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values | title stringlengths 14 86 | content stringlengths 203 553 | key_equations stringclasses 23
values | prerequisites stringclasses 29
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1,901 | physics | mechanics | kinematics_1d | 2 | worked_example | One-dimensional motion with constant acceleration (v0=8.494 m/s, a=1.147 m/s²) | An object starts with initial velocity 8.494 m/s and experiences constant acceleration 1.147 m/s² for 7.118 s. Final velocity: v = v0 + a t = 8.494 + (1.147)(7.118) = 16.66 m/s. Displacement: s = v0 t + (1/2) a t² = 89.53 m. These relations follow directly from the definitions of average velocity and constant accelerat... | v = v_0 + a t; s = v_0 t + (1/2) a t^2; v^2 = v_0^2 + 2 a s | definition of velocity and acceleration | Apply the three kinematic equations for constant acceleration in one dimension. |
1,902 | physics | mechanics | kinematics_1d | 2 | worked_example | One-dimensional motion with constant acceleration (v0=2.888 m/s, a=4.883 m/s²) | An object starts with initial velocity 2.888 m/s and experiences constant acceleration 4.883 m/s² for 1.832 s. Final velocity: v = v0 + a t = 2.888 + (4.883)(1.832) = 11.83 m/s. Displacement: s = v0 t + (1/2) a t² = 13.48 m. These relations follow directly from the definitions of average velocity and constant accelerat... | v = v_0 + a t; s = v_0 t + (1/2) a t^2; v^2 = v_0^2 + 2 a s | definition of velocity and acceleration | Apply the three kinematic equations for constant acceleration in one dimension. |
1,903 | physics | mechanics | kinematics_1d | 2 | worked_example | One-dimensional motion with constant acceleration (v0=28.13 m/s, a=-2.32 m/s²) | An object starts with initial velocity 28.13 m/s and experiences constant acceleration -2.32 m/s² for 1.594 s. Final velocity: v = v0 + a t = 28.13 + (-2.32)(1.594) = 24.43 m/s. Displacement: s = v0 t + (1/2) a t² = 41.9 m. These relations follow directly from the definitions of average velocity and constant accelerati... | v = v_0 + a t; s = v_0 t + (1/2) a t^2; v^2 = v_0^2 + 2 a s | definition of velocity and acceleration | Apply the three kinematic equations for constant acceleration in one dimension. |
1,904 | physics | mechanics | kinematics_1d | 2 | worked_example | One-dimensional motion with constant acceleration (v0=3.212 m/s, a=-2.126 m/s²) | An object starts with initial velocity 3.212 m/s and experiences constant acceleration -2.126 m/s² for 1.605 s. Final velocity: v = v0 + a t = 3.212 + (-2.126)(1.605) = -0.1992 m/s. Displacement: s = v0 t + (1/2) a t² = 2.417 m. These relations follow directly from the definitions of average velocity and constant accel... | v = v_0 + a t; s = v_0 t + (1/2) a t^2; v^2 = v_0^2 + 2 a s | definition of velocity and acceleration | Apply the three kinematic equations for constant acceleration in one dimension. |
1,905 | physics | mechanics | kinematics_1d | 2 | worked_example | One-dimensional motion with constant acceleration (v0=19.25 m/s, a=0.829 m/s²) | An object starts with initial velocity 19.25 m/s and experiences constant acceleration 0.829 m/s² for 10.31 s. Final velocity: v = v0 + a t = 19.25 + (0.829)(10.31) = 27.8 m/s. Displacement: s = v0 t + (1/2) a t² = 242.5 m. These relations follow directly from the definitions of average velocity and constant accelerati... | v = v_0 + a t; s = v_0 t + (1/2) a t^2; v^2 = v_0^2 + 2 a s | definition of velocity and acceleration | Apply the three kinematic equations for constant acceleration in one dimension. |
1,906 | physics | mechanics | kinematics_1d | 2 | worked_example | One-dimensional motion with constant acceleration (v0=18.72 m/s, a=3.855 m/s²) | An object starts with initial velocity 18.72 m/s and experiences constant acceleration 3.855 m/s² for 1.933 s. Final velocity: v = v0 + a t = 18.72 + (3.855)(1.933) = 26.17 m/s. Displacement: s = v0 t + (1/2) a t² = 43.38 m. These relations follow directly from the definitions of average velocity and constant accelerat... | v = v_0 + a t; s = v_0 t + (1/2) a t^2; v^2 = v_0^2 + 2 a s | definition of velocity and acceleration | Apply the three kinematic equations for constant acceleration in one dimension. |
1,907 | physics | mechanics | kinematics_1d | 2 | worked_example | One-dimensional motion with constant acceleration (v0=6.622 m/s, a=5.768 m/s²) | An object starts with initial velocity 6.622 m/s and experiences constant acceleration 5.768 m/s² for 1.249 s. Final velocity: v = v0 + a t = 6.622 + (5.768)(1.249) = 13.82 m/s. Displacement: s = v0 t + (1/2) a t² = 12.76 m. These relations follow directly from the definitions of average velocity and constant accelerat... | v = v_0 + a t; s = v_0 t + (1/2) a t^2; v^2 = v_0^2 + 2 a s | definition of velocity and acceleration | Apply the three kinematic equations for constant acceleration in one dimension. |
1,908 | physics | mechanics | kinematics_1d | 2 | worked_example | One-dimensional motion with constant acceleration (v0=27.63 m/s, a=4.836 m/s²) | An object starts with initial velocity 27.63 m/s and experiences constant acceleration 4.836 m/s² for 18.64 s. Final velocity: v = v0 + a t = 27.63 + (4.836)(18.64) = 117.8 m/s. Displacement: s = v0 t + (1/2) a t² = 1355 m. These relations follow directly from the definitions of average velocity and constant accelerati... | v = v_0 + a t; s = v_0 t + (1/2) a t^2; v^2 = v_0^2 + 2 a s | definition of velocity and acceleration | Apply the three kinematic equations for constant acceleration in one dimension. |
1,909 | physics | mechanics | kinematics_1d | 2 | worked_example | One-dimensional motion with constant acceleration (v0=24.96 m/s, a=2.021 m/s²) | An object starts with initial velocity 24.96 m/s and experiences constant acceleration 2.021 m/s² for 13.14 s. Final velocity: v = v0 + a t = 24.96 + (2.021)(13.14) = 51.52 m/s. Displacement: s = v0 t + (1/2) a t² = 502.4 m. These relations follow directly from the definitions of average velocity and constant accelerat... | v = v_0 + a t; s = v_0 t + (1/2) a t^2; v^2 = v_0^2 + 2 a s | definition of velocity and acceleration | Apply the three kinematic equations for constant acceleration in one dimension. |
1,910 | physics | mechanics | kinematics_1d | 2 | worked_example | One-dimensional motion with constant acceleration (v0=10.83 m/s, a=6.152 m/s²) | An object starts with initial velocity 10.83 m/s and experiences constant acceleration 6.152 m/s² for 17.92 s. Final velocity: v = v0 + a t = 10.83 + (6.152)(17.92) = 121.1 m/s. Displacement: s = v0 t + (1/2) a t² = 1182 m. These relations follow directly from the definitions of average velocity and constant accelerati... | v = v_0 + a t; s = v_0 t + (1/2) a t^2; v^2 = v_0^2 + 2 a s | definition of velocity and acceleration | Apply the three kinematic equations for constant acceleration in one dimension. |
1,911 | physics | mechanics | kinematics_1d | 2 | worked_example | One-dimensional motion with constant acceleration (v0=24.66 m/s, a=6.914 m/s²) | An object starts with initial velocity 24.66 m/s and experiences constant acceleration 6.914 m/s² for 2.108 s. Final velocity: v = v0 + a t = 24.66 + (6.914)(2.108) = 39.23 m/s. Displacement: s = v0 t + (1/2) a t² = 67.33 m. These relations follow directly from the definitions of average velocity and constant accelerat... | v = v_0 + a t; s = v_0 t + (1/2) a t^2; v^2 = v_0^2 + 2 a s | definition of velocity and acceleration | Apply the three kinematic equations for constant acceleration in one dimension. |
