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3,101 | 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. |
3,102 | 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. |
3,103 | 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. |
3,104 | 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. |
3,105 | 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. |
3,106 | 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. |
3,107 | 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. |
3,108 | 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. |
3,109 | 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. |
3,110 | 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. |
3,111 | 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. |
3,112 | 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. |
3,113 | 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. |
3,114 | 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. |
3,115 | 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. |
3,116 | 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. |
3,117 | 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. |
3,118 | 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. |
3,119 | 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. |
3,120 | 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. |
3,121 | 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. |
3,122 | 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. |
3,123 | 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. |
3,124 | 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. |
3,125 | 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. |
3,126 | 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. |
3,127 | 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. |
3,128 | 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. |
3,129 | 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. |
3,130 | 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. |
3,131 | 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. |
3,132 | 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. |
3,133 | 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. |
3,134 | 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. |
3,135 | earth_space | astronomy | kepler_third_law | 5 | worked_example | Orbital period for semi-major axis 15.73 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 = 15.73 AU one obtains T = 62.41 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. |
3,136 | earth_space | astronomy | kepler_third_law | 5 | worked_example | Orbital period for semi-major axis 36.05 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 = 36.05 AU one obtains T = 216.4 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. |
3,137 | earth_space | astronomy | kepler_third_law | 5 | worked_example | Orbital period for semi-major axis 13.82 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 = 13.82 AU one obtains T = 51.4 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. |
3,138 | earth_space | astronomy | kepler_third_law | 5 | worked_example | Orbital period for semi-major axis 2.214 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 = 2.214 AU one obtains T = 3.295 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. |
3,139 | earth_space | astronomy | kepler_third_law | 5 | worked_example | Orbital period for semi-major axis 26.95 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 = 26.95 AU one obtains T = 139.9 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. |
3,140 | earth_space | astronomy | kepler_third_law | 5 | worked_example | Orbital period for semi-major axis 17.95 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 = 17.95 AU one obtains T = 76.03 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. |
3,141 | earth_space | astronomy | kepler_third_law | 5 | worked_example | Orbital period for semi-major axis 11.38 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 = 11.38 AU one obtains T = 38.41 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. |
3,142 | earth_space | astronomy | kepler_third_law | 5 | worked_example | Orbital period for semi-major axis 3.906 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 = 3.906 AU one obtains T = 7.719 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. |
3,143 | earth_space | astronomy | kepler_third_law | 5 | worked_example | Orbital period for semi-major axis 6.958 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 = 6.958 AU one obtains T = 18.36 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. |
3,144 | earth_space | astronomy | kepler_third_law | 5 | worked_example | Orbital period for semi-major axis 1.382 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 = 1.382 AU one obtains T = 1.625 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. |
3,145 | earth_space | astronomy | kepler_third_law | 5 | worked_example | Orbital period for semi-major axis 25.35 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 = 25.35 AU one obtains T = 127.7 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. |
3,146 | earth_space | astronomy | kepler_third_law | 5 | worked_example | Orbital period for semi-major axis 30.96 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 = 30.96 AU one obtains T = 172.2 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. |
3,147 | earth_space | astronomy | kepler_third_law | 5 | worked_example | Orbital period for semi-major axis 12.63 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 = 12.63 AU one obtains T = 44.89 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. |
3,148 | earth_space | astronomy | kepler_third_law | 5 | worked_example | Orbital period for semi-major axis 25.95 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 = 25.95 AU one obtains T = 132.2 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. |
3,149 | earth_space | astronomy | kepler_third_law | 5 | worked_example | Orbital period for semi-major axis 25.9 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 = 25.9 AU one obtains T = 131.8 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. |
3,150 | earth_space | astronomy | kepler_third_law | 5 | worked_example | Orbital period for semi-major axis 3.772 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 = 3.772 AU one obtains T = 7.326 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. |
3,151 | earth_space | astronomy | kepler_third_law | 5 | worked_example | Orbital period for semi-major axis 38.22 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 = 38.22 AU one obtains T = 236.3 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. |
