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54 Pages

### ism_chapter_40

Course: PHYSICS 11, Spring 2007
School: Lehigh
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Word Count: 9662

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40 Chapter Nuclear Physics Conceptual Problems 1 Determine the Concept Two or more nuclides with the same atomic number Z but different N and A numbers are called isotopes. (a) Two other isotopes of 14N are: (b) Two other isotopes of 56Fe are: (c) Two other isotopes of 118Sn are: 15 N, 16N Fe, 55Fe Fe, 55Fe 54 54 2 Determine the Concept The parent of that series, 237Np, has a half-life of 2106 y that is much...

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Lehigh - PHYSICS - 11
Chapter 36 AtomsConceptual Problems*1 Determine the Concept Examination of Figure 35-4 indicates that as n increases, the spacing of adjacent energy levels decreases. 2 Picture the Problem The energy of an atom of atomic number Z, with exactly on
Lehigh - PHYS - 11 and 21
4.1: a) For the magnitude of the sum to be the sum of the magnitudes, the forces must be parallel, and the angle between them is zero. b) The forces form the sides of a right isosceles triangle, and the angle between them is 90 . Alternatively, the l
Lehigh - MECH - 003
COSMOS: Complete Online Solutions Manual Organization SystemChapter 6, Solution 52.FBD Truss:FX = 0: By symmetry,A x + 4 kips - 4 kips = 0 Ax = 0 Ay = Ny = 0FBD Section ABDC:M D = 0:(18 ft )( 4 kips ) - ( 9 ft )4 FCE = 0 5FCE = 10.00
Lehigh - MECH - 003
COSMOS: Complete Online Solutions Manual Organization SystemChapter 9, Solution 32.HaveA = A1 - A2 - A3= (12 )( 8 ) - ( 5 )( 4 ) - ( 2 )( 6 ) in 2 = 64 in 2I x = ( I x )1 - ( I x )2 - ( I x )3Have 1 3 3 3 1 1 = (12 )( 8 ) - ( 5 )(
Lehigh - MECH - 003
COSMOS: Complete Online Solutions Manual Organization SystemChapter 7, Solution 25.FBD Rod:Fy = 0:Ay - W = 0Ay = WM B = 0:r ( Ax - W ) +2rW =02 A x = 1 - W FBD AJ:W = 2 W = W 2sinr =r2 = 2r sin 2 2M J = 0:2
Lehigh - PHYS - 11 and 21
6-1Chapter 6 THE SECOND LAW OF THERMODYNAMICSThe Second Law of Thermodynamics and Thermal Energy Reservoirs6-1C Water is not a fuel; thus the claim is false.6-2C Transferring 5 kWh of heat to an electric resistance wire and producing 5 kWh of
Lehigh - PHYS - 11 and 21
36.1:y1x a x ay1 a x x y1(1.35103m) (7.50 2.00 m104m)5.06107m.36.2:y1a(0.600 m) (5.46 10.2 10310 m7m)3.21105m.36.3: The angle to the first dark fringe is simply:arctan a= arctan633 0.24
Lehigh - PHYS - 11 and 21
11.1:Take the origin to be at the center of the small ball; then,x cm (1 . 00 kg)(0) ( 2 . 00 kg )( 0 . 580 m ) 3 . 00 kg 0 . 387 mfrom the center of the small ball.11.2:The calculation of Exercise 11.1 becomesx cm (1 . 00 kg )( 0 ) (1 . 50
Lehigh - PHYS - 11 and 21
Note: to obtain the numerical results given in this chapter, the following numerical values of certain physical quantities have been used;G 6 . 673 1011N m2kg , g29 . 80 m s and m E25 . 971024kg.Use of other tabulated values f
Lehigh - PHYS - 11 and 21
