Texas A&M - CVEN - 221

Texas A&M - CVEN - 221

Texas A&M - CVEN - 221

Texas A&M - CVEN - 221

Texas A&M - CVEN - 221

Texas A&M - CVEN - 221

Texas A&M - CVEN - 221

Texas A&M - CVEN - 221

Texas A&M - CVEN - 221

Texas A&M - CVEN - 221

Texas A&M - CVEN - 221

Texas A&M - CVEN - 221

Texas A&M - ENGR - 221

TEXAS A&M UNIVERSITY ENGR221/502 Introduction to Engineering Mechanics Fall 2005Final ExamDecember 9, 3-5 pm, 2005Closed books and closed notes. Needed formula sheets are provided in the exam. This exam has 5 questions for a total of 100 points.

Texas A&M - PHYS - 208

Chapter 2CHAPTER 2 - Describing Motion: Kinematics in One Dimension 1. We find the time from average speed = d/t; 15 km/h = (75 km)/t , which givest = 5.0 h. 88 km/h.2. 3.We find the average speed from average speed = d/t = (280 km)/(3.2 h) =

Texas A&M - PHYS - 208

Chapter 3CHAPTER 3 - Kinematics in Two Dimensions; Vectors 1. We choose the west and south coordinate system shown. D 2x For the components of the resultant we have W RW = D1 + D2 cos 45 D 2y D 2 = (200 km) + (80 km) cos 45 = 257 km; RS = 0 + D2 =

Texas A&M - PHYS - 208

Chapter 41CHAPTER 4 - Dynamics: Newton's Laws of Motion 1. We convert the units: # lb = (0.25 lb)(4.45 N/lb) ~ 1 N. If we select the bike and rider as the object, we apply Newton's second law to find the mass: ?F = ma; 255 N = m(2.20 m/s2), which

Texas A&M - PHYS - 208

Chapter 5CHAPTER 5 - Further Applications of Newton's Laws 1. The friction is kinetic, so Ffr = kFN. With constant velocity, the acceleration is zero. Using the force diagram for the crate, we can write ?F = ma: x-component: F kFN = 0; y-component

Texas A&M - PHYS - 208

Chapter 6CHAPTER 6 - Gravitation and Newton's Synthesis 1. Because the spacecraft is 2 Earth radii above the surface, it is 3 Earth radii from the center. The gravitational force on the spacecraft is F = GMEM/r2 = (6.67 1011 N m2/kg2)(5.98 1024 kg

Texas A&M - PHYS - 208

Chapter 7CHAPTER 7 - Work and Energy 1. 2. 3. The displacement is in the direction of the gravitational force, thus W = Fh cos 0 = mgh = (250 kg)(9.80 m/s2)(2.80 m) = 6.86 103 J. The displacement is opposite to the direction of the retarding force,

Texas A&M - PHYS - 208

Chapter 8CHAPTER 8 - Conservation of Energy 1. The potential energy of the spring is zero when the spring is not compressed (x= 0). For the stored potential energy, we have U = ! kxf2 0; 35.0 J = ! (82.0 N/m)xf2 0, which gives xf = 2. 3. 4. For t

Texas A&M - PHYS - 208

Chapter 9CHAPTER 9 - Linear Momentum and Collisions 1. We find the force on the expelled gases from F = ?p/?t = (?m/?t)v = (1200 kg/s)(50,000 m/s) = 6.0 107 N. An equal, but opposite, force will be exerted on the rocket: 6.0 107 N, up. For the mome

Texas A&M - PHYS - 208

Chapter 10 1CHAPTER 10 - Rotational Motion About a Fixed Axis 1. (a) (b) (c) (d) (e) 30 = (30)( rad/180) = p/6 rad = 0.524 rad; 57 = (57)(p rad/180) = 19p/60 = 0.995 rad; 90 = (90)(p rad/180) = p/2 = 1.571 rad; 360 = (360)(p rad/180) = 2p = 6.283 r

