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June 2023 Earth Science Regents — Answer Key & Worked Solutions

Every question from the June 2023 Earth Science Regents exam, with the correct answer and a full step-by-step solution. Check your work, learn the method, then convert your raw score.

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These are real questions from the June 2023 Earth Science New York State Regents examination, published by NYSED. Worked solutions are written by RegentsLab.

Question 1 · 2 pts

Which two characteristics classify Jupiter as a Jovian planet?

Worked solution
$\text{Jovian (gas giant) planets are large in diameter but composed mostly of light gases (H, He), giving them low density.}$
Answer: low density and large diameter

Jovian planets like Jupiter are large-diameter gas giants with low overall density, unlike small, dense terrestrial planets.

Question 2 · 2 pts

As observed from Earth, the wavelengths of light from the most distant galaxies are usually

Worked solution
$\text{Distant galaxies are moving away from Earth, stretching light waves to longer (red) wavelengths.}$
Answer: red shifted due to an expanding universe

Redshift of distant galaxies is evidence that the universe is expanding, per the Doppler effect applied to light.

Question 3 · 2 pts

The rate of Earth's revolution around the Sun is approximately

Worked solution
$\frac{360°}{365 \text{ days}} \approx 1°/\text{day}$
Answer: 1°/day

Earth completes 360° of orbit in about 365 days, roughly 1° per day.

Question 4 · 2 pts

Different star constellations are observed from Earth at different times during the year because

Worked solution
$\text{As Earth revolves, the night side faces different portions of the background stars throughout the year.}$
Answer: Earth travels in an orbit around the Sun

Earth's revolution around the Sun changes which constellations are visible in the night sky at different times of year.

Question 5 · 2 pts

The apparent change in the direction of swing of a Foucault pendulum provides evidence of Earth's

Worked solution
$\text{The pendulum's swing plane stays fixed in space while Earth turns beneath it, so the swing appears to change direction.}$
Answer: rotation

A Foucault pendulum's apparent shift in swing direction is classic evidence of Earth's rotation on its axis.

Question 6 · 2 pts

The Milky Way can best be described as

Worked solution
$\text{The Milky Way is our home spiral galaxy among an estimated billions of galaxies in the universe.}$
Answer: one of billions of galaxies in the universe

The Milky Way is one galaxy among billions that make up the observable universe.

Question 7 · 2 pts

How much heat energy is required to change five grams of ice to liquid water at 0°C?

Worked solution
$\text{Heat of fusion of water} = 334\ \text{J/g (Reference Tables)}$
$5\ \text{g} \times 334\ \text{J/g} = 1670\ \text{J}$
Answer: 1670 joules

Melting ice at 0°C requires the heat of fusion (334 J/g); multiplying by 5 grams gives 1670 joules.

Question 8 · 2 pts

The diagram below represents the change in mass of a potted plant over time. Bottom of potted plant wrapped in plastic. Starting mass just after plastic wrap is applied; mass after 7 days. What is the cause and its effect on the change in mass of the potted plant over time?

earth question 8 figure
Worked solution
$\text{Because the pot's bottom is sealed in plastic, no water can be added; water vapor lost through the leaves (transpiration) escapes the system.}$
$\text{Loss of water vapor through transpiration lowers the overall mass measured on the balance after 7 days.}$
Answer: Transpiration caused a decrease in mass.

With the soil sealed off, water lost from the plant via transpiration through its leaves is not replaced, so the plant's mass decreases over the 7 days.

Question 9 · 2 pts

Air temperatures are recorded as a weather balloon rises from Earth's surface. At first, the air temperature constantly decreases for 12 kilometers, then remains constant briefly before increasing for another 12 kilometers before the balloon bursts. In which layer did the balloon burst?

Worked solution
$\text{Decreasing temperature for the first 12 km = troposphere.}$
$\text{Constant temperature briefly = tropopause.}$
$\text{Increasing temperature for the next 12 km (due to ozone absorption of UV) = stratosphere, where the balloon bursts.}$
Answer: stratosphere

Temperature decreases through the troposphere, levels off at the tropopause, then increases through the stratosphere due to ozone absorbing UV radiation, which is where the balloon bursts.

Question 10 · 2 pts

The diagram below represents the Moon passing through the darker part of Earth's shadow called the umbra. Letters A and B represent locations on Earth's surface. The diagram shows parallel Sun's rays arrows striking Earth from the left; Earth is a circle with location A on the left (sunlit) side and location B on the right (shadowed) side; the umbra extends to the right of Earth as a narrowing shaded cone, with the Moon shown at the tip of the umbra, far to the right. Which statement best identifies this event and at which location it would be viewed?

earth question 10 figure
Worked solution
$\text{The Moon sits inside Earth's umbra (Earth's shadow falling on the Moon) = a lunar eclipse, not a solar eclipse.}$
$\text{Location B is on the night side of Earth (facing away from the Sun, toward the Moon/umbra), so observers there can see the eclipsed Moon.}$
Answer: A lunar eclipse is occurring and can be seen from location B.

Earth's shadow (umbra) falling on the Moon defines a lunar eclipse, and only the night-side location B, facing the Moon, could observe it.

Question 11 · 2 pts

The photograph below shows an impact crater located in the southwestern United States. This crater most likely formed as a result of

earth question 11 figure
Worked solution
$\text{A large, roughly circular bowl-shaped depression with raised rim in an area lacking volcanic or tectonic activity is characteristic of a meteorite/asteroid impact.}$
Answer: an asteroid colliding with Earth

The bowl-shaped crater with an elevated rim, shown in a stable continental area, is characteristic of an asteroid impact rather than volcanic or tectonic processes.

Question 12 · 2 pts

The diagrams below represent three containers, A, B, and C, which were filled with equal volumes of uniformly sorted plastic beads of different sizes (small beads in A, medium beads in B, large beads in C). To determine porosity, water was poured into each container until the water level rose to the top of the beads. The porosity was found to be

earth question 12 figure
Worked solution
$\text{Porosity (percent pore space) depends on sorting and packing arrangement, not on the size of uniformly sorted spherical particles.}$
$\text{Since all three containers hold uniformly sorted beads packed the same way, the percentage of open space (porosity) is equal in A, B, and C.}$
Answer: the same in all three containers

For uniformly sorted spherical particles packed the same way, porosity is independent of particle size, so all three containers have the same porosity.

Question 13 · 2 pts

Which cross section represents the cloud patterns and the direction of air movement at a warm front? Four labeled cross-section diagrams (1) through (4) each show cool air and warm air arrows meeting at a sloped boundary at Earth's surface, with cloud shapes above the boundary.

earth question 13 figure
Worked solution
$\text{At a warm front, warm air gently rises over retreating cooler air along a gently sloping boundary.}$
$\text{This produces a broad sequence of layered clouds (thin, high clouds far ahead, thickening toward the front), matching diagram (3).}$
Answer: Diagram (3)

A warm front has a low-angle slope where warm air overrides cooler air, producing a wide swath of progressively thickening layered clouds ahead of the surface front, matching the pattern in diagram 3.

Question 14 · 2 pts

Which map shows the most likely path of a low pressure system (L) in the United States if it follows a normal storm track?

earth question 14 figure
Worked solution
$\text{In the continental United States, prevailing westerly winds and the jet stream typically steer low pressure systems from the southwest toward the northeast.}$
Answer: Map (1): L moves from southwest to northeast

Normal storm tracks in the United States move from southwest to northeast, following the prevailing westerlies, matching map (1).

Question 15 · 2 pts

Which weather instrument is an anemometer?

earth question 15 figure
Worked solution
$\text{An anemometer measures wind speed using spinning cups.}$
Answer: Image (1): rotating cup instrument

The rotating-cup instrument (image 1) is an anemometer, used to measure wind speed; the others are a barometer, thermometer, and wind vane respectively.

Question 16 · 2 pts

The station model below represents some weather information for a location in New York State. The station model shows the number 73 above and to the left of the station circle, a wind barb extending to the upper right from the circle with a present-weather symbol and the number 2 near it, the station circle is completely filled in (solid black), and the number 72 is shown below and to the left of the circle. Based on the information on the station model, the weather at this location is

earth question 16 figure
Worked solution
$\text{A completely solid black station circle indicates overcast (cloudy) sky cover.}$
$\text{The wind barb attached to the circle indicates wind is present (windy), and the present-weather symbol shown indicates thunderstorm activity.}$
$\text{73 and 72 represent the air temperature and dew point (°F), respectively, not the determining factor for sky/weather condition here.}$
Answer: cloudy and windy with thunderstorms

The fully shaded station circle shows overcast sky, the attached wind barb shows wind, and the plotted present-weather symbol indicates thunderstorms, matching choice 2.

Question 17 · 2 pts

Electromagnetic energy travels through space from the Sun to the top of Earth's atmosphere by

Worked solution
$\text{Radiation is the only heat transfer method that can travel through the vacuum of space, since it does not require a medium.}$
Answer: radiation

Radiation is electromagnetic energy transfer that can travel through the vacuum of space, unlike conduction or convection which require matter.

