t31b - Physics 101 FORM 0 [1] A child is riding a...

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Unformatted text preview: Physics 101 FORM 0 [1] A child is riding a merrymgo-round which has an instantaneous angular velocity of 1.25 rad/s and an angular acceleration of 0.745 rad / s2. rThe child is standing 4.05 In from the center of the merry—go—ronnd. What is the magnitude of the acceleration of the child? [A] 2.58 111/52 [B] 3.45 m/s2 [0] 7.27 31/52 @805 m/s2 [E] 13.4 111/52 [2] A solid cylinder (I m: M R2 /2) is rolling without slipping. What fraction of its kinetic energy is rot onal? [A] 1/4 1/3 [C] 1/2 [D] 2/3 3/4; [3] A 3.50~g bullet, traveling at 250 m/s, hits a hinged, light board at right angles, at a distance of 55.0 cm from its hinges, and emerges onthe other side of the board with a speed of 150'111/5. rI‘he board is initially stationary but is free to rotate about its hinges and has a moment of inertia of 0.275 lag-m2. What is the angular speed of the board after the bullet emerges? . [A] 0.0212 rad/s @ 0.700 rad/s [C] 3.05 rad/s [D] 6.10 rad/s [E] 22.6 rad/s [4] The mass of Muto is 1.29 X 1022 kg and its radius is 1.15 X 106 no. What is the value of g . the surface of Pluto? 0.651 m/s2 [B] 1.05 m/s2 [0] 3.72 m/s2 [D] 6.23 Ira/$2 [s] 9.81 m/s2 [5] In simple harmonic motion, the speed is greatest at that point in the cycle when [A] the magnitude of the acceleration is a maximum. [B] the displacement is a maximum. ©the magnitude of the acceleration is a minimum. D] the potential energy is a maximum. [E] the kinetic energy is a minimum. [6] When a mass of 0.350 kg is attached to a vertical spring and lowered slowly, the spring stretches 12.0 cm. The mass is now displaced from its equilibrium position and undergoes simple harmonic oscillations. What is the period of the oscillations? [A] 0.0769 s [B] 0.286 s [o] 0.483 8 0.695 s [E] 1.44 s [7] A thin hoop is supported in a vertical plane by a nail. What should the radius of the hoop be in order for it to have a period of oscillation of 1.00 s? The moment of inertia of a hoop of mass M and radius R about a po' t on its rim is ZMRz. [A] 00154 m [B] 0.0621 111' @0124. m [D] 0.18% m [E] 0.248 m [8] The vertical displacement of astring is given by y(m,t) m (6.00 min) cos — 3‘3”). What is the speed of propagation of the were? ‘ [A] 0.225 m/s [B] 0.422 m/s [0] 0.750 m/s [D] 0.962 rn/s @111 m/s Physics 101 FORM 0 [9] A cubical box 25.00 cm or}. each side is immersed in a fluid. The pressure at the top surface of the box is 109.4 kPa and the pressure on the bottom surface is 112 kPa. What is the ‘ density of the fluid? {A} 1000 kg/m3 [B] 1030 kg/m3 @1060 kg/m3 [D] 1090 kg/m3 [E] 1120 lag/1113 > [10] A person who weighs 550 N empties her lungs as much as possible and is theh completely immersed in water while suspended from a harness. Her apparent weight is now 21.2 N. What is her density? [A] 56.1 kg/m3 [B] 255 Egg/1113 [C] 960 kg/m3 {D} 1030 kg/m3 @1040 kg/m3 ‘ [11] To determine the force a. person’s triceps muscle can exert, a doctor uses the procedure shown in the figure, where the patient pushes down with the palm of his hand on a force meter. Given that the weight of the lower arm is M g = 16.5 N, and that the force meter reads F = 93.3 N, find: (a) the force Ftp exerted vertically upward by the triceps, (b) the magnitude and the direction of the force FA exerted on the tower arm at the axis (see figure). 91:91}? W‘: ff“ restart-- Mflw: Iszsw \ f5: 01;:- Hiéo Cw 2: Nugzr W‘- (Ct) 1:13? I 3:2?