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values | title stringlengths 14 86 | content stringlengths 203 553 | key_equations stringclasses 23
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7,101 | physics | mechanics | kinematics_1d | 2 | worked_example | One-dimensional motion with constant acceleration (v0=9.883 m/s, a=6.133 m/s²) | An object starts with initial velocity 9.883 m/s and experiences constant acceleration 6.133 m/s² for 8.205 s. Final velocity: v = v0 + a t = 9.883 + (6.133)(8.205) = 60.2 m/s. Displacement: s = v0 t + (1/2) a t² = 287.5 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. |
7,102 | physics | mechanics | kinematics_1d | 2 | worked_example | One-dimensional motion with constant acceleration (v0=25.02 m/s, a=-0.02889 m/s²) | An object starts with initial velocity 25.02 m/s and experiences constant acceleration -0.02889 m/s² for 18.16 s. Final velocity: v = v0 + a t = 25.02 + (-0.02889)(18.16) = 24.5 m/s. Displacement: s = v0 t + (1/2) a t² = 449.5 m. These relations follow directly from the definitions of average velocity and constant acce... | 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. |
7,103 | physics | mechanics | kinematics_1d | 2 | worked_example | One-dimensional motion with constant acceleration (v0=1.382 m/s, a=1.493 m/s²) | An object starts with initial velocity 1.382 m/s and experiences constant acceleration 1.493 m/s² for 18.04 s. Final velocity: v = v0 + a t = 1.382 + (1.493)(18.04) = 28.31 m/s. Displacement: s = v0 t + (1/2) a t² = 267.9 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. |
7,104 | physics | mechanics | kinematics_1d | 2 | worked_example | One-dimensional motion with constant acceleration (v0=11.7 m/s, a=-0.9699 m/s²) | An object starts with initial velocity 11.7 m/s and experiences constant acceleration -0.9699 m/s² for 2.403 s. Final velocity: v = v0 + a t = 11.7 + (-0.9699)(2.403) = 9.368 m/s. Displacement: s = v0 t + (1/2) a t² = 25.31 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. |
7,105 | physics | mechanics | kinematics_1d | 2 | worked_example | One-dimensional motion with constant acceleration (v0=27.15 m/s, a=0.8948 m/s²) | An object starts with initial velocity 27.15 m/s and experiences constant acceleration 0.8948 m/s² for 8.048 s. Final velocity: v = v0 + a t = 27.15 + (0.8948)(8.048) = 34.35 m/s. Displacement: s = v0 t + (1/2) a t² = 247.5 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. |
7,106 | physics | mechanics | kinematics_1d | 2 | worked_example | One-dimensional motion with constant acceleration (v0=26.03 m/s, a=5.634 m/s²) | An object starts with initial velocity 26.03 m/s and experiences constant acceleration 5.634 m/s² for 14.38 s. Final velocity: v = v0 + a t = 26.03 + (5.634)(14.38) = 107 m/s. Displacement: s = v0 t + (1/2) a t² = 956.5 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. |
7,107 | physics | mechanics | kinematics_1d | 2 | worked_example | One-dimensional motion with constant acceleration (v0=26.75 m/s, a=9.139 m/s²) | An object starts with initial velocity 26.75 m/s and experiences constant acceleration 9.139 m/s² for 19.75 s. Final velocity: v = v0 + a t = 26.75 + (9.139)(19.75) = 207.2 m/s. Displacement: s = v0 t + (1/2) a t² = 2311 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. |
7,108 | physics | mechanics | kinematics_1d | 2 | worked_example | One-dimensional motion with constant acceleration (v0=0.04757 m/s, a=1.705 m/s²) | An object starts with initial velocity 0.04757 m/s and experiences constant acceleration 1.705 m/s² for 7.155 s. Final velocity: v = v0 + a t = 0.04757 + (1.705)(7.155) = 12.25 m/s. Displacement: s = v0 t + (1/2) a t² = 43.99 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. |
7,109 | physics | mechanics | kinematics_1d | 2 | worked_example | One-dimensional motion with constant acceleration (v0=11.96 m/s, a=7.269 m/s²) | An object starts with initial velocity 11.96 m/s and experiences constant acceleration 7.269 m/s² for 18.41 s. Final velocity: v = v0 + a t = 11.96 + (7.269)(18.41) = 145.8 m/s. Displacement: s = v0 t + (1/2) a t² = 1452 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. |
7,110 | physics | mechanics | kinematics_1d | 2 | worked_example | One-dimensional motion with constant acceleration (v0=8.066 m/s, a=4.279 m/s²) | An object starts with initial velocity 8.066 m/s and experiences constant acceleration 4.279 m/s² for 1.797 s. Final velocity: v = v0 + a t = 8.066 + (4.279)(1.797) = 15.76 m/s. Displacement: s = v0 t + (1/2) a t² = 21.41 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. |
