs.
- D.
Eliminate damage to aircraft, equipment, and to control tool assets.
Correct Answer
D. Eliminate damage to aircraft, equipment, and to control tool assets.Explanation
The correct answer is to eliminate damage to aircraft, equipment, and to control tool assets. The CTK program aims to prevent any harm or damage to aircraft and equipment by implementing effective measures and controls. By doing so, it helps to minimize costs associated with repairing or replacing damaged aircraft and tools. Additionally, the program focuses on maintaining and controlling tool assets, ensuring their proper usage and preventing any loss or misuse.Rate this question:
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- 5.
Which method of tool layout prevents a quick, accurate inventory of a composite tool kit (CTK)?
- A.
Foam cutouts in a tool tray.
- B.
Tools set loosely in a drawer.
- C.
Contrasting foam depressions.
- D.
Tools hung on a shadow board.
Correct Answer
B. Tools set loosely in a drawer.Explanation
Setting tools loosely in a drawer prevents a quick, accurate inventory of a composite tool kit (CTK) because it makes it difficult to visually assess which tools are present and which are missing. When tools are not securely organized or held in place, they can easily shift or become jumbled together, making it challenging to identify and count them accurately. This method of tool layout also increases the risk of misplacing or losing tools, leading to inefficiency and potential safety hazards.Rate this question:
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- 6.
How are small tools, which cannot be marked, controlled in a composite tool kit (CTK)?
- A.
In a way determined by the wing commander.
- B.
Tools that cannot be marked will not be placed in CTKs.
- C.
Create a designated position in the CTK for the item and listing it on an inventory.
- D.
In a container marked with the identification number and number of tools contained within.
Correct Answer
D. In a container marked with the identification number and number of tools contained within. -
- 7.
What helps to dissipate and lessen the effects of electrostatic discharges?
- A.
Using nonconducting packaging materials.
- B.
Using proper insulating materials.
- C.
Dehumidified air.
- D.
Humid air.
Correct Answer
D. Humid air.Explanation
Humid air helps to dissipate and lessen the effects of electrostatic discharges because it increases the conductivity of the surrounding environment. When the air is humid, the moisture in the air allows for the free movement of ions, which helps to neutralize any charges that may build up on objects. This reduces the likelihood of electrostatic discharges occurring and minimizes their effects.Rate this question:
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- 8.
At what voltage level can electrostatic discharges (ESD) not be seen, felt, or heard?
- A.
15,000.
- B.
9,000.
- C.
6,000.
- D.
4,000.
Correct Answer
D. 4,000.Explanation
Electrostatic discharges (ESD) are typically not visible, felt, or heard at voltage levels below 4,000. This means that any voltage level below 4,000 is too low to generate a noticeable ESD.Rate this question:
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- 9.
What material is designed to prevent the generation of a static charge, but is not a static shield?
- A.
Antistatic.
- B.
Conductive.
- C.
Nonconductive.
- D.
Static dissipative.
Correct Answer
A. Antistatic.Explanation
Antistatic material is designed to prevent the generation of a static charge. It is different from a static shield, which is specifically used to block or redirect static electricity. Antistatic materials work by allowing the static charge to flow through them, effectively neutralizing it. This helps to prevent damage to sensitive electronic components or equipment that may be affected by static electricity. Conductive materials, on the other hand, allow the flow of electricity and are often used for grounding purposes. Nonconductive materials do not allow the flow of electricity, while static dissipative materials allow the controlled dissipation of static charges.Rate this question:
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- 10.
What precaution must you take first before installing an electrostatic discharge (ESD) line replaceable unit (LRU)?
- A.
Remove protective caps or covers.
- B.
Touch the LRU case with your hands.
- C.
Ensure the aircraft is properly grounded.
- D.
Touch the ends of aircraft wiring to the LRU case.
Correct Answer
C. Ensure the aircraft is properly grounded.Explanation
Before installing an electrostatic discharge (ESD) line replaceable unit (LRU), the first precaution that must be taken is to ensure that the aircraft is properly grounded. This is important because grounding the aircraft helps to prevent the buildup of static electricity, which could potentially damage the LRU during installation. By ensuring that the aircraft is properly grounded, the risk of electrostatic discharge and potential damage to the LRU is minimized.Rate this question:
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- 11.
