Devoir 1

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‫ا و ر م ر ‪1‬‬ ‫‪2008 /2009‬‬

‫  ا او‬ ‫ ح ا ‬

‫ ض وس ر‪1‬‬ ‫اء و اء‬

‫اـــــــــــــــء‬ ‫ا"! ا ول‬ ‫ ن ا!  ا" ‪ #‬اـ' ا  &‪:%‬‬ ‫ ‪+‬ة ‪ P‬ذات &!‪ -. %+‬ه  ا ا ‪ 4 R=10cm,r=2cm‬واران ‪3‬ل &‪1‬ر∆‬‫‪+& %& + 75‬آه‪ .‬م ‪94‬ره =< ;‪:‬ا ا‪1‬ر ه ∆‪. J‬‬ ‫ ?<‪ S1 %@ %‬و ‪ S2‬آ  ه  ا ا ‪:‬‬‫‪& M =5kg, m =3kg‬ود‪ & E %4 +D %BC %‬اد آ  ه &; ن )ا‪ +G‬ا'(‬ ‫‪D‬‬ ‫‪+1‬را!  ون ‪ = I  +J‬ا‪ LB=# t1G1‬ا!<‪S2 K‬‬ ‫&‪ %‬ا‪ '9 B N‬إ ا=‪ = A BP‬ا‪VA=0.3m/s  +< t2 Q1‬‬ ‫‪ 1 S1 'P = %3 #‬ا‪ B’ %& 'JR‬إ ’‪ + -) A‬ا ‪ 3‬آت &;(‬ ‫‪α=30° ; g=10Nkg-1 S‬‬ ‫‪ BA %‬ا ‪P‬ل ‪ S2‬و’‪ B’A‬ا ‪P‬ل ‪BA=40cm 73 S1‬‬ ‫‪1‬ن‬ ‫‪1‬ن‬ ‫‪1‬ن‬ ‫‪1.5‬ن‬ ‫‪1‬ن‬ ‫‪1‬ن‬ ‫‪1‬ن‬ ‫‪1‬ن‬

‫‪1‬ن‬ ‫‪1‬ن‬

‫‪S2‬‬ ‫‪B‬‬

‫‪A‬‬

‫‪α‬‬

‫‪1-1‬ا?‪+‬د ا‪P‬ى ا‪  PB‬آ' &‪ %‬ا‪+‬ة ‪ P‬و ‪ S2‬و ‪S1‬‬ ‫‪ 1-2‬او? ا‪ % 4E-‬ا<‪  +‬ا‪ S1 K
‫ا"! ا"‪#$‬‬ ‫‪ + 1‬أ‪BJ‬ا & !<‪ r=0.5m ; -. ،‬و آ  ; ‪ 4 M=20kg‬وران ‪3‬ل &‪1‬ر‪ ، 75‬ا‪+1& BJ‬ك ‪4‬ر] ‪p=2kw‬‬ ‫‪1.5‬ن ‪& 1-1‬ه اة ا&= ا‪E‬ز&  =‪ 'P‬ا‪BJR‬ا &‪ %‬ا<ن ا ا<‪  +‬ااو ‪ω=21rad/s‬‬ ‫‪1‬ن ‪ 1-2‬ا‪ Z<3‬ا‪ 'f‬ا=!&‪+a %‬ف ا‪+1‬ك ‪ES‬ل ه‪ k:‬اة‬

