Tài liệu Physics exercises solution: Chapter 04 pdf

Ngày tải lên: 10/12/2013, 17:15

... investors and potential investors, once floated, is key. As has been emphasised already in this chapter, an investor is looking at two things in considering any investment: risk and return. Once ... also has a less onerous listing procedure for innovative high-growth companies. The LSE introduced Chapter 25 (Innovative High-Growth Companies) into their listing rules at the beginning of 200...

Ngày tải lên: 14/12/2013, 15:15

... m.4.4)s025.0)(s/m100.8()s025.0)(s/m100.9()s025.0( 334223 x b) Differentiating, the velocity as a function of time is so,)s/m 1040 .2()s/m1080.1()( 23524 tttv  23524 )s025.0)(s/m 1040 .2()s025.0)(s/m1080.1()s025.0( v s./m100.3 2  c) The acceleration ... are N32323cos)N50(233cos)N120(  x R .N124323sin)N50(233sin)N120()N250(  y R b) ,N128 22  yx RRR    ...

Ngày tải lên: 10/12/2013, 12:15

... 6.1 107.2 103.11 31 3 3 31 a 1 1 a                      ρ ρ r r 14.6: a)   327 30 3 8 3 4 30 sun sun m 1041 2.1 kg1099.1 m1096.6 kg1099.1       π V M D 33 mkg 1040 9.1  b)   317 313 30 3 4 3 4 30 mkg10594.0 m10351.3 kg1099.1 m1000.2 kg1099.1        π D ...  .mkg1002.7 3 3 mm0.300.150.5 kg0158.0 3   V m ρ You were cheated. 14.4: The length...

Ngày tải lên: 10/12/2013, 12:15

... 12.00 ________________________________________________________________ y(cm) 0.176 0.296 0.243 0 .047 0.176 0.296 0.243 0 .047 0.176 ________________________________________________________________ 15.50: ... TT Hz10.4s106.911 2   Tf m50.0cm0.50  L  (b) Second harmonic (c) sm5.2m)Hz)(0. 504. 10(  fv (d) (i)Maximum displacement, so )sinsin cm5.1((ii)0 ωtkxvv tt y y ...

Ngày tải lên: 10/12/2013, 12:15

... W080.0 2    .m W11 2  17.68: a) From Eq.   ,21.17         W,196 m100.4 K140 m 1040 Km W040.0 2 2     H or W200 to two figures. b) The result of part (a) is the needed power ... .222, 0 22 ALLLALA L L L L   But ,Tα L L   and so   .22 00 TAαTAαA  b)         .m 104. 1C5.12)m275.())C( 104. 2(22 24 2 15   πTAαA  17.27: a)...

Ngày tải lên: 10/12/2013, 12:15

... back to minrev must be made. c) 10.49: a) P MR P K 22 ))2/1)((2/1(        s,1021.2 W 1046 .7 rev/min)(500m)kg)(2.00000,60)(2/1()2/1( 3 4 2 rev/min rad/s 30 2     or 36.8 min. m.N1010.1 360 rad2 /s)00.1( rev/min rad/s 30 rev/min)500(m)kg)(2.00000,60)(2/1( b) 5 2                  ππ Iτ 10.50: ... proportional to .||/1 t Either way,   m.N0.080 s12...

Ngày tải lên: 21/12/2013, 03:15

... ε d A εC b) .kV2.13 F1029.3 C1035.4 12 8       C Q V c) .mV1002.4 m00328.0 V102.13 6 3    d V E 24.3: a) .V 604 F 1045 .2 C10148.0 10 6       C Q V b) .m0091.0 m)103280F)( 1045 .2( 2 0 310 0     ε . ε Cd A c) .mV1084.1 m10328.0 V 604 6 3     d V E d) .C/m1063.1)mV1084.1( 256 00 0   ... 120 b) dAεC 0  C602and2means2 μQQCCdd  c) C4804and,4,4means...

Ngày tải lên: 21/12/2013, 03:16

... 29.35: a) .mA7.55 )m0400.0( A280.0 2 2 0 00     A i A i dt dE j cc D b) .smV1029.6 mA7.55 12 0 2 0   D j dt dE c) Using Ampere’s Law .T100.7)A280.0( )m0400.0( m0200.0 22 : 7 2 0 2 0       D i R r BRr ... b) s.m38)4000( m101.2 m100.2 26 8                VsA dt dI A ρ A ρI dt d dt dE c) .m 104. 3s)sV38( 2 10 00 A dt dE j D    d) A1014....

Ngày tải lên: 17/01/2014, 04:20