6.1: a) ( 2 .40 N ) (1 .5 m ) 3 .60 J b) ( 0 .600 N )(1 .50 m ) c) 3 . 60 J 0 .720 J 2 .70 J .0 . 900 J6.2: a) &quot;Pulling slowly&quot; can be taken to mean that the bucket rises at constant speed, so the tension in the rope may be taken to be the bucket
Lehigh - MECH - 003
COSMOS: Complete Online Solutions Manual Organization SystemChapter 6, Solution 74.Structure (a): No. of membersm = 12 n=8r =4No. of joints No. of react. comps.m + r = 16 = 2n unks = eqnsFBD of EH:M H = 0 FDE ; Fx = 0 FGH ; Fy = 0 H
Lehigh - MECH - 003
COSMOS: Complete Online Solutions Manual Organization SystemChapter 5, Solution 84.Free-Body Diagram:Force from water pressure:1 ApB where A is the rectangular cross sectional area through line BD, and pB is the pressure at 2 point B. Thus 1 1
Lehigh - MECH - 003
COSMOS: Complete Online Solutions Manual Organization SystemChapter 2, Solution 53.From Similar Triangles we have:L2 - ( 2.5 m ) = ( 8 - L ) - ( 5.45 m )2 2 2- 6.25 = 64 - 16 L - 29.7025or Andcos =L = 2.5342 m5.45 m 8 m - 2.5342 mor
Lehigh - MECH - 003
COSMOS: Complete Online Solutions Manual Organization SystemChapter 4, Solution 98.Free-Body Diagram:Fx = 0:Dx = 0M D = 0:( - 7 in.) i C + ( 2 in.) i + ( 3 in.) k ( 530 lb ) j + ( - 192 lb ) k + ( - 3 in.) i + ( 6 in.) j ( 265
Lehigh - MECH - 003
COSMOS: Complete Online Solutions Manual Organization SystemChapter 2, Solution 69.Free-Body Diagram of Pulley and Crate(b)Fy = 0: 3T - ( 280 kg ) 9.81 m/s 2 = 0()T =1 ( 2746.8 N ) 3T = 916 N(d)Fy = 0: 4T - ( 280 kg ) 9.81 m/s 2 =
Lehigh - MECH - 003
Lehigh - MECH - 003
COSMOS: Complete Online Solutions Manual Organization SystemChapter 2, Solution 39.We have:Rx = Fx = - 84 12 3 TBC + (156 lb ) - (100 lb ) 116 13 5or andRx = -0.72414TBC + 84 lbR y = Fy =80 5 4 TBC - (156 lb ) - (100 lb ) 116 13 5Ry =
Lehigh - MECH - 003
COSMOS: Complete Online Solutions Manual Organization SystemChapter 3, Solution 72.Based on M = M1 + M 2M1 = (18 N m ) k M 2 = ( 7.5 N m ) i M = ( 7.5 N m ) i + (18 N m ) kandM =( 7.5 N m )2 + (18 N m )2or M = 19.50 N m= 19.5 N
Lehigh - MECH - 003
COSMOS: Complete Online Solutions Manual Organization SystemChapter 7, Solution 43.(a) FBD Beam:Fy = 0:(8 m ) w - 2 ( 6 kN ) - ( 4 m )( 5 kN/m ) = 0w = 4 kN/mAlong AC:Fy = 0:( 4 kN/m ) x - V=0V = ( 4 kN/m ) x M J = 0: M - x ( 4 kN/
Lehigh - MECH - 003
COSMOS: Complete Online Solutions Manual Organization SystemChapter 7, Solution 21.(a)FBD Rod:Fx = 0: M D = 0:Ax = 0 Ay = P 2aP - 2aAy = 0FBD AJ:Fx = 0: V = 0 Fy = 0:P -F =0 2F=P 2M J = 0: M = 0(b)FBD Rod:M A = 0 4 3 2a
Lehigh - PHYS - 11 and 21
40.1: a) E nE1n h 8 mL22 2E1h2 2( 6 . 631034J s)2 28 mL8 ( 0 . 20 kg)(1.5 m)1 . 2221067J.2 (1 . 2 1067b) Et1 2d vmvv1 .5 m2E m33J)1 . 1 1033m s0.20 kg1 .4 1033s.67 671.110m sc
Lehigh - PHYS - 11 and 21
38.1:fp Ec h pc3.00 5.206.63 5.20 (1.2810 m s 1010 10 10734 785.771.281014Hz27mJ s m2710kg m s 10 m s )8kg m s) (3.003.84161019J2.40 eV.38.2: a) Pt b) hf c)Pt hfhc (0.600 W) (20.03.05 101910
Lehigh - PHYS - 11 and 21
31.1: a) V rmsV 245 . 0 V 231 . 8 V.b) Since the voltage is sinusoidal, the average is zero.31.2: a) I b) I rav2 I2 I rms 2 ( 2 . 10 A ) 2 . 97 A.2( 2 . 97 A )1 . 89 A.c) The root-mean-square voltage is always greater than the rect
Lehigh - PHYS - 11 and 21
