Physics - Metric Flashcards
Water and Heat
Water is able to conduct heat far more efficiently that air because it is:
a) less dense than air
b) more dense than air
c) more fluid than air
d) less fluid than air
b) more dense than air
Water and Heat
Divers are most affected by what form of heat transmission?
a) Conduction
b) Convection
c) Radiation
d) All of the above are equally important
a) Conduction
Water and Heat
Water is able to conduct heat about ______ times faster than air.
a) 3200
b) 775
c) 100
d) 20
d) 20
Water and Light
Refraction is caused by the process of:
a) Light traveling at different speeds as it passes through different substances.
b) Water absorbing various wave lengths of light beginning at the red end of the spectrum.
c) The changing speed of light due to sunspot activity.
d) Light traveling at the same speed as sound once it enters water and encounters resistance.
a) Light traveling at different speeds as it passes through different substances.
Water and Light
When viewed underwater, objects normally appear _______ by a ratio of about _____ (actual to apparent distance).
a) closer/2:1
b) further away/4:3
c) closer/4:3
d) further away/2:1
c) closer/4:3
Water and Light
When viewed underwater, objects tend to be magnified by a factor of about:
a) 10%
b) 33%
c) 50%
d) 74%
b) 33%
Water and Light
“Visual reversal” refers to an object’s tendency to appear:
a) as a mirror image of itself.
b) upside down, as though viewed through a magnifying lens.
c) like a photographic negative.
d) further away than its actual distance.
d) further away than its actual distance.
Water and Light
The single most important factor affecting the “visual reversal” phenomenon is:
a) Depth
b) Turbidity
c) Time of day
d) All of the above are equally important.
b) Turbidity
Water and Sound
Light waves contain _______, while sound waves are comprised of _______.
a) heat energy/air
b) electromagnetic energy/mechanical energy
c) infrared energy/untraviolet energy
d) kinetic energy/potential energy
b) electromagnetic energy/mechanical energy
Water and Sound
The density and elasticity of a medium has what effect upon the transmission of sound?
a) The denser and more elastic the medium; the better sound is transmitted.
b) The denser and more elastic the medium; the poorer sound is transmitted.
c) The denser and more elastic the medium; the less sound can be transmitted.
d) Density and elasticity of a medium has no effect upon the transmission of sound.
a) The denser and more elastic the medium; the better sound is transmitted.
Water and Sound
Sound travels approximately ____ times faster in water than it does in air.
a) two
b) four
c) ten
d) twenty
b) four
Water and Sound
Divers have difficulty determining the direction of sound underwater because:
a) There is an insufficient delay between the sound striking one ear before the other.
b) Water filling the ear canal reduces the ear’s sensitivity to sound.
c) The wet suit hood makes it difficult to perceive sound as accurately as on land.
d) Sound waves travel less efficiently underwater.
a) There is an insufficient delay between the sound striking one ear before the other.
Buoyancy and the ‘Weightless’ World
According to Archimedes’ Principle, “Any object wholly or partially immersed in a fluid is buoyed up by a force ___________.”
a) equal to the weight of the object
b) equal to the weight of the fluid displaced by the object
c) equal to the weight of both the object and the weight of the fluid displaced by the object
d) slightly less than the weight of the object
b) equal to the weight of the fluid displaced by the object
Buoyancy and the ‘Weightless’ World
The specific gravity of pure water is:
a) 0.0
b) 1.0
c) It varies according to where on earth it is measured.
d) It cannot be determined except in a vacuum.
b) 1.0
Buoyancy and the ‘Weightless’ World
Approximately how much air must be added to a lifting device to bring a 600 kilogram object to the surface? The object lies in 30 metres of fresh water.
a) 620 litres
b) 600 litres
c) 580 litres
d) The answer cannot be determined from the data provided.
d) The answer cannot be determined from the data provided.
Buoyancy and the ‘Weightless’ World
Approximately how much water must be displaced to bring a 500 kilogram object to the surface if the object displaced 300 litres? The object lies in 40 metres of seawater.
a) Slightly more than 300 litres.
b) Slightly more than 185 litres.
c) Slightly more than 29 litres.
d) The answer cannot be determined from the data provided.
b) Slightly more than 185 litres.
