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20,133,201 | biology | cell_biology | organelle_function | 3 | practice_problem | Primary function of the Golgi apparatus | Question: What is the primary function of the Golgi apparatus in a eukaryotic cell? Answer: modification, sorting and packaging of proteins. Organelles compartmentalize incompatible biochemical processes and increase efficiency by concentrating enzymes and substrates. | null | prokaryote_eukaryote | Identify the principal function of major eukaryotic organelles. |
20,133,202 | biology | cell_biology | organelle_function | 3 | practice_problem | Primary function of the vacuole | Question: What is the primary function of the vacuole in a eukaryotic cell? Answer: storage and turgor maintenance in plant cells. Organelles compartmentalize incompatible biochemical processes and increase efficiency by concentrating enzymes and substrates. | null | prokaryote_eukaryote | Identify the principal function of major eukaryotic organelles. |
20,133,203 | biology | cell_biology | organelle_function | 3 | practice_problem | Primary function of the endoplasmic reticulum | Question: What is the primary function of the endoplasmic reticulum in a eukaryotic cell? Answer: protein and lipid synthesis. Organelles compartmentalize incompatible biochemical processes and increase efficiency by concentrating enzymes and substrates. | null | prokaryote_eukaryote | Identify the principal function of major eukaryotic organelles. |
20,133,204 | biology | cell_biology | organelle_function | 3 | practice_problem | Primary function of the vacuole | Question: What is the primary function of the vacuole in a eukaryotic cell? Answer: storage and turgor maintenance in plant cells. Organelles compartmentalize incompatible biochemical processes and increase efficiency by concentrating enzymes and substrates. | null | prokaryote_eukaryote | Identify the principal function of major eukaryotic organelles. |
20,133,205 | biology | cell_biology | organelle_function | 3 | practice_problem | Primary function of the nucleus | Question: What is the primary function of the nucleus in a eukaryotic cell? Answer: storage and protection of genomic DNA. Organelles compartmentalize incompatible biochemical processes and increase efficiency by concentrating enzymes and substrates. | null | prokaryote_eukaryote | Identify the principal function of major eukaryotic organelles. |
20,133,206 | biology | cell_biology | organelle_function | 3 | practice_problem | Primary function of the lysosome | Question: What is the primary function of the lysosome in a eukaryotic cell? Answer: degradation of macromolecules. Organelles compartmentalize incompatible biochemical processes and increase efficiency by concentrating enzymes and substrates. | null | prokaryote_eukaryote | Identify the principal function of major eukaryotic organelles. |
20,133,207 | biology | cell_biology | organelle_function | 3 | practice_problem | Primary function of the nucleus | Question: What is the primary function of the nucleus in a eukaryotic cell? Answer: storage and protection of genomic DNA. Organelles compartmentalize incompatible biochemical processes and increase efficiency by concentrating enzymes and substrates. | null | prokaryote_eukaryote | Identify the principal function of major eukaryotic organelles. |
20,133,208 | biology | cell_biology | organelle_function | 3 | practice_problem | Primary function of the ribosome | Question: What is the primary function of the ribosome in a eukaryotic cell? Answer: protein synthesis. Organelles compartmentalize incompatible biochemical processes and increase efficiency by concentrating enzymes and substrates. | null | prokaryote_eukaryote | Identify the principal function of major eukaryotic organelles. |
20,133,209 | biology | cell_biology | organelle_function | 3 | practice_problem | Primary function of the endoplasmic reticulum | Question: What is the primary function of the endoplasmic reticulum in a eukaryotic cell? Answer: protein and lipid synthesis. Organelles compartmentalize incompatible biochemical processes and increase efficiency by concentrating enzymes and substrates. | null | prokaryote_eukaryote | Identify the principal function of major eukaryotic organelles. |
20,133,210 | biology | cell_biology | organelle_function | 3 | practice_problem | Primary function of the mitochondrion | Question: What is the primary function of the mitochondrion in a eukaryotic cell? Answer: ATP synthesis via oxidative phosphorylation. Organelles compartmentalize incompatible biochemical processes and increase efficiency by concentrating enzymes and substrates. | null | prokaryote_eukaryote | Identify the principal function of major eukaryotic organelles. |
20,133,211 | biology | cell_biology | organelle_function | 3 | practice_problem | Primary function of the endoplasmic reticulum | Question: What is the primary function of the endoplasmic reticulum in a eukaryotic cell? Answer: protein and lipid synthesis. Organelles compartmentalize incompatible biochemical processes and increase efficiency by concentrating enzymes and substrates. | null | prokaryote_eukaryote | Identify the principal function of major eukaryotic organelles. |
20,133,212 | biology | cell_biology | organelle_function | 3 | practice_problem | Primary function of the chloroplast | Question: What is the primary function of the chloroplast in a eukaryotic cell? Answer: photosynthesis. Organelles compartmentalize incompatible biochemical processes and increase efficiency by concentrating enzymes and substrates. | null | prokaryote_eukaryote | Identify the principal function of major eukaryotic organelles. |
20,133,213 | biology | cell_biology | organelle_function | 3 | practice_problem | Primary function of the ribosome | Question: What is the primary function of the ribosome in a eukaryotic cell? Answer: protein synthesis. Organelles compartmentalize incompatible biochemical processes and increase efficiency by concentrating enzymes and substrates. | null | prokaryote_eukaryote | Identify the principal function of major eukaryotic organelles. |
20,133,214 | biology | cell_biology | organelle_function | 3 | practice_problem | Primary function of the chloroplast | Question: What is the primary function of the chloroplast in a eukaryotic cell? Answer: photosynthesis. Organelles compartmentalize incompatible biochemical processes and increase efficiency by concentrating enzymes and substrates. | null | prokaryote_eukaryote | Identify the principal function of major eukaryotic organelles. |