7,111 | physics | mechanics | kinematics_1d | 2 | worked_example | One-dimensional motion with constant acceleration (v0=18.46 m/s, a=0.5093 m/s²) | An object starts with initial velocity 18.46 m/s and experiences constant acceleration 0.5093 m/s² for 11.67 s. Final velocity: v = v0 + a t = 18.46 + (0.5093)(11.67) = 24.4 m/s. Displacement: s = v0 t + (1/2) a t² = 250.1 m. These relations follow directly from the definitions of average velocity and constant accelera... | 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. |
7,112 | physics | mechanics | kinematics_1d | 2 | worked_example | One-dimensional motion with constant acceleration (v0=5.839 m/s, a=1.295 m/s²) | An object starts with initial velocity 5.839 m/s and experiences constant acceleration 1.295 m/s² for 4.844 s. Final velocity: v = v0 + a t = 5.839 + (1.295)(4.844) = 12.11 m/s. Displacement: s = v0 t + (1/2) a t² = 43.48 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. |
7,113 | physics | mechanics | kinematics_1d | 2 | worked_example | One-dimensional motion with constant acceleration (v0=16.62 m/s, a=0.6688 m/s²) | An object starts with initial velocity 16.62 m/s and experiences constant acceleration 0.6688 m/s² for 14.79 s. Final velocity: v = v0 + a t = 16.62 + (0.6688)(14.79) = 26.51 m/s. Displacement: s = v0 t + (1/2) a t² = 318.9 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. |
7,114 | physics | mechanics | kinematics_1d | 2 | worked_example | One-dimensional motion with constant acceleration (v0=16.94 m/s, a=1.079 m/s²) | An object starts with initial velocity 16.94 m/s and experiences constant acceleration 1.079 m/s² for 8.38 s. Final velocity: v = v0 + a t = 16.94 + (1.079)(8.38) = 25.98 m/s. Displacement: s = v0 t + (1/2) a t² = 179.8 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. |
7,115 | physics | mechanics | kinematics_1d | 2 | worked_example | One-dimensional motion with constant acceleration (v0=13.77 m/s, a=-4.769 m/s²) | An object starts with initial velocity 13.77 m/s and experiences constant acceleration -4.769 m/s² for 5.448 s. Final velocity: v = v0 + a t = 13.77 + (-4.769)(5.448) = -12.2 m/s. Displacement: s = v0 t + (1/2) a t² = 4.276 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. |
7,116 | physics | mechanics | kinematics_1d | 2 | worked_example | One-dimensional motion with constant acceleration (v0=13 m/s, a=3.331 m/s²) | An object starts with initial velocity 13 m/s and experiences constant acceleration 3.331 m/s² for 18.06 s. Final velocity: v = v0 + a t = 13 + (3.331)(18.06) = 73.16 m/s. Displacement: s = v0 t + (1/2) a t² = 777.9 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. |
7,117 | physics | mechanics | kinematics_1d | 2 | worked_example | One-dimensional motion with constant acceleration (v0=16.97 m/s, a=2.087 m/s²) | An object starts with initial velocity 16.97 m/s and experiences constant acceleration 2.087 m/s² for 19.91 s. Final velocity: v = v0 + a t = 16.97 + (2.087)(19.91) = 58.52 m/s. Displacement: s = v0 t + (1/2) a t² = 751.3 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. |
7,118 | physics | mechanics | kinematics_1d | 2 | worked_example | One-dimensional motion with constant acceleration (v0=19.81 m/s, a=2.279 m/s²) | An object starts with initial velocity 19.81 m/s and experiences constant acceleration 2.279 m/s² for 12.67 s. Final velocity: v = v0 + a t = 19.81 + (2.279)(12.67) = 48.7 m/s. Displacement: s = v0 t + (1/2) a t² = 434.1 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. |
7,119 | physics | mechanics | kinematics_1d | 2 | worked_example | One-dimensional motion with constant acceleration (v0=10.92 m/s, a=0.5649 m/s²) | An object starts with initial velocity 10.92 m/s and experiences constant acceleration 0.5649 m/s² for 12.24 s. Final velocity: v = v0 + a t = 10.92 + (0.5649)(12.24) = 17.84 m/s. Displacement: s = v0 t + (1/2) a t² = 175.9 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. |
7,120 | physics | mechanics | kinematics_1d | 2 | worked_example | One-dimensional motion with constant acceleration (v0=5.418 m/s, a=1.803 m/s²) | An object starts with initial velocity 5.418 m/s and experiences constant acceleration 1.803 m/s² for 2.473 s. Final velocity: v = v0 + a t = 5.418 + (1.803)(2.473) = 9.877 m/s. Displacement: s = v0 t + (1/2) a t² = 18.91 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. |
7,121 | physics | mechanics | kinematics_1d | 2 | worked_example | One-dimensional motion with constant acceleration (v0=13.63 m/s, a=0.6558 m/s²) | An object starts with initial velocity 13.63 m/s and experiences constant acceleration 0.6558 m/s² for 14.17 s. Final velocity: v = v0 + a t = 13.63 + (0.6558)(14.17) = 22.93 m/s. Displacement: s = v0 t + (1/2) a t² = 259.1 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. |
7,122 | physics | mechanics | kinematics_1d | 2 | worked_example | One-dimensional motion with constant acceleration (v0=7.102 m/s, a=-4.666 m/s²) | An object starts with initial velocity 7.102 m/s and experiences constant acceleration -4.666 m/s² for 2.653 s. Final velocity: v = v0 + a t = 7.102 + (-4.666)(2.653) = -5.274 m/s. Displacement: s = v0 t + (1/2) a t² = 2.425 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. |