Buoyancy and the ‘Weightless’ World
An object weighing 350 kilograms and displacing 300 litres is lying in 15 metres of fresh water. If a drum is to be used to lift the object to the surface, how much water must be displaced from the drum?
a) 41 litres
b) 50 litres
c) For any inflexible drum to be used, it must always be completely filled.
d) The answer cannot be determined from the data provided.
b) 50 litres
Under Pressure
The absolute pressure at a depth of 90 metres of seawater is:
a) 8 ata
b) 9 ata
c) 10 ata
d) The answer cannot be determined from the data provided.
c) 10 ata
Under Pressure
The ambient pressure at 30 metres of seawater is:
a) 3 ata
b) 4 ata
c) 5 ata
d) The answer cannot be determined from the data provided.
b) 4 ata
Under Pressure
The gauge pressure at 23 metres of depth in fresh water is:
a) 2.23 atm gauge
b) 3.23 atm gauge
c) 4.23 atm gauge
d) The answer cannot be determined from the data provided.
a) 2.23 atm gauge
Under Pressure
At a depth of 34 metres in fresh water the absolute pressure is __________, the gauge pressure is __________ and the ambient pressure is __________.
a) 4.5 ata/3.5 atm gauge/4.5 ata
b) 3.4 ata/4.4 atm gauge/3.4 ata
c) 4.3 ata/3.3 atm gauge/4.3 ata
d) 4.3 ata/3.3 atm gauge/3.3 ata
c) 4.3 ata/3.3 atm gauge/4.3 ata
Boyle’s Law
A balloon containing 30 litres is released from 10 ata. If it does not explode, how much air will be in the balloon upon reaching the surface?
a) 900 litres
b) 600 litres
c) 300 litres
d) The answer cannot be determined from the data provided.
c) 300 litres
Boyle’s Law
What would the volume of a balloon containing 30 litres at 10 ata be if it is taken to 50 metres of seawater?
a) 60 litres
b) 50 litres
c) 285 litres
d) Unchanged
b) 50 litres
Boyle’s Law
A balloon is filled with 60 litres of air at 30 metres of seawater. What will be the approximate volume of the balloon if it is taken to a depth 90 metres?
a) 24 litres
b) 20 litres
c) 90 litres
d) The answer cannot be determined by the data provided.
a) 24 litres
Boyle’s Law
A balloon containing 300 litres of air at 7 metres of seawater is taken to a depth of 26 metres. What will be the exact volume of the balloon upon reaching 26 metres?
a) 141.66 litres
b) 88.20 litres
c) 58.50 litres
d) The answer cannot be determined from the data provided.
a) 141.66 litres
Boyle’s Law
A scuba cylinder is filled to capacity at the surface. When this cylinder is used at a depth of 30 metres in the sea, the air within the cylinder is four times more dense than it was at the surface.
True
False
False
Boyle’s Law
Because a diver’s lung volume must remain constant regardless of the depth at which he breathes, the density of the air in the diver’s lungs does not change even as he changes depth.
True
False
False
Boyle’s Law
The air that a diver breathes from a scuba cylinder at 50 metres of seawater is ______ as dense as the air breathed from the same cylinder at the surface.
a) 10 times
b) 6 times
c) 5 times
d) exactly
b) 6 times
Boyle’s Law
A diver has an air consumption rate of 2 bar per minute at the surface. If all other factors but depth remain unchanged, what will his consumption rate be at 40 metres of seawater?
a) 3 bar/minute
b) 5 bar/minute
c) 7 bar/minute
d) 10 bar/minute
d) 10 bar/minute
Boyle’s Law
A diver has an air consumption rate of 60 litres per minute at 10 metres of seawater. If all other factors but depth remain unchanged, what will his consumption rate be at 30 metres?
a) 15 litres
b) 30 litres
c) 90 litres
d) 120 litres
d) 120 litres
Boyle’s Law
A diver has an air consumption rate of 90 litres per minute at 20 metres of seawater. If all factors, but depth remain unchanged, what will his consumption rate be at 60 metres?
a) 180 litres per minute
b) 210 litres per minute
c) 240 litres per minute
d) The answer cannot be determined from the data provided.
b) 210 litres per minute
Boyle’s Law
A diver has an air consumption rate of 60 litres per minute at 10 metres of seawater. If all other factors but depth remain unchanged, what will his consumption rate be, in bar per minute, at 30 metres?
a) 3 bar per minute
b) 7 bar per minute
c) 10 bar per minute
d) The answer cannot be provided from the data provided.
d) The answer cannot be provided from the data provided.