41.1:Ll (l1) l (l1)L 24.716 1 . 05410 1034 342J s J sl (l1)20 . 0l4.2 .41.2:Lz La) m lm ax2 , so L z m axb)l (l1) 62 . 45 .c) The angle is arccosarccosml 6, and the angles are, for m l2
Lehigh - PHYS - 11 and 21
39.1: a) b) K 39.2:h p ph p2p ( 2 . 37h 2 mEh 10( 6 . 63 ( 2 . 8024 3110 1034 10 2J s) m) 3 . 082 . 37181024kg m s .kg m s ) kg )10J19 . 3 eV.2m2 ( 9 . 11 10( 6 . 63 2 ( 6 . 64 102710 1034 6J s) eV )
Lehigh - MECH - 003
Lehigh - MECH - 003
COSMOS: Complete Online Solutions Manual Organization SystemChapter 3, Solution 50.Based on M x = ( P cos ) ( 0.225 m ) sin - ( P sin ) ( 0.225 m ) cos M y = - ( P cos )( 0.125 m ) M z = - ( P sin )( 0.125 m ) By Equation ( 3) M z - (
Lehigh - MECH - 003
COSMOS: Complete Online Solutions Manual Organization SystemChapter 3, Solution 60.Have whereM DI = DI rG/I TEG DI = DI = DI()( 4.8 ft ) i - (1.2 ft ) j ( 4.8 ft )2 + ( -1.2 ft )2= 0.97014 i - 0.24254 j rG/I = - ( 35.1 ft ) kTEG =
Lehigh - MECH - 003
COSMOS: Complete Online Solutions Manual Organization SystemChapter 2, Solution 58.Free-Body Diagram At A:First Note . With LAB =( 22 in.)2 + (16.5 in.)2LAB = 27.5 in. LAD =( 30 in.)2 + (16 in.)2LAD = 34 in.Then FAB = k AB ( LAB - LO )
Lehigh - PHYS - 11 and 21
8.1: a) (10 , 000 kg)(12.0 m s )1 . 2010 kg m s.5b) (i) Five times the speed, 60 . 0 m s.(ii)5 12 . 0 m s26 . 8 m s.8.2: See Exercise 8.3 (a); the iceboats have the same kinetic energy, so the boat with the larger mass has the larger
Lehigh - PHYS - 11 and 21
9.1: a)1 . 50 m 2 .50 m0 . 60 rad34.4 .b) c)(14 . 0 cm) (128 )( rad 180 )6 . 27 cm.(1 . 50 m)(0.70 rad)1.05 m.9.2: a) b)1900rev min2 rad rev1 min 60 s199 rad s.(35rad 180 ) (199 rad s)3.07103s.9.3: a) zd
Lehigh - PHYS - 11 and 21
7.1: From Eq. (7.2),mgy (800 kg) (9.80 m s ) (440 m)23.4510 J63.45 MJ .7.2: a) For constant speed, the net force is zero, so the required force is the sack's weight, ( 5 .00 kg)(9.80 m s 2 ) 49 N. b) The lifting force acts in the same dir
Lehigh - PHYS - 11 and 21
42.1: a) KT 6 .1 K3 2kTT2K 3k2 ( 7 .9104eV )(1 . 60 102310 J K)19J eV )3 (1 . 3819b) T2 ( 4 . 48 eV ) (1 . 60 3 (1 . 38 101023J eV )34 , 600 K.J K)c) The thermal energy associated with room temperature (300 K
Lehigh - MECH - 003
COSMOS: Complete Online Solutions Manual Organization SystemChapter 6, Solution 10.FBD Truss:Fx = 0: H x = 0 By symmetry: A y = H y = 4 kips by inspection of joints C and G : FAC = FCE and FBC = 0 FEG = FGH and FFG = 0 also, by symmetry FAB = FFH
Lehigh - PHYS - 11 and 21
30.1: a) 2 M ( di 1 / dt ) ( 3 . 25 10 4 H) (830 A /s) 0.270 V, and is constant. b) If the second coil has the same changing current, then the induced voltage is the same and 1 0 . 270 V.30.2: For a toroidal solenoid, M So, M0N2B2/ i1 , and
Lehigh - PHYS - 11 and 21
4-1Chapter 4 ENERGY ANALYSIS OF CLOSED SYSTEMSMoving Boundary Work 4-1C It represents the boundary work for quasi-equilibrium processes. 4-2C Yes. 4-3C The area under the process curve, and thus the boundary work done, is greater in the constant p
Lehigh - MECH - 003
COSMOS: Complete Online Solutions Manual Organization SystemChapter 2, Solution 35.Cable BC Force:Fx = - (145 lb ) Fy = (145 lb ) 84 = -105 lb 11680 = 100 lb 116 3 = -60 lb 5 4 = -80 lb 5100-lb Force:Fx = - (100 lb ) Fy = - (100 lb )156-l