7,123 | physics | mechanics | kinematics_1d | 2 | worked_example | One-dimensional motion with constant acceleration (v0=12.95 m/s, a=5.523 m/s²) | An object starts with initial velocity 12.95 m/s and experiences constant acceleration 5.523 m/s² for 19.78 s. Final velocity: v = v0 + a t = 12.95 + (5.523)(19.78) = 122.2 m/s. Displacement: s = v0 t + (1/2) a t² = 1336 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. |
7,124 | physics | mechanics | kinematics_1d | 2 | worked_example | One-dimensional motion with constant acceleration (v0=13.64 m/s, a=9.451 m/s²) | An object starts with initial velocity 13.64 m/s and experiences constant acceleration 9.451 m/s² for 10.92 s. Final velocity: v = v0 + a t = 13.64 + (9.451)(10.92) = 116.8 m/s. Displacement: s = v0 t + (1/2) a t² = 712.1 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. |
7,125 | physics | mechanics | kinematics_1d | 2 | worked_example | One-dimensional motion with constant acceleration (v0=16.77 m/s, a=8.09 m/s²) | An object starts with initial velocity 16.77 m/s and experiences constant acceleration 8.09 m/s² for 8.223 s. Final velocity: v = v0 + a t = 16.77 + (8.09)(8.223) = 83.3 m/s. Displacement: s = v0 t + (1/2) a t² = 411.4 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. |
7,126 | physics | mechanics | kinematics_1d | 2 | worked_example | One-dimensional motion with constant acceleration (v0=9.082 m/s, a=-3.82 m/s²) | An object starts with initial velocity 9.082 m/s and experiences constant acceleration -3.82 m/s² for 17.24 s. Final velocity: v = v0 + a t = 9.082 + (-3.82)(17.24) = -56.78 m/s. Displacement: s = v0 t + (1/2) a t² = -411.2 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. |
7,127 | physics | mechanics | kinematics_1d | 2 | worked_example | One-dimensional motion with constant acceleration (v0=22.83 m/s, a=9.706 m/s²) | An object starts with initial velocity 22.83 m/s and experiences constant acceleration 9.706 m/s² for 11.86 s. Final velocity: v = v0 + a t = 22.83 + (9.706)(11.86) = 138 m/s. Displacement: s = v0 t + (1/2) a t² = 953.6 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. |
7,128 | physics | mechanics | kinematics_1d | 2 | worked_example | One-dimensional motion with constant acceleration (v0=28.04 m/s, a=9.111 m/s²) | An object starts with initial velocity 28.04 m/s and experiences constant acceleration 9.111 m/s² for 15.9 s. Final velocity: v = v0 + a t = 28.04 + (9.111)(15.9) = 172.9 m/s. Displacement: s = v0 t + (1/2) a t² = 1597 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. |
7,129 | physics | mechanics | kinematics_1d | 2 | worked_example | One-dimensional motion with constant acceleration (v0=10.84 m/s, a=2.657 m/s²) | An object starts with initial velocity 10.84 m/s and experiences constant acceleration 2.657 m/s² for 19.46 s. Final velocity: v = v0 + a t = 10.84 + (2.657)(19.46) = 62.55 m/s. Displacement: s = v0 t + (1/2) a t² = 714 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. |
7,130 | physics | mechanics | kinematics_1d | 2 | worked_example | One-dimensional motion with constant acceleration (v0=3.533 m/s, a=2.3 m/s²) | An object starts with initial velocity 3.533 m/s and experiences constant acceleration 2.3 m/s² for 6.525 s. Final velocity: v = v0 + a t = 3.533 + (2.3)(6.525) = 18.54 m/s. Displacement: s = v0 t + (1/2) a t² = 72.01 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. |
7,131 | physics | mechanics | kinematics_1d | 2 | worked_example | One-dimensional motion with constant acceleration (v0=2.706 m/s, a=6.219 m/s²) | An object starts with initial velocity 2.706 m/s and experiences constant acceleration 6.219 m/s² for 10.28 s. Final velocity: v = v0 + a t = 2.706 + (6.219)(10.28) = 66.62 m/s. Displacement: s = v0 t + (1/2) a t² = 356.3 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. |
7,132 | physics | mechanics | kinematics_1d | 2 | worked_example | One-dimensional motion with constant acceleration (v0=11.55 m/s, a=-2.227 m/s²) | An object starts with initial velocity 11.55 m/s and experiences constant acceleration -2.227 m/s² for 2.418 s. Final velocity: v = v0 + a t = 11.55 + (-2.227)(2.418) = 6.169 m/s. Displacement: s = v0 t + (1/2) a t² = 21.42 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. |
7,133 | physics | mechanics | kinematics_1d | 2 | worked_example | One-dimensional motion with constant acceleration (v0=16.25 m/s, a=6.959 m/s²) | An object starts with initial velocity 16.25 m/s and experiences constant acceleration 6.959 m/s² for 13.25 s. Final velocity: v = v0 + a t = 16.25 + (6.959)(13.25) = 108.5 m/s. Displacement: s = v0 t + (1/2) a t² = 826.7 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. |