Charles’ Law
A balloon is filled with 30 litres of air at room temperature. Describe what would happen to that balloon if it were put into a freezer at a constant ambient pressure.
a) The volume would increase.
b) The volume would decrease.
c) The volume would remain unchanged but the pressure would decrease.
d) The volume and the pressure would remain unchanged.
b) The volume would decrease.
Charles’ Law
A scuba cylinder is filled to capacity at room temperature. Describe what would happen to that cylinder if it was taken on an ice dive (water at or near freezing).
a) The volume would increase.
b) The volume would decrease.
c) The volume would remain unchanged but the pressure would decrease.
d) The volume and the pressure would remain unchanged.
c) The volume would remain unchanged but the pressure would decrease.
Charles’ Law
A 12 litre scuba cylinder is filled to 200 bar at an ambient temperature of 27° C. If the cylinder is then used in water temperature of 4° C, what would be the approximate cylinder pressure?
a) Unchanged
b) 186 bar
c) 214 bar
d) 116 bar
b) 186 bar
P1 x V1/T1 = P2 x V2/T2
Charles’ Law
A 12 litre scuba cylinder is filled to 200 bar at an ambient temperature of 26°C. What will the cylinder pressure be if the cylinder is used in water temperature of 7°C?
a) 188 bar
b) 187 bar
c) 192 bar
d) The answer cannot be determined from the data provided.
b) 187
First, let’s look at the equation: P1 x V1/T1 = P2 x V2/T2 Where P = pressure, V = volume and T = temperature As we are dealing with a scuba cylinder, its volume will remain unchanged. Therefore, we can ignore the volume variable altogether, restructuring the equation as follows: P1/T1 = P2/T2 Next, let’s determine what we know from the problem: The pressure the cylinder is filled to is 200 bar (P1), and it asks us to determine what the pressure will become (P2). But, these are not yet proper variables to plug into the equation. To predict the behavior of gases we must always work in absolute terms. So, we must add 1 atm (or bar) to the 200 bar for 201bar. This then becomes the quantity to insert into the equation. The temperature of the cylinder is 26°C, and it will be used in 7°C water. But, as stated before, we must work in absolute terms. With temperature, this means working in terms of absolute zero. Readings in degrees Celsius must be converted into degrees Kelvin. To convert a Celsius reading into a Kelvin reading, just add 273.° Therefore, 26°C, becomes 299°K (T1); and 7°C, becomes 280°K (T2). Now we can plug the values into the formula: 201 ÷ 299 = X ÷ 280, where “X” is the unknown. This can be rearranged to become: (280 x 201) ÷ 299 = X and by doing the simple arithmetic, we determine that: 188 bar = X. But this still is not the correct answer since 188 bar is absolute pressure. We need to subtract 1 bar from 188 bar to arrive at the cylinder’s real pressure. Therefore, 188 - 1 = 187 bar.
Dalton’s Law
In a mixture of air comprised of 20% oxygen and 80% nitrogen, at an ambient pressure of 1 ata what is the partial pressure of oxygen?
a) 0.2 ata
b) 0.4 ata
c) 0.9 ata
d) 1 ata
a) 0.2 ata
Dalton’s Law
A gas mixture is comprised of 21% oxygen, 78% nitrogen and 1% carbon dioxide. At a depth of 20 metres, what is the partial pressure of the nitrogen?
a) 1.67 ata
b) 2.34 ata
c) 5.69 ata
d) The answer cannot be determined from the data provided.
b) 2.34 ata
Dalton’s Law
A gas mixture is comprised of 21% oxygen, 78% nitrogen and 1% carbon dioxide. At a depth of 24 metres, what is the partial pressure of the oxygen?
a) 3.6 ata
b) 0.7 ata
c) 0.5 ata
d) The answer cannot be determined from the data provided.
b) 0.7 ata
Dalton’s Law
A scuba cylinder is accidentally filled with 1% carbon monoxide. If a diver breathes air from this cylinder at a depth of 30 metres of seawater, approximately what percentage of carbon monoxide will he be breathing?