Question 18 · 2 pts

The diagram below represents the position of Earth with respect to the Sun on March 21. A and B are two locations on Earth's surface, with A at 30° N latitude and B at 30° S latitude; parallel Sun's rays arrows strike Earth from the right side of the diagram. The intensity of insolation over the next three months will

earth question 18 figure
Worked solution
$\text{March 21 is the spring equinox for the Northern Hemisphere; over the next three months Earth moves toward the June solstice.}$
$\text{As the Northern Hemisphere (location A, 30°N) tilts increasingly toward the Sun, insolation there increases.}$
$\text{Simultaneously, the Southern Hemisphere (location B, 30°S) tilts away from the Sun, so insolation there decreases.}$
Answer: increase for A and decrease for B

Moving from the March equinox toward the June solstice, the Northern Hemisphere (A) receives increasing insolation while the Southern Hemisphere (B) receives decreasing insolation.

Question 19 · 2 pts

Which event is primarily responsible for the production of oxygen that first entered into Earth's early atmosphere?

Worked solution
$\text{Early photosynthetic organisms (e.g., cyanobacteria) released oxygen as a byproduct of photosynthesis, gradually enriching Earth's atmosphere with free oxygen.}$
Answer: evolution of oxygen-producing organisms

Free oxygen first accumulated in Earth's atmosphere as a byproduct of photosynthesis carried out by early oxygen-producing organisms.

Question 20 · 2 pts

Since the break up of Pangaea, it has been inferred that the continent of North America has been generally moving toward the

Worked solution
$\text{Plate tectonic reconstructions since the breakup of Pangaea show the North American plate has drifted generally in a northwesterly direction.}$
Answer: northwest

Geologic and paleomagnetic evidence indicates the North American plate has generally drifted northwest since Pangaea broke apart.

Question 21 · 2 pts

The fossil Mucrospirifer is considered to be an index fossil because it existed over a

Worked solution
$\text{A good index fossil must be widespread (large geographic range) but existed only briefly in geologic time, so it can pinpoint the age of rock layers wherever it is found.}$
Answer: large geographic region for a short period of time

Index fossils, such as Mucrospirifer, are useful for dating rock layers because they were widespread geographically but existed only for a short geologic time span.

Question 22 · 2 pts

Compared to Earth's outer core, the inner core properties are inferred to be

Worked solution
$\text{Seismic wave behavior (S-waves traveling through the inner core but not the outer core) plus immense pressure indicate the inner core is solid.}$
$\text{Greater depth and pressure in the inner core result in greater density than the liquid outer core.}$
Answer: more dense and a solid

The inner core is inferred to be solid due to extreme pressure, and denser than the liquid outer core because of its greater depth and compression.

Question 23 · 2 pts

A seismic station 4000 kilometers from the epicenter of an earthquake records the arrival time of the first S-wave at 1:05:40 p.m. At what time did the first P-wave arrive at this station?

Worked solution
$\text{Using the Earthquake P-S travel time graph, at 4000 km the P-S time interval} = 5\ \text{min } 40\ \text{s}$
$1:05:40\ \text{p.m.} - 5\ \text{min } 40\ \text{s} = 1:00:00\ \text{p.m.}$
Answer: 1:00:00 p.m.

Using the Reference Tables travel-time graph, the P-S interval at 4000 km is about 5 min 40 s, so subtracting this from the S-wave arrival time gives the P-wave arrival time.

Question 24 · 2 pts

The photograph below shows fossilized ammonoids in a rock outcrop. The presence of these fossils in the outcrop indicates that in the past, this region was most likely covered by

earth question 24 figure
Worked solution
$\text{Ammonoids were marine (ocean-dwelling) organisms}$
Answer: ocean water

Ammonoids are extinct marine cephalopods, so their fossils indicate the region was once covered by ocean water.

Question 25 · 2 pts

The cross section below represents rock units within Earth's crust. Sedimentary rock layers are labeled 1 through 6, from bottom (1) to top (6). An igneous intrusion cuts through layers 1, 2, and 3, and layer 4 shows contact metamorphism directly above the intrusion but layers 5 and 6 do not. The age of the igneous rock is

earth question 25 figure
Worked solution
$\text{Contact metamorphism appears only where the intrusion cuts layers 1 and 2, so the intrusion is younger than layers 1 and 2}$
$\text{Layer 3 is neither cut nor baked, so it was deposited after the intrusion solidified, making the intrusion older than layer 3}$
Answer: older than rock layer 3

Contact metamorphism, the baked zone shown in the key, appears only along the narrow neck of the intrusion where it cuts through rock layers 1 and 2, showing those layers already existed when the magma intruded and making the intrusion younger than layers 1 and 2. Layer 3, however, is neither cut through nor baked anywhere along the rounded top of the intrusion, so layer 3 could only have been deposited after the intrusion had already solidified and cooled. Since a rock layer must already be present to be cut or baked by magma, and layer 3 shows neither effect, the intrusion must predate layer 3, making it older.

Question 26 · 2 pts

The maps below show different stream drainage patterns (dendritic, radial, and trellis-like patterns shown as three separate boxes). What factor most likely causes these different stream drainage patterns?

earth question 26 figure
Worked solution
$\text{Drainage pattern shape reflects the underlying rock structure (fractures, folds, dips, resistance)}$
Answer: bedrock structure

Stream drainage patterns develop in response to the underlying bedrock structure, such as jointing, folding, or differences in rock resistance to erosion.

Question 27 · 2 pts

The photograph below shows boulders and other sediments that slid down a hillside. This rock slide is most likely an example of

earth question 27 figure
Worked solution
$\text{Rock/landslides down a hillside are a form of mass movement (mass wasting), commonly triggered by saturation from heavy rain}$
Answer: mass movement following prolonged periods of heavy rain

A rock slide down a hillside is a form of mass movement, typically triggered by saturation of the slope material after prolonged heavy rain reducing internal friction.

Question 28 · 2 pts

The Mariana Trench is Earth's deepest oceanic trench. This trench formed as the Pacific Plate was

Worked solution
$\text{Deep ocean trenches form where a denser oceanic plate subducts beneath an adjacent plate}$
Answer: subducting under the Philippine Plate

The Mariana Trench formed where the denser Pacific Plate subducts beneath the Philippine Plate, creating the deepest oceanic trench on Earth.

Question 29 · 2 pts

The diagram below represents patterns of normal and reverse magnetic polarity in seafloor bedrock, symmetric about a central ridge. In which bedrock type and location can this magnetic pattern most likely be observed?

earth question 29 figure
Worked solution
$\text{Symmetric magnetic striping forms in igneous (basaltic) seafloor rock at a mid-ocean spreading ridge}$
Answer: igneous bedrock at the East Pacific Ridge

Symmetric bands of normal and reversed magnetic polarity form in newly formed igneous seafloor rock at spreading mid-ocean ridges, such as the East Pacific Ridge.

Question 30 · 2 pts

Compared to the thickness and density of the continental crust, the relative thickness and density of Earth's oceanic crust are

Worked solution
$\text{Oceanic crust (basaltic) is thinner and denser than continental crust (granitic)}$
Answer: thinner and more dense

Oceanic crust is composed of denser basaltic rock and is thinner than continental crust, which is thicker and composed of less dense granitic rock.

Question 31 · 2 pts

Iron found in some minerals in basalt can react with oxygen and break down into iron oxide particles. This change is an example of

Worked solution
$\text{Reaction of iron with oxygen (oxidation) forming a new mineral (iron oxide) is a chemical change}$
Answer: chemical weathering

The reaction of iron with oxygen to form iron oxide is a chemical change that alters the mineral composition, which is chemical weathering (oxidation).

Question 32 · 2 pts

Which feature provides the best evidence that thick glacial ice sheets once advanced southward over New York State?

Worked solution
$\text{Glacial ice moving over bedrock carries embedded rock debris that scratches parallel grooves (striations) in the direction of ice movement}$
Answer: parallel grooves in bedrock

Parallel grooves (glacial striations) scratched into bedrock by rock debris embedded in moving ice sheets directly indicate the direction and presence of past glacial advance.

Question 33 · 2 pts

The photograph below shows the metamorphic rock gneiss. Three minerals that would most likely be found in this rock are

earth question 33 figure
Worked solution
$\text{Gneiss is a foliated metamorphic rock typically formed from granite, retaining granitic mineral composition}$
Answer: quartz, mica, and feldspar

Gneiss is a high-grade foliated metamorphic rock typically derived from granite, so it commonly contains quartz, mica, and feldspar, matching the Reference Tables scheme for gneiss.

Question 34 · 2 pts

Which igneous rock has a fine-grained texture and would have a mineral composition of 57% plagioclase feldspar, 28% amphibole, and 15% biotite?

Worked solution
$\text{On the Scheme for Igneous Rock Identification, this mineral composition (plagioclase feldspar, amphibole, biotite) corresponds to the intermediate compositional column}$
$\text{Fine-grained texture in the intermediate composition column} = \text{andesite}$
Answer: andesite

This mineral composition (plagioclase feldspar, amphibole, biotite) corresponds to the intermediate compositional column on the Igneous Rock Identification chart; with fine-grained texture, this is andesite.