7,134 | physics | mechanics | kinematics_1d | 2 | worked_example | One-dimensional motion with constant acceleration (v0=4.34 m/s, a=-3.621 m/s²) | An object starts with initial velocity 4.34 m/s and experiences constant acceleration -3.621 m/s² for 14.54 s. Final velocity: v = v0 + a t = 4.34 + (-3.621)(14.54) = -48.33 m/s. Displacement: s = v0 t + (1/2) a t² = -319.8 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. |
7,135 | physics | mechanics | kinematics_1d | 2 | worked_example | One-dimensional motion with constant acceleration (v0=12.08 m/s, a=-2.933 m/s²) | An object starts with initial velocity 12.08 m/s and experiences constant acceleration -2.933 m/s² for 6.059 s. Final velocity: v = v0 + a t = 12.08 + (-2.933)(6.059) = -5.69 m/s. Displacement: s = v0 t + (1/2) a t² = 19.37 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. |
7,136 | physics | mechanics | kinematics_1d | 2 | worked_example | One-dimensional motion with constant acceleration (v0=4.882 m/s, a=4.047 m/s²) | An object starts with initial velocity 4.882 m/s and experiences constant acceleration 4.047 m/s² for 10.07 s. Final velocity: v = v0 + a t = 4.882 + (4.047)(10.07) = 45.65 m/s. Displacement: s = v0 t + (1/2) a t² = 254.5 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. |
7,137 | physics | mechanics | kinematics_1d | 2 | worked_example | One-dimensional motion with constant acceleration (v0=15.37 m/s, a=-3.883 m/s²) | An object starts with initial velocity 15.37 m/s and experiences constant acceleration -3.883 m/s² for 2.014 s. Final velocity: v = v0 + a t = 15.37 + (-3.883)(2.014) = 7.544 m/s. Displacement: s = v0 t + (1/2) a t² = 23.08 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. |
7,138 | physics | mechanics | kinematics_1d | 2 | worked_example | One-dimensional motion with constant acceleration (v0=21.21 m/s, a=9.936 m/s²) | An object starts with initial velocity 21.21 m/s and experiences constant acceleration 9.936 m/s² for 17.79 s. Final velocity: v = v0 + a t = 21.21 + (9.936)(17.79) = 198 m/s. Displacement: s = v0 t + (1/2) a t² = 1950 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. |
7,139 | physics | mechanics | kinematics_1d | 2 | worked_example | One-dimensional motion with constant acceleration (v0=21.63 m/s, a=0.4966 m/s²) | An object starts with initial velocity 21.63 m/s and experiences constant acceleration 0.4966 m/s² for 18.41 s. Final velocity: v = v0 + a t = 21.63 + (0.4966)(18.41) = 30.77 m/s. Displacement: s = v0 t + (1/2) a t² = 482.3 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. |
7,140 | physics | mechanics | kinematics_1d | 2 | worked_example | One-dimensional motion with constant acceleration (v0=3.157 m/s, a=3.303 m/s²) | An object starts with initial velocity 3.157 m/s and experiences constant acceleration 3.303 m/s² for 13.94 s. Final velocity: v = v0 + a t = 3.157 + (3.303)(13.94) = 49.2 m/s. Displacement: s = v0 t + (1/2) a t² = 364.9 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. |
7,141 | physics | mechanics | kinematics_1d | 2 | worked_example | One-dimensional motion with constant acceleration (v0=11.19 m/s, a=7.502 m/s²) | An object starts with initial velocity 11.19 m/s and experiences constant acceleration 7.502 m/s² for 1.862 s. Final velocity: v = v0 + a t = 11.19 + (7.502)(1.862) = 25.15 m/s. Displacement: s = v0 t + (1/2) a t² = 33.83 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. |
7,142 | physics | mechanics | kinematics_1d | 2 | worked_example | One-dimensional motion with constant acceleration (v0=13.05 m/s, a=2.587 m/s²) | An object starts with initial velocity 13.05 m/s and experiences constant acceleration 2.587 m/s² for 1.055 s. Final velocity: v = v0 + a t = 13.05 + (2.587)(1.055) = 15.78 m/s. Displacement: s = v0 t + (1/2) a t² = 15.21 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. |
7,143 | physics | mechanics | kinematics_1d | 2 | worked_example | One-dimensional motion with constant acceleration (v0=17.34 m/s, a=-3.105 m/s²) | An object starts with initial velocity 17.34 m/s and experiences constant acceleration -3.105 m/s² for 17.52 s. Final velocity: v = v0 + a t = 17.34 + (-3.105)(17.52) = -37.07 m/s. Displacement: s = v0 t + (1/2) a t² = -172.9 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. |
7,144 | physics | mechanics | kinematics_1d | 2 | worked_example | One-dimensional motion with constant acceleration (v0=14.91 m/s, a=-4.999 m/s²) | An object starts with initial velocity 14.91 m/s and experiences constant acceleration -4.999 m/s² for 11.92 s. Final velocity: v = v0 + a t = 14.91 + (-4.999)(11.92) = -44.66 m/s. Displacement: s = v0 t + (1/2) a t² = -177.3 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. |