20,133,215 | biology | cell_biology | organelle_function | 3 | practice_problem | Primary function of the endoplasmic reticulum | Question: What is the primary function of the endoplasmic reticulum in a eukaryotic cell? Answer: protein and lipid synthesis. Organelles compartmentalize incompatible biochemical processes and increase efficiency by concentrating enzymes and substrates. | null | prokaryote_eukaryote | Identify the principal function of major eukaryotic organelles. |
20,133,216 | biology | cell_biology | organelle_function | 3 | practice_problem | Primary function of the chloroplast | Question: What is the primary function of the chloroplast in a eukaryotic cell? Answer: photosynthesis. Organelles compartmentalize incompatible biochemical processes and increase efficiency by concentrating enzymes and substrates. | null | prokaryote_eukaryote | Identify the principal function of major eukaryotic organelles. |
20,133,217 | biology | cell_biology | organelle_function | 3 | practice_problem | Primary function of the vacuole | Question: What is the primary function of the vacuole in a eukaryotic cell? Answer: storage and turgor maintenance in plant cells. Organelles compartmentalize incompatible biochemical processes and increase efficiency by concentrating enzymes and substrates. | null | prokaryote_eukaryote | Identify the principal function of major eukaryotic organelles. |
20,133,218 | biology | cell_biology | organelle_function | 3 | practice_problem | Primary function of the mitochondrion | Question: What is the primary function of the mitochondrion in a eukaryotic cell? Answer: ATP synthesis via oxidative phosphorylation. Organelles compartmentalize incompatible biochemical processes and increase efficiency by concentrating enzymes and substrates. | null | prokaryote_eukaryote | Identify the principal function of major eukaryotic organelles. |
20,133,219 | biology | cell_biology | organelle_function | 3 | practice_problem | Primary function of the Golgi apparatus | Question: What is the primary function of the Golgi apparatus in a eukaryotic cell? Answer: modification, sorting and packaging of proteins. Organelles compartmentalize incompatible biochemical processes and increase efficiency by concentrating enzymes and substrates. | null | prokaryote_eukaryote | Identify the principal function of major eukaryotic organelles. |
20,133,220 | biology | cell_biology | organelle_function | 3 | practice_problem | Primary function of the vacuole | Question: What is the primary function of the vacuole in a eukaryotic cell? Answer: storage and turgor maintenance in plant cells. Organelles compartmentalize incompatible biochemical processes and increase efficiency by concentrating enzymes and substrates. | null | prokaryote_eukaryote | Identify the principal function of major eukaryotic organelles. |
20,133,221 | biology | cell_biology | organelle_function | 3 | practice_problem | Primary function of the ribosome | Question: What is the primary function of the ribosome in a eukaryotic cell? Answer: protein synthesis. Organelles compartmentalize incompatible biochemical processes and increase efficiency by concentrating enzymes and substrates. | null | prokaryote_eukaryote | Identify the principal function of major eukaryotic organelles. |
20,133,222 | biology | cell_biology | organelle_function | 3 | practice_problem | Primary function of the lysosome | Question: What is the primary function of the lysosome in a eukaryotic cell? Answer: degradation of macromolecules. Organelles compartmentalize incompatible biochemical processes and increase efficiency by concentrating enzymes and substrates. | null | prokaryote_eukaryote | Identify the principal function of major eukaryotic organelles. |
20,133,223 | earth_space | astronomy | solar_system_scale | 2 | explanation | Relative Scales in the Solar System | The Sun contains 99.8 % of the mass of the Solar System. The terrestrial planets (Mercury, Venus, Earth, Mars) are small, rocky, and close to the Sun; the Jovian planets (Jupiter, Saturn, Uranus, Neptune) are large, volatile-rich, and farther out. Distances are conveniently measured in astronomical units (1 AU ≈ 1.496 ... | 1 AU ≈ 1.496e11 m | null | Describe the mass distribution and orbital architecture of the Solar System. |
20,133,224 | earth_space | astronomy | stellar_parallax | 5 | explanation | Stellar Parallax and Distance Measurement | The apparent shift in position of a nearby star against the background of distant stars, measured from opposite sides of Earth's orbit, is the trigonometric parallax. Distance in parsecs is the reciprocal of the parallax angle in arcseconds: d (pc) = 1 / p ("). One parsec equals 3.0857 × 10¹⁶ m ≈ 3.26 light-years. Para... | d (pc) = 1 / p (") | basic trigonometry | Explain how trigonometric parallax yields stellar distances. |
20,133,225 | earth_space | geology | plate_tectonics | 4 | explanation | Plate Tectonics | Earth's lithosphere is divided into rigid plates that move relative to one another over the ductile asthenosphere. Divergent boundaries create new crust (mid-ocean ridges); convergent boundaries recycle crust (subduction zones) or build mountain belts; transform boundaries accommodate lateral slip. Mantle convection, s... | null | null | Summarize the types of plate boundaries and the forces that drive plate motion. |
20,133,226 | earth_space | geology | rock_cycle | 3 | explanation | The Rock Cycle | Igneous rocks form by solidification of magma or lava. Sedimentary rocks form by weathering, erosion, deposition, and lithification of pre-existing material. Metamorphic rocks form by recrystallization of existing rocks under elevated temperature and pressure without wholesale melting. Any rock type may be transformed ... | null | null | Describe the three major rock classes and the processes that convert one into another. |
20,133,227 | earth_space | atmospheric_science | greenhouse_effect | 4 | explanation | The Greenhouse Effect | Short-wave solar radiation reaches Earth's surface and is partly absorbed. The surface emits long-wave infrared radiation. Greenhouse gases (H₂O, CO₂, CH₄, etc.) absorb a fraction of this infrared radiation and re-emit it in all directions, including back toward the surface. The result is a higher equilibrium surface t... | null | basic radiation balance | Explain the physical mechanism of the greenhouse effect. |