Shielded cable consists of
- A.
A metallic braid over the insulation.
- B.
A metal conductor covered with a dielectric.
- C.
An inner conductor separated by an insulation dielectric.
- D.
A compact wire bundle with self-contained wire assemblies.
Correct Answer
A. A metallic braid over the insulation.Explanation
Shielded cable consists of a metallic braid over the insulation. This metallic braid acts as a shield to protect the inner conductor from external electromagnetic interference. It helps to reduce noise and signal degradation by providing a barrier against electromagnetic radiation. The insulation layer underneath the metallic braid provides electrical insulation and further protection to the inner conductor. Overall, the metallic braid over the insulation enhances the performance and reliability of the shielded cable.Rate this question:
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- 12.
Which type of cable is used to carry radio frequency (RF) power from one point to another?
- A.
Electromagnetic pulse (EMP).
- B.
Insulated wire.
- C.
Shielded.
- D.
Coaxial.
Correct Answer
D. Coaxial.Explanation
Coaxial cables are specifically designed to carry high-frequency signals, including radio frequency (RF) power, from one point to another. They consist of a central conductor surrounded by an insulating layer, a metallic shield, and an outer insulating layer. The shield helps to minimize interference and signal loss, making coaxial cables ideal for transmitting RF power over long distances with minimal distortion. Electromagnetic pulse (EMP) refers to a burst of electromagnetic energy, not a type of cable. Insulated wire and shielded cables may be used for other purposes, but they are not specifically designed for carrying RF power.Rate this question:
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- 13.
What type of cable is required for greater shielding against electromagnetic interference?
- A.
Twisted pair.
- B.
Coaxial.
- C.
Triaxial.
- D.
Shielded.
Correct Answer
C. Triaxial.Explanation
Triaxial cable provides the highest level of shielding against electromagnetic interference compared to twisted pair, coaxial, and shielded cables. It consists of three layers of shielding, including an inner conductor, an insulating layer, and an outer conductor, which helps to minimize interference from external sources. This design makes triaxial cable ideal for applications where strong shielding is necessary, such as in high-frequency transmission systems or environments with high levels of electromagnetic interference.Rate this question:
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- 14.
What advantage of fiber optic cable allows for a greatly increased information-carrying capacity within a single fiber?
- A.
Smaller size.
- B.
Wider bandwidth.
- C.
Lower signal attenuation.
- D.
Immunity to electromagnetic interference.
Correct Answer
B. Wider bandwidth.Explanation
The advantage of fiber optic cable that allows for a greatly increased information-carrying capacity within a single fiber is its wider bandwidth. Unlike traditional copper cables, fiber optic cables have a much larger bandwidth, which means they can transmit a greater amount of data at a faster rate. This wider bandwidth allows for more information to be carried within a single fiber, making fiber optic cables ideal for high-speed data transmission over long distances.Rate this question:
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- 15.
In a fiber optic cable, what effect does increasing the frequency of a signal have on signal attenuation?
- A.
None.
- B.
Attenuation always increases.
- C.
Attenuation always decreases.
- D.
Attenuation increases or decreases depending on frequency.
Correct Answer
A. None.Explanation
Increasing the frequency of a signal in a fiber optic cable does not have any effect on signal attenuation. Attenuation refers to the loss of signal strength as it travels through the cable. In fiber optic cables, attenuation is primarily caused by factors such as scattering and absorption of light, which are independent of the signal frequency. Therefore, regardless of the frequency, the attenuation remains the same.Rate this question:
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- 16.
What component of a fiber optic communication system changes electrical energy in the form of a current into optical energy (light)?
- A.
Source.
- B.
Emitter.
- C.
Detector.
- D.
Converter.
Correct Answer
A. Source.Explanation
The component of a fiber optic communication system that changes electrical energy into optical energy is the source. The source generates the light signal that is then transmitted through the fiber optic cables. This can be done using a laser diode or a light-emitting diode (LED), which convert electrical energy into light energy. The emitted light signal carries the information to be transmitted through the fiber optic system.Rate this question:
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- 17.