Question 35 · 2 pts

What is a common use of the mineral composed of $Fe_3Al_2Si_3O_{12}$?

Worked solution
$Fe_3Al_2Si_3O_{12} \text{ is the chemical formula for garnet, listed in the Reference Tables Properties of Common Minerals}$
$\text{Garnet's common use listed is jewelry}$
Answer: jewelry

The formula Fe3Al2Si3O12 corresponds to garnet on the Reference Tables, and garnet's common use listed there is jewelry.

Question 36 · 2 pts

Base your answer on the passage below. King Tides and Climate Change Spring tides are extremely high tides that occur when Earth, the Moon, and the Sun are aligned. A king tide is the common name given to an extremely high spring tide that occurs at perigee (when the Moon is closest to Earth in its orbit). King tides occur once or twice every year. Normal high tides are now reaching higher and extending farther inland than in the past because of global climate change and rising sea levels. This causes low-lying shorelines to be at an increased risk of flooding. If climate continues to change and sea levels rise, spring tides and king tides will have a greater effect on shorelines and coastal flooding. Which diagram best illustrates the positions of Earth (E), the Sun (S), and the Moon (M), and the relative distance between Earth and the Moon that results in a king tide?

earth question 36 figure
Worked solution
$\text{A king tide requires syzygy (Sun-Earth-Moon aligned) plus the Moon at perigee (closest distance to Earth)}$
Answer: Diagram 3

A king tide requires the Sun, Earth, and Moon to be aligned in a straight line (syzygy) with the Moon at its closest orbital distance to Earth (perigee), which is shown in diagram 3.

Question 37 · 2 pts

A coastal community can make advanced preparations for a king tide event because the motions of Earth and the Moon are

Worked solution
$\text{Earth's rotation, revolution, and the Moon's orbit are regular, repeating (cyclic) motions that can be calculated in advance}$
Answer: cyclic and predictable

The orbital motions of Earth and the Moon are regular and repeating (cyclic), which allows their positions, and therefore tide events, to be predicted well in advance.

Question 38 · 2 pts

Which factor is primarily causing global average sea levels to rise over time?

Worked solution
$\text{Melting land-based glacial ice adds water volume to the oceans, and warmer ocean water also expands (thermal expansion)}$
Answer: increased melting of glacial ice

The primary cause of global average sea level rise is the increased melting of glacial ice (and ice sheets), which adds additional water to the oceans.

Question 39 · 2 pts

Two major greenhouse gases that may be linked to global warming are

Worked solution
$\text{Carbon dioxide and methane are well-known greenhouse gases that trap heat in the atmosphere}$
Answer: carbon dioxide and methane

Carbon dioxide and methane are two major greenhouse gases that trap heat in Earth's atmosphere and are linked to global warming, unlike nitrogen and oxygen which are not significant greenhouse gases.

Question 40 · 2 pts

Base your answers on the map showing the four time zones and some latitude and longitude lines across the continental United States. Buffalo is in the Eastern time zone; Omaha is in the Central time zone. What time is it in Buffalo, New York, when it is 5:00 p.m. in Omaha, Nebraska?

earth question 40 figure
Worked solution
$\text{Eastern time zone is one hour ahead of Central time zone}$
$5:00\ \text{p.m. (Central)} + 1\ \text{hour} = 6:00\ \text{p.m. (Eastern)}$
Answer: 6:00 p.m.

Buffalo is in the Eastern time zone, which is one hour ahead of the Central time zone where Omaha is located, so 5:00 p.m. Central time is 6:00 p.m. Eastern time.

Question 41 · 2 pts

Base your answers on the map showing the four time zones and some latitude and longitude lines across the continental United States. Reno, Nevada is located at approximately 40 degrees N latitude. The altitude of Polaris, as measured by an observer in Reno, Nevada, is closest to

earth question 41 figure
Worked solution
$\text{The altitude of Polaris above the horizon equals the observer's latitude}$
$\text{Reno is located at approximately } 40°\text{N latitude, so altitude of Polaris} \approx 39°-40°$
Answer: 39°

The altitude of Polaris above the northern horizon is numerically equal to the observer's latitude; Reno's latitude on the map is approximately 39-40 degrees N.

Question 42 · 2 pts

Base your answers on the map showing the four time zones and some latitude and longitude lines across the continental United States, with cities Mobile, Omaha, Miami, and Seattle labeled at different latitudes. Which city has the shortest duration of insolation on December 21?

earth question 42 figure
Worked solution
$\text{On December 21 (winter solstice), duration of daylight in the Northern Hemisphere decreases with increasing latitude}$
$\text{Seattle is the northernmost city listed, so it has the shortest day length}$
Answer: Seattle

On the December solstice, Northern Hemisphere locations farther north have shorter daylight duration; Seattle, being the northernmost of the listed cities, has the shortest period of insolation.

Question 43 · 2 pts

Base your answers on the map showing the four time zones and some latitude and longitude lines across the continental United States. The four time zones shown on the map are based on Earth's

earth question 43 figure
Worked solution
$\text{Earth rotates } 360°\ \text{every 24 hours} = 15°/\text{hour}$
$\text{Time zones are based on 15-degree longitude divisions corresponding to Earth's rotation rate}$
Answer: rate of rotation

Time zones are established based on Earth's rotation rate of 15 degrees of longitude per hour, dividing the globe into 24 standard time zones.

Question 44 · 2 pts

Base your answers on the information and pie chart below. The pie chart shows percentages of different rock types used to make crushed stone in the United States during 2017. Crushed Stone Crushed stone is the world's most basic mineral commodity. It is abundant, widely available, and inexpensive. In 2017, the United States produced 1.3 billion tons of crushed stone, most of which is used in highway and building construction. United States Crushed Stone Rock Types (pie chart): Limestone and Dolostone 70%, Granite 13%, Trap Rock (usually dark colored mafic igneous rock) 6%, Sandstone/Quartzite 4%, Other 7%. Which two elements are common to both rocks that make up the 4% section of the United States crushed stone rock types?

earth question 44 figure
Worked solution
$\text{The 4% section is Sandstone/Quartzite}$
$\text{Sandstone is composed largely of quartz (SiO}_2\text{) grains, and quartzite is metamorphosed quartz sandstone}$
$\text{Both rocks are dominated by quartz, made of silicon and oxygen}$
Answer: silicon and oxygen

The 4% section represents sandstone and quartzite, both of which are composed primarily of the mineral quartz (SiO2), so silicon and oxygen are common to both rocks.

Question 45 · 2 pts

Which rock would most likely be included in trap rock?

Worked solution
$\text{Trap rock is defined in the passage as a dark-colored mafic igneous rock.}$
$\text{Diabase is a dark, mafic (Fe/Mg-rich) intrusive igneous rock, matching this description.}$
$\text{Phyllite is metamorphic; obsidian and rhyolite are light-colored felsic extrusive igneous rocks, so they don't fit.}$
Answer: diabase

Trap rock is a dark, mafic igneous rock, and diabase is the mafic intrusive rock among the choices that matches that composition and color.

Question 46 · 2 pts

Which crushed rock would most likely weather the fastest in areas that receive high levels of acidic rain?

Worked solution
$\text{Limestone is composed mainly of calcite (CaCO}_3\text{).}$
$\text{Calcite reacts chemically with acids, dissolving relatively quickly (chemical weathering).}$
$\text{Trap rock, granite, and quartzite are silicate-rich and far more resistant to acid attack.}$
Answer: limestone

Limestone's calcite mineral content reacts readily with acids, so it chemically weathers much faster than silicate rocks like granite, trap rock, or quartzite when exposed to acid rain.

Question 47 · 2 pts

When the driver of a dump truck raises the bed of the dump truck to release its load, quartzite tends to scratch the bottom and sides of the bed easier than limestone. Which property of the minerals in quartzite causes them to more easily scratch the bed of the dump truck?

Worked solution
$\text{Quartzite is composed largely of quartz, hardness} \approx 7 \text{ on the Mohs scale.}$
$\text{Limestone's calcite has hardness} \approx 3.}$
$\text{A harder mineral scratches a softer one, so quartz's greater hardness causes the scratching.}$
Answer: hardness

Hardness measures a mineral's resistance to scratching; quartz in quartzite is much harder than the truck bed material or limestone's calcite, so it scratches more easily.

Question 48 · 2 pts

One reason why carbon-14 was used to determine the age of the woolly rhinoceros tooth was because carbon-14

Worked solution
$\text{Carbon-14 is incorporated into all living organisms through carbon uptake.}$
$\text{It remains in organic remains (teeth, bone, wood) after death, making it useful for dating such material.}$
$\text{Its half-life (5730 years) is actually short, and it does fully decay over time, ruling out (1) and (2); it is unrelated to Carboniferous bedrock, ruling out (4).}$
Answer: can be found in most organic remains

Carbon-14 is present in all organic material because living things continuously take in carbon, so it is well-suited for dating organic remains like a tooth.

