12 practice questions from the 237 in this chapter of StudyWren Aviation. Pick an answer and check it to see the explanation and its FAA source.
Noise & Altitude PhysiologyPAR-12-228
A new student worries that wearing earplugs under her headset will make radio calls harder to understand. What does the FAA CAMI say?
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The answer is B: Earplugs cut background noise, which can make speech clearer
CAMI emphasizes that hearing protection devices do not interfere with speech communication during flight; by reducing background noise, they make speech signals clearer and more comprehensible.
Why not ?
Source: FAA CAMI brochure 'Hearing and Noise in Aviation'·ACS PA.I.H
The answer is B: Earplugs cut background noise, which can make speech clearer
CAMI emphasizes that hearing protection devices do not interfere with speech communication during flight; by reducing background noise, they make speech signals clearer and more comprehensible.
Source: FAA CAMI brochure 'Hearing and Noise in Aviation'·ACS PA.I.H
Besides hearing damage, which effect can prolonged loud cockpit noise have on a pilot?
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The answer is C: Fatigue, distraction, and more errors on demanding tasks
CAMI notes that annoying high-intensity noise can cause distraction, fatigue, irritability, headache, and impaired concentration and memory, and that tasks requiring vigilance, calculations, and time judgments can be degraded by noise above 90 dB.
Why not ?
Source: FAA CAMI brochure 'Hearing and Noise in Aviation'·ACS PA.I.H
The answer is C: Fatigue, distraction, and more errors on demanding tasks
CAMI notes that annoying high-intensity noise can cause distraction, fatigue, irritability, headache, and impaired concentration and memory, and that tasks requiring vigilance, calculations, and time judgments can be degraded by noise above 90 dB.
Source: FAA CAMI brochure 'Hearing and Noise in Aviation'·ACS PA.I.H
A passenger who ate a large, gassy meal before the flight develops increasing abdominal pain as you climb to 14,000 feet in an unpressurized airplane. What is the most likely cause?
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The answer is C: Gas trapped in the digestive tract expanding as pressure decreases
Per Boyle's law, a gas's volume is inversely proportional to the pressure on it. AC 61-107B notes GI gas expands on ascent and, if it cannot escape, can cause pain and even fainting; the remedy is to pass the gas or, if there is no relief, descend.
The answer is C: Gas trapped in the digestive tract expanding as pressure decreases
Per Boyle's law, a gas's volume is inversely proportional to the pressure on it. AC 61-107B notes GI gas expands on ascent and, if it cannot escape, can cause pain and even fainting; the remedy is to pass the gas or, if there is no relief, descend.
Atmospheric pressure at 18,000 feet is about half of sea-level pressure. By Boyle's law, about how much will a volume of trapped dry gas expand if it is carried from sea level to 18,000 feet?
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The answer is B: To about twice its sea-level volume
Boyle's law: volume varies inversely with pressure. Halving the pressure doubles the volume (V2 = V1 x P1/P2 = V1 x 1/0.5 = 2 x V1). Quadrupling the volume would take a pressure of one quarter of the sea-level value.
Why not ?
Source: AC 61-107B·FAA CAMI brochure 'Hypoxia: The Higher You Fly...'·ACS PA.I.H
The answer is B: To about twice its sea-level volume
Boyle's law: volume varies inversely with pressure. Halving the pressure doubles the volume (V2 = V1 x P1/P2 = V1 x 1/0.5 = 2 x V1). Quadrupling the volume would take a pressure of one quarter of the sea-level value.
Source: AC 61-107B·FAA CAMI brochure 'Hypoxia: The Higher You Fly...'·ACS PA.I.H
Standard atmospheric pressure is about 760 mmHg at sea level and about 282 mmHg at 25,000 feet. Using Boyle's law, about how many times larger would a pocket of trapped dry gas become at 25,000 feet?
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The answer is C: About 2.7 times
V2/V1 = P1/P2 = 760 / 282 = 2.7. Dividing the other way (282 / 760 = 0.37) inverts Boyle's law; gas expands as pressure falls. A factor of about 1.4 corresponds to roughly 10,000 feet.
V2/V1 = P1/P2 = 760 / 282 = 2.7. Dividing the other way (282 / 760 = 0.37) inverts Boyle's law; gas expands as pressure falls. A factor of about 1.4 corresponds to roughly 10,000 feet.
Sea-level atmospheric pressure is about 760 mmHg, and oxygen makes up about 21 percent of the atmosphere. What is the approximate partial pressure of oxygen at sea level?