1,912 | physics | mechanics | kinematics_1d | 2 | worked_example | One-dimensional motion with constant acceleration (v0=18.51 m/s, a=2.935 m/s²) | An object starts with initial velocity 18.51 m/s and experiences constant acceleration 2.935 m/s² for 14.95 s. Final velocity: v = v0 + a t = 18.51 + (2.935)(14.95) = 62.4 m/s. Displacement: s = v0 t + (1/2) a t² = 604.8 m. These relations follow directly from the definitions of average velocity and constant accelerati... | v = v_0 + a t; s = v_0 t + (1/2) a t^2; v^2 = v_0^2 + 2 a s | definition of velocity and acceleration | Apply the three kinematic equations for constant acceleration in one dimension. |
1,913 | physics | mechanics | kinematics_1d | 2 | worked_example | One-dimensional motion with constant acceleration (v0=25.03 m/s, a=-3.831 m/s²) | An object starts with initial velocity 25.03 m/s and experiences constant acceleration -3.831 m/s² for 5.102 s. Final velocity: v = v0 + a t = 25.03 + (-3.831)(5.102) = 5.488 m/s. Displacement: s = v0 t + (1/2) a t² = 77.86 m. These relations follow directly from the definitions of average velocity and constant acceler... | v = v_0 + a t; s = v_0 t + (1/2) a t^2; v^2 = v_0^2 + 2 a s | definition of velocity and acceleration | Apply the three kinematic equations for constant acceleration in one dimension. |
1,914 | physics | mechanics | kinematics_1d | 2 | worked_example | One-dimensional motion with constant acceleration (v0=15.58 m/s, a=-3.482 m/s²) | An object starts with initial velocity 15.58 m/s and experiences constant acceleration -3.482 m/s² for 5.389 s. Final velocity: v = v0 + a t = 15.58 + (-3.482)(5.389) = -3.179 m/s. Displacement: s = v0 t + (1/2) a t² = 33.43 m. These relations follow directly from the definitions of average velocity and constant accele... | v = v_0 + a t; s = v_0 t + (1/2) a t^2; v^2 = v_0^2 + 2 a s | definition of velocity and acceleration | Apply the three kinematic equations for constant acceleration in one dimension. |
1,915 | physics | mechanics | kinematics_1d | 2 | worked_example | One-dimensional motion with constant acceleration (v0=11.56 m/s, a=-2.103 m/s²) | An object starts with initial velocity 11.56 m/s and experiences constant acceleration -2.103 m/s² for 13.48 s. Final velocity: v = v0 + a t = 11.56 + (-2.103)(13.48) = -16.79 m/s. Displacement: s = v0 t + (1/2) a t² = -35.27 m. These relations follow directly from the definitions of average velocity and constant accel... | v = v_0 + a t; s = v_0 t + (1/2) a t^2; v^2 = v_0^2 + 2 a s | definition of velocity and acceleration | Apply the three kinematic equations for constant acceleration in one dimension. |
1,916 | physics | mechanics | kinematics_1d | 2 | worked_example | One-dimensional motion with constant acceleration (v0=17.1 m/s, a=9.213 m/s²) | An object starts with initial velocity 17.1 m/s and experiences constant acceleration 9.213 m/s² for 11.61 s. Final velocity: v = v0 + a t = 17.1 + (9.213)(11.61) = 124.1 m/s. Displacement: s = v0 t + (1/2) a t² = 819.4 m. These relations follow directly from the definitions of average velocity and constant acceleratio... | v = v_0 + a t; s = v_0 t + (1/2) a t^2; v^2 = v_0^2 + 2 a s | definition of velocity and acceleration | Apply the three kinematic equations for constant acceleration in one dimension. |
1,917 | physics | mechanics | kinematics_1d | 2 | worked_example | One-dimensional motion with constant acceleration (v0=16.74 m/s, a=6.192 m/s²) | An object starts with initial velocity 16.74 m/s and experiences constant acceleration 6.192 m/s² for 15.56 s. Final velocity: v = v0 + a t = 16.74 + (6.192)(15.56) = 113.1 m/s. Displacement: s = v0 t + (1/2) a t² = 1010 m. These relations follow directly from the definitions of average velocity and constant accelerati... | v = v_0 + a t; s = v_0 t + (1/2) a t^2; v^2 = v_0^2 + 2 a s | definition of velocity and acceleration | Apply the three kinematic equations for constant acceleration in one dimension. |
1,918 | physics | mechanics | kinematics_1d | 2 | worked_example | One-dimensional motion with constant acceleration (v0=17.3 m/s, a=5.117 m/s²) | An object starts with initial velocity 17.3 m/s and experiences constant acceleration 5.117 m/s² for 14.35 s. Final velocity: v = v0 + a t = 17.3 + (5.117)(14.35) = 90.72 m/s. Displacement: s = v0 t + (1/2) a t² = 775 m. These relations follow directly from the definitions of average velocity and constant acceleration. | v = v_0 + a t; s = v_0 t + (1/2) a t^2; v^2 = v_0^2 + 2 a s | definition of velocity and acceleration | Apply the three kinematic equations for constant acceleration in one dimension. |
1,919 | physics | mechanics | kinematics_1d | 2 | worked_example | One-dimensional motion with constant acceleration (v0=12.81 m/s, a=-3.438 m/s²) | An object starts with initial velocity 12.81 m/s and experiences constant acceleration -3.438 m/s² for 16.06 s. Final velocity: v = v0 + a t = 12.81 + (-3.438)(16.06) = -42.4 m/s. Displacement: s = v0 t + (1/2) a t² = -237.5 m. These relations follow directly from the definitions of average velocity and constant accele... | v = v_0 + a t; s = v_0 t + (1/2) a t^2; v^2 = v_0^2 + 2 a s | definition of velocity and acceleration | Apply the three kinematic equations for constant acceleration in one dimension. |
1,920 | physics | mechanics | kinematics_1d | 2 | worked_example | One-dimensional motion with constant acceleration (v0=16.5 m/s, a=2.57 m/s²) | An object starts with initial velocity 16.5 m/s and experiences constant acceleration 2.57 m/s² for 13.49 s. Final velocity: v = v0 + a t = 16.5 + (2.57)(13.49) = 51.18 m/s. Displacement: s = v0 t + (1/2) a t² = 456.6 m. These relations follow directly from the definitions of average velocity and constant acceleration. | v = v_0 + a t; s = v_0 t + (1/2) a t^2; v^2 = v_0^2 + 2 a s | definition of velocity and acceleration | Apply the three kinematic equations for constant acceleration in one dimension. |
1,921 | physics | mechanics | kinematics_1d | 2 | worked_example | One-dimensional motion with constant acceleration (v0=16.02 m/s, a=3.999 m/s²) | An object starts with initial velocity 16.02 m/s and experiences constant acceleration 3.999 m/s² for 18.74 s. Final velocity: v = v0 + a t = 16.02 + (3.999)(18.74) = 90.94 m/s. Displacement: s = v0 t + (1/2) a t² = 1002 m. These relations follow directly from the definitions of average velocity and constant accelerati... | v = v_0 + a t; s = v_0 t + (1/2) a t^2; v^2 = v_0^2 + 2 a s | definition of velocity and acceleration | Apply the three kinematic equations for constant acceleration in one dimension. |
1,922 | physics | mechanics | kinematics_1d | 2 | worked_example | One-dimensional motion with constant acceleration (v0=27.07 m/s, a=3.251 m/s²) | An object starts with initial velocity 27.07 m/s and experiences constant acceleration 3.251 m/s² for 19.94 s. Final velocity: v = v0 + a t = 27.07 + (3.251)(19.94) = 91.9 m/s. Displacement: s = v0 t + (1/2) a t² = 1186 m. These relations follow directly from the definitions of average velocity and constant acceleratio... | v = v_0 + a t; s = v_0 t + (1/2) a t^2; v^2 = v_0^2 + 2 a s | definition of velocity and acceleration | Apply the three kinematic equations for constant acceleration in one dimension. |
