exam_1_eqsheet_Fall06 - t = 0, respectively. Freely falling...

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Some Equations for Exam 1 sin 30 = 1 2 =0 . 50; cos 30 = 3 / 2=0 . 87 sin 60 = 3 / 2=0 . 87; cos 60 = 1 2 =0 . 50 sin 45 = 2 / 2=0 . 71; cos 45 = 2 / 2=0 . 71 For vectors A = A x ˆ i + A y ˆ j + A z ˆ k and B = B x ˆ i + B y ˆ j + B z ˆ k : | A | = A = ± A 2 x + A 2 y + A 2 z ; A · B = AB cos θ = A x B x + A y B y + A z B z ; A × B =( A y B z - A z B y ) ˆ i +( A z B x - A x B z ) ˆ j +( A x B y - A y B x ) ˆ k ; | A × B | = AB sin θ . For a function of the form f ( x )= ax n , where n is an integer, df dx = nax n - 1 and ² f ( x ) dx = a n +1 x n +1 + C . If a particle’s position is represented by r = r ( t ), v = d r dt and a = d v dt = d 2 r dt 2 . Therefore, v x ( t )= ³ a x ( t ) dt + C and x ( t )= ³ v x ( t ) dt + C , and similarly for y - and z -components. For a constant acceleration a , r ( t )= r 0 + v 0 t + 1 2 a t 2 and v ( t )= v 0 + a t where r 0 and v 0 are the position and velocity at time
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Unformatted text preview: t = 0, respectively. Freely falling objects near the Earths surface accelerate downwards with g = 9 . 8 m/s 2 . For this exam , use the approximation that g = 10 m / s 2 . Newtons Second Law: F = m a . Static friction: f s s N ; kinetic friction: f k = k N (where N is the magnitude of the normal force)....
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This note was uploaded on 04/30/2008 for the course PHY 317k taught by Professor Kopp during the Fall '07 term at University of Texas at Austin.

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