7,145 | physics | mechanics | kinematics_1d | 2 | worked_example | One-dimensional motion with constant acceleration (v0=28.14 m/s, a=-4.253 m/s²) | An object starts with initial velocity 28.14 m/s and experiences constant acceleration -4.253 m/s² for 13.01 s. Final velocity: v = v0 + a t = 28.14 + (-4.253)(13.01) = -27.19 m/s. Displacement: s = v0 t + (1/2) a t² = 6.177 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. |
7,146 | physics | mechanics | kinematics_1d | 2 | worked_example | One-dimensional motion with constant acceleration (v0=6.587 m/s, a=5.562 m/s²) | An object starts with initial velocity 6.587 m/s and experiences constant acceleration 5.562 m/s² for 7.501 s. Final velocity: v = v0 + a t = 6.587 + (5.562)(7.501) = 48.31 m/s. Displacement: s = v0 t + (1/2) a t² = 205.9 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. |
7,147 | physics | mechanics | kinematics_1d | 2 | worked_example | One-dimensional motion with constant acceleration (v0=8.253 m/s, a=-4.207 m/s²) | An object starts with initial velocity 8.253 m/s and experiences constant acceleration -4.207 m/s² for 4.262 s. Final velocity: v = v0 + a t = 8.253 + (-4.207)(4.262) = -9.678 m/s. Displacement: s = v0 t + (1/2) a t² = -3.036 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. |
7,148 | physics | mechanics | kinematics_1d | 2 | worked_example | One-dimensional motion with constant acceleration (v0=24.82 m/s, a=-0.8233 m/s²) | An object starts with initial velocity 24.82 m/s and experiences constant acceleration -0.8233 m/s² for 17.43 s. Final velocity: v = v0 + a t = 24.82 + (-0.8233)(17.43) = 10.47 m/s. Displacement: s = v0 t + (1/2) a t² = 307.5 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. |
7,149 | physics | mechanics | kinematics_1d | 2 | worked_example | One-dimensional motion with constant acceleration (v0=29.76 m/s, a=2.546 m/s²) | An object starts with initial velocity 29.76 m/s and experiences constant acceleration 2.546 m/s² for 18.47 s. Final velocity: v = v0 + a t = 29.76 + (2.546)(18.47) = 76.79 m/s. Displacement: s = v0 t + (1/2) a t² = 984.1 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. |
7,150 | physics | mechanics | kinematics_1d | 2 | worked_example | One-dimensional motion with constant acceleration (v0=18.33 m/s, a=8.311 m/s²) | An object starts with initial velocity 18.33 m/s and experiences constant acceleration 8.311 m/s² for 17.82 s. Final velocity: v = v0 + a t = 18.33 + (8.311)(17.82) = 166.5 m/s. Displacement: s = v0 t + (1/2) a t² = 1647 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. |
7,151 | physics | mechanics | kinematics_1d | 2 | worked_example | One-dimensional motion with constant acceleration (v0=8.4 m/s, a=-4.124 m/s²) | An object starts with initial velocity 8.4 m/s and experiences constant acceleration -4.124 m/s² for 10.22 s. Final velocity: v = v0 + a t = 8.4 + (-4.124)(10.22) = -33.76 m/s. Displacement: s = v0 t + (1/2) a t² = -129.7 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. |
7,152 | physics | mechanics | kinematics_1d | 2 | worked_example | One-dimensional motion with constant acceleration (v0=19.56 m/s, a=8.686 m/s²) | An object starts with initial velocity 19.56 m/s and experiences constant acceleration 8.686 m/s² for 10.82 s. Final velocity: v = v0 + a t = 19.56 + (8.686)(10.82) = 113.5 m/s. Displacement: s = v0 t + (1/2) a t² = 719.8 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. |
7,153 | physics | mechanics | kinematics_1d | 2 | worked_example | One-dimensional motion with constant acceleration (v0=4.828 m/s, a=-1.073 m/s²) | An object starts with initial velocity 4.828 m/s and experiences constant acceleration -1.073 m/s² for 9.6 s. Final velocity: v = v0 + a t = 4.828 + (-1.073)(9.6) = -5.471 m/s. Displacement: s = v0 t + (1/2) a t² = -3.087 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. |
7,154 | physics | mechanics | newton_second_law | 3 | worked_example | Newton's second law: mass 4.959 kg, acceleration 14.91 m/s² | A net force acting on a mass of 4.959 kg produces an acceleration of 14.91 m/s². By Newton's second law, F_net = m a = 4.959 × 14.91 = 73.92 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. |
7,155 | physics | mechanics | newton_second_law | 3 | worked_example | Newton's second law: mass 4.088 kg, acceleration 14.13 m/s² | A net force acting on a mass of 4.088 kg produces an acceleration of 14.13 m/s². By Newton's second law, F_net = m a = 4.088 × 14.13 = 57.76 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. |
7,156 | physics | mechanics | newton_second_law | 3 | worked_example | Newton's second law: mass 27.34 kg, acceleration 13.98 m/s² | A net force acting on a mass of 27.34 kg produces an acceleration of 13.98 m/s². By Newton's second law, F_net = m a = 27.34 × 13.98 = 382.1 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. |