What transmitter device changes electrical signals into light signals?
- A.
Zener diode.
- B.
Pin photodiode.
- C.
Injection laser diode.
- D.
Avalanche photodiode
Correct Answer
C. Injection laser diode.Explanation
The injection laser diode is a transmitter device that converts electrical signals into light signals. It achieves this through a process called injection laser diode, where an electrical current is injected into a semiconductor material, causing it to emit light. This device is commonly used in various applications such as telecommunications, optical fiber communication, and laser printing.Rate this question:
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- 18.
Which receiver device produces current in response to incident light?
- A.
Zener diode.
- B.
Pin photodiode.
- C.
Injection laser diode.
- D.
Light emitting diode.
Correct Answer
B. Pin photodiode.Explanation
A pin photodiode is a type of semiconductor device that generates current when exposed to light. It consists of a p-type layer, an intrinsic layer, and an n-type layer. When light photons strike the intrinsic layer, they create electron-hole pairs, which then generate a current. This current can be used as a signal in various applications such as optical communication, light sensing, and imaging. The other options, Zener diode, injection laser diode, and light emitting diode, do not produce current in response to incident light.Rate this question:
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- 19.
What is the reflective surface that propagates light along the optical transmission path?
- A.
Core.
- B.
Buffer.
- C.
Cladding.
- D.
Shielding.
Correct Answer
C. Cladding.Explanation
The correct answer is cladding. In optical fibers, cladding is the outer layer that surrounds the core. It has a lower refractive index than the core, which allows it to act as a reflective surface. This reflection helps to keep the light within the core and propagate it along the optical transmission path. The cladding serves to protect the core and maintain the integrity of the transmitted light signal.Rate this question:
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- 20.
When wavelength equals rate of travel/frequency, the wavelength for a sine wave with a frequency of 2MHz is
- A.
100 meters.
- B.
150 meters.
- C.
300 meters.
- D.
450 meters.
Correct Answer
B. 150 meters.Explanation
The wavelength of a wave is calculated by dividing the speed of the wave by its frequency. In this question, the frequency is given as 2MHz (2 million cycles per second). Since the speed of the wave is not given, we cannot calculate the exact wavelength. However, we can determine that the wavelength will be inversely proportional to the frequency. This means that as the frequency increases, the wavelength decreases. Therefore, out of the given options, the wavelength of 150 meters is the most reasonable choice for a wave with a frequency of 2MHz.Rate this question:
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- 21.
A transmission line with a simple construction, high radiation loss, and high noise pickup describes
- A.
A shielded line.
- B.
An air coaxial line.
- C.
A flexible coaxial line.
- D.
A parallel two-wire open line.
Correct Answer
D. A parallel two-wire open line.Explanation
A transmission line with a simple construction, high radiation loss, and high noise pickup is likely a parallel two-wire open line. This type of transmission line consists of two parallel conductors that are not shielded, which makes it susceptible to radiation loss and noise pickup. The simple construction refers to the basic design of the parallel wires, and the absence of shielding contributes to the high radiation loss and noise pickup.Rate this question:
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- 22.
How do you ensure that a coaxial cable transmission line is outputting maximum signal strength?
- A.
Check the standing-wave-ratio of the cable.
- B.
Check for an open between the inner and outer conductors.
- C.
Check for an open outer conductor between the connectors on each end of the cable.
- D.
Check for a shorted outer conductor between the connectors on each end of the cable.
Correct Answer
A. Check the standing-wave-ratio of the cable.Explanation
To ensure that a coaxial cable transmission line is outputting maximum signal strength, it is necessary to check the standing-wave-ratio of the cable. The standing-wave-ratio (SWR) is a measure of how well the impedance of the cable matches the impedance of the connected devices. A high SWR indicates a mismatch, which can result in signal loss and reduced performance. By checking the SWR, one can identify any impedance mismatches and take corrective measures such as adjusting the cable length or using impedance matching devices to optimize the signal strength.Rate this question:
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- 23.
What type of loss occurs as a result of electrical current flowing through a conductor?
- A.
Copper.
- B.
Radiation.
- C.
Induction.
- D.
Dielectric.