‫ا‪%‬ــــ!ـــــــــــء‬

‫ا"! ا ول‬ ‫‪ + -‬ا‪+‬آ‪ %‬ا‪ :%R‬آ‪ +‬ت ا‪=&R‬م ا] @‪ ( Al2(SO4 )3,7H2O ) ] f‬و آرر ا &=م @‪( AlCl3)] f‬‬ ‫‪1‬ن ‪ -1‬اآ ‪ Z‬ا‪-‬د ا‪: I‬وان هذ‪ %‬ا ّ‪+‬آ‪%‬‬ ‫‪ + S 1& +Q1- 2‬ت ا‪=&R‬م ا] وذ‪o n‬ذا آ  ‪ %& m‬ه‪:‬ا ا‪+‬آ‪ # Z‬اء ا‪91 XC‬ل  ‬ ‫&‪1‬ل ‪ V=150mL ]!3‬و ‪+‬آ‪CM=7.4.10-2mol/L k‬‬ ‫‪ 2-1 1.5‬ا‪ Z<3‬آ  ا‪+‬آ‪ K Z‬ا‪ 4 V = J‬ا ‪+‬آا  ‪1‬ل‬ ‫‪2-2 1.5‬ا‪ Z<3‬ا ‪+‬اآ ا‪ -‬ا?دة ‪ #‬ا‪1‬ل‬ ‫‪ `Q2-3 1.5‬ا ا‪1‬ل ‪ S‬آ  ‪ %& m’=50g‬آرورا &=م و‪ + -‬ان ا‪ .+f  V K!1‬ا‪ ? %& Z<3‬ا ‪+‬اآ‬ ‫ا ا‪ -‬ا?دة ‪ #‬ا‪1‬ل‬ ‫‪M(Al)=27g/mol ; M(Cl)=35.5g/mol ; M(O)=16g/mol ; M(H)=1g/mol ; M(S)=32g/mol : B-‬‬

‫‪S1‬‬

‫ا"! ا&‪#$‬‬ ‫‪0.5‬ن‬ ‫‪1‬ن‬

‫‪ 1‬ي ‪ # ==4‬ذ ‪  60mL; -J‬آ &‪ %‬ا;اء ‪1‬ـ‪15bar Yf 7‬‬ ‫‪ -1‬ذآ‪P +‬ن ' &رط‬ ‫‪ K!3 & -2‬ا;اءا‪:‬ي ‪ %‬ا‪ %& ]@EC J‬ا‪ r = ==P‬در? ا‪+1‬ارة و‪1barYf 71‬‬

‫ا*)ــــــــــــــــ(ــــــــــ'‬

‫اء‬

‫ا"!ـــــــــــ ا*ول‪:‬‬ ‫‪-1‬‬ ‫‪) -1-1‬د ا‪(-‬ى ا!‪'-./‬‬ ‫ا‪%.‬ة ‪P‬‬ ‫‬ ‫‬ ‫‬ ‫'‪ R‬‬ ‫• وزن ا‪%.‬ة ‪P‬‬ ‫‬ ‫‪T '2‬‬ ‫‪T1‬‬ ‫• ‪ 7(7‬ا‪T 1 56‬‬ ‫‪1‬‬ ‫‪A‬‬ ‫‪S2‬‬ ‫• ‪ 7(7‬ا‪T2 2 56‬‬ ‫‬ ‫‬ ‫‪R‬‬ ‫• ‪ 9:7‬ا!(ر‬ ‫‬ ‫‪P2 α‬‬ ‫ا; ) ‪( S1‬‬ ‫‪T '1‬‬ ‫‪B‬‬ ‫‬ ‫‪S1‬‬ ‫• ‪ 7(7‬ا‪T ' 1 56‬‬ ‫‪ 1‬‬ ‫• وزن ا; ) ‪P1 ( S1‬‬ ‫ا; ) ‪( S 2‬‬ ‫‬ ‫‬ ‫‪P1‬‬ ‫• ‪ 7(7‬ا‪T ' 2 56‬‬ ‫‪ 2‬‬ ‫• وزن ا; ) ‪P2 ( S 2‬‬ ‫‬ ‫• ‪ 9:7‬ا‪R '
‫‬ ‫‪T2‬‬

BA ⊥ R’