3,152 | earth_space | astronomy | kepler_third_law | 5 | worked_example | Orbital period for semi-major axis 21.86 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 = 21.86 AU one obtains T = 102.2 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. |
3,153 | earth_space | astronomy | kepler_third_law | 5 | worked_example | Orbital period for semi-major axis 30.4 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 = 30.4 AU one obtains T = 167.6 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. |
3,154 | earth_space | astronomy | kepler_third_law | 5 | worked_example | Orbital period for semi-major axis 38.31 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 = 38.31 AU one obtains T = 237.1 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. |
3,155 | 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 = 96.66 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. |
3,156 | earth_space | astronomy | kepler_third_law | 5 | worked_example | Orbital period for semi-major axis 30.12 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 = 30.12 AU one obtains T = 165.3 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. |
3,157 | earth_space | astronomy | kepler_third_law | 5 | worked_example | Orbital period for semi-major axis 20.3 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 = 20.3 AU one obtains T = 91.49 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. |
3,158 | earth_space | astronomy | kepler_third_law | 5 | worked_example | Orbital period for semi-major axis 1.139 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 = 1.139 AU one obtains T = 1.216 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. |
3,159 | 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. |
3,160 | earth_space | astronomy | kepler_third_law | 5 | worked_example | Orbital period for semi-major axis 21.46 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 = 21.46 AU one obtains T = 99.42 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. |
3,161 | earth_space | astronomy | kepler_third_law | 5 | worked_example | Orbital period for semi-major axis 38.51 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 = 38.51 AU one obtains T = 239 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. |
3,162 | earth_space | astronomy | kepler_third_law | 5 | worked_example | Orbital period for semi-major axis 11.29 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 = 11.29 AU one obtains T = 37.93 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. |
3,163 | earth_space | astronomy | kepler_third_law | 5 | worked_example | Orbital period for semi-major axis 8.55 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 = 8.55 AU one obtains T = 25 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. |
3,164 | earth_space | astronomy | kepler_third_law | 5 | worked_example | Orbital period for semi-major axis 33.26 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 = 33.26 AU one obtains T = 191.8 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. |
3,165 | earth_space | astronomy | kepler_third_law | 5 | worked_example | Orbital period for semi-major axis 33.62 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 = 33.62 AU one obtains T = 194.9 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. |
3,166 | earth_space | astronomy | kepler_third_law | 5 | worked_example | Orbital period for semi-major axis 23.35 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 = 23.35 AU one obtains T = 112.8 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. |
3,167 | earth_space | astronomy | kepler_third_law | 5 | worked_example | Orbital period for semi-major axis 3.752 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 = 3.752 AU one obtains T = 7.268 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. |
3,168 | earth_space | astronomy | kepler_third_law | 5 | worked_example | Orbital period for semi-major axis 21.53 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 = 21.53 AU one obtains T = 99.89 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. |
3,169 | earth_space | astronomy | kepler_third_law | 5 | worked_example | Orbital period for semi-major axis 13.09 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 = 13.09 AU one obtains T = 47.35 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. |
3,170 | earth_space | astronomy | kepler_third_law | 5 | worked_example | Orbital period for semi-major axis 23.72 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 = 23.72 AU one obtains T = 115.5 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. |
3,171 | earth_space | astronomy | kepler_third_law | 5 | worked_example | Orbital period for semi-major axis 19.86 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 = 19.86 AU one obtains T = 88.52 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. |
3,172 | earth_space | astronomy | kepler_third_law | 5 | worked_example | Orbital period for semi-major axis 37.38 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 = 37.38 AU one obtains T = 228.5 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. |
3,173 | earth_space | astronomy | kepler_third_law | 5 | worked_example | Orbital period for semi-major axis 13.86 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 = 13.86 AU one obtains T = 51.61 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. |
3,174 | earth_space | astronomy | kepler_third_law | 5 | worked_example | Orbital period for semi-major axis 7.548 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 = 7.548 AU one obtains T = 20.74 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. |
3,175 | earth_space | astronomy | kepler_third_law | 5 | worked_example | Orbital period for semi-major axis 1.353 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 = 1.353 AU one obtains T = 1.573 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. |
3,176 | earth_space | astronomy | kepler_third_law | 5 | worked_example | Orbital period for semi-major axis 21.3 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 = 21.3 AU one obtains T = 98.34 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. |
3,177 | earth_space | astronomy | kepler_third_law | 5 | worked_example | Orbital period for semi-major axis 15.53 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 = 15.53 AU one obtains T = 61.19 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. |
3,178 | earth_space | astronomy | kepler_third_law | 5 | worked_example | Orbital period for semi-major axis 4.005 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 = 4.005 AU one obtains T = 8.014 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. |