Texas A&M - PHYS - 208

Chapter 11CHAPTER 11 General Rotation 1.z (a) For the magnitudes of the vector products we have i i = i i sin 0 = 0; k j j = j j sin 0 = 0; j k k = k k sin 0 = 0. y (b) For the magnitudes of the vector products we have i j = i j sin 90 = (1)(1)(1

Texas A&M - PHYS - 208

Chapter 12CHAPTER 12 Static Equilibrium; Elasticity and Fracture 1. From the force diagram for the sapling we can write ?Fx = F1 F2 sin 20 F3 cos = 0; 380 N (255 N) sin 20 F3 cos = 0, or F3 cos = 293 N. ?Fy = F2 cos 20 F3 sin = 0; F3 sin

Texas A&M - MEEN - 221

Texas A&M - MEEN - 221

Texas A&M - MEEN - 221

Texas A&M - MEEN - 221

Texas A&M - MEEN - 221

Texas A&M - CHEN - 323

PROBLEM 1.1 KNOWN: Heat rate, q, through one-dimensional wall of area A, thickness L, thermalconductivity k and inner temperature, T1. FIND: The outer temperature of the wall, T2. SCHEMATIC:ASSUMPTIONS: (1) One-dimensional conduction in the x-dire

Texas A&M - CHEN - 323

PROBLEM 1.41KNOWN: Hot plate-type wafer thermal processing tool based upon heat transfer modes by conduction through gas within the gap and by radiation exchange across gap. FIND: (a) Radiative and conduction heat fluxes across gap for specified hot

Texas A&M - ELEN - 214

Circuit Variables1Assessment ProblemsAP 1.1 To solve this problem we use a product of ratios to change units from dollars/year to dollars/millisecond. We begin by expressing $10 billion in scientific notation: $100 billion = $100 109 Now we det

Texas A&M - ELEN - 214

Circuit Elements2Assessment ProblemsAP 2.1[a] To find vg write a KVL equation clockwise around the left loop, starting below the dependent source: ib ib so vg = + vg = 0 4 4 To find ib write a KCL equation at the upper right node. Sum the curr

Texas A&M - ELEN - 214

Simple Resistive Circuits3Assessment ProblemsAP 3.1Start from the right hand side of the circuit and make series and parallel combinations of the resistors until one equivalent resistor remains. Begin by combining the 6 resistor and the 10 r

Texas A&M - ELEN - 214

4 Techniques of Circuit AnalysisAssessment ProblemsAP 4.1 [a] Redraw the circuit, labeling the reference node and the two node voltages:The two node voltage equations are v1 v1 - v2 v1 + + = 0 -15 + 60 15 5 v2 v2 - v1 5+ + = 0 2 5 Place these equ

Texas A&M - ELEN - 214

The Operational Amplifier5Assessment ProblemsAP 5.1 [a] This is an inverting amplifier, so vo = (-Rf /Ri )vs = (-80/16)vs , vs ( V) 0.4 2.0 so vo = -5vs3.5 -0.6 -1.6 -2.4vo ( V) -2.0 -10.0 -15.0 3.0 8.0 10.0 Two of the vs values, 3.5 V and -

Texas A&M - ELEN - 214

6 Inductance, Capacitance, and MutualInductanceAssessment ProblemsAP 6.1 [a] ig = 8e-300t - 8e-1200t A v=L dig = -9.6e-300t + 38.4e-1200t V, dt 38.4e-1200t = 9.6e-300t t > 0+v(0+ ) = -9.6 + 38.4 = 28.8 V [b] v = 0 when or t = (ln 4)/900 = 1.54 m

Texas A&M - ELEN - 214

7 Response of First-Order RL and RCCircuitsAssessment ProblemsAP 7.1 [a] The circuit for t < 0 is shown below. Note that the inductor behaves like a short circuit, effectively eliminating the 2 resistor from the circuit.First combine the 30 an

Texas A&M - ELEN - 214

8 Natural and Step Responses of RLCCircuitsAssessment ProblemsAP 8.1 [a] 1 1 = , (2RC)2 LC 1 , [b] = 5000 = 2RC s1,2 = -5000 [c] 1 = 20,000, LC therefore C = 500 nF C = 1 Ftherefore 25 106 - therefore(103 )(106 ) = (-5000 j5000) rad/s 20