7,157 | physics | mechanics | newton_second_law | 3 | worked_example | Newton's second law: mass 38.74 kg, acceleration 0.1931 m/s² | A net force acting on a mass of 38.74 kg produces an acceleration of 0.1931 m/s². By Newton's second law, F_net = m a = 38.74 × 0.1931 = 7.479 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. |
7,158 | physics | mechanics | newton_second_law | 3 | worked_example | Newton's second law: mass 3.763 kg, acceleration 11.82 m/s² | A net force acting on a mass of 3.763 kg produces an acceleration of 11.82 m/s². By Newton's second law, F_net = m a = 3.763 × 11.82 = 44.46 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. |
7,159 | physics | mechanics | newton_second_law | 3 | worked_example | Newton's second law: mass 47.18 kg, acceleration 9.658 m/s² | A net force acting on a mass of 47.18 kg produces an acceleration of 9.658 m/s². By Newton's second law, F_net = m a = 47.18 × 9.658 = 455.6 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. |
7,160 | physics | mechanics | newton_second_law | 3 | worked_example | Newton's second law: mass 37.78 kg, acceleration 0.9791 m/s² | A net force acting on a mass of 37.78 kg produces an acceleration of 0.9791 m/s². By Newton's second law, F_net = m a = 37.78 × 0.9791 = 36.99 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. |
7,161 | physics | mechanics | newton_second_law | 3 | worked_example | Newton's second law: mass 6.25 kg, acceleration 5.64 m/s² | A net force acting on a mass of 6.25 kg produces an acceleration of 5.64 m/s². By Newton's second law, F_net = m a = 6.25 × 5.64 = 35.25 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. |
7,162 | physics | mechanics | newton_second_law | 3 | worked_example | Newton's second law: mass 0.5508 kg, acceleration 2.752 m/s² | A net force acting on a mass of 0.5508 kg produces an acceleration of 2.752 m/s². By Newton's second law, F_net = m a = 0.5508 × 2.752 = 1.516 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. |
7,163 | physics | mechanics | newton_second_law | 3 | worked_example | Newton's second law: mass 23.46 kg, acceleration 12.13 m/s² | A net force acting on a mass of 23.46 kg produces an acceleration of 12.13 m/s². By Newton's second law, F_net = m a = 23.46 × 12.13 = 284.5 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. |
7,164 | physics | mechanics | newton_second_law | 3 | worked_example | Newton's second law: mass 49.06 kg, acceleration 9.888 m/s² | A net force acting on a mass of 49.06 kg produces an acceleration of 9.888 m/s². By Newton's second law, F_net = m a = 49.06 × 9.888 = 485.1 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. |
7,165 | physics | mechanics | newton_second_law | 3 | worked_example | Newton's second law: mass 16.09 kg, acceleration 6.468 m/s² | A net force acting on a mass of 16.09 kg produces an acceleration of 6.468 m/s². By Newton's second law, F_net = m a = 16.09 × 6.468 = 104.1 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. |
7,166 | physics | mechanics | newton_second_law | 3 | worked_example | Newton's second law: mass 37.5 kg, acceleration 0.7499 m/s² | A net force acting on a mass of 37.5 kg produces an acceleration of 0.7499 m/s². By Newton's second law, F_net = m a = 37.5 × 0.7499 = 28.12 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. |
7,167 | physics | mechanics | newton_second_law | 3 | worked_example | Newton's second law: mass 18.46 kg, acceleration 14.21 m/s² | A net force acting on a mass of 18.46 kg produces an acceleration of 14.21 m/s². By Newton's second law, F_net = m a = 18.46 × 14.21 = 262.3 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. |
7,168 | physics | mechanics | newton_second_law | 3 | worked_example | Newton's second law: mass 43.18 kg, acceleration 6.665 m/s² | A net force acting on a mass of 43.18 kg produces an acceleration of 6.665 m/s². By Newton's second law, F_net = m a = 43.18 × 6.665 = 287.8 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. |
7,169 | physics | mechanics | newton_second_law | 3 | worked_example | Newton's second law: mass 41.13 kg, acceleration 2.129 m/s² | A net force acting on a mass of 41.13 kg produces an acceleration of 2.129 m/s². By Newton's second law, F_net = m a = 41.13 × 2.129 = 87.56 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. |
7,170 | physics | mechanics | newton_second_law | 3 | worked_example | Newton's second law: mass 2.671 kg, acceleration 6.989 m/s² | A net force acting on a mass of 2.671 kg produces an acceleration of 6.989 m/s². By Newton's second law, F_net = m a = 2.671 × 6.989 = 18.67 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. |
7,171 | physics | mechanics | newton_second_law | 3 | worked_example | Newton's second law: mass 40.58 kg, acceleration 8.945 m/s² | A net force acting on a mass of 40.58 kg produces an acceleration of 8.945 m/s². By Newton's second law, F_net = m a = 40.58 × 8.945 = 363 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. |