Question 49 · 2 pts

Approximately how much of the original uranium-238 is left in the sample of zircon crystals?

The sample of zircon crystals is approximately 4.4 billion years old (one uranium-238 half-life).

Worked solution
$\text{Half-life of U-238} \approx 4.5 \text{ billion years (per Reference Tables).}$
$\text{Sample age} = 4.4 \text{ billion years} \approx 1 \text{ half-life.}$
$\text{After one half-life, 50\% of the original parent isotope remains.}$
Answer: 50%

Since the zircon's age of 4.4 billion years is approximately equal to one half-life of uranium-238 (4.5 billion years), about half of the original U-238 remains.

Question 50 · 2 pts

The iridium-enriched sediment layer was deposited at the end of which geological period?

Worked solution
$\text{Sample age} = 65{,}500{,}000 \text{ years, matching the end of the Cretaceous Period on the Geologic Time Scale.}$
$\text{An iridium-rich layer at this age is the well-known K-Pg boundary marker from the asteroid impact mass extinction.}$
Answer: Cretaceous

An iridium-enriched layer dated to about 65.5 million years corresponds to the Cretaceous-Paleogene boundary, marking the end of the Cretaceous Period.

Question 51 · 1 pt · constructed response

Base your answers to questions 51 through 54 on the satellite image below and on your knowledge of Earth science. The satellite image shows the Horton River flowing into Franklin Bay in northwestern Canada. Large abandoned meanders, now lakes, indicate where the Horton River once flowed before it changed its path over time. Identify the name of the large depositional feature formed by the sediment deposited where the Horton River flows into Franklin Bay.

earth question 51 figure
Model answer: Delta
Worked solution
$\text{As the river enters the standing water of the bay, its velocity drops sharply.}$
$\text{Loss of velocity causes loss of carrying capacity, so sediment is deposited at the mouth.}$
$\text{This buildup of deposited sediment at a river's mouth forms a delta.}$
Answer: Delta

A delta forms where a river's velocity decreases as it enters a body of standing water, causing it to deposit its sediment load at its mouth.

Question 52 · 1 pt · constructed response

State the general relationship between the distance from the shoreline and the average diameters of the rock particles deposited in Franklin Bay.

earth question 52 figure
Model answer: As the distance from the shoreline increases, the average diameter of the deposited rock particles decreases.
Worked solution
$\text{Larger, heavier particles have higher settling velocities and drop out of the water quickly near the river mouth/shore.}$
$\text{Smaller, lighter particles remain suspended longer and are carried farther out into the bay before settling.}$
$\text{Therefore, particle size decreases with increasing distance from the shoreline.}$
Answer: As the distance from the shoreline increases, the average diameter of the deposited rock particles decreases.

Sediment sorts by settling velocity: coarser particles settle first near shore, while finer particles are carried farther out, so grain size decreases with distance from shore.

Question 53 · 1 pt · constructed response

Identify the dominant process that occurs along the outside of a meander curve and the dominant process that occurs along the inside of a meander curve that caused this meandering river to change its path.

The satellite image shows the Horton River flowing into Franklin Bay in northwestern Canada. Large abandoned meanders, now lakes, indicate where the Horton River once flowed before it changed its path over time. Identify the name of the large depositional feature formed by the sediment deposited where the Horton River flows into Franklin Bay.

Model answer: Erosion is dominant along the outside of a meander curve, and deposition is dominant along the inside of a meander curve.
Worked solution
$\text{Stream velocity is greatest on the outside of a meander curve, causing erosion of the bank.}$
$\text{Stream velocity is slowest on the inside of a meander curve, causing deposition of sediment (point bar).}$
$\text{Over time, this erosion/deposition pattern shifts the meander's position, changing the river's path.}$
Answer: Erosion is dominant along the outside of a meander curve, and deposition is dominant along the inside of a meander curve.

Faster current on the outside of a meander erodes the bank, while slower current on the inside deposits sediment, gradually shifting the meander and changing the river's course.

Question 54 · 1 pt · constructed response

Explain how abrasion causes rounding of rock particles as they are transported by the Horton River.

Model answer: As rock particles are carried along by the river's current, they repeatedly collide and rub against each other and against the streambed; these impacts wear away sharp corners and edges over time, progressively rounding the particles the farther/longer they are transported.
Worked solution
$\text{Moving water carries particles that collide with each other and the streambed.}$
$\text{Repeated collisions chip and grind away sharp edges and corners.}$
$\text{Over distance and time, particles become progressively smoother and more rounded.}$
Answer: As rock particles are carried along by the river's current, they repeatedly collide and rub against each other and against the streambed; these impacts wear away sharp corners and edges over time, progressively rounding the particles the farther/longer they are transported.

Abrasion is the mechanical wearing down of particles through collision and friction during transport, which removes sharp edges and rounds the particles.

Question 55 · 1 pt · constructed response

Base your answers to questions 55 through 57 on the passage and map below and on your knowledge of Earth science. The map shows the area of a portion of Africa that includes the Ethiopian Rift and the East African Rift. The large arrows show the directions that the Arabian Plate and the two new plates forming from the African Plate, labeled X and Y, are moving. Letter A represents a plate boundary. East African Rift: The East African Rift System begins at a location where two tectonic plates meet in the Afar region of Africa. The two plates that meet at this location are the Arabian Plate to the north and the African Plate to the south. The African Plate is rifting to form two new plates. The Nubian Plate is moving westward and the Somalian Plate is moving eastward relative to the rift. The rift system began approximately 25 million years ago and now stretches over 3000 kilometers from the Red Sea and Gulf of Aden down toward the southern tip of Africa. The process of rifting is believed to occur due to heat rising from magma below Earth's surface. In 2005, a part of the crust in the Afar region gave way, spreading the rift area by an additional 26 feet. This crack was filled with 600 billion gallons of molten rock, which solidified into basalt. Identify the type of plate boundary at location A.

earth question 55 figure
Model answer: Divergent plate boundary
Worked solution
$\text{Location A lies along the Red Sea, where the Arabian and African Plates are pulling apart (rifting).}$
$\text{Plates moving apart, with magma rising to fill the gap and form new crust, define a divergent boundary.}$
Answer: Divergent plate boundary

Location A is where the Red Sea is widening as the Arabian and African Plates move apart, which is the definition of a divergent plate boundary.

Question 56 · 1 pt · constructed response

Identify the names of the new plates, labeled plate X and plate Y, that are forming from the African Plate.

earth question 56 figure
Model answer: Plate X is the Nubian Plate (moving westward) and Plate Y is the Somalian Plate (moving eastward).
Worked solution
$\text{Passage states the Nubian Plate moves westward and the Somalian Plate moves eastward relative to the rift.}$
$\text{On the map, Plate X is positioned to the west of the East African Rift and Plate Y to the east.}$
$\text{Matching directions: Plate X = Nubian Plate, Plate Y = Somalian Plate.}$
Answer: Plate X is the Nubian Plate (moving westward) and Plate Y is the Somalian Plate (moving eastward).

The passage identifies the two new plates by their motion (Nubian moving west, Somalian moving east), which matches the map positions of Plate X (west) and Plate Y (east).

Question 57 · 1 pt · constructed response

Identify one tectonic event, other than volcanic activity, that is associated with rifting in Africa.

Model answer: Earthquakes (seismic activity)
Worked solution
$\text{Rifting stretches and fractures Earth's crust along fault lines.}$
$\text{Stress buildup and release along these faults generates earthquakes.}$
Answer: Earthquakes (seismic activity)

As tectonic plates pull apart during rifting, crustal faulting releases stress as earthquakes, in addition to the volcanic activity from rising magma.

Question 58 · 1 pt · constructed response

Base your answers to questions 58 and 59 on the isoline map on your answer booklet and on your knowledge of Earth science. The map shows the snowfall totals recorded in inches from a lake effect snowstorm that occurred as a result of cold air passing over the warmer water of Lake Michigan in January 2012. The 2-inch and 8-inch snowfall isolines are shown. On the map in your answer booklet, draw the 4-inch and 6-inch snowfall isolines. Extend both isolines to the top edge of the map.

Model answer: The 4-inch isoline should be drawn between the given 2-inch and 8-inch isolines (closer to the 2-inch line), and the 6-inch isoline drawn between the 4-inch isoline and the 8-inch isoline. Both new isolines should follow the same general curved, roughly concentric pattern as the given isolines (not crossing any other isoline) and should be extended so they reach the top edge of the map.
Worked solution
$\text{Isolines of a continuous field (snowfall totals) never cross and are spaced in order of value.}$
$\text{The 4-inch isoline must lie between the 2-inch and 8-inch isolines, and the 6-inch isoline between the 4-inch and 8-inch isolines.}$
$\text{Both should mimic the shape/curvature of the existing isolines and extend to the map's top edge as instructed.}$
Answer: The 4-inch isoline should be drawn between the given 2-inch and 8-inch isolines (closer to the 2-inch line), and the 6-inch isoline drawn between the 4-inch isoline and the 8-inch isoline. Both new isolines should follow the same general curved, roughly concentric pattern as the given isolines (not crossing any other isoline) and should be extended so they reach the top edge of the map.

