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Unformatted text preview: April 15, 2011 Exam 2 Physics 1025 MDC—IAC Name K614) c. Calculate its maximum cross—radial velocity, U1. :1” — r:- u. A- .. _ (- L685. T Emu: "‘ “’ wr'lli 1 0L 1- 3 r .._—.=-.-u 5'; - feign“. (Liter) \ . d. Calculate its centripetal force, FE, at a height of0.1m. :3. Fe, : (4%)(15/75%3im is J 3) (30pts.) The two masses are each attached by a 1.0m long weightless string. The charge of mass m1 = 1.0"10‘5 C and that of mass m = 1.17"10'4 C, they repel. The mass m; = 1.7kg and the tension (T) on the cable holding m1 is 12(\/2)N. A free-body diagram of all of the forces acting on m has been drawn for you (you’re welcome). (Hint: Create x and y equilibrium vector equations. Cos45° = Sin45° = 1N (2) = 0.707) a. Solve for Fe,( and H and use the Pythagorean Theorem to calculate the Coulomb force (Ft) between the two masses. ( x ( R] Q S— , _. .1 .— 0“, “‘ - 7— _ CZ («0w TCon + my : mt; F9; '* “‘95) J ‘3 "2 “PC-3 : )7 N — [QM : 5k) m “WHEN—“ma I I . . f ‘ H X'Pﬁm : inc; :— T {Jim 1* Fax 3 Q1 - iQN) )1. \$95“) Fe. =' (W7 M L W i b. Use Coulomb’s Law to determine the displacement (r) between the two masses. (Hint: 1.17/13 = 0.09, (W181) = 0.9) _)(‘-?.I()—Xili)qrr:;3/c2)(l am" 5a)( 1. Who Q3) I a \ . IN“ .— April 15, 2011 Exam 2 Physics 1025 MDC—IAC Name ‘ 3M c. TanGC = Fey/F“ = 5/12, making the directional angle 'ofthe Coulomb force: E3C = 22.6” above or below the horizontal. Use the distance r above, and basic trigonometry to determine the length b. [Hint: You can calculate Ah of ml or use r, Cos(22.6°) = 0.92] Qt'lQ-T O-qm(<_otg‘li7-.UD) :9 O, giR‘QDv‘A 1.“ of] M lW‘ [It-Jo ‘— 5 o C1-— Ort‘lo a :; (Maya/X : 0.707% 19 S oﬁaﬁm-Oﬂmm d. Use the value of b and basic trigonometry to determine the angle 02. (Hint: Sin'1(0.121) = 7°) ' (_ o_ 11.1 ’ : Sy-h-l (04:21 92 lg 3M 9; '3 "-lEL‘ - 7 91 > [m ‘0 l e. Draw a free-body diagram for mass m;. f. Create two equations for the tension on the cable attached to mass m;, as you did for mass m1. Solve for T and set the equations Equal to one another. The only unknown should now be m2, solve for this mass. (Hint: the values of F0. and Fay have not changed from part a above. Sin(7°) = 0.12, Cos(7°) = 1.0) ...
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## This note was uploaded on 12/04/2011 for the course PHYSICS 1025 taught by Professor Ramos during the Summer '11 term at Miami Dade College, Miami.

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Pages from Exam3key - April 15, 2011 Exam 2 Physics 1025...

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