7,172 | physics | mechanics | newton_second_law | 3 | worked_example | Newton's second law: mass 34.18 kg, acceleration 0.1686 m/s² | A net force acting on a mass of 34.18 kg produces an acceleration of 0.1686 m/s². By Newton's second law, F_net = m a = 34.18 × 0.1686 = 5.763 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. |
7,173 | physics | mechanics | newton_second_law | 3 | worked_example | Newton's second law: mass 24.45 kg, acceleration 13.6 m/s² | A net force acting on a mass of 24.45 kg produces an acceleration of 13.6 m/s². By Newton's second law, F_net = m a = 24.45 × 13.6 = 332.5 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. |
7,174 | physics | mechanics | newton_second_law | 3 | worked_example | Newton's second law: mass 41.58 kg, acceleration 9.839 m/s² | A net force acting on a mass of 41.58 kg produces an acceleration of 9.839 m/s². By Newton's second law, F_net = m a = 41.58 × 9.839 = 409.2 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. |
7,175 | physics | mechanics | newton_second_law | 3 | worked_example | Newton's second law: mass 43.92 kg, acceleration 8.653 m/s² | A net force acting on a mass of 43.92 kg produces an acceleration of 8.653 m/s². By Newton's second law, F_net = m a = 43.92 × 8.653 = 380.1 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. |
7,176 | physics | mechanics | newton_second_law | 3 | worked_example | Newton's second law: mass 15.96 kg, acceleration 10.7 m/s² | A net force acting on a mass of 15.96 kg produces an acceleration of 10.7 m/s². By Newton's second law, F_net = m a = 15.96 × 10.7 = 170.7 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. |
7,177 | physics | mechanics | newton_second_law | 3 | worked_example | Newton's second law: mass 35.11 kg, acceleration 3.482 m/s² | A net force acting on a mass of 35.11 kg produces an acceleration of 3.482 m/s². By Newton's second law, F_net = m a = 35.11 × 3.482 = 122.3 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. |
7,178 | physics | mechanics | newton_second_law | 3 | worked_example | Newton's second law: mass 32.47 kg, acceleration 11.98 m/s² | A net force acting on a mass of 32.47 kg produces an acceleration of 11.98 m/s². By Newton's second law, F_net = m a = 32.47 × 11.98 = 389.1 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. |
7,179 | physics | mechanics | newton_second_law | 3 | worked_example | Newton's second law: mass 26.33 kg, acceleration 6.257 m/s² | A net force acting on a mass of 26.33 kg produces an acceleration of 6.257 m/s². By Newton's second law, F_net = m a = 26.33 × 6.257 = 164.8 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. |
7,180 | physics | mechanics | newton_second_law | 3 | worked_example | Newton's second law: mass 27.99 kg, acceleration 3.279 m/s² | A net force acting on a mass of 27.99 kg produces an acceleration of 3.279 m/s². By Newton's second law, F_net = m a = 27.99 × 3.279 = 91.77 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. |
7,181 | physics | mechanics | newton_second_law | 3 | worked_example | Newton's second law: mass 42.94 kg, acceleration 5.235 m/s² | A net force acting on a mass of 42.94 kg produces an acceleration of 5.235 m/s². By Newton's second law, F_net = m a = 42.94 × 5.235 = 224.8 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. |
7,182 | physics | mechanics | newton_second_law | 3 | worked_example | Newton's second law: mass 29.96 kg, acceleration 4.258 m/s² | A net force acting on a mass of 29.96 kg produces an acceleration of 4.258 m/s². By Newton's second law, F_net = m a = 29.96 × 4.258 = 127.6 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. |
7,183 | physics | mechanics | newton_second_law | 3 | worked_example | Newton's second law: mass 23.87 kg, acceleration 1.822 m/s² | A net force acting on a mass of 23.87 kg produces an acceleration of 1.822 m/s². By Newton's second law, F_net = m a = 23.87 × 1.822 = 43.5 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. |
7,184 | physics | mechanics | newton_second_law | 3 | worked_example | Newton's second law: mass 41.9 kg, acceleration 1.413 m/s² | A net force acting on a mass of 41.9 kg produces an acceleration of 1.413 m/s². By Newton's second law, F_net = m a = 41.9 × 1.413 = 59.18 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. |
7,185 | physics | mechanics | newton_second_law | 3 | worked_example | Newton's second law: mass 31.41 kg, acceleration 10.84 m/s² | A net force acting on a mass of 31.41 kg produces an acceleration of 10.84 m/s². By Newton's second law, F_net = m a = 31.41 × 10.84 = 340.5 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. |
7,186 | physics | mechanics | newton_second_law | 3 | worked_example | Newton's second law: mass 46.32 kg, acceleration 14.66 m/s² | A net force acting on a mass of 46.32 kg produces an acceleration of 14.66 m/s². By Newton's second law, F_net = m a = 46.32 × 14.66 = 679.2 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. |