1,923 | physics | mechanics | kinematics_1d | 2 | worked_example | One-dimensional motion with constant acceleration (v0=25.35 m/s, a=1.185 m/s²) | An object starts with initial velocity 25.35 m/s and experiences constant acceleration 1.185 m/s² for 13.03 s. Final velocity: v = v0 + a t = 25.35 + (1.185)(13.03) = 40.79 m/s. Displacement: s = v0 t + (1/2) a t² = 430.9 m. These relations follow directly from the definitions of average velocity and constant accelerat... | v = v_0 + a t; s = v_0 t + (1/2) a t^2; v^2 = v_0^2 + 2 a s | definition of velocity and acceleration | Apply the three kinematic equations for constant acceleration in one dimension. |
1,924 | physics | mechanics | kinematics_1d | 2 | worked_example | One-dimensional motion with constant acceleration (v0=24.78 m/s, a=-0.1482 m/s²) | An object starts with initial velocity 24.78 m/s and experiences constant acceleration -0.1482 m/s² for 5.574 s. Final velocity: v = v0 + a t = 24.78 + (-0.1482)(5.574) = 23.96 m/s. Displacement: s = v0 t + (1/2) a t² = 135.8 m. These relations follow directly from the definitions of average velocity and constant accel... | v = v_0 + a t; s = v_0 t + (1/2) a t^2; v^2 = v_0^2 + 2 a s | definition of velocity and acceleration | Apply the three kinematic equations for constant acceleration in one dimension. |
1,925 | physics | mechanics | kinematics_1d | 2 | worked_example | One-dimensional motion with constant acceleration (v0=16.37 m/s, a=0.6259 m/s²) | An object starts with initial velocity 16.37 m/s and experiences constant acceleration 0.6259 m/s² for 11.45 s. Final velocity: v = v0 + a t = 16.37 + (0.6259)(11.45) = 23.54 m/s. Displacement: s = v0 t + (1/2) a t² = 228.6 m. These relations follow directly from the definitions of average velocity and constant acceler... | v = v_0 + a t; s = v_0 t + (1/2) a t^2; v^2 = v_0^2 + 2 a s | definition of velocity and acceleration | Apply the three kinematic equations for constant acceleration in one dimension. |
1,926 | physics | mechanics | kinematics_1d | 2 | worked_example | One-dimensional motion with constant acceleration (v0=22.34 m/s, a=0.4001 m/s²) | An object starts with initial velocity 22.34 m/s and experiences constant acceleration 0.4001 m/s² for 9.596 s. Final velocity: v = v0 + a t = 22.34 + (0.4001)(9.596) = 26.18 m/s. Displacement: s = v0 t + (1/2) a t² = 232.8 m. These relations follow directly from the definitions of average velocity and constant acceler... | v = v_0 + a t; s = v_0 t + (1/2) a t^2; v^2 = v_0^2 + 2 a s | definition of velocity and acceleration | Apply the three kinematic equations for constant acceleration in one dimension. |
1,927 | physics | mechanics | kinematics_1d | 2 | worked_example | One-dimensional motion with constant acceleration (v0=28.79 m/s, a=-4.997 m/s²) | An object starts with initial velocity 28.79 m/s and experiences constant acceleration -4.997 m/s² for 1.659 s. Final velocity: v = v0 + a t = 28.79 + (-4.997)(1.659) = 20.51 m/s. Displacement: s = v0 t + (1/2) a t² = 40.89 m. These relations follow directly from the definitions of average velocity and constant acceler... | v = v_0 + a t; s = v_0 t + (1/2) a t^2; v^2 = v_0^2 + 2 a s | definition of velocity and acceleration | Apply the three kinematic equations for constant acceleration in one dimension. |
1,928 | physics | mechanics | kinematics_1d | 2 | worked_example | One-dimensional motion with constant acceleration (v0=8.534 m/s, a=-4.755 m/s²) | An object starts with initial velocity 8.534 m/s and experiences constant acceleration -4.755 m/s² for 3.567 s. Final velocity: v = v0 + a t = 8.534 + (-4.755)(3.567) = -8.427 m/s. Displacement: s = v0 t + (1/2) a t² = 0.1925 m. These relations follow directly from the definitions of average velocity and constant accel... | v = v_0 + a t; s = v_0 t + (1/2) a t^2; v^2 = v_0^2 + 2 a s | definition of velocity and acceleration | Apply the three kinematic equations for constant acceleration in one dimension. |
1,929 | physics | mechanics | kinematics_1d | 2 | worked_example | One-dimensional motion with constant acceleration (v0=26.71 m/s, a=-4.901 m/s²) | An object starts with initial velocity 26.71 m/s and experiences constant acceleration -4.901 m/s² for 19.7 s. Final velocity: v = v0 + a t = 26.71 + (-4.901)(19.7) = -69.87 m/s. Displacement: s = v0 t + (1/2) a t² = -425.2 m. These relations follow directly from the definitions of average velocity and constant acceler... | v = v_0 + a t; s = v_0 t + (1/2) a t^2; v^2 = v_0^2 + 2 a s | definition of velocity and acceleration | Apply the three kinematic equations for constant acceleration in one dimension. |
1,930 | physics | mechanics | kinematics_1d | 2 | worked_example | One-dimensional motion with constant acceleration (v0=6.445 m/s, a=-0.2764 m/s²) | An object starts with initial velocity 6.445 m/s and experiences constant acceleration -0.2764 m/s² for 15.66 s. Final velocity: v = v0 + a t = 6.445 + (-0.2764)(15.66) = 2.116 m/s. Displacement: s = v0 t + (1/2) a t² = 67.03 m. These relations follow directly from the definitions of average velocity and constant accel... | v = v_0 + a t; s = v_0 t + (1/2) a t^2; v^2 = v_0^2 + 2 a s | definition of velocity and acceleration | Apply the three kinematic equations for constant acceleration in one dimension. |
1,931 | physics | mechanics | kinematics_1d | 2 | worked_example | One-dimensional motion with constant acceleration (v0=28.3 m/s, a=7.335 m/s²) | An object starts with initial velocity 28.3 m/s and experiences constant acceleration 7.335 m/s² for 18.46 s. Final velocity: v = v0 + a t = 28.3 + (7.335)(18.46) = 163.7 m/s. Displacement: s = v0 t + (1/2) a t² = 1773 m. These relations follow directly from the definitions of average velocity and constant acceleration... | v = v_0 + a t; s = v_0 t + (1/2) a t^2; v^2 = v_0^2 + 2 a s | definition of velocity and acceleration | Apply the three kinematic equations for constant acceleration in one dimension. |
1,932 | physics | mechanics | kinematics_1d | 2 | worked_example | One-dimensional motion with constant acceleration (v0=15.34 m/s, a=3.56 m/s²) | An object starts with initial velocity 15.34 m/s and experiences constant acceleration 3.56 m/s² for 11.86 s. Final velocity: v = v0 + a t = 15.34 + (3.56)(11.86) = 57.56 m/s. Displacement: s = v0 t + (1/2) a t² = 432.3 m. These relations follow directly from the definitions of average velocity and constant acceleratio... | v = v_0 + a t; s = v_0 t + (1/2) a t^2; v^2 = v_0^2 + 2 a s | definition of velocity and acceleration | Apply the three kinematic equations for constant acceleration in one dimension. |
1,933 | physics | mechanics | kinematics_1d | 2 | worked_example | One-dimensional motion with constant acceleration (v0=2.829 m/s, a=1.906 m/s²) | An object starts with initial velocity 2.829 m/s and experiences constant acceleration 1.906 m/s² for 3.966 s. Final velocity: v = v0 + a t = 2.829 + (1.906)(3.966) = 10.39 m/s. Displacement: s = v0 t + (1/2) a t² = 26.2 m. These relations follow directly from the definitions of average velocity and constant accelerati... | v = v_0 + a t; s = v_0 t + (1/2) a t^2; v^2 = v_0^2 + 2 a s | definition of velocity and acceleration | Apply the three kinematic equations for constant acceleration in one dimension. |