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The answer is B: About 160 mmHg
Partial pressure = total pressure × the gas's fraction: 760 × 0.21 ≈ 160 mmHg (AC 61-107B rounds oxygen to 20 percent and gets about 152 mmHg). About 600 mmHg is roughly the nitrogen share (79 percent), and 36 mmHg comes from dividing by 21 instead of multiplying by 0.21.
Partial pressure = total pressure × the gas's fraction: 760 × 0.21 ≈ 160 mmHg (AC 61-107B rounds oxygen to 20 percent and gets about 152 mmHg). About 600 mmHg is roughly the nitrogen share (79 percent), and 36 mmHg comes from dividing by 21 instead of multiplying by 0.21.
At 18,000 feet, atmospheric pressure is about 380 mmHg, and oxygen is still about 21 percent of the air. What is the approximate partial pressure of oxygen there?
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The answer is C: About 80 mmHg
380 x 0.21 = about 80 mmHg, half the sea-level value of about 160 mmHg, because the oxygen percentage is unchanged while the total pressure is halved. That reduced partial pressure is what causes hypoxic hypoxia; 300 mmHg is the nitrogen share.
380 x 0.21 = about 80 mmHg, half the sea-level value of about 160 mmHg, because the oxygen percentage is unchanged while the total pressure is halved. That reduced partial pressure is what causes hypoxic hypoxia; 300 mmHg is the nitrogen share.
Atmospheric pressure at 10,000 feet is about 523 mmHg. With oxygen at about 21 percent of the air, what is the approximate partial pressure of oxygen?
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The answer is A: About 110 mmHg
Partial pressure = total pressure × the gas's fraction: 523 × 0.21 ≈ 110 mmHg, compared with about 160 mmHg at sea level. About 413 mmHg is the nitrogen share (79 percent).
Partial pressure = total pressure × the gas's fraction: 523 × 0.21 ≈ 110 mmHg, compared with about 160 mmHg at sea level. About 413 mmHg is the nitrogen share (79 percent).
A pilot who had a cavity filled two days ago feels a sharp pain in that one tooth that grows worse as the airplane climbs. What is the likely cause?
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The answer is B: Air trapped beneath the filling expanding as pressure drops
AC 61-107B describes a tooth block as rare but possible after a recent filling: any air trapped between the filling and the tooth pulp expands on ascent and causes pain. Level off; if there is no relief, land as soon as practical and see a dentist.
The answer is B: Air trapped beneath the filling expanding as pressure drops
AC 61-107B describes a tooth block as rare but possible after a recent filling: any air trapped between the filling and the tooth pulp expands on ascent and causes pain. Level off; if there is no relief, land as soon as practical and see a dentist.
Why does the PHAK caution that certain antacids can cause problems at altitude?
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The answer is B: They can release gas that expands and causes abdominal pain
The PHAK's table of drug effects notes that antacids can liberate carbon dioxide at altitude; the resulting distension may cause acute abdominal pain and may mask other medical problems.
The answer is B: They can release gas that expands and causes abdominal pain
The PHAK's table of drug effects notes that antacids can liberate carbon dioxide at altitude; the resulting distension may cause acute abdominal pain and may mask other medical problems.
According to the FAA CAMI sunglasses brochure, exposure to ultraviolet radiation increases by approximately how much for every 1,000 feet of altitude?
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The answer is A: About 5 percent
CAMI notes UV exposure increases approximately 5 percent for every 1,000 feet of altitude, one reason pilots are advised to wear sunglasses with 99 to 100 percent UVA and UVB protection.
Why not ?
Source: FAA CAMI brochure 'Sunglasses for Pilots: Beyond the Image'·ACS PA.I.H
CAMI notes UV exposure increases approximately 5 percent for every 1,000 feet of altitude, one reason pilots are advised to wear sunglasses with 99 to 100 percent UVA and UVB protection.
Source: FAA CAMI brochure 'Sunglasses for Pilots: Beyond the Image'·ACS PA.I.H
According to the FAA CAMI, at roughly what intensity can impulse or blast noise rupture the eardrum?
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The answer is A: About 140 dB
CAMI lists ear discomfort at about 120 dB, ear pain at about 130 dB, and eardrum rupture at about 140 dB. Normal conversation is about 60 to 65 dB, and 90 dB is the OSHA 8-hour limit.
Why not ?
Source: FAA CAMI brochure 'Hearing and Noise in Aviation'·ACS PA.I.H
CAMI lists ear discomfort at about 120 dB, ear pain at about 130 dB, and eardrum rupture at about 140 dB. Normal conversation is about 60 to 65 dB, and 90 dB is the OSHA 8-hour limit.
Source: FAA CAMI brochure 'Hearing and Noise in Aviation'·ACS PA.I.H
