Tài liệu Target score chapter 06 docx

... for 2 v , .s/m1041.2 kg) 1067 .1( J) 1065 .2(2 s)/m1000.3( 2 5 27 17 252 12       m W vv b) With 12 ,0 KWK  . Using 2 x W   , .m1082.2 )s/m1000.3)(kg 1067 .1( )mN1012.2(22 10 2527 226 2 11 2            mvK x c) ... so .s/m93.6)m060.0( kg0300.0 m/N400 0  x m k v b) tot W must now include friction, so 0 2 0 2 1 tot 2 2 1 fxkxWmv  , where f is the magn...

Ngày tải lên: 10/12/2013, 12:15

... energy is the work fyW  other ). Eliminating y in favor of k by k y N 1066 .3 4   leads to kk )N 1066 .3)(N1070.1()N 1066 .3( 2 1 4424    . N)10N)(3.6610(1.96 J105.62 44 4 k   This ... done by gravity and friction, h,hhfmg )N600,36()N000,17)sm80.9)(kg2000(()( 2  so m.3.17 J 1066 .3 J1033.6 4 5    h c) The net work done on the elevator between the highest poin...

Ngày tải lên: 21/12/2013, 03:15

... N2700 20sin 40cos 040cos 2 )20(sin 2 1             mg T L mgLT b) Take +y upward. N137260cosgives0 N6.73so060sin gives0 sx   TfF nTwnF y 19 N73.6 N1372 , s sss  n f μnμf The floor must ... force that the pivot exerts is (300 N) –(132 N) sin N1830.62  and the horizontal force is N6 2062 cos)N132( . , for a magnitude of 193 N and an angle of 71 ab...

Ngày tải lên: 21/12/2013, 03:15

... need the flux through the parallelepiped: CmN5.3760cos)CN1050.2)(m0600.0)(m0500.0(60cos 24 11  AE CmN10560cos)CN1000.7)(m0600.0)(m0500.0(120cos 24 22  AE So the total flux is ,CmN5.67CmN)1055.37( 22 21  and ... the surface, not its distribution within the surface. 22.7: a) C.Nm1007.4C) 1060 .3( 25 0 6 0   εεq b) C. 106. 90)CNm780( 92 000   εεqεq...

Ngày tải lên: 21/12/2013, 03:16

... 11 22 9 2222                         kV ax a kQV ax kQ V But s.m 1067 .1 2 1 7 kg109.11 V)(796C) 1060 .1(2 2 31 19      vmvVqW 23.34: Energy is conserved: .V0117.0 C) 1060 .1(2 )sm1500()kg 1067 .1( 2 1 19 227 2       VVqmv But: m.158.0 m/C1000.5 )V0117.0(2 exp)m180.0( λ 2 exp λ 2 exp)ln( 2 λ 12 0 0 0 0 00 0               ...

Ngày tải lên: 21/12/2013, 03:16

... a) .J1032.4J/eV) 106. 1(eV)102.7(MeV7.2 13196  K m .068 .0 T)5.3(C) 106. 1( )m/s1027.2(kg) 1067 .1( m/s.1027.2 kg 1067 .1 J)1032.4(22 19 727 7 27 13             qB mv R m K v rad/s.1034.3 m068.0 m/s1027.2 Also, 8 7    R v ω b) ... . 2222 22 mV eBD mV qBD d  b) cm.6.7m0 .067 V)kg)(750102(9.11 C)10(1.6 2 T)10(5.0m)(0.50 31 1952       d ,of%13 Dd  wh...

Ngày tải lên: 21/12/2013, 03:16

... PRINCIPLES AND APPLICATIONS, Third Edition, by Allan R. Hambley, ©2005 Pearson Education, Inc. Chapter 14 Operational Amplifiers ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Third Edition, ... PRINCIPLES AND APPLICATIONS, Third Edition, by Allan R. Hambley, ©2005 Pearson Education, Inc. Chapter 14 Operational Amplifiers 1. List the characteristics of ideal op amps. 2. Identify n...

Ngày tải lên: 23/12/2013, 01:18

... PRINCIPLES AND APPLICATIONS, Third Edition, by Allan R. Hambley, ©2005 Pearson Education, Inc. Chapter 11 Amplifiers: Specifications and External Characteristics ELECTRICAL ENGINEERING: PRINCIPLES ... PRINCIPLES AND APPLICATIONS, Third Edition, by Allan R. Hambley, ©2005 Pearson Education, Inc. Chapter 11 Amplifiers: Specifications and External Characteristics 1. Use various amplifier...

Ngày tải lên: 23/12/2013, 01:18

...    200 V2.21 Z ,V2.21 2 V0.30 2 rmsrms rmsrms1 R VV I V VV source A 106. 0              ,V4.27 F1000.6srad645.5 A 106. 0 .V4.27H400.0srad5.645A 106. 0 2 6 0 rms 3 0rms2 V C ω I V L ωIV        ... 62.31ωLX L ;    906. 71 ωCX C       A10108.271.23V1000.5 71.23 43 2 2    ZVI XXXXRZ CLCL V; 1066 7.1 3 CC   IXV this is the maximum v...

Ngày tải lên: 17/01/2014, 04:20

... interference. λ 2 1 sin   d m 1064 .1 0.11sin2 m 1062 4 sin2 λ 6 9        d (b) Bright fringes: λsin maxmax mθd  The largest that θ can be is mdm Since6.2λ/so,90 m 1062 4 m101.64 max 9 6      ... are .3,2,1,0for2.54,4.35,3.20,65.6            m 35.9: .m105.90 m2.20 m)10(2.82m)10(4.60 λ λ 7 34        R yd d R y 35.10: For bright fringes: .mm1.14m1...

Ngày tải lên: 17/01/2014, 04:20