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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 freebody 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 “PC3 : )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'lQT Oqm(<_otg‘li7.UD) :9 O, giR‘QDv‘A 1.“ of] M lW‘ [ItJo ‘— 5 o C1— Ort‘lo a :; (Maya/X : 0.707% 19 S oﬁaﬁmOﬂmm d. Use the value of b and basic trigonometry to determine the angle 02. (Hint: Sin'1(0.121) = 7°) '
(_ o_ 11.1 ’ : Syhl (04:21
92 lg 3M 9; '3 "lEL‘  7 91 > [m ‘0 l e. Draw a freebody 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.
 Summer '11
 Ramos

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