20,133,228 | earth_space | astronomy | kepler_third_law | 5 | worked_example | Orbital period for semi-major axis 23.54 AU | For a planet orbiting the Sun, Kepler's third law states that the square of the sidereal orbital period T (in years) equals the cube of the semi-major axis a (in AU): T² = a³. With a = 23.54 AU one obtains T = 114.2 years. The law is a direct consequence of Newtonian gravity for a central inverse-square force. | T^2 = a^3 (solar units) | newtonian gravity | Apply Kepler's third law to relate orbital period and semi-major axis. |
20,133,229 | earth_space | astronomy | kepler_third_law | 5 | worked_example | Orbital period for semi-major axis 24.61 AU | For a planet orbiting the Sun, Kepler's third law states that the square of the sidereal orbital period T (in years) equals the cube of the semi-major axis a (in AU): T² = a³. With a = 24.61 AU one obtains T = 122.1 years. The law is a direct consequence of Newtonian gravity for a central inverse-square force. | T^2 = a^3 (solar units) | newtonian gravity | Apply Kepler's third law to relate orbital period and semi-major axis. |
20,133,230 | earth_space | astronomy | kepler_third_law | 5 | worked_example | Orbital period for semi-major axis 31.92 AU | For a planet orbiting the Sun, Kepler's third law states that the square of the sidereal orbital period T (in years) equals the cube of the semi-major axis a (in AU): T² = a³. With a = 31.92 AU one obtains T = 180.3 years. The law is a direct consequence of Newtonian gravity for a central inverse-square force. | T^2 = a^3 (solar units) | newtonian gravity | Apply Kepler's third law to relate orbital period and semi-major axis. |
20,133,231 | earth_space | astronomy | kepler_third_law | 5 | worked_example | Orbital period for semi-major axis 11.53 AU | For a planet orbiting the Sun, Kepler's third law states that the square of the sidereal orbital period T (in years) equals the cube of the semi-major axis a (in AU): T² = a³. With a = 11.53 AU one obtains T = 39.13 years. The law is a direct consequence of Newtonian gravity for a central inverse-square force. | T^2 = a^3 (solar units) | newtonian gravity | Apply Kepler's third law to relate orbital period and semi-major axis. |
20,133,232 | earth_space | astronomy | kepler_third_law | 5 | worked_example | Orbital period for semi-major axis 31.33 AU | For a planet orbiting the Sun, Kepler's third law states that the square of the sidereal orbital period T (in years) equals the cube of the semi-major axis a (in AU): T² = a³. With a = 31.33 AU one obtains T = 175.4 years. The law is a direct consequence of Newtonian gravity for a central inverse-square force. | T^2 = a^3 (solar units) | newtonian gravity | Apply Kepler's third law to relate orbital period and semi-major axis. |
20,133,233 | earth_space | astronomy | kepler_third_law | 5 | worked_example | Orbital period for semi-major axis 21.06 AU | For a planet orbiting the Sun, Kepler's third law states that the square of the sidereal orbital period T (in years) equals the cube of the semi-major axis a (in AU): T² = a³. With a = 21.06 AU one obtains T = 96.63 years. The law is a direct consequence of Newtonian gravity for a central inverse-square force. | T^2 = a^3 (solar units) | newtonian gravity | Apply Kepler's third law to relate orbital period and semi-major axis. |
20,133,234 | earth_space | astronomy | kepler_third_law | 5 | worked_example | Orbital period for semi-major axis 14.88 AU | For a planet orbiting the Sun, Kepler's third law states that the square of the sidereal orbital period T (in years) equals the cube of the semi-major axis a (in AU): T² = a³. With a = 14.88 AU one obtains T = 57.38 years. The law is a direct consequence of Newtonian gravity for a central inverse-square force. | T^2 = a^3 (solar units) | newtonian gravity | Apply Kepler's third law to relate orbital period and semi-major axis. |
20,133,235 | earth_space | astronomy | kepler_third_law | 5 | worked_example | Orbital period for semi-major axis 17.63 AU | For a planet orbiting the Sun, Kepler's third law states that the square of the sidereal orbital period T (in years) equals the cube of the semi-major axis a (in AU): T² = a³. With a = 17.63 AU one obtains T = 74 years. The law is a direct consequence of Newtonian gravity for a central inverse-square force. | T^2 = a^3 (solar units) | newtonian gravity | Apply Kepler's third law to relate orbital period and semi-major axis. |
20,133,236 | earth_space | astronomy | kepler_third_law | 5 | worked_example | Orbital period for semi-major axis 14.91 AU | For a planet orbiting the Sun, Kepler's third law states that the square of the sidereal orbital period T (in years) equals the cube of the semi-major axis a (in AU): T² = a³. With a = 14.91 AU one obtains T = 57.57 years. The law is a direct consequence of Newtonian gravity for a central inverse-square force. | T^2 = a^3 (solar units) | newtonian gravity | Apply Kepler's third law to relate orbital period and semi-major axis. |
20,133,237 | earth_space | astronomy | kepler_third_law | 5 | worked_example | Orbital period for semi-major axis 26.47 AU | For a planet orbiting the Sun, Kepler's third law states that the square of the sidereal orbital period T (in years) equals the cube of the semi-major axis a (in AU): T² = a³. With a = 26.47 AU one obtains T = 136.2 years. The law is a direct consequence of Newtonian gravity for a central inverse-square force. | T^2 = a^3 (solar units) | newtonian gravity | Apply Kepler's third law to relate orbital period and semi-major axis. |
20,133,238 | earth_space | astronomy | kepler_third_law | 5 | worked_example | Orbital period for semi-major axis 18.21 AU | For a planet orbiting the Sun, Kepler's third law states that the square of the sidereal orbital period T (in years) equals the cube of the semi-major axis a (in AU): T² = a³. With a = 18.21 AU one obtains T = 77.69 years. The law is a direct consequence of Newtonian gravity for a central inverse-square force. | T^2 = a^3 (solar units) | newtonian gravity | Apply Kepler's third law to relate orbital period and semi-major axis. |