1,934 | physics | mechanics | kinematics_1d | 2 | worked_example | One-dimensional motion with constant acceleration (v0=7.95 m/s, a=8.344 m/s²) | An object starts with initial velocity 7.95 m/s and experiences constant acceleration 8.344 m/s² for 5.747 s. Final velocity: v = v0 + a t = 7.95 + (8.344)(5.747) = 55.91 m/s. Displacement: s = v0 t + (1/2) a t² = 183.5 m. These relations follow directly from the definitions of average velocity and constant acceleratio... | v = v_0 + a t; s = v_0 t + (1/2) a t^2; v^2 = v_0^2 + 2 a s | definition of velocity and acceleration | Apply the three kinematic equations for constant acceleration in one dimension. |
1,935 | physics | mechanics | kinematics_1d | 2 | worked_example | One-dimensional motion with constant acceleration (v0=7.687 m/s, a=-0.4955 m/s²) | An object starts with initial velocity 7.687 m/s and experiences constant acceleration -0.4955 m/s² for 11.84 s. Final velocity: v = v0 + a t = 7.687 + (-0.4955)(11.84) = 1.821 m/s. Displacement: s = v0 t + (1/2) a t² = 56.29 m. These relations follow directly from the definitions of average velocity and constant accel... | v = v_0 + a t; s = v_0 t + (1/2) a t^2; v^2 = v_0^2 + 2 a s | definition of velocity and acceleration | Apply the three kinematic equations for constant acceleration in one dimension. |
1,936 | physics | mechanics | kinematics_1d | 2 | worked_example | One-dimensional motion with constant acceleration (v0=17.07 m/s, a=-1.738 m/s²) | An object starts with initial velocity 17.07 m/s and experiences constant acceleration -1.738 m/s² for 2.467 s. Final velocity: v = v0 + a t = 17.07 + (-1.738)(2.467) = 12.78 m/s. Displacement: s = v0 t + (1/2) a t² = 36.82 m. These relations follow directly from the definitions of average velocity and constant acceler... | v = v_0 + a t; s = v_0 t + (1/2) a t^2; v^2 = v_0^2 + 2 a s | definition of velocity and acceleration | Apply the three kinematic equations for constant acceleration in one dimension. |
1,937 | physics | mechanics | kinematics_1d | 2 | worked_example | One-dimensional motion with constant acceleration (v0=9.741 m/s, a=2.238 m/s²) | An object starts with initial velocity 9.741 m/s and experiences constant acceleration 2.238 m/s² for 10.28 s. Final velocity: v = v0 + a t = 9.741 + (2.238)(10.28) = 32.74 m/s. Displacement: s = v0 t + (1/2) a t² = 218.3 m. These relations follow directly from the definitions of average velocity and constant accelerat... | v = v_0 + a t; s = v_0 t + (1/2) a t^2; v^2 = v_0^2 + 2 a s | definition of velocity and acceleration | Apply the three kinematic equations for constant acceleration in one dimension. |
1,938 | physics | mechanics | kinematics_1d | 2 | worked_example | One-dimensional motion with constant acceleration (v0=26.97 m/s, a=5.543 m/s²) | An object starts with initial velocity 26.97 m/s and experiences constant acceleration 5.543 m/s² for 16.57 s. Final velocity: v = v0 + a t = 26.97 + (5.543)(16.57) = 118.8 m/s. Displacement: s = v0 t + (1/2) a t² = 1208 m. These relations follow directly from the definitions of average velocity and constant accelerati... | v = v_0 + a t; s = v_0 t + (1/2) a t^2; v^2 = v_0^2 + 2 a s | definition of velocity and acceleration | Apply the three kinematic equations for constant acceleration in one dimension. |
1,939 | physics | mechanics | kinematics_1d | 2 | worked_example | One-dimensional motion with constant acceleration (v0=2.861 m/s, a=4.634 m/s²) | An object starts with initial velocity 2.861 m/s and experiences constant acceleration 4.634 m/s² for 1.71 s. Final velocity: v = v0 + a t = 2.861 + (4.634)(1.71) = 10.79 m/s. Displacement: s = v0 t + (1/2) a t² = 11.67 m. These relations follow directly from the definitions of average velocity and constant acceleratio... | v = v_0 + a t; s = v_0 t + (1/2) a t^2; v^2 = v_0^2 + 2 a s | definition of velocity and acceleration | Apply the three kinematic equations for constant acceleration in one dimension. |
1,940 | physics | mechanics | kinematics_1d | 2 | worked_example | One-dimensional motion with constant acceleration (v0=24.25 m/s, a=0.7232 m/s²) | An object starts with initial velocity 24.25 m/s and experiences constant acceleration 0.7232 m/s² for 15.82 s. Final velocity: v = v0 + a t = 24.25 + (0.7232)(15.82) = 35.69 m/s. Displacement: s = v0 t + (1/2) a t² = 474.2 m. These relations follow directly from the definitions of average velocity and constant acceler... | v = v_0 + a t; s = v_0 t + (1/2) a t^2; v^2 = v_0^2 + 2 a s | definition of velocity and acceleration | Apply the three kinematic equations for constant acceleration in one dimension. |
1,941 | physics | mechanics | kinematics_1d | 2 | worked_example | One-dimensional motion with constant acceleration (v0=27.66 m/s, a=0.01584 m/s²) | An object starts with initial velocity 27.66 m/s and experiences constant acceleration 0.01584 m/s² for 17.72 s. Final velocity: v = v0 + a t = 27.66 + (0.01584)(17.72) = 27.94 m/s. Displacement: s = v0 t + (1/2) a t² = 492.4 m. These relations follow directly from the definitions of average velocity and constant accel... | v = v_0 + a t; s = v_0 t + (1/2) a t^2; v^2 = v_0^2 + 2 a s | definition of velocity and acceleration | Apply the three kinematic equations for constant acceleration in one dimension. |
1,942 | physics | mechanics | kinematics_1d | 2 | worked_example | One-dimensional motion with constant acceleration (v0=13.03 m/s, a=8.954 m/s²) | An object starts with initial velocity 13.03 m/s and experiences constant acceleration 8.954 m/s² for 3.29 s. Final velocity: v = v0 + a t = 13.03 + (8.954)(3.29) = 42.49 m/s. Displacement: s = v0 t + (1/2) a t² = 91.34 m. These relations follow directly from the definitions of average velocity and constant acceleratio... | v = v_0 + a t; s = v_0 t + (1/2) a t^2; v^2 = v_0^2 + 2 a s | definition of velocity and acceleration | Apply the three kinematic equations for constant acceleration in one dimension. |
1,943 | physics | mechanics | newton_second_law | 3 | worked_example | Newton's second law: mass 38.63 kg, acceleration 11.24 m/s² | A net force acting on a mass of 38.63 kg produces an acceleration of 11.24 m/s². By Newton's second law, F_net = m a = 38.63 × 11.24 = 434.3 N. Direction of F_net is the same as the direction of the acceleration. This relation defines the inertial mass and is the foundation of classical dynamics. | F_net = m a | kinematics_1d | Compute net force from mass and acceleration using Newton's second law. |
1,944 | physics | mechanics | newton_second_law | 3 | worked_example | Newton's second law: mass 42.55 kg, acceleration 13.94 m/s² | A net force acting on a mass of 42.55 kg produces an acceleration of 13.94 m/s². By Newton's second law, F_net = m a = 42.55 × 13.94 = 593 N. Direction of F_net is the same as the direction of the acceleration. This relation defines the inertial mass and is the foundation of classical dynamics. | F_net = m a | kinematics_1d | Compute net force from mass and acceleration using Newton's second law. |
1,945 | physics | mechanics | newton_second_law | 3 | worked_example | Newton's second law: mass 31.88 kg, acceleration 9.52 m/s² | A net force acting on a mass of 31.88 kg produces an acceleration of 9.52 m/s². By Newton's second law, F_net = m a = 31.88 × 9.52 = 303.5 N. Direction of F_net is the same as the direction of the acceleration. This relation defines the inertial mass and is the foundation of classical dynamics. | F_net = m a | kinematics_1d | Compute net force from mass and acceleration using Newton's second law. |
