{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

205_test2_spring10

# 205_test2_spring10 - Dr Gundersen Signature Idnumber Phy...

This preview shows pages 1–3. Sign up to view the full content.

Dr. Gundersen Phy 205DJ Test 2 22 March 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 For constant acceleration: x = x 0 + v 0 t + 1 2 at 2 , v 2 = v 2 0 + 2 a ( x - x 0 ) , x = x 0 + 1 2 ( v 0 + v ) t, x = x 0 + vt - 1 2 at 2 v = v 0 + at, v av = 1 2 ( v 0 + v ) = v 0 + 1 2 at 2 , vector r ( t ) = vector r 0 + vector v 0 t + 1 2 vector a t 2 , vector v ( t ) = vector v 0 + vector a t For nonconstant acceleration: vector v = dvector r dt , vector v av = vector r 1 - vector r 0 t 1 - t 0 , x 1 - x 0 = integraldisplay t 1 t 0 v x ( t ) dt vector a = dvector v dt , vector a av = vector v 1 - vector v 0 t 1 - t 0 , v x 1 - v x 0 = integraldisplay t 1 t 0 a x ( t ) dt Physics 205DJ TEST 2 22 March 2010

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
Dr. Gundersen Phy 205DJ Test 2 22 March 2010 Forces: vector F net = mvector a , vector F BA = - vector F AB , | vector F g | = mg, 0 ≤ | vector f s | ≤ μ s F N , | vector f k | = μ k F N vector F = - mv 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 Uniform Circular Motion: vector r ( t ) = r (cos θ ˆ i + sin θ ˆ j ) = r ˆ r , vector v ( t ) = ωr ( - sin θ ˆ i + cos θ ˆ j ) = ωr ˆ θ, θ ( t ) = ωt + θ 0 vector a ( t ) = - ω 2 r (cos θ ˆ i + sin θ ˆ j ) = - ω 2 r ˆ r = - v 2 r ˆ r , ω = dt = 2 πf = 2
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

### Page1 / 12

205_test2_spring10 - Dr Gundersen Signature Idnumber Phy...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online