20,133,239 | earth_space | astronomy | kepler_third_law | 5 | worked_example | Orbital period for semi-major axis 4.401 AU | For a planet orbiting the Sun, Kepler's third law states that the square of the sidereal orbital period T (in years) equals the cube of the semi-major axis a (in AU): T² = a³. With a = 4.401 AU one obtains T = 9.233 years. The law is a direct consequence of Newtonian gravity for a central inverse-square force. | T^2 = a^3 (solar units) | newtonian gravity | Apply Kepler's third law to relate orbital period and semi-major axis. |
20,133,240 | earth_space | astronomy | kepler_third_law | 5 | worked_example | Orbital period for semi-major axis 14.18 AU | For a planet orbiting the Sun, Kepler's third law states that the square of the sidereal orbital period T (in years) equals the cube of the semi-major axis a (in AU): T² = a³. With a = 14.18 AU one obtains T = 53.41 years. The law is a direct consequence of Newtonian gravity for a central inverse-square force. | T^2 = a^3 (solar units) | newtonian gravity | Apply Kepler's third law to relate orbital period and semi-major axis. |
20,133,241 | earth_space | astronomy | kepler_third_law | 5 | worked_example | Orbital period for semi-major axis 28.68 AU | For a planet orbiting the Sun, Kepler's third law states that the square of the sidereal orbital period T (in years) equals the cube of the semi-major axis a (in AU): T² = a³. With a = 28.68 AU one obtains T = 153.6 years. The law is a direct consequence of Newtonian gravity for a central inverse-square force. | T^2 = a^3 (solar units) | newtonian gravity | Apply Kepler's third law to relate orbital period and semi-major axis. |
20,133,242 | earth_space | astronomy | kepler_third_law | 5 | worked_example | Orbital period for semi-major axis 8.867 AU | For a planet orbiting the Sun, Kepler's third law states that the square of the sidereal orbital period T (in years) equals the cube of the semi-major axis a (in AU): T² = a³. With a = 8.867 AU one obtains T = 26.41 years. The law is a direct consequence of Newtonian gravity for a central inverse-square force. | T^2 = a^3 (solar units) | newtonian gravity | Apply Kepler's third law to relate orbital period and semi-major axis. |
20,133,243 | earth_space | astronomy | kepler_third_law | 5 | worked_example | Orbital period for semi-major axis 33.32 AU | For a planet orbiting the Sun, Kepler's third law states that the square of the sidereal orbital period T (in years) equals the cube of the semi-major axis a (in AU): T² = a³. With a = 33.32 AU one obtains T = 192.3 years. The law is a direct consequence of Newtonian gravity for a central inverse-square force. | T^2 = a^3 (solar units) | newtonian gravity | Apply Kepler's third law to relate orbital period and semi-major axis. |
20,133,244 | earth_space | astronomy | kepler_third_law | 5 | worked_example | Orbital period for semi-major axis 32.3 AU | For a planet orbiting the Sun, Kepler's third law states that the square of the sidereal orbital period T (in years) equals the cube of the semi-major axis a (in AU): T² = a³. With a = 32.3 AU one obtains T = 183.6 years. The law is a direct consequence of Newtonian gravity for a central inverse-square force. | T^2 = a^3 (solar units) | newtonian gravity | Apply Kepler's third law to relate orbital period and semi-major axis. |
20,133,245 | earth_space | astronomy | kepler_third_law | 5 | worked_example | Orbital period for semi-major axis 29.19 AU | For a planet orbiting the Sun, Kepler's third law states that the square of the sidereal orbital period T (in years) equals the cube of the semi-major axis a (in AU): T² = a³. With a = 29.19 AU one obtains T = 157.7 years. The law is a direct consequence of Newtonian gravity for a central inverse-square force. | T^2 = a^3 (solar units) | newtonian gravity | Apply Kepler's third law to relate orbital period and semi-major axis. |
20,133,246 | earth_space | astronomy | kepler_third_law | 5 | worked_example | Orbital period for semi-major axis 3.209 AU | For a planet orbiting the Sun, Kepler's third law states that the square of the sidereal orbital period T (in years) equals the cube of the semi-major axis a (in AU): T² = a³. With a = 3.209 AU one obtains T = 5.747 years. The law is a direct consequence of Newtonian gravity for a central inverse-square force. | T^2 = a^3 (solar units) | newtonian gravity | Apply Kepler's third law to relate orbital period and semi-major axis. |
20,133,247 | earth_space | astronomy | kepler_third_law | 5 | worked_example | Orbital period for semi-major axis 8.133 AU | For a planet orbiting the Sun, Kepler's third law states that the square of the sidereal orbital period T (in years) equals the cube of the semi-major axis a (in AU): T² = a³. With a = 8.133 AU one obtains T = 23.19 years. The law is a direct consequence of Newtonian gravity for a central inverse-square force. | T^2 = a^3 (solar units) | newtonian gravity | Apply Kepler's third law to relate orbital period and semi-major axis. |
20,133,248 | earth_space | astronomy | kepler_third_law | 5 | worked_example | Orbital period for semi-major axis 10.4 AU | For a planet orbiting the Sun, Kepler's third law states that the square of the sidereal orbital period T (in years) equals the cube of the semi-major axis a (in AU): T² = a³. With a = 10.4 AU one obtains T = 33.52 years. The law is a direct consequence of Newtonian gravity for a central inverse-square force. | T^2 = a^3 (solar units) | newtonian gravity | Apply Kepler's third law to relate orbital period and semi-major axis. |
20,133,249 | earth_space | astronomy | kepler_third_law | 5 | worked_example | Orbital period for semi-major axis 2.304 AU | For a planet orbiting the Sun, Kepler's third law states that the square of the sidereal orbital period T (in years) equals the cube of the semi-major axis a (in AU): T² = a³. With a = 2.304 AU one obtains T = 3.496 years. The law is a direct consequence of Newtonian gravity for a central inverse-square force. | T^2 = a^3 (solar units) | newtonian gravity | Apply Kepler's third law to relate orbital period and semi-major axis. |
20,133,250 | earth_space | astronomy | kepler_third_law | 5 | worked_example | Orbital period for semi-major axis 9.849 AU | For a planet orbiting the Sun, Kepler's third law states that the square of the sidereal orbital period T (in years) equals the cube of the semi-major axis a (in AU): T² = a³. With a = 9.849 AU one obtains T = 30.91 years. The law is a direct consequence of Newtonian gravity for a central inverse-square force. | T^2 = a^3 (solar units) | newtonian gravity | Apply Kepler's third law to relate orbital period and semi-major axis. |