 U '?!S ‫آت‬%"K ‫ *ن ا‬W ( R ' ) = 0 :‫ن‬V B ‫و‬

1 Mv A2 = T ' 2 . AB − MgAB sin α 2 Mv A2 T '2 = + Mg sin α = 25,56 N 2 AB

T '1 ‫ب‬K :Y? Z[$ S1 ;‫' آ' ا‬. '‫' اآ‬/‫ه' ا‬. X./"   1 2 1 2 mv A' − mv B ' = W (T '1 ) + W ( P1 ) 2 2 1 2 mv A' = −T '1 . A' B'+ mg . A' B' 2 mv A2 ' mRv A2 T '1 = mg − = mg − = 28,31N 2. A' B' 2 AB.r :‫ن‬V (‫' اوران‬K) '‫' اآ‬/‫ه' ا‬. NK -1-5   ∆E c = W (T1 ) + (T2 ) 1 1 J ∆ ω A2 ' − J ∆ ω B2 ' = T1 R.∆θ − T2 .r.∆θ 2 2 1 J ∆ ω A2 ' = T1 R.∆θ − T2 .r.∆θ 2 2(T1 R − T2 .r ).∆θ 2.r. AB(T1 .R − T2 .r ) J∆ = = = 0,41kg .m 2 2 2 ω A' vA

-2 '‫' اآ‬/‫ه' ا‬. NK -2-1   1 1 MvC2 − Mv A2 = W ( P2 ) + W ( R ' ) 2 2 1 − Mv A2 = − Mg ( Z A − Z C ) = − MgAC sin α = 2 v A2 AC = = 9.10 −3 m 2 g . sin α : '‫' اآ‬/‫ه' ا‬. NK -2-2  1 1 Mv B2 − MvC2 = W ( P2 ) = MgBC sin α 2 2 v B = 2 gBC sin α = 2,09m.s −1 :;$ '‫' اآ‬/‫ه' ا‬. X./" -2-3 1 1 Mv E2 − Mv B2 = Mg ( Z B − Z E ) = − Mgh 2 2 2 v h = B = 0.202m 2g

‫‪-3‬‬ ‫‪ ! -3-1‬أن دوران ا‪%.‬ة "‪ (ω = '"9 ) I‬اذن‪:‬‬ ‫ ;!(ع وم ا‪(-‬ى ا!‪ Y? '-./‬ا‪%.‬ة `م‪:‬‬ ‫‪"6$‬ر ا!‪ Y‬ا!()‪ N‬ه( ا!‪ Y‬ا!`آ‪ Y! a‬دوران ‪-‬رب ا'‪:‬‬ ‫‪M C + T1 .R = 0‬‬ ‫‪M C = −T1 .R = − mg .R = −3 N .m‬‬ ‫‪ UK‬أن م آ‪ Z‬وزن ا‪%.‬ة و ‪ 97‬ا ‪` Z‬م ‪(%‬ن أن ‪ 9:7 #/b‬ه‪ 7‬ا‪`d-" 7(-‬ن ‪( c‬ر‬ ‫اوران‪ ،‬آ! أن ‪ 7(7‬ا‪ 1 56‬وي وزن ا; ) ‪. NK ( S1‬أ ا‪([-‬ر‪.‬‬ ‫‪. X./7 -3-2‬ه' ا‪ '/‬اآ'‪:‬‬ ‫ ‪ #‬ه‪ hi‬ا' !دو)' (ى ا ‪%"K‬ك ه‪ #‬و‪K‬ه ا"‪` f ZOP S #‬م ‪ a‬ا*=‪.‬ب ا(اردة ‪ #‬ا‪j‬ال‬ ‫ا‪X‬‬

‫‪1‬‬ ‫‪J ∆ ω 2 = M C .∆θ = M C .2π .n‬‬ ‫‪2‬‬ ‫‪J ∆ω 2‬‬ ‫‪n=−‬‬ ‫‪= 2,7tr‬‬ ‫‪4πM C‬‬