Isolines of increasing snowfall value must be nested in numerical order without crossing, following the shape of the existing lines, and extended per the instructions.

Question 59 · 1 pt · constructed response

The air mass that moved over Lake Michigan producing the snowstorm originated in Central Canada. Write the two-letter air mass symbol for this air mass that originated over Canada.

Model answer: cP
Worked solution
$\text{Air mass source region: Central Canada = continental (dry) source, high-latitude (polar) temperature.}$
$\text{Continental + polar} \Rightarrow \text{symbol cP.}$
Answer: cP

An air mass originating over land in Central Canada is continental and polar in character, giving it the two-letter symbol cP.

Question 60 · 1 pt · constructed response

Base your answers to questions 60 through 62 on the diagram on your answer booklet and on your knowledge of Earth science. The diagram represents the apparent path of the Sun for a location in New York State on October 21st. Three positions of the Sun are labeled A, B, and C. The length and direction of the shadow of a vertical stick are represented at three different times during the day. Compass directions along the horizon are labeled. On the diagram in your answer booklet, write the letter in each box labeled "Sun position" that caused each of the three shadows.

Sun-path diagram for a New York State location on October 21: The horizon is labeled with compass directions N, E, S, and W. The Sun's apparent path is an arc that begins at the horizon south of due east, reaches its highest point in the southern sky, and ends at the horizon south of due west. Three Sun positions are marked on the arc: A is low in the sky in the southeast (shortly after sunrise), B is at the top of the arc due south (midday), and C is low in the sky in the southwest (shortly before sunset). A vertical stick at the observer's location casts three shadows at three different times, each with a box labeled "Sun position": | Shadow | Length | Direction it points | |---|---|---| | 1 | long | toward the northwest | | 2 | short | due north | | 3 | long | toward the northeast | A shadow points directly away from the Sun, so match each shadow to the Sun position (A, B, or C) that produced it.

Model answer: Match each shadow box to the Sun position whose location in the sky is geometrically consistent with that shadow's length and direction: the shortest shadow, pointing most nearly north, corresponds to the Sun position highest in the sky (nearest due south, close to solar noon); the longer shadows pointing toward the west and toward the east correspond to the lower morning (Sun in the east) and afternoon (Sun in the west) Sun positions, respectively.
Worked solution
$\text{A shadow points opposite the direction of the Sun and lengthens as the Sun's altitude decreases.}$
$\text{The Sun position with the greatest altitude (nearest solar noon, due south) casts the shortest, most northward-pointing shadow.}$
$\text{Lower Sun positions (morning in the east, afternoon in the west) cast longer shadows pointing west and east, respectively.}$
Answer: Match each shadow box to the Sun position whose location in the sky is geometrically consistent with that shadow's length and direction: the shortest shadow, pointing most nearly north, corresponds to the Sun position highest in the sky (nearest due south, close to solar noon); the longer shadows pointing toward the west and toward the east correspond to the lower morning (Sun in the east) and afternoon (Sun in the west) Sun positions, respectively.

Shadow length and direction are determined by the Sun's altitude and azimuth, so each shadow can be matched to the Sun position consistent with that geometry.

Question 61 · 1 pt · constructed response

On the diagram in your answer booklet, draw the apparent path of the Sun from sunrise to sunset on March 21.

The diagram shows the Sun's apparent path on October 21 for a New York State location: an arc rising at the horizon south of due east, peaking in the southern sky, and setting south of due west. Compass directions N, E, S, and W are labeled on the horizon. Draw (or describe) the March 21 path relative to this October 21 path: where it rises, where it sets, and how high it arcs compared to the October path.

Model answer: On March 21 (the spring equinox), the Sun rises due east and sets due west, tracing an arc across the sky with the Sun's noon altitude midway between the October 21 path shown and the summer solstice path, giving about 12 hours of daylight.
Worked solution
$\text{March 21 is the vernal equinox, when the Sun's direct rays strike the equator.}$
$\text{At the equinox, the Sun rises due east and sets due west everywhere.}$
$\text{The arc's maximum altitude at solar noon is intermediate between the winter and summer extremes, giving equal day and night (12 hours each).}$
Answer: On March 21 (the spring equinox), the Sun rises due east and sets due west, tracing an arc across the sky with the Sun's noon altitude midway between the October 21 path shown and the summer solstice path, giving about 12 hours of daylight.

At an equinox the Sun rises due east, sets due west, and follows an arc with noon altitude midway between the solstice extremes, producing 12 hours of daylight.

Question 62 · 1 pt · constructed response

Identify the letter of the Sun's position where it is closest to solar noon. State one piece of evidence shown in the diagram that supports your answer.

Sun-path diagram for New York State on October 21: Sun position A is low in the southeastern sky (morning), position B is at the highest point of the arc due south, and position C is low in the southwestern sky (afternoon). The vertical stick's three shadows are: a long shadow pointing northwest, the shortest shadow pointing due north, and a long shadow pointing northeast.

Model answer: The Sun's position closest to solar noon is the one that produced the shortest shadow (pointing most directly north), since at solar noon the Sun reaches its highest altitude and is due south, casting the day's shortest shadow.
Worked solution
$\text{Solar noon = the moment the Sun is highest in the sky and directly on the observer's meridian (due south in NY State).}$
$\text{The highest Sun altitude produces the shortest shadow of the day.}$
$\text{Therefore the Sun position associated with the shortest, most-northward shadow marks solar noon.}$
Answer: The Sun's position closest to solar noon is the one that produced the shortest shadow (pointing most directly north), since at solar noon the Sun reaches its highest altitude and is due south, casting the day's shortest shadow.

Solar noon corresponds to the Sun's maximum daily altitude, which produces the shortest shadow of the day, pointing due north in the Northern Hemisphere midlatitudes.

Question 63 · 1 pt · constructed response

Base your answers to questions 63 through 65 on the passage below and on your knowledge of Earth science. The passage describes the life cycle of a Sun-like star. The Life Cycle of a Sun-like Star: Stars like the Sun begin as a nebula, which is a large cloud of gas and dust in space. Under the right conditions, a clump in the nebula begins to rotate, contract, and heat up, causing the gas (mainly hydrogen) and dust to glow more brightly. This protostar, or early star, may reach a surface temperature of 3000 K with a luminosity 10 times greater than the Sun. As the protostar continues to collapse, heat and pressure in its core become so great that nuclear reactions begin to use the hydrogen as fuel to produce energy, and the protostar becomes a star. The pressure created by nuclear reactions in the star's core pushes outward and exactly balances the inward pull of gravity. This balance between forces allows the star to maintain a certain size and brightness for about 80% of its life. The star may reach a surface temperature of 5500 K and a luminosity equal to the Sun. After most of the hydrogen in the core is used up, a series of changes first causes the star to collapse, then greatly expand beyond its original size as the surface temperature lowers to 3000 K and the luminosity increases to 2000 times greater than the Sun. As the star reaches the end of its supply of nuclear fuel, its size changes to an equatorial diameter similar to the size of Earth. The surface temperature will be approximately 13,000 K and the luminosity will decrease to 0.001 that of the Sun. In your answer booklet, complete the table by matching the stage of star development listed below to its description. Each stage will be used only once. (Giant, Main Sequence, Nebula, Protostar, White Dwarf)

Complete the table by writing one stage next to each description (each stage used once: Giant, Main Sequence, Nebula, Protostar, White Dwarf): | Description | Stage of Star Development | |---|---| | Large cloud of gas and dust in space | ____ | | Surface temperature of 3000 K and luminosity 10 times greater than the Sun | ____ | | Surface temperature of 5500 K and luminosity equal to the Sun | ____ | | Surface temperature of 3000 K and luminosity 2000 times greater than the Sun | ____ | | Surface temperature of about 13,000 K, luminosity 0.001 that of the Sun, and equatorial diameter similar to Earth | ____ |

Model answer: Nebula: a large cloud of gas and dust in space, before contraction begins. Protostar: a contracting, rotating clump that heats up, reaching about 3000 K surface temperature with a luminosity 10 times the Sun's. Main Sequence: nuclear fusion pressure balances gravity, keeping the star's size/brightness steady for about 80% of its life, at about 5500 K surface temperature with luminosity equal to the Sun's. Giant: after core hydrogen is used up the star expands greatly, surface cooling to about 3000 K while luminosity rises to 2000 times the Sun's. White Dwarf: after nuclear fuel is exhausted, the star collapses to about Earth's diameter, with surface temperature about 13,000 K and luminosity only 0.001 times the Sun's.
Worked solution
$\text{Match each stage's description directly to the sequence given in the passage, in order: Nebula} \to \text{Protostar} \to \text{Main Sequence} \to \text{Giant} \to \text{White Dwarf}.$
$\text{Each stage's key data (temperature, luminosity, size) from the passage identifies it uniquely.}$
Answer: Nebula: a large cloud of gas and dust in space, before contraction begins. Protostar: a contracting, rotating clump that heats up, reaching about 3000 K surface temperature with a luminosity 10 times the Sun's. Main Sequence: nuclear fusion pressure balances gravity, keeping the star's size/brightness steady for about 80% of its life, at about 5500 K surface temperature with luminosity equal to the Sun's. Giant: after core hydrogen is used up the star expands greatly, surface cooling to about 3000 K while luminosity rises to 2000 times the Sun's. White Dwarf: after nuclear fuel is exhausted, the star collapses to about Earth's diameter, with surface temperature about 13,000 K and luminosity only 0.001 times the Sun's.