Correct Answer
A. Copper.Explanation
The correct answer is Copper. When electrical current flows through a conductor, there is a certain amount of resistance that causes the conductor to heat up. This heating effect is known as copper loss, as copper is commonly used as a conductor material. The resistance of the copper conductor leads to energy loss in the form of heat, which is why it is important to consider this factor when designing electrical systems.Rate this question:
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- 24.
When performing electrical measurements of radio frequency (RF) lines, which of the following will not load the line and upset its normal characteristics?
- A.
An ammeter.
- B.
A voltmeter.
- C.
A neon bulb.
- D.
A slotted line section.
Correct Answer
D. A slotted line section.Explanation
A slotted line section will not load the RF line and upset its normal characteristics because it is specifically designed to allow the RF signal to pass through without impedance or reflection. A slotted line section is used for measuring the standing wave ratio (SWR) of the RF line and does not introduce any additional impedance or disturbance to the line. In contrast, an ammeter, voltmeter, or neon bulb can all introduce additional impedance or load to the RF line, which can affect its normal characteristics.Rate this question:
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- 25.
The disadvantage of using a wave guide is its
- A.
High power loss.
- B.
Size restrictions.
- C.
Low power handling capability.
- D.
Use is limited to low frequencies.
Correct Answer
B. Size restrictions.Explanation
Waveguides are structures that guide electromagnetic waves along a specific path. One of the disadvantages of using a waveguide is its size restrictions. Waveguides are designed to operate at specific frequencies, and their dimensions are determined by the wavelength of the waves they are intended to guide. This means that waveguides can only be used for certain frequency ranges, and their size cannot be easily adjusted to accommodate different frequencies. Therefore, the size restrictions of waveguides limit their versatility and applicability in various applications.Rate this question:
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- 26.
The most common device used to connect the sections of a wave guide is the
- A.
Isolator.
- B.
Magic tee.
- C.
Choke joint.
- D.
Directional coupler.
Correct Answer
C. Choke joint.Explanation
A choke joint is commonly used to connect the sections of a waveguide. It is designed to prevent the leakage of electromagnetic waves and to provide a smooth transition between the sections. The choke joint consists of a flange with a series of concentric ridges that create a high impedance barrier, effectively blocking the propagation of unwanted waves. This helps to maintain the integrity and efficiency of the waveguide system by minimizing signal loss and reflections. The other options, such as the isolator, magic tee, and directional coupler, serve different purposes and are not primarily used for connecting waveguide sections.Rate this question:
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- 27.
Which is not a type of coupling device normally found in an electrical connector?
- A.
Friction coupling.
- B.
Threaded coupling ring.
- C.
Bayonet lock.
- D.
Insertion.
Correct Answer
D. Insertion.Explanation
The given options describe different types of coupling devices found in electrical connectors. Friction coupling, threaded coupling ring, and bayonet lock are all commonly used types of coupling devices. However, "insertion" is not a type of coupling device. It is a general action that refers to the process of inserting a connector into a corresponding socket or receptacle. Therefore, "insertion" is the correct answer as it does not represent a specific type of coupling device.Rate this question:
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- 28.
The minimum bend radius for a coaxial cable is
- A.
Twice the inside diameter.
- B.
Double the outside diameter.
- C.
Six times the inside diameter.
- D.
Six times the outside diameter.
Correct Answer
D. Six times the outside diameter.Explanation
The minimum bend radius for a coaxial cable is six times the outside diameter. This means that the cable should not be bent in a radius smaller than six times the size of its outer diameter. Bending the cable in a smaller radius can cause signal loss, interference, or even damage to the cable. Therefore, it is important to adhere to the recommended minimum bend radius to ensure optimal performance and longevity of the coaxial cable.Rate this question:
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- 29.
Which of the following is not a proper procedure for soldering a center pin on a coax cable center conductor?
- A.
Use rosin-core 60/40 tin/lead solder.
- B.
Use a small soldering pencil (less than 100 watts).
- C.
Apply heat at the dielectric end of the center pin body.
- D.
Feed solder through the hole in the side of the center pin.