Texas A&M - ELEN - 214

Sinusoidal Steady State Analysis9Assessment ProblemsAP 9.1 [a] V = 170/-40 V [b] 10 sin(1000t + 20 ) = 10 cos(1000t - 70 ) . I = 10/-70 A[c] I = 5/36.87 + 10/-53.13 = 4 + j3 + 6 - j8 = 10 - j5 = 11.18/-26.57 A [d] sin(20,000t + 30 ) = cos(20,0

Texas A&M - ELEN - 214

10 Sinusoidal Steady State PowerCalculationsAssessment ProblemsAP 10.1 [a] V = 100/- 45 V, I = 20/15 A Therefore 1 P = (100)(20) cos[-45 - (15)] = 500 W, 2 Q = 1000 sin -60 = -866.03 VAR, [b] V = 100/- 45 , I = 20/165 BA ABABBAP = 1000 cos(-

Texas A&M - ELEN - 214

11 Balanced Three-Phase CircuitsAssessment ProblemsAP 11.1 Make a sketch:We know VAN and wish to find VBC . To do this, write a KVL equation to find VAB , and use the known phase angle relationship between VAB and VBC to find VBC . VAB = VAN + VN

Texas A&M - ELEN - 214

12 Introduction to the Laplace TransformAssessment ProblemsAP 12.1 [a] cosh t = et + e-t 2 Therefore, 1 -(s-)t [e + e-(s+)t ]dt L{cosh t} = 2 0- = = [b] sinh t = 1 e-(s-)t 2 -(s - ) 1 2 0-+e-(s+)t -(s + ) = s2 0-1 1 + s- s+s - 2et - e

Texas A&M - ELEN - 214

13 The Laplace Transform in CircuitAnalysisAssessment ProblemsAP 13.1 [a] Y = 1 1 C[s2 + (1/RC)s + (1/LC) + + sC = R sL s 1 = 25 108 LC106 1 = = 80,000; RC (500)(0.025) Therefore Y = [b] -z1,225 10-9 (s2 + 80,000s + 25 108 ) s = -40,000 1

Texas A&M - ELEN - 214

14Introduction to Frequency-Selective CircuitsAssessment ProblemsAP 14.1 fc = 8 kHz, c = 1 ; RC c = 2fc = 16 krad/s R = 10 k;. C =1 1 = = 1.99 nF c R (16 103 )(104 )AP 14.2 [a] c = 2fc = 2(2000) = 4 krad/s L= 5000 R = = 0.40 H c 4000 4000

Texas A&M - ELEN - 214

Active Filter Circuits15Assessment ProblemsAP 15.1 H(s) = -(R2 /R1 )s s + (1/R1 C) R1 = 1 , . C = 1 F1 = 1 rad/s; R1 C R2 = 1, R1 .. R2 = R1 = 1 -s s+1Hprototype (s) =AP 15.2 H(s) =-20,000 -(1/R1 C) = s + (1/R2 C) s + 5000 C = 5 F1

Texas A&M - ELEN - 214

Fourier Series16Assessment ProblemsAP 16.1 av = 1 T 2 T2T /3 0Vm dt +1 TT 2T /3Vm 3T7 dt = Vm = 7 V 9 Vm cos k0 t dt 3ak = = bk = =2T /3 0Vm cos k0 t dt + sin 4k 3 =2T /34Vm 3k0 T 2 T2T /3 06 4k sin k 3T 2T /3Vm sin k

Texas A&M - ELEN - 214

The Fourier Transform17Assessment ProblemsAP 17.1 [a] F () = =0 - /2(-Ae-jt ) dt + /2 0Ae-jt dtA [2 - ej /2 - e-j /2 ] j ej /2 + e-j /2 2A 1- = j 2 -j2A [1 - cos ] = 2 [b] F () = AP 17.2 f (t) = = = = 1 2 0te-at e-jt dt = 4ejt d +

Texas A&M - PHYS - 218

44.2: The total energy of the positron is E K mc 2 5.00 MeV 0.511 MeV 5.51 MeV. We can calculate the speed of the positron from Eq. 37.38Emc 2 1v2 c2v c1mc 2 E210.511 MeV 5.51 MeV20.996.

