Question 15, Physics - equation sheet attached Physics 114 Equation SheetConstants and ConversionsKinematics Continuedg = 9.80 m/sFree-fall accelerationΔνInstantaneousainst. = limAt-o AtAcceleration1N = 1 kg m/s?NewtonUniform motion(v) = (v); = constantPosition in uniformX = x + (v)AtMathematics, Scaling and Vectorsb = a* + loga (b) = xmotionLogarithmConstant(v); = (v,); + azAtacceleration:1log(ab) = log(a) + log (b)x, = x, + (v,),At +a, (at)?log Ax" = n log x + log A(v,); = (v,)} + 2a,AxVolume of a sphereV =Surface area of aA = 4ar?ForcessphereNewton's second lawFnet = EF = mã%3DVolume of a cylinderV = arlNewton's second lawFnetx = EF = ma,%3DSurface area of aA = 2ar? + 2rlComponent formFnety = ER, = maycylinderMass densityp = m/VNewton's Third LawFA en =-ton AA, = A cos e (rel. to x-axis)Weightw = mgx -component of avector ÅApparent weightWapp = magnitude of supporting forcesy -component of aAy = A sin 8 (rel to x-axis)vector ÅKinetic frictionfk = HanMagnitude of vector ẢStatic frictionA = JA + A,Reynolds numberRe = pvl/nDirection of A relative8 = tan-(Ay/A,)1Drag (high Reynoldsnumber)=CopAv?to x-axisAddition of two vectors If = Å + B, thenC, = A, + B,D = 6nyrvDrag (low Reynoldsnumber)Cy = Ay + ByCircular MotionKinematicsCentripetal accelerationa =DisplacementAx = x - XAverage VelocityAxFrequency1VargT2arAtAxVinst. = limInstantaneous VelocityAt+0 AtAvAverage AccelerationdavgΔε An object is thrown horizontally off the top of a 110 m high cliff. The object travels 170 mhorizontally. At what initial speed was the object thrown?+y110 m170 mO 58 m/sO Om/sCorrect answer is not listed.63 m/s46 m/s36 m/s

Given Angle of projection θ = 00 x component of initial velocity ux = u cosθ = u y component of…

Constants and Conversions 9- 9.80 m/s 1N = 1 kg m/s* Kinematics Continued Free-fall acceleration Av Instantaneous Acceleration Ainst. = lim At0 At Newton Uniform motion (v)r = (v), = constant Position in uniform X = x + (v),At Mathematics, Scaling and Vectors Logarithm motion b= a" ++ loga (b) = x (v,), = (v,), + a,At Constant acceleration: *, - x, + (v,),at +a, (At)? log(ab) = log(a) + tog (b) log Ax" = n log x + log A (v,); = (v,)} + 2a,Ax Volume of a sphere v = ar Surface area of a A = 4ar? Forces sphere Newton's second law Fnet = EF = må V = rl A = 2ar? + 2arl Volume of a cylinder Newton's second law Fnetx = ER = maz Surface area of a Component form Fnety = Ek, = ma, cylinder Mass density FA on =-fu on A Newton's Third Law p = m/V x -component of a vector Å A, - A cos 6 (rel. to x-axis) Weight W = mg Apparent weight Wapp - magnitude of supporting forces A, - A sin 6 (rel. to x- axis) y -component of a vector Å fa = Han Kinetic friction Magnitude of vector Ä Static friction A= A + A

Constants and Conversions 9- 9.80 m/s 1N = 1 kg m/s* Kinematics Continued Free-fall acceleration Av Instantaneous Acceleration Ainst. = lim At0 At Newton Uniform motion (v)r = (v), = constant Position in uniform X = x + (v),At Mathematics, Scaling and Vectors Logarithm motion b= a" ++ loga (b) = x (v,), = (v,), + a,At Constant acceleration: *, - x, + (v,),at +a, (At)? log(ab) = log(a) + tog (b) log Ax" = n log x + log A (v,); = (v,)} + 2a,Ax Volume of a sphere v = ar Surface area of a A = 4ar? Forces sphere Newton's second law Fnet = EF = må V = rl A = 2ar? + 2arl Volume of a cylinder Newton's second law Fnetx = ER = maz Surface area of a Component form Fnety = Ek, = ma, cylinder Mass density FA on =-fu on A Newton's Third Law p = m/V x -component of a vector Å A, - A cos 6 (rel. to x-axis) Weight W = mg Apparent weight Wapp - magnitude of supporting forces A, - A sin 6 (rel. to x- axis) y -component of a vector Å fa = Han Kinetic friction Magnitude of vector Ä Static friction A= A + A

College Physics

11th Edition

ISBN:9781305952300

Author:Raymond A. Serway, Chris Vuille

Publisher:Raymond A. Serway, Chris Vuille

Chapter1: Units, Trigonometry. And Vectors

Section: Chapter Questions

Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...

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