20,133,251 | earth_space | astronomy | kepler_third_law | 5 | worked_example | Orbital period for semi-major axis 24.82 AU | For a planet orbiting the Sun, Kepler's third law states that the square of the sidereal orbital period T (in years) equals the cube of the semi-major axis a (in AU): T² = a³. With a = 24.82 AU one obtains T = 123.6 years. The law is a direct consequence of Newtonian gravity for a central inverse-square force. | T^2 = a^3 (solar units) | newtonian gravity | Apply Kepler's third law to relate orbital period and semi-major axis. |
20,133,252 | earth_space | astronomy | kepler_third_law | 5 | worked_example | Orbital period for semi-major axis 13.48 AU | For a planet orbiting the Sun, Kepler's third law states that the square of the sidereal orbital period T (in years) equals the cube of the semi-major axis a (in AU): T² = a³. With a = 13.48 AU one obtains T = 49.48 years. The law is a direct consequence of Newtonian gravity for a central inverse-square force. | T^2 = a^3 (solar units) | newtonian gravity | Apply Kepler's third law to relate orbital period and semi-major axis. |
20,133,253 | earth_space | astronomy | kepler_third_law | 5 | worked_example | Orbital period for semi-major axis 22.25 AU | For a planet orbiting the Sun, Kepler's third law states that the square of the sidereal orbital period T (in years) equals the cube of the semi-major axis a (in AU): T² = a³. With a = 22.25 AU one obtains T = 104.9 years. The law is a direct consequence of Newtonian gravity for a central inverse-square force. | T^2 = a^3 (solar units) | newtonian gravity | Apply Kepler's third law to relate orbital period and semi-major axis. |
20,133,254 | earth_space | astronomy | kepler_third_law | 5 | worked_example | Orbital period for semi-major axis 34.6 AU | For a planet orbiting the Sun, Kepler's third law states that the square of the sidereal orbital period T (in years) equals the cube of the semi-major axis a (in AU): T² = a³. With a = 34.6 AU one obtains T = 203.5 years. The law is a direct consequence of Newtonian gravity for a central inverse-square force. | T^2 = a^3 (solar units) | newtonian gravity | Apply Kepler's third law to relate orbital period and semi-major axis. |
20,133,255 | earth_space | astronomy | kepler_third_law | 5 | worked_example | Orbital period for semi-major axis 32.8 AU | For a planet orbiting the Sun, Kepler's third law states that the square of the sidereal orbital period T (in years) equals the cube of the semi-major axis a (in AU): T² = a³. With a = 32.8 AU one obtains T = 187.9 years. The law is a direct consequence of Newtonian gravity for a central inverse-square force. | T^2 = a^3 (solar units) | newtonian gravity | Apply Kepler's third law to relate orbital period and semi-major axis. |
20,133,256 | earth_space | astronomy | kepler_third_law | 5 | worked_example | Orbital period for semi-major axis 28.92 AU | For a planet orbiting the Sun, Kepler's third law states that the square of the sidereal orbital period T (in years) equals the cube of the semi-major axis a (in AU): T² = a³. With a = 28.92 AU one obtains T = 155.5 years. The law is a direct consequence of Newtonian gravity for a central inverse-square force. | T^2 = a^3 (solar units) | newtonian gravity | Apply Kepler's third law to relate orbital period and semi-major axis. |
20,133,257 | earth_space | astronomy | kepler_third_law | 5 | worked_example | Orbital period for semi-major axis 14.67 AU | For a planet orbiting the Sun, Kepler's third law states that the square of the sidereal orbital period T (in years) equals the cube of the semi-major axis a (in AU): T² = a³. With a = 14.67 AU one obtains T = 56.16 years. The law is a direct consequence of Newtonian gravity for a central inverse-square force. | T^2 = a^3 (solar units) | newtonian gravity | Apply Kepler's third law to relate orbital period and semi-major axis. |
20,133,258 | earth_space | astronomy | kepler_third_law | 5 | worked_example | Orbital period for semi-major axis 23.16 AU | For a planet orbiting the Sun, Kepler's third law states that the square of the sidereal orbital period T (in years) equals the cube of the semi-major axis a (in AU): T² = a³. With a = 23.16 AU one obtains T = 111.4 years. The law is a direct consequence of Newtonian gravity for a central inverse-square force. | T^2 = a^3 (solar units) | newtonian gravity | Apply Kepler's third law to relate orbital period and semi-major axis. |
20,133,259 | earth_space | astronomy | kepler_third_law | 5 | worked_example | Orbital period for semi-major axis 9.283 AU | For a planet orbiting the Sun, Kepler's third law states that the square of the sidereal orbital period T (in years) equals the cube of the semi-major axis a (in AU): T² = a³. With a = 9.283 AU one obtains T = 28.29 years. The law is a direct consequence of Newtonian gravity for a central inverse-square force. | T^2 = a^3 (solar units) | newtonian gravity | Apply Kepler's third law to relate orbital period and semi-major axis. |
20,133,260 | earth_space | astronomy | kepler_third_law | 5 | worked_example | Orbital period for semi-major axis 20.53 AU | For a planet orbiting the Sun, Kepler's third law states that the square of the sidereal orbital period T (in years) equals the cube of the semi-major axis a (in AU): T² = a³. With a = 20.53 AU one obtains T = 93.03 years. The law is a direct consequence of Newtonian gravity for a central inverse-square force. | T^2 = a^3 (solar units) | newtonian gravity | Apply Kepler's third law to relate orbital period and semi-major axis. |
20,133,261 | earth_space | astronomy | kepler_third_law | 5 | worked_example | Orbital period for semi-major axis 24.39 AU | For a planet orbiting the Sun, Kepler's third law states that the square of the sidereal orbital period T (in years) equals the cube of the semi-major axis a (in AU): T² = a³. With a = 24.39 AU one obtains T = 120.5 years. The law is a direct consequence of Newtonian gravity for a central inverse-square force. | T^2 = a^3 (solar units) | newtonian gravity | Apply Kepler's third law to relate orbital period and semi-major axis. |
