30163788
Descripción
Fichas por Adriana Vincelli-Joma, actualizado hace más de 1 año
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Creado por Adriana Vincelli-Joma
hace más de 3 años
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| Pregunta | Respuesta |
| uniform circular motion | motion of object in circular path at constant speed |
| rotational angle | -ratio of arc length to radius curvature on circular path -Δθ = Δs/r |
| arc length | -distance traveled by object along circular path -Δs |
| radius of curvature | -radius of circular path -r |
| revolution | -2π radians |
| angular velocity | -rate of change of angle with which object moves on a circular path -ω = Δθ/Δt -ω = v/r |
| linear velocity | -v = rω |
| centripetal acceleration | -acceleration of object moving in circle directed toward center -a_c = v^2/r -a_c = rω^2 |
| centripetal force | -any net force causing uniform circular motion -perpendicular to path and pointing to center |
| centripetal force formulas | -F_c = ma_c -F_c = m (v^2/r) -F_c = mrω^2 |
| centrifugal force | -fictitious force that tends to throw an object off when the object is rotating in a non-inertial frame of reference |
| banked curve | -curve in road that is sloping in a manner that helps a vehicle negotiate the curve |
| static coefficient of friction for banked curve | -μ_s = v^2/rg |
| angle of ideally banked curve | -θ = tan^-1 (v^2/rg) |
| Newton's Universal Law of Gravitation | -every particle in universe attracts every other particle with force along line joining them -force is directly proportional to product of masses and inversely proportional to square of distance between them |
| Newton's Universal Law of Gravitation formulas | -F = G mM/r^2 -G = 6.674 x 10^-11 N x m^2/kg^2 -m and M: masses of two bodies -r: distance between centers of mass -F: magnitude of gravitational force -mg = G mM/r^2 -g = GM/r^2 |
| Kepler's First Law | orbit of each planet about Sun is ellipse with Sun at one focus |
| Kepler's Second Law | each planet moves so that imaginary line drawn from Sun to planet sweeps out equal areas in equal times |
| Kepler's Third Law | -ratio of squares of period of any two planets about Sun is equal to ratio of cubes of their average distances from Sun -(T_1)^2/ (T_2 )^2 = (r_1)^3/(r_2)^3 |
| Period and radius of a satellite’s orbit about a larger body M | -T^2 = 4π^2/GM x r^3 -r^3 / T^2 = GM/4π^2 |
| altitude of satellite above Earth | h = r - R_E -r: distance between satellite and Earth -R_E: radius of Earth |
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