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6,601 | 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. |
6,602 | 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. |
6,603 | 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. |
6,604 | 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. |
6,605 | 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. |
6,606 | 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. |
6,607 | 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. |
6,608 | 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. |
6,609 | earth_space | astronomy | kepler_third_law | 5 | worked_example | Orbital period for semi-major axis 12.04 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 = 12.04 AU one obtains T = 41.79 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. |
6,610 | earth_space | astronomy | kepler_third_law | 5 | worked_example | Orbital period for semi-major axis 2.103 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 = 2.103 AU one obtains T = 3.049 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. |
6,611 | earth_space | astronomy | kepler_third_law | 5 | worked_example | Orbital period for semi-major axis 32.77 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 = 32.77 AU one obtains T = 187.6 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. |
6,612 | earth_space | astronomy | kepler_third_law | 5 | worked_example | Orbital period for semi-major axis 17.47 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 = 17.47 AU one obtains T = 73.04 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. |
6,613 | earth_space | astronomy | kepler_third_law | 5 | worked_example | Orbital period for semi-major axis 11.68 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 = 11.68 AU one obtains T = 39.9 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. |
6,614 | earth_space | astronomy | kepler_third_law | 5 | worked_example | Orbital period for semi-major axis 2.495 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 = 2.495 AU one obtains T = 3.941 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. |
6,615 | earth_space | astronomy | kepler_third_law | 5 | worked_example | Orbital period for semi-major axis 18.29 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 = 18.29 AU one obtains T = 78.21 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. |
6,616 | earth_space | astronomy | kepler_third_law | 5 | worked_example | Orbital period for semi-major axis 27.41 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 = 27.41 AU one obtains T = 143.5 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. |
6,617 | earth_space | astronomy | kepler_third_law | 5 | worked_example | Orbital period for semi-major axis 33.35 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 = 33.35 AU one obtains T = 192.6 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. |
6,618 | earth_space | astronomy | kepler_third_law | 5 | worked_example | Orbital period for semi-major axis 16.18 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 = 16.18 AU one obtains T = 65.05 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. |
6,619 | earth_space | astronomy | kepler_third_law | 5 | worked_example | Orbital period for semi-major axis 21.55 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 = 21.55 AU one obtains T = 100.1 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. |
6,620 | earth_space | astronomy | kepler_third_law | 5 | worked_example | Orbital period for semi-major axis 37.14 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 = 37.14 AU one obtains T = 226.3 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. |
6,621 | earth_space | astronomy | kepler_third_law | 5 | worked_example | Orbital period for semi-major axis 36.24 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 = 36.24 AU one obtains T = 218.2 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. |
6,622 | earth_space | astronomy | kepler_third_law | 5 | worked_example | Orbital period for semi-major axis 29.74 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 = 29.74 AU one obtains T = 162.2 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. |
6,623 | earth_space | astronomy | kepler_third_law | 5 | worked_example | Orbital period for semi-major axis 20.19 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 = 20.19 AU one obtains T = 90.7 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. |
6,624 | earth_space | astronomy | kepler_third_law | 5 | worked_example | Orbital period for semi-major axis 20.73 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 = 20.73 AU one obtains T = 94.39 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. |
6,625 | earth_space | astronomy | kepler_third_law | 5 | worked_example | Orbital period for semi-major axis 36.64 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 = 36.64 AU one obtains T = 221.8 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. |
6,626 | earth_space | astronomy | kepler_third_law | 5 | worked_example | Orbital period for semi-major axis 3.377 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 = 3.377 AU one obtains T = 6.206 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. |