a) 1%
b) 2%
c) 3%
d) 4%
a) 1%
a percentage is constant
Dalton’s Law
A scuba cylinder is accidentally filled with 1% carbon monoxide, it is determined that at the surface the diver inhales 500,000 molecules of carbon monoxide with each breath. Therefore, when breathing the air at 30 metres of seawater (assuming all other factors but depth are unchanged), he would breathe approximately how many molecules?
a) 500,000
b) 1,000,000
c) 1,500,000
d) 2,000,000
d) 2,000,000
Dalton’s Law
A scuba cylinder is accidentally filled with 1% carbon monoxide, breathing the contaminated air at 30 metres of seawater would have the same effect on the diver as breathing what percentage of carbon monoxide at the surface?
a) 1%
b) 2%
c) 3%
d) 4%
d) 4%
Dalton’s Law
Breathing from a contaminated air source, with 1.5% carbon monoxide, at a depth of 90 metres of seawater, would have the same effect as breathing approximately what percentage of carbon monoxide at the surface?
a) 1.5%
b) 13.5%
c) 15%
d) The answer cannot be determined from the data provided.
c) 15%
Henry’s Law
A glass of water has been placed in a vacuum for several days. It no longer contains any dissolved gas within it. If it is then placed in a pressure pot and pressurized to 2 ata for several days, what will be the gas pressure within the liquid?
a) 1 ata
b) 2 ata
c) 4 ata
d) The answer is impossible to determine.
b) 2 ata
Henry’s Law
If the pressure within a pot of a glass of water that has been placed in a vacuum for several days. It no longer contains any dissolved gas within it. it is then placed in a pressure pot and pressurized to 2 ata for several days and is increased, the pressure of the gas within the liquid will:
a) increase.
b) decrease.
c) remain unchanged.
d) The answer cannot be determined from the data provided.
a) increase.
Henry’s Law
If a vacuum is created within the pressure pot in question - A glass of water has been placed in a vacuum for several days. It no longer contains any dissolved gas within it. it is then placed in a pressure pot and pressurized to 2 ata for several days, the pressure of the gas within the liquid will:
a) increase.
b) decrease.
c) remain unchanged.
d) The answer cannot be determined from the data provided.
b) decrease.
Henry’s Law
Assume that the atmosphere contains 80% nitrogen. At sea level, the nitrogen partial pressure within our tissues would be:
a) 1.0 ata
b) 0.8 ata
c) 0.2 ata
d) The answer cannot be determined from the data provided.
b) 0.8 ata
Henry’s Law
Assume that the atmosphere contains 80% nitrogen, the total gas pressure within our tissues is:
a) 1.0 ata
b) 0.8 ata
c) 0.2 ata
d) The answer cannot be determined from the data provided.
a) 1.0 ata
Henry’s Law
Because our tissues cannot hold any more gas pressure at sea level, our body is referred to as being:
a) pressurized.
b) saturated.
c) supersaturated.
d) unsaturated.
b) saturated.
Henry’s Law
When breathing at depth, the tissues of a diver begin to take on additional gas pressure. If the diver remains at depth long enough, his tissues will again equalize with the ambient pressure.
True
False
True
Henry’s Law
The condition described in the previous question (When breathing at depth, the tissues of a diver begin to take on additional gas pressure. If the diver remains at depth long enough, his tissues will again equalize with the ambient pressure) is referred to as supersaturation.
True
False
False
We discussed previously that at the surface our tissues, or any liquid, can only hold an amount of gas equal to 1 atm. Regardless of how long we remain at 1 atm, our tissues can never achieve a gas tension of more than that amount. However, there is one way more gas can be absorbed — increase the pressure in contact with the tissues. At a greater ambient pressure the tissues will again equalize with the gas in contact with them. Therefore, as the pressure of the gas in contact with the tissues increases, so does the pressure within the tissues. But after time, if the ambient pressure is reduced, the reverse phenomenon will occur. More pressure or gas tension will exist within the tissue than the ambient pressure. In order words, a “pressure gradient” will exist in favor of the tissue. In this case the tissue is referred to as supersaturated, as it contains a higher gas tension that the ambient pressure. (Just as the gas tension within a can of soda is higher than the ambient pressure when the top is initially removed). For the diver this is an extremely important phenomenon to understand.