The passage describes the stages of stellar evolution in chronological order, and each stage's temperature/luminosity/size data uniquely identifies it as Nebula, Protostar, Main Sequence, Giant, or White Dwarf.

Question 64 · 1 pt · constructed response

Identify the force responsible for the contraction of gas and dust within a nebula.

Model answer: Gravity
Worked solution
$\text{Gravitational attraction between gas/dust particles pulls the nebula's clump inward, causing it to contract.}$
Answer: Gravity

Gravity is the attractive force that pulls the gas and dust in a nebula together, causing it to contract and eventually form a protostar.

Question 65 · 1 pt · constructed response

State the name of the gas that fuels the nuclear process in the core of a star for 80% of its life.

Model answer: Hydrogen
Worked solution
$\text{During the main sequence (about 80\% of a star's life), core nuclear fusion converts hydrogen into helium, releasing energy.}$
Answer: Hydrogen

Hydrogen is the fuel used in nuclear fusion reactions in a star's core during its main-sequence stage, which lasts about 80% of its life.

Question 66 · 1 pt · constructed response

Base your answers to questions 66 through 68 on the passage and map below and on your knowledge of Earth science. The lines on the map show the average locations of the Intertropical Convergence Zone (ITCZ) in January and July. ITCZ: The Intertropical Convergence Zone (ITCZ) is a band of low atmospheric pressure where prevailing winds generally blow toward the equator and come together. As the Sun's direct rays move northward from the equator after the Northern Hemisphere's spring equinox, the ITCZ also moves northward. As the Sun's direct rays move southward from the equator after the Northern Hemisphere's autumnal equinox, the ITCZ also moves southward. This northward and southward movement shifts a greater distance over land than over water. Explain why the Intertropical Convergence Zone shifts northward after the Northern Hemisphere's spring equinox.

earth question 66 figure
Model answer: After the spring equinox, Earth's Northern Hemisphere tilts increasingly toward the Sun, so the Sun's most direct rays move north of the equator toward the Tropic of Cancer. Since the ITCZ forms where solar heating and rising air are greatest, this zone of maximum heating (and the associated ITCZ low-pressure band) migrates northward along with the Sun's direct rays.
Worked solution
$\text{After the spring equinox, Earth's axial tilt causes the Sun's direct rays to move north of the equator.}$
$\text{The ITCZ is located where solar heating (and thus rising air/low pressure) is greatest.}$
$\text{As the zone of most intense heating shifts north with the Sun's direct rays, the ITCZ follows it northward.}$
Answer: After the spring equinox, Earth's Northern Hemisphere tilts increasingly toward the Sun, so the Sun's most direct rays move north of the equator toward the Tropic of Cancer. Since the ITCZ forms where solar heating and rising air are greatest, this zone of maximum heating (and the associated ITCZ low-pressure band) migrates northward along with the Sun's direct rays.

The ITCZ tracks the latitude of the Sun's most direct rays because that is where surface heating and rising air are strongest; after the spring equinox those rays move north, so the ITCZ follows.

Question 67 · 1 pt · constructed response

Base your answer on the passage and map below and on your knowledge of Earth science. The lines on the map show the average locations of the Intertropical Convergence Zone (ITCZ) in January and July. ITCZ The Intertropical Convergence Zone (ITCZ) is a band of low atmospheric pressure where prevailing winds generally blow toward the equator and come together. As the Sun's direct rays move northward from the equator after the Northern Hemisphere's spring equinox, the ITCZ also moves northward. As the Sun's direct rays move southward from the equator after the Northern Hemisphere's autumnal equinox, the ITCZ also moves southward. This northward and southward movement shifts a greater distance over land than over water. Use the map to identify the maximum latitude that the Intertropical Convergence Zone reached when it shifted farthest north in July and the maximum latitude that it reached when it shifted farthest south in January.

earth question 67 figure
Model answer: Approximately 25°N in July (over land in Asia/Africa) and approximately 20° to 25°S in January (over land in southern Africa/Australia).
Worked solution
$\text{Locate the highest point of the solid (July) curve on the latitude scale}$ read directly off the map grid lines
$\text{Locate the lowest point of the dashed (January) curve on the latitude scale}$
$\text{Record both readings as the maximum north and south latitudes}$
Answer: Approximately 25°N in July (over land in Asia/Africa) and approximately 20° to 25°S in January (over land in southern Africa/Australia).

The ITCZ migrates with the Sun's direct rays; because land heats and cools faster than water, its northward and southward swing is exaggerated over continents, producing peaks near 25°N in July and around 20-25°S in January.

Question 68 · 1 pt · constructed response

Base your answers to questions 66 through 68 on the passage and map (ITCZ) described above. 68 Explain why air is rising along the Intertropical Convergence Zone.

earth question 68 figure
Model answer: Air is rising along the ITCZ because intense solar heating near the equator warms the surface air, making it less dense; this warm, low-density air expands and rises by convection, creating the band of low pressure.
Worked solution
$\text{Equator receives the most direct/intense insolation}$
$\text{Surface air is heated} \Rightarrow \text{density decreases}$
$\text{Less dense air rises (convection)} \Rightarrow \text{low pressure at surface}$
Answer: Air is rising along the ITCZ because intense solar heating near the equator warms the surface air, making it less dense; this warm, low-density air expands and rises by convection, creating the band of low pressure.

The ITCZ marks where trade winds from both hemispheres converge over strongly heated equatorial land/ocean; the resulting warm air is buoyant and rises, which is why it is a region of low pressure and heavy convective rainfall.

Question 69 · 1 pt · constructed response

Base your answers to questions 69 through 71 on the diagram on your answer booklet and on your knowledge of Earth science. The diagram represents a satellite in four positions, labeled A through D, in its orbit around a celestial object. 69 The celestial object represents one of the foci of this satellite's elliptical orbit. On the diagram in your answer booklet, draw an X to represent the position of the other focus of this orbit.

The diagram shows an elongated elliptical orbit around a celestial object. The celestial object is located at one focus, on the major axis to one side of the ellipse's center. Four satellite positions are marked on the orbit: A is at the end of the major axis nearest the celestial object, C is at the opposite end of the major axis farthest from the celestial object, and B and D are at opposite ends of the minor axis. The second focus lies on the major axis on the other side of the center, the same distance from the center as the celestial object.

Model answer: The X should be placed on the major axis of the ellipse, on the opposite side of the orbit's center from the celestial object, the same distance from the center as the celestial object is.
Worked solution
$\text{An ellipse has two foci located symmetrically about its center along the major axis}$
$\text{Distance from center to focus 1 (celestial object)} = \text{distance from center to focus 2 (X)}$
$\text{Plot X at this mirrored location}$
Answer: The X should be placed on the major axis of the ellipse, on the opposite side of the orbit's center from the celestial object, the same distance from the center as the celestial object is.

Every ellipse has two foci equidistant from its center along the major axis; since the celestial object occupies one focus, the second focus (X) must lie the same distance from the center on the opposite side.

Question 70 · 1 pt · constructed response

Base your answers to questions 69 through 71 on the diagram on your answer booklet described above. 70 Describe what would happen to the shape of the orbit represented in the diagram if the two foci were close together.

Model answer: The orbit would become more circular (less eccentric); as the two foci move closer together and approach the same point, the ellipse approaches the shape of a circle.
Worked solution
$\text{Eccentricity } e = \frac{\text{distance between foci}}{\text{length of major axis}}$
$\text{As distance between foci} \to 0,\ e \to 0$
$e \to 0 \Rightarrow \text{orbit shape} \to \text{circle}$
Answer: The orbit would become more circular (less eccentric); as the two foci move closer together and approach the same point, the ellipse approaches the shape of a circle.

Orbital eccentricity depends on how far apart the two foci are relative to the major axis; foci close together give a low eccentricity, meaning a more circular orbit.

Question 71 · 1 pt · constructed response

Base your answers to questions 69 through 71 on the diagram on your answer booklet described above. 71 Identify the letter of the position where the satellite experiences the greatest amount of gravitational attraction with the celestial object. Explain why the greatest amount of gravitational attraction occurs at this position.

The elliptical orbit diagram shows the celestial object at one focus. Satellite position A is at the end of the major axis nearest the celestial object, C is at the far end of the major axis farthest from the object, and B and D are at opposite ends of the minor axis (intermediate distances).