“ figfl'Mt 2""— F33 (1 t Fig” Mgr)“; Humerus [ Db Triceps FY M Ft3“’aah F?" H 61 g r 1533 132,33.»- its ib. .q .q W; T ” A}? Physics 101 FORM 0 [12] On Apollo Moon missions, the lunar module would blast off fiom the moon’s surface and dock with the command module in lunar orbit. After docking, the lunar module“ would be jettisoned (for a short time) and allowed to crash back onto the lunar surface. Assume that the module is jettisoned from an orbit 130 km above the lunar surface, moving with a speed of1510 m/s. ' (a) Find the kinetic energy and the gravitational potential energy, per kg of the moduie, right after it is jettisoned. What is its total mechanical energy, per kg? (The mass and the radius of the moon are: MM m 7’35 x 102‘2 kg, RM m 1MB km). (b) What is the gravitational potential energy, per kg of the module, when it hits the lunar surface? (c) Determine the impact speed of the lunar module on the moon’s surface. - If U“: 7- lSlOm; FEE 1' flm‘tk *1 (040+ {gallows L g?‘ min“! szfli1=’3?4"mma v S (60 Km (A; E 37%" r J "m “2 )“fimz “Kt”- ? twigs-i 3“"45101'5» («ill Naomi; w» mtg, L)_,2,U:m 2’ Eye-wit Iva) ML _ L (s. Gris?“ )3“ GHM “6 H i0_{i?«3§*1’022' j" l“ 'M 0‘ 3-: WNW L3, _ H. %x~2,gz»€ot~Z—( 15”“) ‘ , (As .w ' “5.3:...” , @461”; ET? "filmil‘iqo algal) m Incl,” 1L”? {01-911, g2 Dies) at... - . (L) M '2 (M 45'; {O‘HMFJOZA ' leis-.. '“w-wflw - 9;; J3 1M.“ .12); °" 1,?J1,{0£ 7:5“- Q'gl, i0 bagegig f0 mgvmfisiifit) :: attmuzssytot it , Physics 101 FORM 0 FORMULAE SHEET _, - - ' —.‘ —» Relatwe veloc1ty. em m um + "930, pm m _UAB —+ Motion at constant acceleration: me E m(t m 0), '00 E *(t m 0), 17m 2 W '03 m 110,; + amt, a: = $0 + fist, a: = 339 + watt + %amt2, 1):: '03: »i- 2%(33 — 339) Free fall and projectile motion: (1' = g(—;&), g m 9.81 m/sz, R = iezsi—ngfz—m Newton’s Laws: fine»; 2 2 F 2 mi, PEA = MFAB Weight: W m mg(—~3}), friction: ‘kinetic: fk m pkN, static: f3 3 ,uSN Springs: F; 2 wake, circular motion: Eel, = %:(——7‘), £1, m m:2(—'F) Work and Energy: W 3 Fun! a moose, K e 2W _—_ AK, P m 311 m file 2 lfVc x: —AU, Ug = mgy, Us Eg—z, E = K + U, Wnc = $13 ll Impuise ancl Momentum: 35': m5, f: FAt m A}? If the net extemal impulse is zero then fitot =_ 215': const TWO particle collision: mlé’li + mzfi’zi m 171.3611: + mg'é’gf, completely inelastic: 6'1; 2 1333: Elastic collision: Kf = Ki, in one dimension: '02; w elf = vh- — 112i . w —v M -» ' __ A Center of mass. mam _ Z mgwi/ 27m, pcm «m ptot, rocket. F ~— 0—K? Rotational Motion: 271' rad : 1 rev m 360°, 3 = r9, (3 z 33%, w w 27”, E: 2 %—:i wmwg+at, 9290+(31'2fl, 9=HD+wgt-i-%at2, w2=wfi+2a(9~—Hn) wt 2 1w, 03,:P : 7102, at at: To, 3 m 5c}, + fit, roiling motion: *0 m m, e = at 2 Rotational kinetic energy: Kmt 2 1—3-3, I = Z miff, K = 32m + W Torque: 'r = 7'in : TFHL s: rFsin t9, Twat m 21* : Ia a %—%, work: Wmt = 7A9 Angular momentum: L 2 T3) Sing 2 ho, if Them,“ = 0 then Ltog = const - "‘ A __ 2 2 2 Grawty: F12 2: 9%(WT12), Q m 6.67 X 10 111E; , g = gfi—Aéf», g»; = 3734, U3 m —G“;M Oscillations: m = Acoswt v : —Aw sinwt a = —Aw2 coswt xx ~w2$ ? I f: e: 5;, w: 5;, Tm27r lg», T=27r\/:~I9;, T='21r gm!” Waves: 6 = 3f = A)”, y(ar;,t) = Acos (gig—E $ 3-5;), string: '0 = “gag Fluids: p m e, pw «£1000 51%, P m EAL, 195,: m 1 am m 101 kPa, Pg = P m Pat P2 = P1 + pgh, E‘- = 53* F}, = pfiléubg, 3—: = At: = const, P + g + pgy m coost ...
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t31b - Physics 101 FORM 0 [1] A child is riding a...

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