* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
Download 1. (a) GPE = KE m gL = ½m
Survey
Document related concepts
Galvanometer wikipedia , lookup
Valve RF amplifier wikipedia , lookup
Oscilloscope history wikipedia , lookup
Spark-gap transmitter wikipedia , lookup
Josephson voltage standard wikipedia , lookup
Power electronics wikipedia , lookup
Schmitt trigger wikipedia , lookup
Integrating ADC wikipedia , lookup
Electrical ballast wikipedia , lookup
Operational amplifier wikipedia , lookup
Power MOSFET wikipedia , lookup
Resistive opto-isolator wikipedia , lookup
Voltage regulator wikipedia , lookup
Surge protector wikipedia , lookup
Opto-isolator wikipedia , lookup
Switched-mode power supply wikipedia , lookup
Current source wikipedia , lookup
Current mirror wikipedia , lookup
Transcript
Mechanics 1. (a) GPE1 = KE2 m1gL = ½m1v2 v = 2004 C 2gL (b) SF = FT - Fg = ma FT - m1g = m1(0 m/s2) FT = m1g (c) p1 + p2 = p1' + p2' m1v1 + m2v2 = m1v1' + m2v2' m1v1 + m2v2 = (m1 + m2)v' Inelastic Collision m1 2gL + 0 = ( m1 + m 2 )v' ö æ m1 ÷÷ 2gL v 2 ' = çç è m1 + m 2 ø KE sys 2 1 2 m1 v1 (d) ratio = = 1 = 2 KE sys ' m + m v' ( ) 1 2 2 ratio = 1 2 m1 ( ) 2 2gL ìïé üï ù m1 1 2gL m + m ( ) í ý 1 2 ê ú 2 ïþ îïë( m1 + m 2 ) û 2 m1 + m 2 m1 (e) From position A to position B the horizontal distance is L The system experiences projectile motion from B to D in which the initial velocity is all horizontal. x y v xo = 2gL v yo = 0 m / s x=? g = -9.8 m/s2 yo = L y=0m t=? y = yo + vyot + ½at2 0 m = L + (0 m/s)t + ½gt2 ½gt2 = L L t = 1 2g x = v xo t = x = 2L so the total horizontal distance from A to D is given by: xTotal = xA-B + xB-D = L + 2L xTotal = 3L ( æ 2gL × ç ç è ) L ö÷ 1 ÷ 2g ø Mechanics 2. (a) y = yo + vot + ½at2 0 = D + 0 + ½at2 a = 2004 C 2D t2 (b) i. D should be graphed as a function of t2 to best determine the acceleration of the block because the relationship between D and t is D a t2, so graphing D vs. t would give a parabola which would be difficult to analyze for acceleration. Graphing D vs. t2 would yield a straight line (direct proportion). The slope of this graph would determine the proportionality constant for the relation between D and t2 (D a t2). From the kinematic equation y = yo + vot + ½at2 which simplifies to D = ½at2 in this circumstance, the proportionality constant is one-half of the acceleration. Therefore, the slope of this graph (D vs. t2) would equal one-half of the acceleration. ii. 2 2.0 D (m) 1.5 1.0 0.5 0.0 0.0 0.5 1.0 t (s2) 1.5 2.0 2 iii. slope, m = Dy Dx = DD Dt 2 = (2.0 m - (1.91 s 2 0.00 m) - 0.00 s 2 ) = 1.05 m / s 2 , but m = ½a 1.05 m/s2 = ½a a = 2.10 m/s2 (c) SF = FT - Fg = ma FT - mg = ma FT = ma - mg = m(a - g) ST = Ia RFT = I a R a R a g gö æ ö 2æ I = mR 2 ç ÷ = mR ç1 - ÷ aø è a ø è Rm(a - g) = I (d) The weight of the string was ignored which would reduce the experimental value of the rotational inertia. As the string unravels, the mass of the falling weight would increase, thus, increasing the measured acceleration, thereby, reducing the experimental rotational inertia. Mechanics 2004 C 3. (a) dI = dmr2 = ldxrx2 I = ò dI 2L 3 -L 3 ò = l = x3 lx 2dx = l 3 2L 3 -L 3 3 éæ 8L3 éæ 2L ö 3 æ -L ö ù ç ÷ ú êçç êç ÷ M 3 3 è ø è ø êè 27 ú = I = lê 3 ú Lê 3 ê 3 ê ê ú ë û