1,946 | physics | mechanics | newton_second_law | 3 | worked_example | Newton's second law: mass 3.179 kg, acceleration 7.913 m/s² | A net force acting on a mass of 3.179 kg produces an acceleration of 7.913 m/s². By Newton's second law, F_net = m a = 3.179 × 7.913 = 25.16 N. Direction of F_net is the same as the direction of the acceleration. This relation defines the inertial mass and is the foundation of classical dynamics. | F_net = m a | kinematics_1d | Compute net force from mass and acceleration using Newton's second law. |
1,947 | physics | mechanics | newton_second_law | 3 | worked_example | Newton's second law: mass 16.31 kg, acceleration 8.517 m/s² | A net force acting on a mass of 16.31 kg produces an acceleration of 8.517 m/s². By Newton's second law, F_net = m a = 16.31 × 8.517 = 138.9 N. Direction of F_net is the same as the direction of the acceleration. This relation defines the inertial mass and is the foundation of classical dynamics. | F_net = m a | kinematics_1d | Compute net force from mass and acceleration using Newton's second law. |
1,948 | physics | mechanics | newton_second_law | 3 | worked_example | Newton's second law: mass 16.68 kg, acceleration 10.5 m/s² | A net force acting on a mass of 16.68 kg produces an acceleration of 10.5 m/s². By Newton's second law, F_net = m a = 16.68 × 10.5 = 175.1 N. Direction of F_net is the same as the direction of the acceleration. This relation defines the inertial mass and is the foundation of classical dynamics. | F_net = m a | kinematics_1d | Compute net force from mass and acceleration using Newton's second law. |
1,949 | physics | mechanics | newton_second_law | 3 | worked_example | Newton's second law: mass 8.083 kg, acceleration 10.14 m/s² | A net force acting on a mass of 8.083 kg produces an acceleration of 10.14 m/s². By Newton's second law, F_net = m a = 8.083 × 10.14 = 81.94 N. Direction of F_net is the same as the direction of the acceleration. This relation defines the inertial mass and is the foundation of classical dynamics. | F_net = m a | kinematics_1d | Compute net force from mass and acceleration using Newton's second law. |
1,950 | physics | mechanics | newton_second_law | 3 | worked_example | Newton's second law: mass 49.87 kg, acceleration 14.57 m/s² | A net force acting on a mass of 49.87 kg produces an acceleration of 14.57 m/s². By Newton's second law, F_net = m a = 49.87 × 14.57 = 726.5 N. Direction of F_net is the same as the direction of the acceleration. This relation defines the inertial mass and is the foundation of classical dynamics. | F_net = m a | kinematics_1d | Compute net force from mass and acceleration using Newton's second law. |
1,951 | physics | mechanics | newton_second_law | 3 | worked_example | Newton's second law: mass 37.77 kg, acceleration 8.671 m/s² | A net force acting on a mass of 37.77 kg produces an acceleration of 8.671 m/s². By Newton's second law, F_net = m a = 37.77 × 8.671 = 327.5 N. Direction of F_net is the same as the direction of the acceleration. This relation defines the inertial mass and is the foundation of classical dynamics. | F_net = m a | kinematics_1d | Compute net force from mass and acceleration using Newton's second law. |
1,952 | physics | mechanics | newton_second_law | 3 | worked_example | Newton's second law: mass 4.959 kg, acceleration 6.536 m/s² | A net force acting on a mass of 4.959 kg produces an acceleration of 6.536 m/s². By Newton's second law, F_net = m a = 4.959 × 6.536 = 32.41 N. Direction of F_net is the same as the direction of the acceleration. This relation defines the inertial mass and is the foundation of classical dynamics. | F_net = m a | kinematics_1d | Compute net force from mass and acceleration using Newton's second law. |
1,953 | physics | mechanics | newton_second_law | 3 | worked_example | Newton's second law: mass 39.43 kg, acceleration 2.305 m/s² | A net force acting on a mass of 39.43 kg produces an acceleration of 2.305 m/s². By Newton's second law, F_net = m a = 39.43 × 2.305 = 90.89 N. Direction of F_net is the same as the direction of the acceleration. This relation defines the inertial mass and is the foundation of classical dynamics. | F_net = m a | kinematics_1d | Compute net force from mass and acceleration using Newton's second law. |
1,954 | physics | mechanics | newton_second_law | 3 | worked_example | Newton's second law: mass 17.14 kg, acceleration 7.19 m/s² | A net force acting on a mass of 17.14 kg produces an acceleration of 7.19 m/s². By Newton's second law, F_net = m a = 17.14 × 7.19 = 123.2 N. Direction of F_net is the same as the direction of the acceleration. This relation defines the inertial mass and is the foundation of classical dynamics. | F_net = m a | kinematics_1d | Compute net force from mass and acceleration using Newton's second law. |
1,955 | physics | mechanics | newton_second_law | 3 | worked_example | Newton's second law: mass 37.39 kg, acceleration 8.267 m/s² | A net force acting on a mass of 37.39 kg produces an acceleration of 8.267 m/s². By Newton's second law, F_net = m a = 37.39 × 8.267 = 309.1 N. Direction of F_net is the same as the direction of the acceleration. This relation defines the inertial mass and is the foundation of classical dynamics. | F_net = m a | kinematics_1d | Compute net force from mass and acceleration using Newton's second law. |
1,956 | physics | mechanics | newton_second_law | 3 | worked_example | Newton's second law: mass 13.5 kg, acceleration 6.038 m/s² | A net force acting on a mass of 13.5 kg produces an acceleration of 6.038 m/s². By Newton's second law, F_net = m a = 13.5 × 6.038 = 81.49 N. Direction of F_net is the same as the direction of the acceleration. This relation defines the inertial mass and is the foundation of classical dynamics. | F_net = m a | kinematics_1d | Compute net force from mass and acceleration using Newton's second law. |
1,957 | physics | mechanics | newton_second_law | 3 | worked_example | Newton's second law: mass 22.01 kg, acceleration 10.5 m/s² | A net force acting on a mass of 22.01 kg produces an acceleration of 10.5 m/s². By Newton's second law, F_net = m a = 22.01 × 10.5 = 231 N. Direction of F_net is the same as the direction of the acceleration. This relation defines the inertial mass and is the foundation of classical dynamics. | F_net = m a | kinematics_1d | Compute net force from mass and acceleration using Newton's second law. |
1,958 | physics | mechanics | newton_second_law | 3 | worked_example | Newton's second law: mass 38.22 kg, acceleration 7.266 m/s² | A net force acting on a mass of 38.22 kg produces an acceleration of 7.266 m/s². By Newton's second law, F_net = m a = 38.22 × 7.266 = 277.7 N. Direction of F_net is the same as the direction of the acceleration. This relation defines the inertial mass and is the foundation of classical dynamics. | F_net = m a | kinematics_1d | Compute net force from mass and acceleration using Newton's second law. |
1,959 | physics | mechanics | newton_second_law | 3 | worked_example | Newton's second law: mass 10.31 kg, acceleration 5.99 m/s² | A net force acting on a mass of 10.31 kg produces an acceleration of 5.99 m/s². By Newton's second law, F_net = m a = 10.31 × 5.99 = 61.74 N. Direction of F_net is the same as the direction of the acceleration. This relation defines the inertial mass and is the foundation of classical dynamics. | F_net = m a | kinematics_1d | Compute net force from mass and acceleration using Newton's second law. |