20,133,262 | earth_space | astronomy | kepler_third_law | 5 | worked_example | Orbital period for semi-major axis 2.247 AU | For a planet orbiting the Sun, Kepler's third law states that the square of the sidereal orbital period T (in years) equals the cube of the semi-major axis a (in AU): T² = a³. With a = 2.247 AU one obtains T = 3.368 years. The law is a direct consequence of Newtonian gravity for a central inverse-square force. | T^2 = a^3 (solar units) | newtonian gravity | Apply Kepler's third law to relate orbital period and semi-major axis. |
20,133,263 | earth_space | astronomy | kepler_third_law | 5 | worked_example | Orbital period for semi-major axis 3.142 AU | For a planet orbiting the Sun, Kepler's third law states that the square of the sidereal orbital period T (in years) equals the cube of the semi-major axis a (in AU): T² = a³. With a = 3.142 AU one obtains T = 5.568 years. The law is a direct consequence of Newtonian gravity for a central inverse-square force. | T^2 = a^3 (solar units) | newtonian gravity | Apply Kepler's third law to relate orbital period and semi-major axis. |
20,133,264 | earth_space | astronomy | kepler_third_law | 5 | worked_example | Orbital period for semi-major axis 34.1 AU | For a planet orbiting the Sun, Kepler's third law states that the square of the sidereal orbital period T (in years) equals the cube of the semi-major axis a (in AU): T² = a³. With a = 34.1 AU one obtains T = 199.1 years. The law is a direct consequence of Newtonian gravity for a central inverse-square force. | T^2 = a^3 (solar units) | newtonian gravity | Apply Kepler's third law to relate orbital period and semi-major axis. |
20,133,265 | earth_space | astronomy | kepler_third_law | 5 | worked_example | Orbital period for semi-major axis 21.5 AU | For a planet orbiting the Sun, Kepler's third law states that the square of the sidereal orbital period T (in years) equals the cube of the semi-major axis a (in AU): T² = a³. With a = 21.5 AU one obtains T = 99.67 years. The law is a direct consequence of Newtonian gravity for a central inverse-square force. | T^2 = a^3 (solar units) | newtonian gravity | Apply Kepler's third law to relate orbital period and semi-major axis. |
20,133,266 | earth_space | astronomy | kepler_third_law | 5 | worked_example | Orbital period for semi-major axis 5.163 AU | For a planet orbiting the Sun, Kepler's third law states that the square of the sidereal orbital period T (in years) equals the cube of the semi-major axis a (in AU): T² = a³. With a = 5.163 AU one obtains T = 11.73 years. The law is a direct consequence of Newtonian gravity for a central inverse-square force. | T^2 = a^3 (solar units) | newtonian gravity | Apply Kepler's third law to relate orbital period and semi-major axis. |
20,133,267 | earth_space | astronomy | kepler_third_law | 5 | worked_example | Orbital period for semi-major axis 17.69 AU | For a planet orbiting the Sun, Kepler's third law states that the square of the sidereal orbital period T (in years) equals the cube of the semi-major axis a (in AU): T² = a³. With a = 17.69 AU one obtains T = 74.43 years. The law is a direct consequence of Newtonian gravity for a central inverse-square force. | T^2 = a^3 (solar units) | newtonian gravity | Apply Kepler's third law to relate orbital period and semi-major axis. |
20,133,268 | earth_space | astronomy | kepler_third_law | 5 | worked_example | Orbital period for semi-major axis 9.747 AU | For a planet orbiting the Sun, Kepler's third law states that the square of the sidereal orbital period T (in years) equals the cube of the semi-major axis a (in AU): T² = a³. With a = 9.747 AU one obtains T = 30.43 years. The law is a direct consequence of Newtonian gravity for a central inverse-square force. | T^2 = a^3 (solar units) | newtonian gravity | Apply Kepler's third law to relate orbital period and semi-major axis. |
20,133,269 | earth_space | astronomy | kepler_third_law | 5 | worked_example | Orbital period for semi-major axis 7.847 AU | For a planet orbiting the Sun, Kepler's third law states that the square of the sidereal orbital period T (in years) equals the cube of the semi-major axis a (in AU): T² = a³. With a = 7.847 AU one obtains T = 21.98 years. The law is a direct consequence of Newtonian gravity for a central inverse-square force. | T^2 = a^3 (solar units) | newtonian gravity | Apply Kepler's third law to relate orbital period and semi-major axis. |
20,133,270 | earth_space | astronomy | kepler_third_law | 5 | worked_example | Orbital period for semi-major axis 22.67 AU | For a planet orbiting the Sun, Kepler's third law states that the square of the sidereal orbital period T (in years) equals the cube of the semi-major axis a (in AU): T² = a³. With a = 22.67 AU one obtains T = 107.9 years. The law is a direct consequence of Newtonian gravity for a central inverse-square force. | T^2 = a^3 (solar units) | newtonian gravity | Apply Kepler's third law to relate orbital period and semi-major axis. |
20,133,271 | earth_space | astronomy | kepler_third_law | 5 | worked_example | Orbital period for semi-major axis 13.53 AU | For a planet orbiting the Sun, Kepler's third law states that the square of the sidereal orbital period T (in years) equals the cube of the semi-major axis a (in AU): T² = a³. With a = 13.53 AU one obtains T = 49.76 years. The law is a direct consequence of Newtonian gravity for a central inverse-square force. | T^2 = a^3 (solar units) | newtonian gravity | Apply Kepler's third law to relate orbital period and semi-major axis. |
20,133,272 | earth_space | astronomy | kepler_third_law | 5 | worked_example | Orbital period for semi-major axis 5.889 AU | For a planet orbiting the Sun, Kepler's third law states that the square of the sidereal orbital period T (in years) equals the cube of the semi-major axis a (in AU): T² = a³. With a = 5.889 AU one obtains T = 14.29 years. The law is a direct consequence of Newtonian gravity for a central inverse-square force. | T^2 = a^3 (solar units) | newtonian gravity | Apply Kepler's third law to relate orbital period and semi-major axis. |
20,133,273 | earth_space | astronomy | kepler_third_law | 5 | worked_example | Orbital period for semi-major axis 16.45 AU | For a planet orbiting the Sun, Kepler's third law states that the square of the sidereal orbital period T (in years) equals the cube of the semi-major axis a (in AU): T² = a³. With a = 16.45 AU one obtains T = 66.75 years. The law is a direct consequence of Newtonian gravity for a central inverse-square force. | T^2 = a^3 (solar units) | newtonian gravity | Apply Kepler's third law to relate orbital period and semi-major axis. |