ë I = 1 9 ö ÷÷ ø - ( ) g = v = æ -L3 çç è 27 3 öù ÷÷ ú 3 3 3 ø ú = M é8L + L ù = M é9L ù ú L êë 81 81 úû L êë 81 úû ú û ML2 (b) GPE1 = RKE2 mgy1 = ½Iw2 1 1 Mg L = Iw 2 6 2 2 1æ1 1 2 öæ v ö Mg L = ç ML ÷ç ÷ 2è9 6 øè r ø 2 1 2 æ v ö gL = L çç 2 ÷÷ 3 è 3 Lø gL = m dm = L dx ( ) 1 2 L 3 v2 ( 23 L)2 1 2 æ 9 v2 L çç 2 3 è4 L ( ) 3 4 Note: Use the lowest point of center of mass (cm) as 0 height, therefore, y1, the initial height of the cm (located in the center of the rod), would be the distance the center of mass is from the pivot point. The cm is at 12 L while the pivot point is at 23 L. The distance between these two points,y1, is given by: y1 = 12 L - 13 L = 36 L - 26 L = 16 L Note: r is the distance from the pivot point to the bottom of the rod when it is in the vertical position. This distance is 23 L. ö ÷÷ ø gL (c) Period, T, for a physical pendulum is given by T = 2p I where D is the distance from the MgD pivot point to the cm. For this rod, D is 16 L. 1 ML I T = 2p = 2p 9 1 MgD Mg 6 L 2L 3g T = 2p D q q Fg Note: A simple pendulum consists of a mass (bob) suspended from a string. A physical pendulum consists of a rigid body suspended body suspended by some point other than its cm such as the rod in this problem. When the body is displaced from its equilibrium position it will oscillate. When the angle of the displacement is small, the physical pendulum's oscillation approximates simple harmonic motion just as a simple pendulum does. The torque created by the physical pendulum is given by: T = -rFsinq = Ia let r = D which represents the distance from the pivot d 2q point to the cm. For small angles, sin q » q -Dmgq = I 2 dt MgD 2p d 2q MgD and the period is given by T = . = q so w 2 = 2 I w I dt E&M 1. (a) i. ii. 2004 C + + - + + - + - - + - + - - + + + (b) _____ 1 Va _____ 3 Vb _____ 5 Vc _____ 1 Vd _____ 3 Ve (c) For a Gaussian surface, use a cylindrical shell with its center at the center of the infinite line charge for each of the following (i., ii., iii.). The electric field through the end faces of the cylindrical shell is 0. The only differences among i., ii., and iii. is the size of the Gaussian surface (shown as dotted lines below). r r ldl Q i. ò E × dA = = eo eo l/l E2prl/ = r2 eo E = 1 l 2pe o r r1 r r ll + rV ll + rAl Q ii. ò E × dA = = = eo eo eo E2prl/ = ( l/l + r pr 2 - pr1 2 eo )/l i. ( 2 1 l + r pr - pr1 E = 2pe o r 2 ) ii. r r ldl + rdV Q iii. ò E × dA = = eo eo E2prl/ = ( l/l + r pr2 - pr1 l + 1 E = 2pe o E = 2 2 eo (pr 1 l + lC 2pe o r 2 )/l lC 2 - pr1 2 r ) iii. Note: r = (pr 2 2 - pr1 2 ) Q = V (pr 2 Q2 2 - pr1 2 )l = (pr 2 lC 2 - pr1 2 ) because l C = Q l E&M 2. (a) 20 V (b) 8V 2004 C Voltage across the capacitor, C, is zero initially. The branch with the capacitor acts as a short circuit since charges will flow directly onto the top plate unimpeded (there is no charge, initially, on this plate so no repulsion to charges flowing onto this plate) and away from the bottom plate. Since the sum ov the voltages in any loop must be zero (Kirchoff's loop rule), there must be a voltage drop of 20 V (equivalent to EMF of battery) across R 2. A long time after the switch is closed, the voltage across the C is 12 V which is the voltage across R1 (parallel). Since the voltage drops in