# Phys1 - Motion

 The flashcards below were created by user mse263 on FreezingBlue Flashcards. Meter Conversions 1 gigameter (G) = 109 m1 megameter (M) = 106 m1 kilometer (km) = 103 m1 centimeter (cm) = 10-2 m1 millimeter (mm) = 10-3 m1 micrometer (µm) = 10-6 m1 nanometer (nm) = 10-9 m1 Ångström (Å) = 10-10 m1 picometer (pm) = 10-12 m1 femptometer (fm) = 10-15 m Other Conversions √2 = 1.4√3 = 1.7sin30 = 1/2cos30 = √3/2 ~ .86sin60 = √3/2cos60 = 1/2sin & cos 45 = √2/2 ~ .7 Milligram Conversions 1 mg = 10-3 g = 10-6 kg Distance a Scalar quantity that refers to how much ground an object has covered during its motionwill only equal 0 if you don’t move at all Displacement a Vector quantity that refers to how far out of place an object is from where it started, aka it’s the object's overall change in positiondisplacement is the quantity that can equal 0 when you end in the same place you started (average) Acceleration change in velocity divided by the elapsed time for that changea = v2 - v1 / Δtaka change in velocity / elapsed time Most basic translational motion equation: d = vtdistance (meters) = velocity (m/s) * time (s) What equation should be used if vf is missing? d = vot + 1/2at2 What equation should be used if d is missing? vf = vo + at What equation should be used if t is missing? vf2 = vo2 + 2ad Equation to find the time at the top of a projectile motion problem? ttop = voy/gsame thing as saying m/s * s2/m Equation to find the height at the top of a projectile motion problem? h = voy2 / 2g What equation can be used to determine an object’s displacement during FREE-FALL? d (Δh) = 5t2from d = vot + 1/2at2 Objects with a ____ contact area-to-mass ratio experience a more significant impact of wind resistance than their denser counterparts. Objects with a LARGE surface area : mass ratio experience MORE wind resistance than their denser (therefore heavier) counterparts Terminal Velocity • vterm occurs when the drag Force upward is equal & opposite to the Force of gravity (Fg) • the object with the greatest terminal velocity will be:1. the DENSEST 2. have the LEAST amount of contact area Projectile Motion vox = vocosθvoy = vosinθ Range Equations (3) r = vo2sin2θ / gr = 2voxvoy / gr45° = 4hmax Which launch angle results in the maximum range & which results in the maximum height? • a 45° launch angle corresponds to the maximum possible RANGE • a 90° launch angle corresponds to the maximum possible HEIGHT • 30° & 60° launch angles correspond to the same range Authormse263 ID285169 Card SetPhys1 - Motion DescriptionChapter 1 Updated2014-10-07T23:56:09Z Show Answers