test2_09FS - LAST NAME: STUDENT NR: PHYS 1010 6.0: CLASS...

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Unformatted text preview: LAST NAME: STUDENT NR: PHYS 1010 6.0: CLASS TEST 2 Time: 50 minutes; Calculators & formulae provided at the end = only aid; Total = 20 points. Emphasis is on problem solving out of chapters 4,5, and concepts from 6.1 FORMULA SHEET ￿ v (tf ) = v (ti ) + ti a(t) dt s(tf ) = s(ti ) + ttif v (t) dt vf = vi + a∆t sf = si + vi ∆t + 1 a∆t2 vf2 = vi2 + 2a∆s 2 ￿ tf g = 9.8 m/s2 f (t) = t f (t) = a = 0 F (t) = f (t) dt = at + C F (t) = anti-derivative = indefinite integral area under the curve f (t) between limits t1 and t2 : F (t2 ) − F (t1 ) x2 + px + q = 0 factored by: x1,2 = − p ± p4 − q 2 uniform circular m. ￿(t) = R(cos ω t ˆ + sin ω t ˆ); ￿ (t) = r i jv exp = exp; sin = cos; cos = − sin. m￿ = Fnet ; a￿ FG = Gm1 m2 ; r2 ￿ ￿ ￿ d [f (g (x))] dx df dt df dt = 1 F (t) = ￿ ￿ f (t) dt = t2 2 +C ￿ 2 = d￿ r dt df dg ; dg dx = ...; ￿ (t) = a ￿ ￿ d￿ v dt = .... (f g ) = f g + f g ￿ 2 g= GME 2; RE fs ≤ µs n; fk = µk n; fr = µr n; µr << µk < µs . FH = −k ∆x = −k (x − x0 ). ￿d ∼ −￿ ; linear: Fd = dv ; quadratic: Fd = 0.5ρAv 2 ; A = cross sectional area F v W = F ∆x = F (∆r) cos θ For F (x) the work is given as area under the Fx vs x curve. RE = 6370 km; G = 6.67 × 10−11 Nm2 ; ME = 6.0 × 1024 kg kg 1 ...
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This note was uploaded on 09/28/2010 for the course MATH 1025 taught by Professor Tammie during the Fall '10 term at York University.

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