the circuit must add to the EMF of the battery (Kirchoff's loop rule), the voltage across R2 would be 8 V (20 V - 12 V). (c) A long time after the switch is closed, when the capacitor is fully charged, the branch with the capacitor acts as an open circuit, so no current flows into that branch. At this time, the current through R 2 is the same as that through R1. Since the voltage across R1 is 12 V, the current can be determined from Ohm's V 12 V law: I = = = 8.0 x 10 -4 A = 0.80 mA. Since this is the same current through R2 R 15,000 W which has a voltage of 8 V across it, the value of the resistance R 2 can be determined from Ohm's law. V2 = IR2 8 V = (8 x 10-4 A)R2 R2 = 10,000 W = 10 kW (d) E = ½QV, but Q = CV, so E = ½CV2 E = ½(20 mF)(12 V)2 E = 1.44 x 10-3 J = 1.44 mJ (e) 2.0 Current in R2 (mA) 1.5 1.0 0.5 0 0 5 10 Time (s) (f) Initially, with no charge on the capacitor, the branch with the capacitor acts as a short circuit with no resistance. So current through R2 can be determined by using Ohm's law. V 20 V Io = = R 10,000 W I o = 2.0 x 10 -3 A I o = 2.0 mA It then would follow an exponential decay with a time constant of 1.5 s (the same as the voltage graph) 15 to 0.80 mA (when the capacitor branch acts as an open circuit). _____ Greater than _____ Less than _____ x The same as The energy stored depends on the physical nature of the capacitor and the voltage across the capacitor which is the same as the battery after a long time. Decreasing the resistance of R 2 will only reduce the time needed to fully charge this capacitor and, thereby, reach full energy storage. E&M 2004 C mo I dr 2p r l l m I mo Iù m æ1 mo æ 4 mo æ 3 ö 1ö 1ö o B = ò dr = = o Iç Iç Iç ÷ ÷ = ÷ = ú 4 l 2p r 2p r û 4 l 2p è l 4l ø 2p è 4l 4l ø 2p è 4l ø é m æ 3 öù f B = BAcosq = ê o Iç ÷ú 12l 2 cos(0) ë 2 p è 4 l øû 3. (a) db = ( fB = ) mo I(9l ) 2p (b) _____ x Counterclockwise _____ Clockwise The current is decreasing exponentially with time causing the flux within the rectangular loop which is out of the page to decrease with time. Lenz's law states that the induced current will be in a direction such that the magnetic field created by this induced current will oppose the change. Since the current in the long straight wire is decreasing (exponentially) resulting in a decrease in flux out of the page, the induced current will travel counterclockwise producing a magnetic field out of the page to oppose the decrease caused by the decreasing current in the long straight wire. ( ) d I oe -kt dI ( t ) (c) = = I oe -kt (-k ) = - kI oe -kt dt dt df B m m = o -kI oe -kt (9l ) = (-9kl ) o I oe -kt e = dt 2p 2p mo -kt (-9kl ) I oe e 2p I = = R R ( ) æ -9kl ö m o -kt I = ç ÷ I oe è R ø 2p 2 æ 81m o 2 k 2 l 2 ö 2 -2kt æ 81k 2 l 2 öæ m o ö 2 -2kt éæ -9kl ö m o -kt ù ÷I o e ÷÷ç ÷ I o e (d) P = I R = êç = çç ÷ I oe ú R = çç 2 ÷ 2 R 2 R p p 4 R p è ø è ø ë û è ø ø è æ 81m o 2 k 2 l 2 ö 2 -2kt dE ÷I o e dt dP = so dE = dP × dt = çç 2 ÷ dt è 4p R ø ¥ 2 2 2 ¥ æ 81m k l ö 2 æ 81m o 2 kl 2 ö 2 -2kt ù æ 81m o 2 kl 2 ö 2 -2k( ¥ ) -2k 0 -2kt o ç ÷ ç ÷I o e ÷ e dt = E = ò çç I e = - e () I ú o o 2 2 2 ç ÷ ç ÷ ÷ 0 úû 0 è 8p R ø è 8p R ø è 4p R ø æ 81m o 2 kl 2 ö 2 ÷I o (0 - 1) E = çç 2 ÷ è 8p R ø 2 2 [ æ -81m o 2 kl 2 E = çç 2 è 8p R ö 2 ÷I o ÷ ø ]