1,960 | physics | mechanics | newton_second_law | 3 | worked_example | Newton's second law: mass 28.78 kg, acceleration 8.461 m/s² | A net force acting on a mass of 28.78 kg produces an acceleration of 8.461 m/s². By Newton's second law, F_net = m a = 28.78 × 8.461 = 243.5 N. Direction of F_net is the same as the direction of the acceleration. This relation defines the inertial mass and is the foundation of classical dynamics. | F_net = m a | kinematics_1d | Compute net force from mass and acceleration using Newton's second law. |
1,961 | physics | mechanics | newton_second_law | 3 | worked_example | Newton's second law: mass 49.86 kg, acceleration 7.911 m/s² | A net force acting on a mass of 49.86 kg produces an acceleration of 7.911 m/s². By Newton's second law, F_net = m a = 49.86 × 7.911 = 394.4 N. Direction of F_net is the same as the direction of the acceleration. This relation defines the inertial mass and is the foundation of classical dynamics. | F_net = m a | kinematics_1d | Compute net force from mass and acceleration using Newton's second law. |
1,962 | physics | mechanics | newton_second_law | 3 | worked_example | Newton's second law: mass 12.49 kg, acceleration 4.067 m/s² | A net force acting on a mass of 12.49 kg produces an acceleration of 4.067 m/s². By Newton's second law, F_net = m a = 12.49 × 4.067 = 50.82 N. Direction of F_net is the same as the direction of the acceleration. This relation defines the inertial mass and is the foundation of classical dynamics. | F_net = m a | kinematics_1d | Compute net force from mass and acceleration using Newton's second law. |
1,963 | physics | mechanics | newton_second_law | 3 | worked_example | Newton's second law: mass 21.79 kg, acceleration 10.3 m/s² | A net force acting on a mass of 21.79 kg produces an acceleration of 10.3 m/s². By Newton's second law, F_net = m a = 21.79 × 10.3 = 224.5 N. Direction of F_net is the same as the direction of the acceleration. This relation defines the inertial mass and is the foundation of classical dynamics. | F_net = m a | kinematics_1d | Compute net force from mass and acceleration using Newton's second law. |
1,964 | physics | mechanics | newton_second_law | 3 | worked_example | Newton's second law: mass 29.01 kg, acceleration 1.777 m/s² | A net force acting on a mass of 29.01 kg produces an acceleration of 1.777 m/s². By Newton's second law, F_net = m a = 29.01 × 1.777 = 51.55 N. Direction of F_net is the same as the direction of the acceleration. This relation defines the inertial mass and is the foundation of classical dynamics. | F_net = m a | kinematics_1d | Compute net force from mass and acceleration using Newton's second law. |
1,965 | physics | mechanics | newton_second_law | 3 | worked_example | Newton's second law: mass 20.97 kg, acceleration 5.542 m/s² | A net force acting on a mass of 20.97 kg produces an acceleration of 5.542 m/s². By Newton's second law, F_net = m a = 20.97 × 5.542 = 116.2 N. Direction of F_net is the same as the direction of the acceleration. This relation defines the inertial mass and is the foundation of classical dynamics. | F_net = m a | kinematics_1d | Compute net force from mass and acceleration using Newton's second law. |
1,966 | physics | mechanics | newton_second_law | 3 | worked_example | Newton's second law: mass 9.929 kg, acceleration 5.233 m/s² | A net force acting on a mass of 9.929 kg produces an acceleration of 5.233 m/s². By Newton's second law, F_net = m a = 9.929 × 5.233 = 51.96 N. Direction of F_net is the same as the direction of the acceleration. This relation defines the inertial mass and is the foundation of classical dynamics. | F_net = m a | kinematics_1d | Compute net force from mass and acceleration using Newton's second law. |
1,967 | physics | mechanics | newton_second_law | 3 | worked_example | Newton's second law: mass 32.94 kg, acceleration 11.57 m/s² | A net force acting on a mass of 32.94 kg produces an acceleration of 11.57 m/s². By Newton's second law, F_net = m a = 32.94 × 11.57 = 381.1 N. Direction of F_net is the same as the direction of the acceleration. This relation defines the inertial mass and is the foundation of classical dynamics. | F_net = m a | kinematics_1d | Compute net force from mass and acceleration using Newton's second law. |
1,968 | physics | mechanics | newton_second_law | 3 | worked_example | Newton's second law: mass 8.878 kg, acceleration 5.311 m/s² | A net force acting on a mass of 8.878 kg produces an acceleration of 5.311 m/s². By Newton's second law, F_net = m a = 8.878 × 5.311 = 47.15 N. Direction of F_net is the same as the direction of the acceleration. This relation defines the inertial mass and is the foundation of classical dynamics. | F_net = m a | kinematics_1d | Compute net force from mass and acceleration using Newton's second law. |
1,969 | physics | mechanics | newton_second_law | 3 | worked_example | Newton's second law: mass 41.17 kg, acceleration 12.06 m/s² | A net force acting on a mass of 41.17 kg produces an acceleration of 12.06 m/s². By Newton's second law, F_net = m a = 41.17 × 12.06 = 496.6 N. Direction of F_net is the same as the direction of the acceleration. This relation defines the inertial mass and is the foundation of classical dynamics. | F_net = m a | kinematics_1d | Compute net force from mass and acceleration using Newton's second law. |
1,970 | physics | mechanics | newton_second_law | 3 | worked_example | Newton's second law: mass 31.35 kg, acceleration 6.663 m/s² | A net force acting on a mass of 31.35 kg produces an acceleration of 6.663 m/s². By Newton's second law, F_net = m a = 31.35 × 6.663 = 208.9 N. Direction of F_net is the same as the direction of the acceleration. This relation defines the inertial mass and is the foundation of classical dynamics. | F_net = m a | kinematics_1d | Compute net force from mass and acceleration using Newton's second law. |
1,971 | physics | mechanics | newton_second_law | 3 | worked_example | Newton's second law: mass 46.25 kg, acceleration 3.579 m/s² | A net force acting on a mass of 46.25 kg produces an acceleration of 3.579 m/s². By Newton's second law, F_net = m a = 46.25 × 3.579 = 165.5 N. Direction of F_net is the same as the direction of the acceleration. This relation defines the inertial mass and is the foundation of classical dynamics. | F_net = m a | kinematics_1d | Compute net force from mass and acceleration using Newton's second law. |
1,972 | physics | mechanics | newton_second_law | 3 | worked_example | Newton's second law: mass 34.43 kg, acceleration 2.424 m/s² | A net force acting on a mass of 34.43 kg produces an acceleration of 2.424 m/s². By Newton's second law, F_net = m a = 34.43 × 2.424 = 83.46 N. Direction of F_net is the same as the direction of the acceleration. This relation defines the inertial mass and is the foundation of classical dynamics. | F_net = m a | kinematics_1d | Compute net force from mass and acceleration using Newton's second law. |
1,973 | physics | mechanics | newton_second_law | 3 | worked_example | Newton's second law: mass 36.56 kg, acceleration 1.379 m/s² | A net force acting on a mass of 36.56 kg produces an acceleration of 1.379 m/s². By Newton's second law, F_net = m a = 36.56 × 1.379 = 50.4 N. Direction of F_net is the same as the direction of the acceleration. This relation defines the inertial mass and is the foundation of classical dynamics. | F_net = m a | kinematics_1d | Compute net force from mass and acceleration using Newton's second law. |