Lehigh - MECH - 003
COSMOS: Complete Online Solutions Manual Organization SystemChapter 7, Solution 39.(a) FBD Beam:by symmetry, C y = Gy =Cx = 0, and1 2 (12 lb/in )(10 in ) + 2 (100 lb ) + (150 lb ) 2C y = G y = 295 lb Along AC:Fy = 0:- (12 lb/in.) x
Lehigh - PHYS - 11 and 21
35.1: Measuring with a ruler from both S 1 and S 2 to there different points in the antinodal line labeled m = 3, we find that the difference in path length is three times the wavelength of the wave, as measured from one crest to the next on the diag
Lehigh - MECH - 003
COSMOS: Complete Online Solutions Manual Organization SystemChapter 2, Solution 126.See Problem 2.125 for the analysis leading to the linear algebraic Equations (1 ) , ( 2 ) , and ( 3 ) below: i component: j component: k component:- 81 TAB + 34
Lehigh - MECH - 003
COSMOS: Complete Online Solutions Manual Organization SystemChapter 7, Solution 32.(a) FBD Beam:Fx = 0:Bx = 0M B = 0: Fy = 0:aP + 2aC - 3.5aP = 0 C = 1.25P - P + By + 1.25P - P = 0 By = 0.75PAlong AB:Fy = 0: M J = 0:Along BC:-P - V
Lehigh - PHYS - 11 and 21
32.1: a) td c3 . 84 3 . 0010 1088m m s1 . 28 s.b) Light travel time is:8 . 61 years ( 8 . 61 years ) ( 365 days ) ( 24 hours ) ( 3600 s ) (1 year ) (1 day )82 . 7210138s(1 hour )dctc t(3 .0(3 .0108m s) (2.72
Lehigh - PHYS - 11 and 21
5.1: a) The tension in the rope must be equal to each suspended weight, 25.0 N. b) If the mass of the light pulley may be neglected, the net force on the pulley is the vector sum of the tension in the chain and the tensions in the two parts of the ro
Lehigh - MECH - 003
COSMOS: Complete Online Solutions Manual Organization SystemChapter 9, Solution 12.x2 y2 + 2 =1 a2 by = b 1-x2 a2dA = 2 ydxdI y = x 2dA = 2 x 2 ydxI y = dI y = 0 2 x 2 ydx = 2b 0 x 2 1 -Set:a ax2 dx a2x = a sin dx = a cos dI
Lehigh - MECH - 003
COSMOS: Complete Online Solutions Manual Organization SystemChapter 9, Solution 19.First note: At x = 0: b = c cos ( 0 ) orc=bAt x = 2a : b = b sin k ( 2a )or2ka = k=24aThen 2a A = a b sin x - b cos x dx 4a 4a 4a 4a =
Lehigh - PHYS - 11 and 21
43.1: a) b) c)85 37 205 8128 14Si has 14 protons and 14 neutrons.Rb Tlhas 37 protons and 48 neutrons. has 81 protons and 124 neutrons.43.2: a) Using R (1 . 2 fm) A 1 3 , the radii are roughly 3.6 fm, 5.3 fm, and 7.1 fm. b) Using 4 R 2 for
Lehigh - PHYS - 11 and 21
37.1: If O sees simultaneous flashes then O will see the A( A ) flash first since O would believe that the A flash must have traveled longer to reach O , and hence started first.37.2: a) 11 ( 0 .9 )22 . 29 .t( 2 . 29 ) ( 2 . 20 ( 0 . 90
Lehigh - PHYS - 11 and 21
1-1Chapter 1 INTRODUCTION AND BASIC CONCEPTSThermodynamics 1-1C Classical thermodynamics is based on experimental observations whereas statistical thermodynamics is based on the average behavior of large groups of particles. 1-2C On a downhill roa
Lehigh - PHYS - 11 and 21
5-1Chapter 5 MASS AND ENERGY ANALYSIS OF CONTROL VOLUMESConservation of Mass 5-1C Mass, energy, momentum, and electric charge are conserved, and volume and entropy are not conserved during a process. 5-2C Mass flow rate is the amount of mass flowi