20,133,274 | earth_space | astronomy | kepler_third_law | 5 | worked_example | Orbital period for semi-major axis 25.83 AU | For a planet orbiting the Sun, Kepler's third law states that the square of the sidereal orbital period T (in years) equals the cube of the semi-major axis a (in AU): T² = a³. With a = 25.83 AU one obtains T = 131.3 years. The law is a direct consequence of Newtonian gravity for a central inverse-square force. | T^2 = a^3 (solar units) | newtonian gravity | Apply Kepler's third law to relate orbital period and semi-major axis. |
20,133,275 | earth_space | astronomy | kepler_third_law | 5 | worked_example | Orbital period for semi-major axis 14.27 AU | For a planet orbiting the Sun, Kepler's third law states that the square of the sidereal orbital period T (in years) equals the cube of the semi-major axis a (in AU): T² = a³. With a = 14.27 AU one obtains T = 53.92 years. The law is a direct consequence of Newtonian gravity for a central inverse-square force. | T^2 = a^3 (solar units) | newtonian gravity | Apply Kepler's third law to relate orbital period and semi-major axis. |
20,133,276 | earth_space | astronomy | kepler_third_law | 5 | worked_example | Orbital period for semi-major axis 18.21 AU | For a planet orbiting the Sun, Kepler's third law states that the square of the sidereal orbital period T (in years) equals the cube of the semi-major axis a (in AU): T² = a³. With a = 18.21 AU one obtains T = 77.69 years. The law is a direct consequence of Newtonian gravity for a central inverse-square force. | T^2 = a^3 (solar units) | newtonian gravity | Apply Kepler's third law to relate orbital period and semi-major axis. |
20,133,277 | earth_space | astronomy | kepler_third_law | 5 | worked_example | Orbital period for semi-major axis 32.14 AU | For a planet orbiting the Sun, Kepler's third law states that the square of the sidereal orbital period T (in years) equals the cube of the semi-major axis a (in AU): T² = a³. With a = 32.14 AU one obtains T = 182.2 years. The law is a direct consequence of Newtonian gravity for a central inverse-square force. | T^2 = a^3 (solar units) | newtonian gravity | Apply Kepler's third law to relate orbital period and semi-major axis. |
20,133,278 | mathematics | algebra | linear_relation | 2 | worked_example | Evaluate linear function y = 0.9546 x + 1.73 at x = -5.091 | The linear relation y = m x + b with slope m = 0.9546 and intercept b = 1.73 evaluated at x = -5.091 yields y = -3.13. Linear models appear throughout science whenever a rate of change is approximately constant. | y = m x + b | basic arithmetic | Evaluate and interpret a linear function in a scientific context. |
20,133,279 | mathematics | algebra | linear_relation | 2 | worked_example | Evaluate linear function y = -4.867 x + 16.94 at x = 0.9954 | The linear relation y = m x + b with slope m = -4.867 and intercept b = 16.94 evaluated at x = 0.9954 yields y = 12.09. Linear models appear throughout science whenever a rate of change is approximately constant. | y = m x + b | basic arithmetic | Evaluate and interpret a linear function in a scientific context. |
20,133,280 | mathematics | algebra | linear_relation | 2 | worked_example | Evaluate linear function y = 4.086 x + -15.36 at x = -0.5356 | The linear relation y = m x + b with slope m = 4.086 and intercept b = -15.36 evaluated at x = -0.5356 yields y = -17.55. Linear models appear throughout science whenever a rate of change is approximately constant. | y = m x + b | basic arithmetic | Evaluate and interpret a linear function in a scientific context. |
20,133,281 | mathematics | algebra | linear_relation | 2 | worked_example | Evaluate linear function y = 2.627 x + -6.269 at x = 1.278 | The linear relation y = m x + b with slope m = 2.627 and intercept b = -6.269 evaluated at x = 1.278 yields y = -2.911. Linear models appear throughout science whenever a rate of change is approximately constant. | y = m x + b | basic arithmetic | Evaluate and interpret a linear function in a scientific context. |
20,133,282 | mathematics | algebra | linear_relation | 2 | worked_example | Evaluate linear function y = 4.826 x + -17.8 at x = -2.566 | The linear relation y = m x + b with slope m = 4.826 and intercept b = -17.8 evaluated at x = -2.566 yields y = -30.18. Linear models appear throughout science whenever a rate of change is approximately constant. | y = m x + b | basic arithmetic | Evaluate and interpret a linear function in a scientific context. |
20,133,283 | mathematics | algebra | linear_relation | 2 | worked_example | Evaluate linear function y = 1.792 x + 3.347 at x = -9.504 | The linear relation y = m x + b with slope m = 1.792 and intercept b = 3.347 evaluated at x = -9.504 yields y = -13.69. Linear models appear throughout science whenever a rate of change is approximately constant. | y = m x + b | basic arithmetic | Evaluate and interpret a linear function in a scientific context. |
20,133,284 | mathematics | algebra | linear_relation | 2 | worked_example | Evaluate linear function y = 0.3698 x + 4.576 at x = 4.011 | The linear relation y = m x + b with slope m = 0.3698 and intercept b = 4.576 evaluated at x = 4.011 yields y = 6.06. Linear models appear throughout science whenever a rate of change is approximately constant. | y = m x + b | basic arithmetic | Evaluate and interpret a linear function in a scientific context. |
20,133,285 | mathematics | algebra | linear_relation | 2 | worked_example | Evaluate linear function y = 2.443 x + -9.088 at x = 5.045 | The linear relation y = m x + b with slope m = 2.443 and intercept b = -9.088 evaluated at x = 5.045 yields y = 3.235. Linear models appear throughout science whenever a rate of change is approximately constant. | y = m x + b | basic arithmetic | Evaluate and interpret a linear function in a scientific context. |
20,133,286 | mathematics | algebra | linear_relation | 2 | worked_example | Evaluate linear function y = -2.712 x + 10.22 at x = -2.389 | The linear relation y = m x + b with slope m = -2.712 and intercept b = 10.22 evaluated at x = -2.389 yields y = 16.7. Linear models appear throughout science whenever a rate of change is approximately constant. | y = m x + b | basic arithmetic | Evaluate and interpret a linear function in a scientific context. |