6,627 | earth_space | astronomy | kepler_third_law | 5 | worked_example | Orbital period for semi-major axis 33.42 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 = 33.42 AU one obtains T = 193.2 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. |
6,628 | earth_space | astronomy | kepler_third_law | 5 | worked_example | Orbital period for semi-major axis 25.71 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 = 25.71 AU one obtains T = 130.3 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. |
6,629 | earth_space | astronomy | kepler_third_law | 5 | worked_example | Orbital period for semi-major axis 35.18 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 = 35.18 AU one obtains T = 208.6 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. |
6,630 | earth_space | astronomy | kepler_third_law | 5 | worked_example | Orbital period for semi-major axis 35.81 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 = 35.81 AU one obtains T = 214.3 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. |
6,631 | 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 = 115.6 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. |
6,632 | earth_space | astronomy | kepler_third_law | 5 | worked_example | Orbital period for semi-major axis 33.16 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 = 33.16 AU one obtains T = 191 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. |
6,633 | earth_space | astronomy | kepler_third_law | 5 | worked_example | Orbital period for semi-major axis 4.453 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 = 4.453 AU one obtains T = 9.395 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. |
6,634 | earth_space | astronomy | kepler_third_law | 5 | worked_example | Orbital period for semi-major axis 8.058 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 = 8.058 AU one obtains T = 22.87 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. |
6,635 | earth_space | astronomy | kepler_third_law | 5 | worked_example | Orbital period for semi-major axis 20.97 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 = 20.97 AU one obtains T = 96.02 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. |
6,636 | earth_space | astronomy | kepler_third_law | 5 | worked_example | Orbital period for semi-major axis 38.96 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 = 38.96 AU one obtains T = 243.2 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. |
6,637 | earth_space | astronomy | kepler_third_law | 5 | worked_example | Orbital period for semi-major axis 7.196 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 = 7.196 AU one obtains T = 19.3 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. |
6,638 | earth_space | astronomy | kepler_third_law | 5 | worked_example | Orbital period for semi-major axis 22.03 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 = 22.03 AU one obtains T = 103.4 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. |
6,639 | earth_space | astronomy | kepler_third_law | 5 | worked_example | Orbital period for semi-major axis 29.49 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 = 29.49 AU one obtains T = 160.1 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. |
6,640 | 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 = 88.48 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. |
6,641 | earth_space | astronomy | kepler_third_law | 5 | worked_example | Orbital period for semi-major axis 29.36 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 = 29.36 AU one obtains T = 159.1 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. |
6,642 | earth_space | astronomy | kepler_third_law | 5 | worked_example | Orbital period for semi-major axis 24.42 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 = 24.42 AU one obtains T = 120.7 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. |
6,643 | earth_space | astronomy | kepler_third_law | 5 | worked_example | Orbital period for semi-major axis 9.661 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 = 9.661 AU one obtains T = 30.03 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. |
6,644 | earth_space | astronomy | kepler_third_law | 5 | worked_example | Orbital period for semi-major axis 17.27 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 = 17.27 AU one obtains T = 71.77 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. |
6,645 | earth_space | astronomy | kepler_third_law | 5 | worked_example | Orbital period for semi-major axis 33.12 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 = 33.12 AU one obtains T = 190.6 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. |
6,646 | earth_space | astronomy | kepler_third_law | 5 | worked_example | Orbital period for semi-major axis 35.04 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 = 35.04 AU one obtains T = 207.4 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. |
6,647 | earth_space | astronomy | kepler_third_law | 5 | worked_example | Orbital period for semi-major axis 21.76 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 = 21.76 AU one obtains T = 101.5 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. |