Texas A&M - PHYS - 218

44.1:ma) Kmc 2311 1 v c2 210.1547mc 2149.109 10kg, so K1.27 10Jb) The total energy of each electron or positron is E K mc2 1.1547mc2 9.46 10 14 J. The total energy of the electron and positron is converted into the total energ

Texas A&M - PHYS - 218

44.4: a) hc Ehc m c 2h m c(6.626 10 34 J s) (207)(9.11 10 31 kg) (3.00 108 m s)1.17 10 14 m 0.0117 pm. In this case, the muons are created at rest (no kinetic energy). b) Shorter wavelengths would mean higher photon energy, and the muons wo

Texas A&M - PHYS - 218

44.5: a)mmm270 me207 me63 meE 63(0.511 MeV) 32 MeV. b) A positive muon has less mass than a positive pion, so if the decay from muon to pion was to happen, you could always find a frame where energy was not conserved. This cannot occur.

Texas A&M - PHYS - 218

44.3: Each photon gets half of the energy of the pion 1 1 1 E m c2 (270 me )c 2 (270)(0.511 MeV) 69 MeV 2 2 2 E (6.9 107 eV)(1.6 10 19 J eV) f 1.7 1022 Hz 34 h (6.63 10 J s)c f3.00 108 m s 1.7 1022 Hz1.8 10 14 m gamma ray.

Texas A&M - PHYS - 218

44.6: a) The energy will be the proton rest energy, 938.3 MeV, corresponding to a frequency of 2.27 10 23 Hz and a wavelength of 1.32 10 15 m. b) The energy of each photon will be 938.3 MeV 830 MeV 1768 MeV, with frequency 42.8 10 22 Hz and wavelengt

Texas A&M - PHYS - 218

44.7:E( m)c 2(400 kg400 kg )(3.00 108 m s) 27.20 1019 J.

Texas A&M - PHYS - 218

44.9:1 0n10 5B4 27 3Li4 2He4.002603u 11.018607 um( n m( Li7 31 010 5B) 1.008665 u 10.012937 u 11.021602u He) 7.016004 um 0.002995 u; (0.002995u) (931.5 MeV u ) 2.79 MeV The mass decreases so energy is released and the rea

Texas A&M - PHYS - 218

1 44.8: 4 He 9 Be 12 C 0 n 2 4 6 We take the masses for these reactants from Table 43.2, and use Eq. 43.23 Q (4.002603 u 9.012182 u 12.000000 u 1.008665 u) (931.5 MeV u )5.701 MeV. This is an exoergic reaction.

Texas A&M - PHYS - 218

44.10: a) The energy is so high that the total energy of each particle is half of the available energy, 50 GeV. b) Equation (44.11) is applicable, and Ea 226 MeV.

Texas A&M - PHYS - 218

44.11:a) BqB mBm 2mf q q2 (2.01 u) (1.66 10 27 kg u )(9.00 106 Hz) 1.60 10 19 C B 1.18 Tq 2 B 2 R 2 (1.60 1019 C) 2 (1.18 T) 2 (0.32 m) 2 b) K 5.47 1013 J 27 2m 2(2.01 u )(1.66 10 kg u ) 3.42 106 eV 3.42 MeV and v 2K 2(

Texas A&M - PHYS - 218

eB eBR 3.97 107 s. b) R 3.12 107 m s. c) For threem m figure precision, the relativistic form of the kinetic energy must be used, ( 1 )mc 2 eV ( 1 )mc 2 , so eV ( 1 )mc 2 , so V 5.11 106 V. e44.12: a) 2 f

Texas A&M - PHYS - 218

1000 103 MeV 1065.8, so v 938.3 MeV b) Nonrelativistic: eB 3.83 108 rad s . m Relativistic: eB 1 3.59 105 rad s . m 44.14: a) 0.999999559 . c