20,133,287 | mathematics | algebra | linear_relation | 2 | worked_example | Evaluate linear function y = 0.848 x + 16.73 at x = -9.78 | The linear relation y = m x + b with slope m = 0.848 and intercept b = 16.73 evaluated at x = -9.78 yields y = 8.434. Linear models appear throughout science whenever a rate of change is approximately constant. | y = m x + b | basic arithmetic | Evaluate and interpret a linear function in a scientific context. |
20,133,288 | mathematics | algebra | linear_relation | 2 | worked_example | Evaluate linear function y = -2.755 x + -11.75 at x = 2.355 | The linear relation y = m x + b with slope m = -2.755 and intercept b = -11.75 evaluated at x = 2.355 yields y = -18.23. Linear models appear throughout science whenever a rate of change is approximately constant. | y = m x + b | basic arithmetic | Evaluate and interpret a linear function in a scientific context. |
20,133,289 | mathematics | algebra | linear_relation | 2 | worked_example | Evaluate linear function y = 1.701 x + 15.93 at x = 6.009 | The linear relation y = m x + b with slope m = 1.701 and intercept b = 15.93 evaluated at x = 6.009 yields y = 26.15. Linear models appear throughout science whenever a rate of change is approximately constant. | y = m x + b | basic arithmetic | Evaluate and interpret a linear function in a scientific context. |
20,133,290 | mathematics | algebra | linear_relation | 2 | worked_example | Evaluate linear function y = 1.293 x + 16.25 at x = 9.635 | The linear relation y = m x + b with slope m = 1.293 and intercept b = 16.25 evaluated at x = 9.635 yields y = 28.7. Linear models appear throughout science whenever a rate of change is approximately constant. | y = m x + b | basic arithmetic | Evaluate and interpret a linear function in a scientific context. |
20,133,291 | mathematics | algebra | linear_relation | 2 | worked_example | Evaluate linear function y = 2.571 x + 2.383 at x = 0.841 | The linear relation y = m x + b with slope m = 2.571 and intercept b = 2.383 evaluated at x = 0.841 yields y = 4.546. Linear models appear throughout science whenever a rate of change is approximately constant. | y = m x + b | basic arithmetic | Evaluate and interpret a linear function in a scientific context. |
20,133,292 | mathematics | algebra | linear_relation | 2 | worked_example | Evaluate linear function y = -2.631 x + 0.9067 at x = -3.835 | The linear relation y = m x + b with slope m = -2.631 and intercept b = 0.9067 evaluated at x = -3.835 yields y = 11. Linear models appear throughout science whenever a rate of change is approximately constant. | y = m x + b | basic arithmetic | Evaluate and interpret a linear function in a scientific context. |
20,133,293 | mathematics | algebra | linear_relation | 2 | worked_example | Evaluate linear function y = -3.305 x + 2.151 at x = 5.167 | The linear relation y = m x + b with slope m = -3.305 and intercept b = 2.151 evaluated at x = 5.167 yields y = -14.93. Linear models appear throughout science whenever a rate of change is approximately constant. | y = m x + b | basic arithmetic | Evaluate and interpret a linear function in a scientific context. |
20,133,294 | mathematics | algebra | linear_relation | 2 | worked_example | Evaluate linear function y = -3.35 x + -19.79 at x = -7.259 | The linear relation y = m x + b with slope m = -3.35 and intercept b = -19.79 evaluated at x = -7.259 yields y = 4.523. Linear models appear throughout science whenever a rate of change is approximately constant. | y = m x + b | basic arithmetic | Evaluate and interpret a linear function in a scientific context. |
20,133,295 | mathematics | algebra | linear_relation | 2 | worked_example | Evaluate linear function y = 0.1819 x + 3.591 at x = 6.208 | The linear relation y = m x + b with slope m = 0.1819 and intercept b = 3.591 evaluated at x = 6.208 yields y = 4.72. Linear models appear throughout science whenever a rate of change is approximately constant. | y = m x + b | basic arithmetic | Evaluate and interpret a linear function in a scientific context. |
20,133,296 | mathematics | algebra | linear_relation | 2 | worked_example | Evaluate linear function y = 4.318 x + 7.886 at x = -0.8195 | The linear relation y = m x + b with slope m = 4.318 and intercept b = 7.886 evaluated at x = -0.8195 yields y = 4.348. Linear models appear throughout science whenever a rate of change is approximately constant. | y = m x + b | basic arithmetic | Evaluate and interpret a linear function in a scientific context. |
20,133,297 | mathematics | algebra | linear_relation | 2 | worked_example | Evaluate linear function y = 1.165 x + -14.98 at x = -5.298 | The linear relation y = m x + b with slope m = 1.165 and intercept b = -14.98 evaluated at x = -5.298 yields y = -21.15. Linear models appear throughout science whenever a rate of change is approximately constant. | y = m x + b | basic arithmetic | Evaluate and interpret a linear function in a scientific context. |
20,133,298 | mathematics | algebra | linear_relation | 2 | worked_example | Evaluate linear function y = 4.976 x + 13.87 at x = -6.288 | The linear relation y = m x + b with slope m = 4.976 and intercept b = 13.87 evaluated at x = -6.288 yields y = -17.42. Linear models appear throughout science whenever a rate of change is approximately constant. | y = m x + b | basic arithmetic | Evaluate and interpret a linear function in a scientific context. |
20,133,299 | mathematics | algebra | linear_relation | 2 | worked_example | Evaluate linear function y = -1.616 x + 16.8 at x = -1.861 | The linear relation y = m x + b with slope m = -1.616 and intercept b = 16.8 evaluated at x = -1.861 yields y = 19.81. Linear models appear throughout science whenever a rate of change is approximately constant. | y = m x + b | basic arithmetic | Evaluate and interpret a linear function in a scientific context. |
20,133,300 | mathematics | algebra | linear_relation | 2 | worked_example | Evaluate linear function y = -3.678 x + 0.4993 at x = 6.533 | The linear relation y = m x + b with slope m = -3.678 and intercept b = 0.4993 evaluated at x = 6.533 yields y = -23.53. Linear models appear throughout science whenever a rate of change is approximately constant. | y = m x + b | basic arithmetic | Evaluate and interpret a linear function in a scientific context. |
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