6,648 | earth_space | astronomy | kepler_third_law | 5 | worked_example | Orbital period for semi-major axis 35.12 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 = 35.12 AU one obtains T = 208.1 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. |
6,649 | earth_space | astronomy | kepler_third_law | 5 | worked_example | Orbital period for semi-major axis 33.98 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 = 33.98 AU one obtains T = 198.1 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. |
6,650 | earth_space | astronomy | kepler_third_law | 5 | worked_example | Orbital period for semi-major axis 7.418 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 = 7.418 AU one obtains T = 20.2 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. |
6,651 | earth_space | astronomy | kepler_third_law | 5 | worked_example | Orbital period for semi-major axis 27.39 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 = 27.39 AU one obtains T = 143.3 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. |
6,652 | earth_space | astronomy | kepler_third_law | 5 | worked_example | Orbital period for semi-major axis 13.01 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 = 13.01 AU one obtains T = 46.93 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. |
6,653 | earth_space | astronomy | kepler_third_law | 5 | worked_example | Orbital period for semi-major axis 1.555 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 = 1.555 AU one obtains T = 1.938 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. |
6,654 | earth_space | astronomy | kepler_third_law | 5 | worked_example | Orbital period for semi-major axis 9.465 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 = 9.465 AU one obtains T = 29.12 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. |
6,655 | earth_space | astronomy | kepler_third_law | 5 | worked_example | Orbital period for semi-major axis 30.11 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 = 30.11 AU one obtains T = 165.2 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. |
6,656 | earth_space | astronomy | kepler_third_law | 5 | worked_example | Orbital period for semi-major axis 38.41 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 = 38.41 AU one obtains T = 238 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. |
6,657 | earth_space | astronomy | kepler_third_law | 5 | worked_example | Orbital period for semi-major axis 1.308 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 = 1.308 AU one obtains T = 1.495 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. |
6,658 | earth_space | astronomy | kepler_third_law | 5 | worked_example | Orbital period for semi-major axis 31.51 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 = 31.51 AU one obtains T = 176.9 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. |
6,659 | mathematics | algebra | linear_relation | 2 | worked_example | Evaluate linear function y = 0.8861 x + 8.767 at x = -1.426 | The linear relation y = m x + b with slope m = 0.8861 and intercept b = 8.767 evaluated at x = -1.426 yields y = 7.504. 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. |
6,660 | mathematics | algebra | linear_relation | 2 | worked_example | Evaluate linear function y = 1.807 x + 19.78 at x = -0.9285 | The linear relation y = m x + b with slope m = 1.807 and intercept b = 19.78 evaluated at x = -0.9285 yields y = 18.1. 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. |
6,661 | mathematics | algebra | linear_relation | 2 | worked_example | Evaluate linear function y = 2.001 x + -3.894 at x = 8.764 | The linear relation y = m x + b with slope m = 2.001 and intercept b = -3.894 evaluated at x = 8.764 yields y = 13.64. 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. |
6,662 | mathematics | algebra | linear_relation | 2 | worked_example | Evaluate linear function y = -2.641 x + -3.797 at x = 5.598 | The linear relation y = m x + b with slope m = -2.641 and intercept b = -3.797 evaluated at x = 5.598 yields y = -18.58. 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. |
6,663 | mathematics | algebra | linear_relation | 2 | worked_example | Evaluate linear function y = -3.535 x + 11.7 at x = -2.698 | The linear relation y = m x + b with slope m = -3.535 and intercept b = 11.7 evaluated at x = -2.698 yields y = 21.23. 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. |
6,664 | mathematics | algebra | linear_relation | 2 | worked_example | Evaluate linear function y = -4.977 x + 1.625 at x = -8.313 | The linear relation y = m x + b with slope m = -4.977 and intercept b = 1.625 evaluated at x = -8.313 yields y = 43. 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. |
6,665 | mathematics | algebra | linear_relation | 2 | worked_example | Evaluate linear function y = 3.407 x + -12.78 at x = 2.583 | The linear relation y = m x + b with slope m = 3.407 and intercept b = -12.78 evaluated at x = 2.583 yields y = -3.981. 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. |
6,666 | mathematics | algebra | linear_relation | 2 | worked_example | Evaluate linear function y = -1.721 x + -13.63 at x = 9.843 | The linear relation y = m x + b with slope m = -1.721 and intercept b = -13.63 evaluated at x = 9.843 yields y = -30.57. 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. |