Correct Answer
C. Apply heat at the dielectric end of the center pin body.Explanation
Applying heat at the dielectric end of the center pin body is not a proper procedure for soldering a center pin on a coax cable center conductor. Heat should be applied to the metal part of the center pin, not the dielectric end, in order to properly melt the solder and create a secure connection.Rate this question:
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- 30.
What length of lacing cord is required to properly lace a cable bunch?
- A.
One and a half times the length of the cable.
- B.
Two times the length of the cable.
- C.
Two and a half times the length of the finished bundle.
- D.
Three times the length of the cable.
Correct Answer
C. Two and a half times the length of the finished bundle.Explanation
To properly lace a cable bunch, two and a half times the length of the finished bundle is required. This is because when lacing a cable bunch, the cord needs to go around the cables multiple times to secure them properly. By using two and a half times the length of the finished bundle, there will be enough cord to wrap around the cables and tie them securely without running out of cord. This ensures that the cables are properly laced and organized.Rate this question:
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- 31.
Which condition of an EMP hardened cable is considered nonrepairable by local personnel?
- A.
Frayed to cable jacket.
- B.
Damage to internal wires.
- C.
Loose outer shield.
- D.
Broken pins at the connector.
Correct Answer
B. Damage to internal wires.Explanation
Damage to internal wires is considered nonrepairable by local personnel because it requires specialized knowledge and equipment to repair. Local personnel may not have the necessary training or tools to fix internal wire damage in an EMP hardened cable. Frayed cable jacket, loose outer shield, and broken pins at the connector can be repaired or replaced by local personnel with basic training and tools.Rate this question:
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- 32.
What percentage of minor shield damage can the flight line technician repair?
- A.
Up to 20 percent of the inner shield circumference.
- B.
Up to 20 percent of the outer shield circumference.
- C.
Up to 25 percent of the inner shield circumference.
- D.
Up to 25 percent of the outer shield circumference.
Correct Answer
D. Up to 25 percent of the outer shield circumference.Explanation
The flight line technician can repair up to 25 percent of the outer shield circumference. This means that any damage to the outer shield that is within this percentage can be repaired by the technician. The inner shield circumference is not mentioned in relation to the repair percentage, so it is not relevant to the answer.Rate this question:
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- 33.
What precaution must you adhere to prior to applying heat shrink on a wire repair job?
- A.
Ensure proper lighting.
- B.
Inventory required tools.
- C.
Read all procedures in TO 1–1A–14.
- D.
Warnings concerning use of heat tools on aircraft.
Correct Answer
D. Warnings concerning use of heat tools on aircraft.Explanation
Prior to applying heat shrink on a wire repair job, it is important to adhere to the precaution of reading all procedures in TO 1-1A-14. This ensures that the repair job is done correctly and in accordance with the proper guidelines. Additionally, it is important to be aware of the warnings concerning the use of heat tools on aircraft to prevent any potential damage or safety hazards.Rate this question:
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- 34.
What type of fiber optic loss occurs when part of the light being transmitted is blocked?
- A.
End separation.
- B.
Lateral displacement.
- C.
Angular misalignment.
- D.
Perpendicular displacement.
Correct Answer
B. Lateral displacement.Explanation
Lateral displacement refers to the phenomenon where part of the light being transmitted through a fiber optic cable is blocked or diverted from its intended path. This can happen due to factors such as bends, twists, or misalignments in the fiber optic cable. As a result, the blocked or diverted light causes loss in signal strength and quality, leading to a decrease in overall transmission efficiency.Rate this question:
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- 35.
What method, used to produce correct end finishes during a fiber optic cable spice, is more commonly used with connectors?
- A.
Shine.
- B.
Polish.
- C.
Etch-and-bend.
- D.
Scribe-and-break.
Correct Answer
D. Scribe-and-break.Explanation
The method commonly used with connectors to produce correct end finishes during a fiber optic cable splice is scribe-and-break. This technique involves using a diamond scribe to create a small scratch or scribe mark on the fiber, and then breaking the fiber along the scribe mark. This results in a clean and flat end finish, which is essential for proper alignment and optimal performance of the connector.Rate this question:
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- 36.
What fiber optic connector is best used in rugged, high vibration areas?
- A.
ST.
- B.
BNC.
- C.
SMA.
- D.
TNC.