Lehigh - MECH - 003
COSMOS: Complete Online Solutions Manual Organization SystemChapter 6, Solution 23.FBD Truss:Fx = 0:Ax = 0M A = 0:(12 m ) (M y - 1 kN)- (2.4 m + 4.8 m + 6 m + 8.4 m + 10.8 m) (1.5 kN) = 0M y = 5.05 kNFy = 0: Ay - 2(1 kN) - 5(1.5 kN) +
Lehigh - MECH - 003
COSMOS: Complete Online Solutions Manual Organization SystemChapter 5, Solution 28.The centroid coincides with the center of gravity because the wire is homogeneous.L1 2 3 r2 rxxL- -r 2-r2 2r sin l 2- 2r 2 sin ll2 2The
Lehigh - MECH - 003
COSMOS: Complete Online Solutions Manual Organization SystemChapter 6, Solution 98. FBD Frame:M A = 0: Fx = 0: Fy = 0:( 0.625 m ) F - ( 0.75 m )( 4 kN ) - (1.25 m )( 3 kN ) = 0Fx = 10.8 kNAx - 10.8 kN = 0, Ay - 4 kN - 3 kN = 0,A x = 10.80 k
Lehigh - MECH - 003
COSMOS: Complete Online Solutions Manual Organization SystemChapter 6, Solution 104. Members FBDs:I:M A = 0:M C = 0:(12.8 ft ) Bx - ( 32 ft ) By - ( 20 ft )(14 kips ) = 0( 7.2 ft ) Bx + ( 24 ft ) By - (12 ft )( 21 kips ) = 0Solving: Bx = 2
Lehigh - MECH - 003
COSMOS: Complete Online Solutions Manual Organization SystemChapter 6, Solution 123.Member FBDs: = tan -1( BC + CD ) sin 60 AF + ( AB + BC - CD ) cos 60( 2.5 in. + 1 in.) sin 60 4.5 in. + (1.5 in. + 2.5 in. - 1 in.) cos 60= tan -1 = 26.80
Lehigh - MECH - 003
COSMOS: Complete Online Solutions Manual Organization SystemChapter 6, Solution 128.(a) FBD BC: M B = 0:( 0.045 m ) FCD sin 45 - 6.00 N m = 0FCD = 188.562 N CFBD Joint D: = tan -138.8 mm = 22.823 92.2 mm 37.5 mm - 18.8 mm = 25.732 38.8 m
Lehigh - MECH - 003
COSMOS: Complete Online Solutions Manual Organization SystemChapter 6, Solution 139. FBD top handle:Note CD and DE are two-force membersM A = 0:( 4 in.)6 5 FCD - (1.5 in.) FCD - (13.2 in.) ( 90 lb ) = 0 61 61FCD = 72 61 lb By symmetry: FDE
Lehigh - MECH - 003
COSMOS: Complete Online Solutions Manual Organization SystemChapter 6, Solution 141.FBD cutter AC:M C = 0:( 32 mm )1.5 KN - ( 28 mm ) Ay - (10 mm ) Ax 11 10 Ax + 28 Ax = 48 kN 13 Ax = 1.42466 kN Ay = 1.20548 kN=0FBD handle AD:M D =
Lehigh - MECH - 003
COSMOS: Complete Online Solutions Manual Organization SystemChapter 6, Solution 150.FBD Bucket (one side):(a)M D = 0:( 0.8 m ) ( 2.4525 kN ) - ( 0.5 m ) FAB = 0FAB = 3.924 kNw = ( 250 kg ) ( 9.81 N/kg ) = 2452.5 N = 2.4525 kNFBD link BE:
Lehigh - MECH - 003
COSMOS: Complete Online Solutions Manual Organization SystemChapter 7, Solution 6.FBD CD:Note: AB is a two-force member C = 0:(18 in.) 5 12 FAB + ( 7.5 in.) FAB 13 13 - ( 24 in.)( 78 lb ) = 0FAB = 135.2 lbFx = 0:Cx -12
Lehigh - MECH - 003
COSMOS: Complete Online Solutions Manual Organization SystemChapter 7, Solution 8.FBD AJ:Note: Cut is just left of contact with ground. AlsoW = ( 9 kg )( 9.81 N/kg ) = 88.29 Nx= 2randr = 0.15 m F=0 V = 44.1 NFx = 0: Fy = 0: M J = 0:
Lehigh - MECH - 003
COSMOS: Complete Online Solutions Manual Organization SystemChapter 7, Solution 56.M A = 0: aDy - ( 5 ft )( 250 lb ) - (10 ft )( 500 lb ) = 0Dy =With a in ft, Dy =6250 lb ft a6250 lb aSince there are no distributed loads, M is piecewise