7,187 | physics | mechanics | newton_second_law | 3 | worked_example | Newton's second law: mass 10.51 kg, acceleration 4.732 m/s² | A net force acting on a mass of 10.51 kg produces an acceleration of 4.732 m/s². By Newton's second law, F_net = m a = 10.51 × 4.732 = 49.72 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. |
7,188 | physics | mechanics | newton_second_law | 3 | worked_example | Newton's second law: mass 27.15 kg, acceleration 3.277 m/s² | A net force acting on a mass of 27.15 kg produces an acceleration of 3.277 m/s². By Newton's second law, F_net = m a = 27.15 × 3.277 = 88.97 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. |
7,189 | physics | mechanics | newton_second_law | 3 | worked_example | Newton's second law: mass 38.64 kg, acceleration 11.62 m/s² | A net force acting on a mass of 38.64 kg produces an acceleration of 11.62 m/s². By Newton's second law, F_net = m a = 38.64 × 11.62 = 449.2 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. |
7,190 | physics | mechanics | newton_second_law | 3 | worked_example | Newton's second law: mass 2.258 kg, acceleration 5.051 m/s² | A net force acting on a mass of 2.258 kg produces an acceleration of 5.051 m/s². By Newton's second law, F_net = m a = 2.258 × 5.051 = 11.41 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. |
7,191 | physics | mechanics | newton_second_law | 3 | worked_example | Newton's second law: mass 8.11 kg, acceleration 11.64 m/s² | A net force acting on a mass of 8.11 kg produces an acceleration of 11.64 m/s². By Newton's second law, F_net = m a = 8.11 × 11.64 = 94.42 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. |
7,192 | physics | mechanics | newton_second_law | 3 | worked_example | Newton's second law: mass 33.37 kg, acceleration 6.109 m/s² | A net force acting on a mass of 33.37 kg produces an acceleration of 6.109 m/s². By Newton's second law, F_net = m a = 33.37 × 6.109 = 203.8 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. |
7,193 | physics | mechanics | newton_second_law | 3 | worked_example | Newton's second law: mass 34.39 kg, acceleration 1.124 m/s² | A net force acting on a mass of 34.39 kg produces an acceleration of 1.124 m/s². By Newton's second law, F_net = m a = 34.39 × 1.124 = 38.65 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. |
7,194 | physics | mechanics | newton_second_law | 3 | worked_example | Newton's second law: mass 44.67 kg, acceleration 3.279 m/s² | A net force acting on a mass of 44.67 kg produces an acceleration of 3.279 m/s². By Newton's second law, F_net = m a = 44.67 × 3.279 = 146.5 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. |
7,195 | physics | mechanics | newton_second_law | 3 | worked_example | Newton's second law: mass 44.14 kg, acceleration 11.88 m/s² | A net force acting on a mass of 44.14 kg produces an acceleration of 11.88 m/s². By Newton's second law, F_net = m a = 44.14 × 11.88 = 524.4 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. |
7,196 | physics | mechanics | newton_second_law | 3 | worked_example | Newton's second law: mass 18.32 kg, acceleration 3.915 m/s² | A net force acting on a mass of 18.32 kg produces an acceleration of 3.915 m/s². By Newton's second law, F_net = m a = 18.32 × 3.915 = 71.71 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. |
7,197 | physics | mechanics | newton_second_law | 3 | worked_example | Newton's second law: mass 38.32 kg, acceleration 12.38 m/s² | A net force acting on a mass of 38.32 kg produces an acceleration of 12.38 m/s². By Newton's second law, F_net = m a = 38.32 × 12.38 = 474.2 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. |
7,198 | physics | mechanics | newton_second_law | 3 | worked_example | Newton's second law: mass 17.2 kg, acceleration 13.78 m/s² | A net force acting on a mass of 17.2 kg produces an acceleration of 13.78 m/s². By Newton's second law, F_net = m a = 17.2 × 13.78 = 237.1 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. |
7,199 | physics | mechanics | newton_second_law | 3 | worked_example | Newton's second law: mass 21.32 kg, acceleration 1.001 m/s² | A net force acting on a mass of 21.32 kg produces an acceleration of 1.001 m/s². By Newton's second law, F_net = m a = 21.32 × 1.001 = 21.35 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. |
7,200 | physics | mechanics | newton_second_law | 3 | worked_example | Newton's second law: mass 43.17 kg, acceleration 14.69 m/s² | A net force acting on a mass of 43.17 kg produces an acceleration of 14.69 m/s². By Newton's second law, F_net = m a = 43.17 × 14.69 = 634.1 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. |
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