Model answer: The greatest gravitational attraction occurs at the lettered position closest to the celestial object (perihelion-like point) because gravitational force increases as the distance between two objects decreases.
Worked solution
$F_g \propto \frac{1}{d^2}$
$\text{Smallest } d \Rightarrow \text{largest } F_g$
$\text{Identify the labeled position (A, B, C, or D) nearest the celestial object on the diagram}$
Answer: The greatest gravitational attraction occurs at the lettered position closest to the celestial object (perihelion-like point) because gravitational force increases as the distance between two objects decreases.

Gravitational force is inversely proportional to the square of the distance between two bodies, so the satellite feels the strongest pull at the point in its orbit where it is physically closest to the celestial object.

Question 72 · 1 pt · constructed response

Base your answers to questions 72 through 74 on the data table below and on your knowledge of Earth science. The data table shows the barometric pressure at the center of Hurricane Katrina and the wind speed recorded at the same time each day from August 23 through August 30, 2005. Hurricane Katrina Data | Date | Barometric Pressure (mb) | Wind Speed (mph) | | --- | --- | --- | | August 23 | 1007 | 35 | | August 24 | 1002 | 45 | | August 25 | 985 | 75 | | August 26 | 965 | 100 | | August 27 | 945 | 115 | | August 28 | 902 | 165 | | August 29 | 960 | 75 | | August 30 | 991 | 30 | 72 On the grid in your answer booklet, construct a line graph by plotting the data for the wind speed for each date shown on the data table. Connect all eight plots with a line.

Model answer: Plot the eight points (Aug 23, 35 mph), (Aug 24, 45 mph), (Aug 25, 75 mph), (Aug 26, 100 mph), (Aug 27, 115 mph), (Aug 28, 165 mph), (Aug 29, 75 mph), (Aug 30, 30 mph) on the wind-speed-vs-date grid and connect them in sequence with straight line segments, producing a curve that rises to a peak on August 28 and then falls sharply.
Worked solution
$(\text{Aug 23},35),(\text{Aug 24},45),(\text{Aug 25},75),(\text{Aug 26},100)$
$(\text{Aug 27},115),(\text{Aug 28},165),(\text{Aug 29},75),(\text{Aug 30},30)$
$\text{Connect all eight points in date order with straight segments}$
Answer: Plot the eight points (Aug 23, 35 mph), (Aug 24, 45 mph), (Aug 25, 75 mph), (Aug 26, 100 mph), (Aug 27, 115 mph), (Aug 28, 165 mph), (Aug 29, 75 mph), (Aug 30, 30 mph) on the wind-speed-vs-date grid and connect them in sequence with straight line segments, producing a curve that rises to a peak on August 28 and then falls sharply.

Plotting wind speed against date for all eight recorded days produces a line that increases steadily to a maximum of 165 mph on August 28, then drops rapidly as the storm weakens.

Question 73 · 1 pt · constructed response

Base your answers to questions 72 through 74 on the Hurricane Katrina data table described above. 73 Hurricane Katrina made landfall on August 29. Describe one piece of evidence shown in the data table that supports this statement.

earth question 73 figure
Model answer: Between August 28 and August 29, the barometric pressure rose sharply (from 902 mb to 960 mb) and the wind speed dropped sharply (from 165 mph to 75 mph), showing the storm rapidly weakened as it moved from open water onto land.
Worked solution
$\Delta P = 960-902 = 58\text{ mb increase (Aug 28} \to \text{Aug 29)}$
$\Delta v = 165-75 = 90\text{ mph decrease (Aug 28} \to \text{Aug 29)}$
$\text{Sudden weakening} \Rightarrow \text{consistent with loss of warm-ocean energy source at landfall}$
Answer: Between August 28 and August 29, the barometric pressure rose sharply (from 902 mb to 960 mb) and the wind speed dropped sharply (from 165 mph to 75 mph), showing the storm rapidly weakened as it moved from open water onto land.

Hurricanes lose strength (pressure rises, winds slow) once they move over land and are cut off from the warm ocean water that fuels them; the abrupt reversal in both variables on August 29 is direct evidence of landfall.

Question 74 · 1 pt · constructed response

Base your answers to questions 72 through 74 on the Hurricane Katrina data table described above. 74 Describe the relationship between barometric pressure and wind speed.

earth question 74 figure
Model answer: Barometric pressure and wind speed have an inverse (negative) relationship: as barometric pressure decreases, wind speed increases, and as barometric pressure increases, wind speed decreases.
Worked solution
$\text{Lowest pressure (902 mb, Aug 28)} \leftrightarrow \text{highest wind speed (165 mph)}$
$\text{Highest pressure (1007 mb, Aug 23)} \leftrightarrow \text{lowest wind speed (35 mph)}$
$\text{Pattern holds across all eight data points} \Rightarrow \text{inverse relationship}$
Answer: Barometric pressure and wind speed have an inverse (negative) relationship: as barometric pressure decreases, wind speed increases, and as barometric pressure increases, wind speed decreases.

A stronger pressure gradient (lower central pressure) drives faster winds toward the storm's center, so pressure and wind speed vary inversely throughout the table.

Question 75 · 1 pt · constructed response

Base your answer on the cross section below and on your knowledge of Earth science. The cross section represents rock units, labeled A through G, and a fault labeled H. Letters X and Y represent unconformities. The rock layers have not been overturned. Identify, by letter, the only rock unit that remains in the original position in which the rock-forming sediments were deposited.

earth question 75 figure
Model answer: Rock unit A.
Worked solution
$\text{Layer A lies flat/horizontal at the surface, above unconformity } X$
$\text{All other layers (B, C, D, E, F, G) are tilted, folded, or cut by fault } H$
$\Rightarrow \text{A is the only layer never disturbed from its original horizontal depositional position}$
Answer: Rock unit A.

Sediments are originally deposited in horizontal layers; layer A is still flat and undisturbed, while the layers beneath it have been tilted and faulted, so A alone preserves its original position.

Question 76 · 1 pt · constructed response

Base your answers to questions 75 through 78 on the cross section described above. 76 Indicate the relative ages of rock layers A, B, C, and D, and fault H by listing the letters from oldest to youngest.

earth question 76 figure
Model answer: C, D, B, H, A (from oldest to youngest).
Worked solution
$\text{Law of superposition: C deposited before D, D before B (bottom to top)}$
$\text{Fault } H \text{ cuts through C, D, and B} \Rightarrow H \text{ younger than all three}$
$\text{Layer A lies above unconformity } X \text{ and is not cut by } H \Rightarrow A \text{ is youngest}$
Answer: C, D, B, H, A (from oldest to youngest).

Using superposition for the tilted sequence (C oldest, then D, then B) and cross-cutting relationships (the fault must be younger than every layer it cuts, and layer A must be younger still since it is undisturbed by the fault), the oldest-to-youngest order is C, D, B, H, A.

Question 77 · 1 pt · constructed response

Base your answers to questions 75 through 78 on the cross section described above. 77 Identify the letter and the name of the rock unit that has a bioclastic texture and is mostly composed of carbon.

earth question 77 figure
Model answer: Rock unit G, coal.
Worked solution
$\text{Bioclastic texture} \Rightarrow \text{formed from remains of once-living organisms}$
$\text{Composition mostly carbon} \Rightarrow \text{coal (from compacted plant remains)}$
$\text{Match this description to rock unit } G \text{ on the cross section}$
Answer: Rock unit G, coal.

According to the Earth Science Reference Tables' rock classification, coal is the organic sedimentary rock with a bioclastic texture that is composed mainly of carbon, matching rock unit G in the diagram.

Question 78 · 1 pt · constructed response

Base your answers to questions 75 through 78 on the cross section described above. 78 Layer C is Ordovician age bedrock. Identify the name of the trilobite index fossil and the name of the nautiloid index fossil that could be found in rock layer C.

earth question 78 figure
Model answer: Trilobite: Triarthrus eatoni; Nautiloid (cephalopod): Endoceras.
Worked solution
$\text{Layer C is Ordovician age}$
$\text{Consult the Reference Tables' Ordovician index fossils}$
$\text{Trilobite example: } \textit{Triarthrus eatoni}; \text{ Nautiloid example: } \textit{Endoceras}$
Answer: Trilobite: Triarthrus eatoni; Nautiloid (cephalopod): Endoceras.

The New York State Earth Science Reference Tables list Triarthrus eatoni as a representative Ordovician trilobite and Endoceras as a representative Ordovician straight-shelled nautiloid (cephalopod), both valid index fossils for Ordovician bedrock.

