# CHEATSHEET - M easurement 1012 109 106 103 102 101 Basic...

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Measurement 10 12 tera- T 10 9 giga- G 10 6 mega- M 10 3 kilo- k 10 2 hecto- h 10 1 deka- da Basic SI unit 10 -1 deci- d 10 -2 centi- c 10 -3 milli- m 10 -6 micro- μ 10 -9 nano- n 10 -12 pico- p 1,230,000 1.23 x 10 6 0.00000123 1.23 x 10 -6 2 /360 π o Density: mass/volume = m/v ρ Vectors Vectors – magnitude and direction Scalars – magnitude ONLY Magnitude is the distance between the original and final position. Direction can be – or + a x = a cos θ a y = a sin θ a 2 = a x 2 + a y 2 tan θ = a y /a x Scalar product i.e. dot product a · b = | a | | b | cos θ Vector product i.e. cross product | a x b | = | a | | b | sin θ Kinematics Velocity (m/s) Average velocity is the slope of an x vs t graph v avg = ( x)/( t) Δ Δ Average speed s avg = (total distance)/( t) Δ The velocity is the first derivative of its position x(t) with respect to time. Acceleration (m/ s 2 ) The acceleration is the second derivative of its position x(t) with respect to time. a avg = ( v)/( t) Δ Δ Equations for motion with constant acceleration Equation missing v = v 0 + at x – x 0 x – x 0 = v 0 t + ½at 2 v v 2 = v 0 2 + 2a(x – x 0 ) t x –x 0 = ½(v 0 + v)t a x – x 0 = vt – 1/2at 2 v 0 Free-Fall acceleration is –g = 9.81 m/s 2 but its magnitude is g = 9.81 m/s 2 Projectile Motion v ox =v o cos θ (zero acceleration) v oy =v o sin θ -gt (contstant acceleration) x – x 0 = (v o cos θ )t y – y 0 = (v o sin θ )t -1/2gt 2 v y 2 = (v o sin θ ) 2 – 2g(y – y 0 ) when v y = 0 the projectile reached maximum height. Projectile’s path (trajectory) Y = (tan θ )x – [(gx 2 )/(2 (v o cos θ ) 2 )] Horizontal Range (R) at the same launching and final height R = (v 0 2 /g) * sin 2 θ 45 deg provides the max range

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## This note was uploaded on 11/15/2010 for the course PHYSICS 1322 taught by Professor Michaelgorman during the Spring '10 term at University of Houston.

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CHEATSHEET - M easurement 1012 109 106 103 102 101 Basic...

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