205_test3_spring10

# 205_test3_spring10 - Dr Gundersen Phy 205DJ Test 3 12 April...

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Unformatted text preview: Dr. Gundersen Phy 205DJ Test 3 12 April 2010 Signature: Name: 1 2 3 4 5 Idnumber: Do only four out of the five problems. The first problem consists of five multiple choice questions. If you do more only your FIRST four answered problems will be graded. Clearly cross out the page and the numbered box of the problem omitted. Do not write in the other boxes. TO GET PARTIAL CREDIT IN PROBLEMS 2 - 5 YOU MUST SHOW GOOD WORK. CHECK DISCUSSION SECTION ATTENDED: [ ] Dr. Gundersen 2O, 9:30 - 10:20 a.m. [ ] Dr. Nepomechie 2P, 11:00 - 11:50a.m. [ ] Dr. Alvarez 2Q, 12:30 - 1:20 p.m. [ ] Dr. Barnes 2R, 2:00 - 2:50 p.m. [ ] Mr. Perez-Veitia 2S, 3:30 - 4:20 p.m. Vector relations: vector a = a x ˆ i + a y ˆ j , a = | vector a | = radicalBig a 2 x + a 2 y , θ = tan- 1 a y a x , a x = a cos θ, a y = a sin θ vector b = b x ˆ i + b y ˆ j + b z ˆ k , b = | vector b | = radicalBig b 2 x + b 2 y + b 2 z , ˆ b = vector b /b, vector v AC = vector v AB + vector v BC vector a · vector b = | vector a || vector b | cos θ = a x b x + a y b y + a z b z vector a × vector 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 , | vector a × vector b | = | vector a || vector b || sin θ | For constant acceleration: x = x + v t + 1 2 at 2 , v 2 = v 2 + 2 a ( x- x ) , x = x + 1 2 ( v + v ) t, x = x + vt- 1 2 at 2 v = v + at, v av = 1 2 ( v + v ) = v + 1 2 at 2 , vector r ( t ) = vector r + vector v t + 1 2 vector a t 2 , vector v ( t ) = vector v + vector a t ω = ω + αt, θ = θ + ω t + 1 2 αt 2 , ω 2 = ω 2 + 2 α ( θ- θ ) For nonconstant acceleration: vector v = dvector r dt , vector v av = vector r 1- vector r t 1- t , x 1- x = integraldisplay t 1 t v x ( t ) dt, vector a = dvector v dt , vector a av = vector v 1- vector v t 1- t , v x 1- v x = integraldisplay t 1 t a x ( t ) dt Physics 205DJ TEST 3 12 April 2010 Dr. Gundersen Phy 205DJ Test 3 12 April 2010 Force, Momentum and Impulse: vector F net = mvector a = dvector p dt = d ( mvector v ) dt , vector F BA =- vector F AB , | vector F g | = mg, ≤ | vector f s | ≤ μ s F N , | vector f k | = μ k F N vector F =- mv 2 r ˆ r =- mω 2 r ˆ r , vector F s =- k vector d , vector F =- dU dx ˆ i- dU dy ˆ j- dU dz ˆ k , vector F =- dU ( r ) dr ˆ r vector p = mvector v , vector p 1 i + vector p 2 i = vector p 1 f + vector p 2 f , vector...
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## This note was uploaded on 01/08/2012 for the course PHYSICS 205 taught by Professor Galeazzi during the Fall '11 term at University of Miami.

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205_test3_spring10 - Dr Gundersen Phy 205DJ Test 3 12 April...

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