3,179 | earth_space | astronomy | kepler_third_law | 5 | worked_example | Orbital period for semi-major axis 39.2 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 = 39.2 AU one obtains T = 245.4 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. |
3,180 | earth_space | astronomy | kepler_third_law | 5 | worked_example | Orbital period for semi-major axis 21.85 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 = 21.85 AU one obtains T = 102.2 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. |
3,181 | earth_space | astronomy | kepler_third_law | 5 | worked_example | Orbital period for semi-major axis 11.7 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 = 11.7 AU one obtains T = 40.02 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. |
3,182 | earth_space | astronomy | kepler_third_law | 5 | worked_example | Orbital period for semi-major axis 7.634 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 = 7.634 AU one obtains T = 21.09 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. |
3,183 | earth_space | astronomy | kepler_third_law | 5 | worked_example | Orbital period for semi-major axis 30.82 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 = 30.82 AU one obtains T = 171.1 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. |
3,184 | earth_space | astronomy | kepler_third_law | 5 | worked_example | Orbital period for semi-major axis 6.507 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 = 6.507 AU one obtains T = 16.6 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. |
3,185 | mathematics | algebra | linear_relation | 2 | worked_example | Evaluate linear function y = 4.522 x + 7.144 at x = -8.822 | The linear relation y = m x + b with slope m = 4.522 and intercept b = 7.144 evaluated at x = -8.822 yields y = -32.74. 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. |
3,186 | mathematics | algebra | linear_relation | 2 | worked_example | Evaluate linear function y = 4.611 x + -12.74 at x = 4.266 | The linear relation y = m x + b with slope m = 4.611 and intercept b = -12.74 evaluated at x = 4.266 yields y = 6.93. 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. |
3,187 | mathematics | algebra | linear_relation | 2 | worked_example | Evaluate linear function y = -2.02 x + -16.05 at x = -7.151 | The linear relation y = m x + b with slope m = -2.02 and intercept b = -16.05 evaluated at x = -7.151 yields y = -1.607. 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. |
3,188 | mathematics | algebra | linear_relation | 2 | worked_example | Evaluate linear function y = 3.93 x + -9.683 at x = 8.832 | The linear relation y = m x + b with slope m = 3.93 and intercept b = -9.683 evaluated at x = 8.832 yields y = 25.03. 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. |
3,189 | mathematics | algebra | linear_relation | 2 | worked_example | Evaluate linear function y = 3.995 x + -17.92 at x = -5.208 | The linear relation y = m x + b with slope m = 3.995 and intercept b = -17.92 evaluated at x = -5.208 yields y = -38.73. 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. |
3,190 | mathematics | algebra | linear_relation | 2 | worked_example | Evaluate linear function y = 0.97 x + -10.54 at x = 0.1724 | The linear relation y = m x + b with slope m = 0.97 and intercept b = -10.54 evaluated at x = 0.1724 yields y = -10.37. 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. |
3,191 | mathematics | algebra | linear_relation | 2 | worked_example | Evaluate linear function y = 0.04421 x + 17.97 at x = -4.614 | The linear relation y = m x + b with slope m = 0.04421 and intercept b = 17.97 evaluated at x = -4.614 yields y = 17.77. 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. |
3,192 | mathematics | algebra | linear_relation | 2 | worked_example | Evaluate linear function y = -1.817 x + 2.125 at x = -8.691 | The linear relation y = m x + b with slope m = -1.817 and intercept b = 2.125 evaluated at x = -8.691 yields y = 17.91. 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. |
3,193 | mathematics | algebra | linear_relation | 2 | worked_example | Evaluate linear function y = -4.711 x + 5.342 at x = 8.275 | The linear relation y = m x + b with slope m = -4.711 and intercept b = 5.342 evaluated at x = 8.275 yields y = -33.64. 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. |
3,194 | mathematics | algebra | linear_relation | 2 | worked_example | Evaluate linear function y = -4.814 x + 2.163 at x = 1.723 | The linear relation y = m x + b with slope m = -4.814 and intercept b = 2.163 evaluated at x = 1.723 yields y = -6.133. 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. |
3,195 | mathematics | algebra | linear_relation | 2 | worked_example | Evaluate linear function y = -3.985 x + 10.52 at x = 3.031 | The linear relation y = m x + b with slope m = -3.985 and intercept b = 10.52 evaluated at x = 3.031 yields y = -1.559. 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. |
3,196 | mathematics | algebra | linear_relation | 2 | worked_example | Evaluate linear function y = -3.744 x + 2.73 at x = -1.08 | The linear relation y = m x + b with slope m = -3.744 and intercept b = 2.73 evaluated at x = -1.08 yields y = 6.775. 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. |
3,197 | mathematics | algebra | linear_relation | 2 | worked_example | Evaluate linear function y = 3.463 x + 0.01729 at x = 5.658 | The linear relation y = m x + b with slope m = 3.463 and intercept b = 0.01729 evaluated at x = 5.658 yields y = 19.61. 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. |
3,198 | mathematics | algebra | linear_relation | 2 | worked_example | Evaluate linear function y = -3.576 x + 1.618 at x = -6.636 | The linear relation y = m x + b with slope m = -3.576 and intercept b = 1.618 evaluated at x = -6.636 yields y = 25.35. 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. |
3,199 | mathematics | algebra | linear_relation | 2 | worked_example | Evaluate linear function y = -2.548 x + 18.1 at x = 9.227 | The linear relation y = m x + b with slope m = -2.548 and intercept b = 18.1 evaluated at x = 9.227 yields y = -5.41. 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. |
3,200 | mathematics | algebra | linear_relation | 2 | worked_example | Evaluate linear function y = 0.5515 x + -18.33 at x = 6.253 | The linear relation y = m x + b with slope m = 0.5515 and intercept b = -18.33 evaluated at x = 6.253 yields y = -14.88. 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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