Question 79 · 1 pt · constructed response

Base your answer on the topographic map below and on your knowledge of Earth science. Letters A, B, C, X, and Y are points on the map. Contour interval is 10 feet. Scale: 0 to 5 miles. Calculate the gradient between points X and Y in feet per mile.

earth question 79 figure
Model answer: Gradient ≈ 50 feet per mile (based on an elevation difference of about 50 ft between the contours at X and Y over a map distance of about 1 mile); exact value depends on direct measurement of the map.
Worked solution
$\text{Gradient} = \frac{\text{change in elevation}}{\text{distance}}$
$\Delta elevation \approx 450\text{ ft} - 400\text{ ft} = 50\text{ ft}$
$\text{distance} \approx 1\text{ mile (measured with map scale)}$
$\text{Gradient} \approx \frac{50\text{ ft}}{1\text{ mi}} = 50\text{ ft/mi}$
Answer: Gradient ≈ 50 feet per mile (based on an elevation difference of about 50 ft between the contours at X and Y over a map distance of about 1 mile); exact value depends on direct measurement of the map.

Gradient is found by dividing the difference in elevation between two points by the straight-line map distance between them, measured using the map's bar scale.

Question 80 · 1 pt · constructed response

Base your answers to questions 79 through 82 on the topographic map described above. 80 Determine one possible elevation of point A.

earth question 80 figure
Model answer: Any elevation between 450 ft and 460 ft (exclusive of 460 ft), for example 455 ft, since point A lies within the closed 450-ft contour with no higher contour line shown inside it.
Worked solution
$\text{Point A is enclosed by the 450-ft contour}$
$\text{Contour interval} = 10\text{ ft} \Rightarrow \text{next possible contour would be 460 ft}$
$450\text{ ft} < \text{elevation of A} < 460\text{ ft}$
Answer: Any elevation between 450 ft and 460 ft (exclusive of 460 ft), for example 455 ft, since point A lies within the closed 450-ft contour with no higher contour line shown inside it.

Because point A sits inside the innermost closed contour line (450 ft) with no further contour drawn inside it, its true elevation must be somewhere between 450 ft and the next contour interval, 460 ft.

Question 81 · 1 pt · constructed response

Base your answers to questions 79 through 82 on the topographic map described above. 81 On the grid in your answer booklet, construct a topographic profile along line BC by plotting the elevation of each contour line that crosses line BC. Points B and C have already been plotted on the grid. Connect all nine plots with a line from B to C to complete the profile.

earth question 81 figure
Model answer: Plot the elevation of each of the nine contour crossings along line BC (in the order they occur from B to C) on the provided elevation-vs-distance grid, then connect all nine points with a smooth line from B to C, producing a profile that rises and falls corresponding to the hills and valleys crossed.
Worked solution
$\text{Identify each of the 9 contour lines line BC crosses, in order}$
$\text{Record the elevation value of each crossing}$
$\text{Plot each elevation at its correct horizontal position on the profile grid}$
$\text{Connect all 9 points, including endpoints B and C, with a line}$
Answer: Plot the elevation of each of the nine contour crossings along line BC (in the order they occur from B to C) on the provided elevation-vs-distance grid, then connect all nine points with a smooth line from B to C, producing a profile that rises and falls corresponding to the hills and valleys crossed.

A topographic profile is built by transferring each contour crossing along the chosen transect to an elevation-vs-distance grid in the same left-to-right order as they occur on the map, then connecting the points to show the shape of the land surface.

Question 82 · 1 pt · constructed response

Base your answers to questions 79 through 82 on the topographic map described above. 82 State the general compass direction in which Lyman Stream flows. Describe the evidence shown by the contour lines on the map that supports this stream flowing downhill in that direction.

earth question 82 figure
Model answer: Lyman Stream flows generally toward the south. The contour lines crossing the stream form V-shapes that point north (upstream, toward higher elevation), which shows the stream is flowing downhill from the higher elevations in the north to lower elevations in the south.
Worked solution
$\text{Rule: contour V's always point upstream (toward higher elevation)}$
$\text{Observed V's along Lyman Stream point toward the north (higher contour values)}$
$\Rightarrow \text{stream flow direction is south, opposite the V's point}$
Answer: Lyman Stream flows generally toward the south. The contour lines crossing the stream form V-shapes that point north (upstream, toward higher elevation), which shows the stream is flowing downhill from the higher elevations in the north to lower elevations in the south.

On a topographic map, contour lines crossing a stream always bend to point in the upstream direction (toward higher ground); since these V-shapes point north along Lyman Stream, the water itself must be flowing south, downhill.

Question 83 · 1 pt · constructed response

Base your answers to questions 83 through 85 on the diagram on your answer booklet and on your knowledge of Earth science. The diagram represents temperatures that occur on either side of a mountain. Elevations are recorded in meters (m) above sea level and air temperature is recorded in degrees Celsius (°C). Letters A and B represent reference lines on the diagram. 83 On the diagram in your answer booklet, draw one arrowhead on reference line A and one arrowhead on reference line B to indicate the direction of air flow on each side of the mountain.

The diagram is a cross-section of a mountain. Prevailing winds move air toward the mountain from the west (left side). Air rises up the west slope, passes over the summit, and descends the east slope. Reference line A runs parallel to the west slope and reference line B runs parallel to the east slope. Air temperatures marked on the diagram decrease with elevation going up the west side; at equal elevations the descending air on the east side is warmer than the air on the west side, because the rising air cools and loses moisture (clouds and precipitation on the windward side) while the sinking air on the leeward side warms by compression.

Model answer: On reference line A (the windward, western side), draw the arrowhead pointing up the slope (air rising); on reference line B (the leeward, eastern side), draw the arrowhead pointing down the slope (air sinking/descending).
Worked solution
$\text{Prevailing wind pushes air up the windward (west) slope} \Rightarrow \text{arrow on A points upward}$
$\text{After crossing the summit, air descends the leeward (east) slope} \Rightarrow \text{arrow on B points downward}$
Answer: On reference line A (the windward, western side), draw the arrowhead pointing up the slope (air rising); on reference line B (the leeward, eastern side), draw the arrowhead pointing down the slope (air sinking/descending).

In an orographic circulation, air is forced to rise on the windward side of a mountain and, having lost most of its moisture, descends and warms on the leeward side, so the arrowheads must point upslope on A and downslope on B.

Question 84 · 1 pt · constructed response

Base your answers to questions 83 through 85 on the mountain diagram described above. 84 Compared to the air temperature and moisture conditions on the western side of the mountain at sea level, describe how the air temperature and moisture conditions on the eastern side of the mountain at sea level are different.

The diagram is a cross-section of a mountain. Prevailing winds move air toward the mountain from the west (left side). Air rises up the west slope, passes over the summit, and descends the east slope. Reference line A runs parallel to the west slope and reference line B runs parallel to the east slope. Air temperatures marked on the diagram decrease with elevation going up the west side; at equal elevations the descending air on the east side is warmer than the air on the west side, because the rising air cools and loses moisture (clouds and precipitation on the windward side) while the sinking air on the leeward side warms by compression.

Model answer: The air on the eastern (leeward) side at sea level is warmer and drier than the air on the western (windward) side at sea level.
Worked solution
$\text{Windward (west) air rises, cools, condenses, and loses moisture as precipitation}$
$\text{Leeward (east) air descends and warms by compression (dry adiabatic warming)}$
$\Rightarrow \text{east-side sea-level air is warmer and has lower humidity (rain-shadow effect)}$
Answer: The air on the eastern (leeward) side at sea level is warmer and drier than the air on the western (windward) side at sea level.

This is the classic orographic/rain-shadow effect: moist air loses its water vapor while rising and cooling on the windward slope, then the now-dry air warms adiabatically as it sinks on the leeward slope, leaving the eastern base warmer and drier than the western base.

Question 85 · 1 pt · constructed response

Base your answers to questions 83 through 85 on the mountain diagram described above. 85 State the most probable air temperature that would occur on the west side of the mountain at 1500 meters and the most probable air temperature on the east side of the mountain at 1500 meters.

Model answer: The west (windward) side at 1500 m would be cooler, having cooled at the wet adiabatic lapse rate of about 0.6°C per 100 m while rising and condensing moisture; the east (leeward) side at 1500 m would be warmer, having warmed at the dry adiabatic lapse rate of about 1°C per 100 m while descending. Exact numeric values depend on the sea-level starting temperatures given in the diagram.
Worked solution
$\text{West side (rising, saturated air): cools at wet adiabatic rate} \approx 0.6^\circ\text{C}/100\text{ m}$
$\text{East side (descending, dry air): warms at dry adiabatic rate} \approx 1^\circ\text{C}/100\text{ m}$
$\text{Apply each rate from sea level to 1500 m using the diagram's given sea-level temperatures}$
Answer: The west (windward) side at 1500 m would be cooler, having cooled at the wet adiabatic lapse rate of about 0.6°C per 100 m while rising and condensing moisture; the east (leeward) side at 1500 m would be warmer, having warmed at the dry adiabatic lapse rate of about 1°C per 100 m while descending. Exact numeric values depend on the sea-level starting temperatures given in the diagram.

Rising saturated air cools more slowly (wet adiabatic rate, about 0.6°C/100 m) because condensation releases latent heat, while descending dry air on the leeward side warms faster (dry adiabatic rate, about 1°C/100 m), so at the same elevation the leeward-side air is warmer than the windward-side air.

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