6,667 | mathematics | algebra | linear_relation | 2 | worked_example | Evaluate linear function y = 1.787 x + 6.382 at x = -9.703 | The linear relation y = m x + b with slope m = 1.787 and intercept b = 6.382 evaluated at x = -9.703 yields y = -10.96. 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. |
6,668 | mathematics | algebra | linear_relation | 2 | worked_example | Evaluate linear function y = -1.649 x + 19.55 at x = 8.741 | The linear relation y = m x + b with slope m = -1.649 and intercept b = 19.55 evaluated at x = 8.741 yields y = 5.136. 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. |
6,669 | mathematics | algebra | linear_relation | 2 | worked_example | Evaluate linear function y = 1.867 x + -3.803 at x = -6.993 | The linear relation y = m x + b with slope m = 1.867 and intercept b = -3.803 evaluated at x = -6.993 yields y = -16.86. 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. |
6,670 | mathematics | algebra | linear_relation | 2 | worked_example | Evaluate linear function y = -4.271 x + 6.052 at x = 8.741 | The linear relation y = m x + b with slope m = -4.271 and intercept b = 6.052 evaluated at x = 8.741 yields y = -31.29. 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. |
6,671 | mathematics | algebra | linear_relation | 2 | worked_example | Evaluate linear function y = -4.272 x + 2.825 at x = -3.452 | The linear relation y = m x + b with slope m = -4.272 and intercept b = 2.825 evaluated at x = -3.452 yields y = 17.57. 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. |
6,672 | mathematics | algebra | linear_relation | 2 | worked_example | Evaluate linear function y = -1.021 x + -1.731 at x = 1.883 | The linear relation y = m x + b with slope m = -1.021 and intercept b = -1.731 evaluated at x = 1.883 yields y = -3.653. 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. |
6,673 | mathematics | algebra | linear_relation | 2 | worked_example | Evaluate linear function y = -4.606 x + 3.625 at x = 5.451 | The linear relation y = m x + b with slope m = -4.606 and intercept b = 3.625 evaluated at x = 5.451 yields y = -21.48. 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. |
6,674 | mathematics | algebra | linear_relation | 2 | worked_example | Evaluate linear function y = -2.315 x + -2.802 at x = 2.179 | The linear relation y = m x + b with slope m = -2.315 and intercept b = -2.802 evaluated at x = 2.179 yields y = -7.846. 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. |
6,675 | mathematics | algebra | linear_relation | 2 | worked_example | Evaluate linear function y = -2.112 x + 1.869 at x = -9.677 | The linear relation y = m x + b with slope m = -2.112 and intercept b = 1.869 evaluated at x = -9.677 yields y = 22.31. 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. |
6,676 | mathematics | algebra | linear_relation | 2 | worked_example | Evaluate linear function y = -3.417 x + 15.69 at x = -9.244 | The linear relation y = m x + b with slope m = -3.417 and intercept b = 15.69 evaluated at x = -9.244 yields y = 47.27. 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. |
6,677 | mathematics | algebra | linear_relation | 2 | worked_example | Evaluate linear function y = -1.691 x + -18.64 at x = 2.103 | The linear relation y = m x + b with slope m = -1.691 and intercept b = -18.64 evaluated at x = 2.103 yields y = -22.2. 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. |
6,678 | mathematics | algebra | linear_relation | 2 | worked_example | Evaluate linear function y = 2.296 x + 6.493 at x = 0.8456 | The linear relation y = m x + b with slope m = 2.296 and intercept b = 6.493 evaluated at x = 0.8456 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. |
6,679 | mathematics | algebra | linear_relation | 2 | worked_example | Evaluate linear function y = 4.75 x + 3.929 at x = 3.828 | The linear relation y = m x + b with slope m = 4.75 and intercept b = 3.929 evaluated at x = 3.828 yields y = 22.11. 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. |
6,680 | mathematics | algebra | linear_relation | 2 | worked_example | Evaluate linear function y = 0.07361 x + -8.325 at x = -7.4 | The linear relation y = m x + b with slope m = 0.07361 and intercept b = -8.325 evaluated at x = -7.4 yields y = -8.87. 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. |
6,681 | mathematics | algebra | linear_relation | 2 | worked_example | Evaluate linear function y = 3.825 x + -6.491 at x = 4.708 | The linear relation y = m x + b with slope m = 3.825 and intercept b = -6.491 evaluated at x = 4.708 yields y = 11.52. 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. |
6,682 | mathematics | algebra | linear_relation | 2 | worked_example | Evaluate linear function y = 0.1221 x + 5.181 at x = 2.945 | The linear relation y = m x + b with slope m = 0.1221 and intercept b = 5.181 evaluated at x = 2.945 yields y = 5.54. 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. |
6,683 | mathematics | algebra | linear_relation | 2 | worked_example | Evaluate linear function y = -2.164 x + -1.925 at x = 0.9138 | The linear relation y = m x + b with slope m = -2.164 and intercept b = -1.925 evaluated at x = 0.9138 yields y = -3.903. 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. |