1,974 | physics | mechanics | newton_second_law | 3 | worked_example | Newton's second law: mass 37.11 kg, acceleration 3.721 m/s² | A net force acting on a mass of 37.11 kg produces an acceleration of 3.721 m/s². By Newton's second law, F_net = m a = 37.11 × 3.721 = 138.1 N. Direction of F_net is the same as the direction of the acceleration. This relation defines the inertial mass and is the foundation of classical dynamics. | F_net = m a | kinematics_1d | Compute net force from mass and acceleration using Newton's second law. |
1,975 | physics | mechanics | newton_second_law | 3 | worked_example | Newton's second law: mass 31.24 kg, acceleration 2.3 m/s² | A net force acting on a mass of 31.24 kg produces an acceleration of 2.3 m/s². By Newton's second law, F_net = m a = 31.24 × 2.3 = 71.86 N. Direction of F_net is the same as the direction of the acceleration. This relation defines the inertial mass and is the foundation of classical dynamics. | F_net = m a | kinematics_1d | Compute net force from mass and acceleration using Newton's second law. |
1,976 | physics | mechanics | newton_second_law | 3 | worked_example | Newton's second law: mass 41.71 kg, acceleration 4.387 m/s² | A net force acting on a mass of 41.71 kg produces an acceleration of 4.387 m/s². By Newton's second law, F_net = m a = 41.71 × 4.387 = 182.9 N. Direction of F_net is the same as the direction of the acceleration. This relation defines the inertial mass and is the foundation of classical dynamics. | F_net = m a | kinematics_1d | Compute net force from mass and acceleration using Newton's second law. |
1,977 | physics | mechanics | newton_second_law | 3 | worked_example | Newton's second law: mass 17.49 kg, acceleration 12.08 m/s² | A net force acting on a mass of 17.49 kg produces an acceleration of 12.08 m/s². By Newton's second law, F_net = m a = 17.49 × 12.08 = 211.4 N. Direction of F_net is the same as the direction of the acceleration. This relation defines the inertial mass and is the foundation of classical dynamics. | F_net = m a | kinematics_1d | Compute net force from mass and acceleration using Newton's second law. |
1,978 | physics | mechanics | newton_second_law | 3 | worked_example | Newton's second law: mass 15.6 kg, acceleration 7.121 m/s² | A net force acting on a mass of 15.6 kg produces an acceleration of 7.121 m/s². By Newton's second law, F_net = m a = 15.6 × 7.121 = 111.1 N. Direction of F_net is the same as the direction of the acceleration. This relation defines the inertial mass and is the foundation of classical dynamics. | F_net = m a | kinematics_1d | Compute net force from mass and acceleration using Newton's second law. |
1,979 | physics | mechanics | newton_second_law | 3 | worked_example | Newton's second law: mass 16.93 kg, acceleration 11.13 m/s² | A net force acting on a mass of 16.93 kg produces an acceleration of 11.13 m/s². By Newton's second law, F_net = m a = 16.93 × 11.13 = 188.3 N. Direction of F_net is the same as the direction of the acceleration. This relation defines the inertial mass and is the foundation of classical dynamics. | F_net = m a | kinematics_1d | Compute net force from mass and acceleration using Newton's second law. |
1,980 | physics | mechanics | newton_second_law | 3 | worked_example | Newton's second law: mass 10.99 kg, acceleration 8.242 m/s² | A net force acting on a mass of 10.99 kg produces an acceleration of 8.242 m/s². By Newton's second law, F_net = m a = 10.99 × 8.242 = 90.56 N. Direction of F_net is the same as the direction of the acceleration. This relation defines the inertial mass and is the foundation of classical dynamics. | F_net = m a | kinematics_1d | Compute net force from mass and acceleration using Newton's second law. |
1,981 | physics | mechanics | newton_second_law | 3 | worked_example | Newton's second law: mass 37.64 kg, acceleration 8.277 m/s² | A net force acting on a mass of 37.64 kg produces an acceleration of 8.277 m/s². By Newton's second law, F_net = m a = 37.64 × 8.277 = 311.6 N. Direction of F_net is the same as the direction of the acceleration. This relation defines the inertial mass and is the foundation of classical dynamics. | F_net = m a | kinematics_1d | Compute net force from mass and acceleration using Newton's second law. |
1,982 | physics | mechanics | newton_second_law | 3 | worked_example | Newton's second law: mass 31.34 kg, acceleration 14.89 m/s² | A net force acting on a mass of 31.34 kg produces an acceleration of 14.89 m/s². By Newton's second law, F_net = m a = 31.34 × 14.89 = 466.5 N. Direction of F_net is the same as the direction of the acceleration. This relation defines the inertial mass and is the foundation of classical dynamics. | F_net = m a | kinematics_1d | Compute net force from mass and acceleration using Newton's second law. |
1,983 | physics | mechanics | newton_second_law | 3 | worked_example | Newton's second law: mass 34.91 kg, acceleration 10.06 m/s² | A net force acting on a mass of 34.91 kg produces an acceleration of 10.06 m/s². By Newton's second law, F_net = m a = 34.91 × 10.06 = 351.1 N. Direction of F_net is the same as the direction of the acceleration. This relation defines the inertial mass and is the foundation of classical dynamics. | F_net = m a | kinematics_1d | Compute net force from mass and acceleration using Newton's second law. |
1,984 | physics | mechanics | newton_second_law | 3 | worked_example | Newton's second law: mass 19.14 kg, acceleration 13.6 m/s² | A net force acting on a mass of 19.14 kg produces an acceleration of 13.6 m/s². By Newton's second law, F_net = m a = 19.14 × 13.6 = 260.3 N. Direction of F_net is the same as the direction of the acceleration. This relation defines the inertial mass and is the foundation of classical dynamics. | F_net = m a | kinematics_1d | Compute net force from mass and acceleration using Newton's second law. |
1,985 | physics | mechanics | newton_second_law | 3 | worked_example | Newton's second law: mass 26.58 kg, acceleration 3.906 m/s² | A net force acting on a mass of 26.58 kg produces an acceleration of 3.906 m/s². By Newton's second law, F_net = m a = 26.58 × 3.906 = 103.8 N. Direction of F_net is the same as the direction of the acceleration. This relation defines the inertial mass and is the foundation of classical dynamics. | F_net = m a | kinematics_1d | Compute net force from mass and acceleration using Newton's second law. |
1,986 | physics | mechanics | newton_second_law | 3 | worked_example | Newton's second law: mass 3.475 kg, acceleration 3.09 m/s² | A net force acting on a mass of 3.475 kg produces an acceleration of 3.09 m/s². By Newton's second law, F_net = m a = 3.475 × 3.09 = 10.74 N. Direction of F_net is the same as the direction of the acceleration. This relation defines the inertial mass and is the foundation of classical dynamics. | F_net = m a | kinematics_1d | Compute net force from mass and acceleration using Newton's second law. |