Correct Answer
C. SMA.Explanation
The SMA fiber optic connector is best used in rugged, high vibration areas because it has a threaded coupling mechanism that provides a secure and reliable connection. The threaded design helps to prevent accidental disconnection due to vibrations or movement. Additionally, the SMA connector is more robust and durable compared to other connectors like ST, BNC, or TNC, making it ideal for harsh environments where there may be frequent vibrations or mechanical stress.Rate this question:
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- 37.
What multiplexing method is used to transmit information over a data bus?
- A.
Code division.
- B.
Time division.
- C.
Frequency division.
- D.
Wavelength division.
Correct Answer
B. Time division.Explanation
Time division multiplexing (TDM) is the multiplexing method used to transmit information over a data bus. In TDM, the available bandwidth is divided into time slots, and each slot is allocated to a different source or channel. This allows multiple sources to share the same transmission medium by taking turns transmitting their data during their allocated time slots. TDM is commonly used in telecommunications systems and is efficient for transmitting data over a data bus as it ensures that each source gets a fair and equal opportunity to transmit its information.Rate this question:
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- 38.
All of the following are advantages of a multiplex bus except
- A.
Increases reliability.
- B.
Decreases available space.
- C.
Reduces aircraft weight.
- D.
Reduces the volume of wires.
Correct Answer
B. Decreases available space.Explanation
A multiplex bus is a communication system that allows multiple signals to be transmitted simultaneously over a single channel. It offers several advantages, including increased reliability, reduced aircraft weight, and a decrease in the volume of wires required for communication. However, it does not decrease available space. In fact, a multiplex bus can help optimize space utilization by reducing the number of individual wires and cables needed for communication, thus freeing up more space for other components or purposes.Rate this question:
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- 39.
How is a serial multiplex bus different than a parallel multiplex bus?
- A.
It is faster and uses less twisted wire pairs.
- B.
It is faster and uses more twisted wire pairs.
- C.
All data bits are received in sequence and uses less twisted wire pairs.
- D.
All data bits are received at the same time and uses more twisted wire pairs.
Correct Answer
C. All data bits are received in sequence and uses less twisted wire pairs.Explanation
A serial multiplex bus is different from a parallel multiplex bus because it receives all data bits in sequence, meaning that each bit is transmitted one after the other. This allows for a more efficient use of the twisted wire pairs, as fewer pairs are needed to transmit the data. In contrast, a parallel multiplex bus receives all data bits at the same time, requiring more twisted wire pairs to transmit the data simultaneously. Therefore, the correct answer is that a serial multiplex bus receives all data bits in sequence and uses less twisted wire pairs.Rate this question:
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- 40.
A 1553 data bus message must contain at least one command word with a maximum of how many data words?
- A.
3.
- B.
16.
- C.
20.
- D.
32.
Correct Answer
D. 32.Explanation
A 1553 data bus message must contain at least one command word, which is used to send instructions or commands. The maximum number of data words that can be included in the message is 32. Data words are used to transmit data or information. Therefore, the correct answer is 32.Rate this question:
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- 41.
The typical 1553 data bus system command that provides a mechanism for an remote terminal (RT) to send data to another RT when the BC does not require the data is called
- A.
Bus controller (BC) to RT.
- B.
RT to BC.
- C.
RT to RT.
- D.
BC to BC.
Correct Answer
C. RT to RT.Explanation
The correct answer is RT to RT. In a typical 1553 data bus system, when the Bus Controller (BC) does not require the data, the Remote Terminal (RT) can still send data to another RT using the RT to RT command. This command allows direct communication between two RTs without involving the BC.Rate this question:
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- 42.
Which of the following is a function of the MIL-STD–1553 data bus mode codes?
- A.
Transferring the subsystem time.
- B.
Obtaining program data from an RT.
- C.
Sending initialization instructions to an RT.
- D.
Commanding an RT to initiate a system self-test.
Correct Answer
D. Commanding an RT to initiate a system self-test.Explanation
The MIL-STD-1553 data bus mode codes are used for various functions in the system. One of these functions is commanding an RT (Remote Terminal) to initiate a system self-test. This means that the mode codes can be used to send specific instructions to an RT to start a self-test of the system. This allows for the testing and verification of the system's functionality and performance.Rate this question:
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- 43.
