P112Prelim1SolS06

# P112Prelim1SolS06 - Solutions Prelim 1 Spring 2006 1(a A(b...

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s d Solutions: Prelim 1, Spring 2006 1.(a) A; (b) A; (c) A 2. F=0.18N 3.(a) C; (b) C; (c) B 4. (a) W=5 N; (b) W=40 N 5. (a) D; (b) E; (c) C; (d) F; (e) A 6. (a) F=500N; (b) v=30 m/s; (c) v=15 m/s 7. (a) so (b) t BI > t IB : You have a headwind from Boston to Ithaca. (c) v ave < v 737 : Your land speed is less than your airspeed because you have to fly into the wind to go straight to your destination. (d) v ave < v 737 : You have a headwind for more time than you have a tailwind. 8. An equivalent configuration is shown below with
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Unformatted text preview: (1/k 1 )=(1/k)+1/(2k)=3/(2k) (1/k 2 )=(1/k)+(1/k)=2/k Combining these in parallel gives K eff =k 1 +k 2 =(5/6)k=83.3 N/m 9. (a) (b) (c) Solving for t, and substituting into the equation for x gives 10.(a) (b) None (c) (d) Use coordinate system parallel and perp to wedge, We then use that the two tensions are equal and the two accelerations are equal, to find θ φ v 737 v wind V ground F F K 1 K 2 T W Before W After...
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• Summer '07
• LECLAIR,A
• Equivalence relation, Coordinate system, Runway, Tailwind

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