1,987 | physics | mechanics | newton_second_law | 3 | worked_example | Newton's second law: mass 34.98 kg, acceleration 4.142 m/s² | A net force acting on a mass of 34.98 kg produces an acceleration of 4.142 m/s². By Newton's second law, F_net = m a = 34.98 × 4.142 = 144.9 N. Direction of F_net is the same as the direction of the acceleration. This relation defines the inertial mass and is the foundation of classical dynamics. | F_net = m a | kinematics_1d | Compute net force from mass and acceleration using Newton's second law. |
1,988 | physics | mechanics | newton_second_law | 3 | worked_example | Newton's second law: mass 23.52 kg, acceleration 14.18 m/s² | A net force acting on a mass of 23.52 kg produces an acceleration of 14.18 m/s². By Newton's second law, F_net = m a = 23.52 × 14.18 = 333.5 N. Direction of F_net is the same as the direction of the acceleration. This relation defines the inertial mass and is the foundation of classical dynamics. | F_net = m a | kinematics_1d | Compute net force from mass and acceleration using Newton's second law. |
1,989 | physics | mechanics | newton_second_law | 3 | worked_example | Newton's second law: mass 3.797 kg, acceleration 9.94 m/s² | A net force acting on a mass of 3.797 kg produces an acceleration of 9.94 m/s². By Newton's second law, F_net = m a = 3.797 × 9.94 = 37.74 N. Direction of F_net is the same as the direction of the acceleration. This relation defines the inertial mass and is the foundation of classical dynamics. | F_net = m a | kinematics_1d | Compute net force from mass and acceleration using Newton's second law. |
1,990 | physics | mechanics | newton_second_law | 3 | worked_example | Newton's second law: mass 33.32 kg, acceleration 13.7 m/s² | A net force acting on a mass of 33.32 kg produces an acceleration of 13.7 m/s². By Newton's second law, F_net = m a = 33.32 × 13.7 = 456.5 N. Direction of F_net is the same as the direction of the acceleration. This relation defines the inertial mass and is the foundation of classical dynamics. | F_net = m a | kinematics_1d | Compute net force from mass and acceleration using Newton's second law. |
1,991 | physics | mechanics | newton_second_law | 3 | worked_example | Newton's second law: mass 18.33 kg, acceleration 1.708 m/s² | A net force acting on a mass of 18.33 kg produces an acceleration of 1.708 m/s². By Newton's second law, F_net = m a = 18.33 × 1.708 = 31.3 N. Direction of F_net is the same as the direction of the acceleration. This relation defines the inertial mass and is the foundation of classical dynamics. | F_net = m a | kinematics_1d | Compute net force from mass and acceleration using Newton's second law. |
1,992 | physics | mechanics | newton_second_law | 3 | worked_example | Newton's second law: mass 14.98 kg, acceleration 13.26 m/s² | A net force acting on a mass of 14.98 kg produces an acceleration of 13.26 m/s². By Newton's second law, F_net = m a = 14.98 × 13.26 = 198.7 N. Direction of F_net is the same as the direction of the acceleration. This relation defines the inertial mass and is the foundation of classical dynamics. | F_net = m a | kinematics_1d | Compute net force from mass and acceleration using Newton's second law. |
1,993 | physics | mechanics | newton_second_law | 3 | worked_example | Newton's second law: mass 19.63 kg, acceleration 14.7 m/s² | A net force acting on a mass of 19.63 kg produces an acceleration of 14.7 m/s². By Newton's second law, F_net = m a = 19.63 × 14.7 = 288.6 N. Direction of F_net is the same as the direction of the acceleration. This relation defines the inertial mass and is the foundation of classical dynamics. | F_net = m a | kinematics_1d | Compute net force from mass and acceleration using Newton's second law. |
1,994 | physics | mechanics | newton_second_law | 3 | worked_example | Newton's second law: mass 25.67 kg, acceleration 10.67 m/s² | A net force acting on a mass of 25.67 kg produces an acceleration of 10.67 m/s². By Newton's second law, F_net = m a = 25.67 × 10.67 = 274 N. Direction of F_net is the same as the direction of the acceleration. This relation defines the inertial mass and is the foundation of classical dynamics. | F_net = m a | kinematics_1d | Compute net force from mass and acceleration using Newton's second law. |
1,995 | physics | mechanics | newton_second_law | 3 | worked_example | Newton's second law: mass 45.24 kg, acceleration 7.321 m/s² | A net force acting on a mass of 45.24 kg produces an acceleration of 7.321 m/s². By Newton's second law, F_net = m a = 45.24 × 7.321 = 331.2 N. Direction of F_net is the same as the direction of the acceleration. This relation defines the inertial mass and is the foundation of classical dynamics. | F_net = m a | kinematics_1d | Compute net force from mass and acceleration using Newton's second law. |
1,996 | physics | mechanics | newton_second_law | 3 | worked_example | Newton's second law: mass 30.06 kg, acceleration 1.412 m/s² | A net force acting on a mass of 30.06 kg produces an acceleration of 1.412 m/s². By Newton's second law, F_net = m a = 30.06 × 1.412 = 42.43 N. Direction of F_net is the same as the direction of the acceleration. This relation defines the inertial mass and is the foundation of classical dynamics. | F_net = m a | kinematics_1d | Compute net force from mass and acceleration using Newton's second law. |
1,997 | physics | mechanics | newton_second_law | 3 | worked_example | Newton's second law: mass 10.34 kg, acceleration 14.51 m/s² | A net force acting on a mass of 10.34 kg produces an acceleration of 14.51 m/s². By Newton's second law, F_net = m a = 10.34 × 14.51 = 150 N. Direction of F_net is the same as the direction of the acceleration. This relation defines the inertial mass and is the foundation of classical dynamics. | F_net = m a | kinematics_1d | Compute net force from mass and acceleration using Newton's second law. |
1,998 | physics | mechanics | newton_second_law | 3 | worked_example | Newton's second law: mass 28.25 kg, acceleration 10.72 m/s² | A net force acting on a mass of 28.25 kg produces an acceleration of 10.72 m/s². By Newton's second law, F_net = m a = 28.25 × 10.72 = 302.8 N. Direction of F_net is the same as the direction of the acceleration. This relation defines the inertial mass and is the foundation of classical dynamics. | F_net = m a | kinematics_1d | Compute net force from mass and acceleration using Newton's second law. |
1,999 | physics | mechanics | newton_second_law | 3 | worked_example | Newton's second law: mass 2.468 kg, acceleration 13.84 m/s² | A net force acting on a mass of 2.468 kg produces an acceleration of 13.84 m/s². By Newton's second law, F_net = m a = 2.468 × 13.84 = 34.16 N. Direction of F_net is the same as the direction of the acceleration. This relation defines the inertial mass and is the foundation of classical dynamics. | F_net = m a | kinematics_1d | Compute net force from mass and acceleration using Newton's second law. |
2,000 | physics | mechanics | newton_second_law | 3 | worked_example | Newton's second law: mass 15.59 kg, acceleration 7.4 m/s² | A net force acting on a mass of 15.59 kg produces an acceleration of 7.4 m/s². By Newton's second law, F_net = m a = 15.59 × 7.4 = 115.4 N. Direction of F_net is the same as the direction of the acceleration. This relation defines the inertial mass and is the foundation of classical dynamics. | F_net = m a | kinematics_1d | Compute net force from mass and acceleration using Newton's second law. |
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