On the B–1B, which of the following is not a major aircraft structural area?
- A.
Fuselage.
- B.
Windows.
- C.
Stabilizers.
- D.
Wing carry-through.
Correct Answer
D. Wing carry-through.Explanation
The B-1B aircraft has several major structural areas, including the fuselage, windows, stabilizers, and wing carry-through. However, the wing carry-through is not considered a major aircraft structural area. This term refers to the structure that connects the wings to the fuselage and helps distribute the load and stress between them. While it is an important component for the overall integrity of the aircraft, it is not classified as a major structural area like the fuselage, windows, and stabilizers.Rate this question:
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- 44.
How many fuselage sections is the B–1B divided into?
- A.
8.
- B.
6.
- C.
5.
- D.
4.
Correct Answer
C. 5.Explanation
The B-1B is divided into five fuselage sections.Rate this question:
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- 45.
The forward intermediate fuselage of the B–1B contains
- A.
Movable fairings for wing sweep mechanism.
- B.
The forward equipment avionics bay.
- C.
A wheel well fuel tank.
- D.
Two weapons bays.
Correct Answer
D. Two weapons bays.Explanation
The forward intermediate fuselage of the B-1B contains two weapons bays. These bays are used to store and deploy various weapons, such as bombs and missiles. They are an essential part of the aircraft's offensive capabilities, allowing it to carry and release weapons during missions.Rate this question:
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- 46.
The B–2A is divided into how many major sections?
- A.
4.
- B.
5.
- C.
6.
- D.
7.
Correct Answer
B. 5.Explanation
The correct answer is 5 because the given options are the number of major sections into which the B-2A is divided. Out of the given options, 5 is the only number that is mentioned. Therefore, the B-2A is divided into 5 major sections.Rate this question:
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- 47.
Which B–2 aircraft section serves as the exhaust ramp surface?
- A.
Forward center.
- B.
Aft center.
- C.
Aft intermediate.
- D.
Outboard aft intermediate.
Correct Answer
C. Aft intermediate.Explanation
The aft intermediate section of a B-2 aircraft serves as the exhaust ramp surface.Rate this question:
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- 48.
Which B–2 doors have blade seals?
- A.
Interior.
- B.
Exterior.
- C.
Intermediate.
- D.
GLAS.
Correct Answer
B. Exterior.Explanation
The correct answer is "Exterior." Blade seals are typically used in exterior doors to provide a tight seal against air and water infiltration. These seals are designed to prevent drafts, noise, and moisture from entering the building. In contrast, interior doors do not usually require blade seals as they are not exposed to outdoor elements. Intermediate doors may or may not have blade seals depending on their location and purpose. GLAS doors are not commonly used, so it is unlikely that they would have blade seals.Rate this question:
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- 49.
Section 41 of B–52H fuselage contains the
- A.
Upper and lower crew compartments.
- B.
Attachment point for the wings to the fuselage.
- C.
Landing gear, the bomb bay, and center wing tank.
- D.
Drag chute compartment, aft electronic countermeasures (ECM) system, and antennas.
Correct Answer
A. Upper and lower crew compartments.Explanation
The correct answer is upper and lower crew compartments. This section of the B-52H fuselage contains the areas where the crew members are located during flight. The upper crew compartment is typically where the pilot and co-pilot sit, while the lower crew compartment is where the rest of the crew, such as the navigator and weapons systems officer, are situated. This section is important for housing the crew and providing them with the necessary space and equipment to perform their duties during flight.Rate this question:
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- 50.
Which engine cowling covers starter cartridges and the engine combustion chamber?
- A.
Firewall cowling.
- B.
Nose cowl.
- C.
Wrap cowl.
- D.
Afterbody cowl.
Correct Answer
D. Afterbody cowl.Explanation
The afterbody cowl is the correct answer because it covers both the starter cartridges and the engine combustion chamber. The other options, such as the firewall cowling, nose cowl, and wrap cowl, do not specifically cover these components.Rate this question:
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Apr 19, 2024Quiz Edited by
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Nov 04, 2011Quiz Created by
Airmancollin
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