6,684 | mathematics | algebra | linear_relation | 2 | worked_example | Evaluate linear function y = -0.4211 x + 12.41 at x = 8.258 | The linear relation y = m x + b with slope m = -0.4211 and intercept b = 12.41 evaluated at x = 8.258 yields y = 8.932. 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. |
6,685 | mathematics | algebra | linear_relation | 2 | worked_example | Evaluate linear function y = 1.953 x + -2.344 at x = -2.861 | The linear relation y = m x + b with slope m = 1.953 and intercept b = -2.344 evaluated at x = -2.861 yields y = -7.931. 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. |
6,686 | mathematics | algebra | linear_relation | 2 | worked_example | Evaluate linear function y = -3.21 x + 15.81 at x = -1.967 | The linear relation y = m x + b with slope m = -3.21 and intercept b = 15.81 evaluated at x = -1.967 yields y = 22.12. 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. |
6,687 | mathematics | algebra | linear_relation | 2 | worked_example | Evaluate linear function y = -1.095 x + -12.16 at x = 5.382 | The linear relation y = m x + b with slope m = -1.095 and intercept b = -12.16 evaluated at x = 5.382 yields y = -18.05. 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. |
6,688 | mathematics | algebra | linear_relation | 2 | worked_example | Evaluate linear function y = 1.567 x + -18.95 at x = 2.312 | The linear relation y = m x + b with slope m = 1.567 and intercept b = -18.95 evaluated at x = 2.312 yields y = -15.33. 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. |
6,689 | mathematics | algebra | linear_relation | 2 | worked_example | Evaluate linear function y = -2.251 x + 11.21 at x = 9.27 | The linear relation y = m x + b with slope m = -2.251 and intercept b = 11.21 evaluated at x = 9.27 yields y = -9.658. 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. |
6,690 | mathematics | algebra | linear_relation | 2 | worked_example | Evaluate linear function y = -4.365 x + 5.113 at x = 9.467 | The linear relation y = m x + b with slope m = -4.365 and intercept b = 5.113 evaluated at x = 9.467 yields y = -36.21. 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. |
6,691 | mathematics | algebra | linear_relation | 2 | worked_example | Evaluate linear function y = -0.05102 x + 1.288 at x = -2.311 | The linear relation y = m x + b with slope m = -0.05102 and intercept b = 1.288 evaluated at x = -2.311 yields y = 1.406. 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. |
6,692 | mathematics | algebra | linear_relation | 2 | worked_example | Evaluate linear function y = -3.501 x + 4.946 at x = 8.306 | The linear relation y = m x + b with slope m = -3.501 and intercept b = 4.946 evaluated at x = 8.306 yields y = -24.13. 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. |
6,693 | mathematics | algebra | linear_relation | 2 | worked_example | Evaluate linear function y = 0.05805 x + 8.742 at x = -6.922 | The linear relation y = m x + b with slope m = 0.05805 and intercept b = 8.742 evaluated at x = -6.922 yields y = 8.34. 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. |
6,694 | mathematics | algebra | linear_relation | 2 | worked_example | Evaluate linear function y = 3.341 x + -7.195 at x = 1.536 | The linear relation y = m x + b with slope m = 3.341 and intercept b = -7.195 evaluated at x = 1.536 yields y = -2.063. 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. |
6,695 | mathematics | algebra | linear_relation | 2 | worked_example | Evaluate linear function y = -0.06093 x + 9.991 at x = 5.811 | The linear relation y = m x + b with slope m = -0.06093 and intercept b = 9.991 evaluated at x = 5.811 yields y = 9.637. 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. |
6,696 | mathematics | algebra | linear_relation | 2 | worked_example | Evaluate linear function y = 0.3214 x + 7.263 at x = -9.987 | The linear relation y = m x + b with slope m = 0.3214 and intercept b = 7.263 evaluated at x = -9.987 yields y = 4.053. 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. |
6,697 | mathematics | algebra | linear_relation | 2 | worked_example | Evaluate linear function y = 1.435 x + 7.294 at x = 5.848 | The linear relation y = m x + b with slope m = 1.435 and intercept b = 7.294 evaluated at x = 5.848 yields y = 15.69. 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. |
6,698 | mathematics | algebra | linear_relation | 2 | worked_example | Evaluate linear function y = -1.313 x + 7.82 at x = 7.66 | The linear relation y = m x + b with slope m = -1.313 and intercept b = 7.82 evaluated at x = 7.66 yields y = -2.24. 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. |
6,699 | mathematics | algebra | linear_relation | 2 | worked_example | Evaluate linear function y = 1.636 x + 14.71 at x = 5.111 | The linear relation y = m x + b with slope m = 1.636 and intercept b = 14.71 evaluated at x = 5.111 yields y = 23.07. 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. |
6,700 | mathematics | algebra | linear_relation | 2 | worked_example | Evaluate linear function y = -1.098 x + 9.108 at x = -3.362 | The linear relation y = m x + b with slope m = -1.098 and intercept b = 9.108 evaluated at x = -3.362 yields y = 12.8. 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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