‫‪0−‬‬

‫ا"!ـــــــــــ ا&‪:#$‬‬ ‫‪-1‬‬ ‫‪ -1-1‬ا!;!(' ا!رو=' ا*=‪(/‬ا‪'$‬‬ ‫)د ا‪(-‬ى‪:‬‬ ‫‬ ‫ دو)' ا‪(-‬ى‬ ‫ا!آ' ‪∑ Fi‬‬ ‫‬ ‫وزن ا*=‪(/‬ا‪P '$‬‬ ‫‬ ‫‪ 9:7‬ا!(ر ‪R‬‬ ‫‪. X./$‬ه' ا‪ '/‬اآ'‪:‬‬ ‫‬ ‫‬ ‫‬ ‫‬ ‫‪1‬‬ ‫) ‪J ∆ ω 2 − 0 = ∑ W ( F ) =W ( P ) + W ( R ) + W (∑ Fi‬‬ ‫‪2‬‬ ‫‬ ‫‬ ‫و ! أن ‪* W ( P ) = W ( R ) = 0 :‬ن ‪ 97 #/b‬ه‪ 7‬ا‪`d-" 7(-‬ن ‪ c‬ا!(ر ∆‬ ‫‬ ‫‪1‬‬ ‫‪J ∆ ω 2 = W (∑ Fi ) = p.∆t‬‬ ‫‪2‬‬ ‫إذن‪:‬‬ ‫‪J ∆ ω 2 Mr 2ω 2‬‬ ‫= ‪∆t‬‬ ‫=‬ ‫‪= 0,27 s‬‬ ‫‪2. p‬‬ ‫‪4p‬‬ ‫‪ -1-2‬ا‪ ZOk‬ا!;‪:‬‬ ‫‬ ‫‪W = W (∑ Fi ) = p.∆t = 540 J‬‬

‫ا‪!%‬ء‬

‫ا"! ا*ول‪:‬‬ ‫‪-1‬‬ ‫‪+ 7H 2O‬‬

‫‪2−‬‬ ‫‪4‬‬ ‫‪aq‬‬

‫) ‪+ 3( SO‬‬

‫‪3+‬‬ ‫‪aq‬‬

‫‪Al 2 ( SO4 ) 3 ,7 H 2 O ‬‬ ‫‪→ 2 Al‬‬ ‫‪H 2O‬‬

‫‪3+‬‬ ‫‪2O‬‬ ‫‪AlCl 3 H‬‬ ‫‪‬‬ ‫‪→ 3Cl aq− + Al aq‬‬

-2 :‫`? أن‬$ -2-1 m n= M n = c M .V :‫إذن‬ m M .V m = C M .M .V = 5,19 g

cM =

:#?"%‫ا"آ ا‬ m = 34,6 g / L .V :‫ أن‬l""$ ‫ون‬i‫ `د' ا‬/$‫ ا‬-2-2 cm =

[Al ] = 2.C [SO ] = 3.C 3+

1

2− 4

M

= 1,48.10 −1 mol.L−1

M

= 2,22.10 −1 mol.L−1 -2-3 :‫ ا!?(ل‬Y‫ ' إ‬m!‫ ا‬AlCl3 ‫د آ!' دة‬

n( AlCl 3 ) =

m( AlCl 3 ) = 3,74.10 −1 mol M ( AlCl 3)

:#‫ ' ه‬m!‫ت ا*( (م ا‬$(‫ن آ!' دة أ‬V AlCl3 ‫ `د' ذون‬NK n 2 ( Al 3+ ) = n( AlCl 3 ) = 3,74.10 −1 mol :#S B!"!‫"ت ا*( (م ا‬.‫;'  ذون آ‬7‫ت ا*( (م ا‬$(‫أ  آ!' دة أ‬ n1 ( Al 3+ ) = Al 3+ 1 .V = 2,22.10 −2 mol :#‫< آ!' دة أ(ن ا*( (م ا!"(ا)ة !?(ل ه‬.[7 ‫ا‬i%‫و ه‬ 3+ n1 ( Al ) + n2 ( Al 3+ ) = 0,396mol :(‫ا ا*(ن ه‬i‫آ ه‬7 <.[ 3+ n ( Al ) + n 2 ( Al 3+ ) Al 3+ = 1 = 2,64mol.L−1 V ‫ ا!?(ل‬Y‫ إ‬AlCl3 ‫ ذون‬Sm  UK ،O7 ‫ أي‬B? ‫أ‬/ ? ‫"ت ا[(د(م‬.‫ت آ‬$(‫[(ص ا‬6  ‫أ‬ :‫أي‬ 2− −1 −1 SO4 = 2,22.10 mol.L Cl– ‫?(رور‬%‫ت ا‬$(‫آ أ‬7 ‫ب‬K :‫; أن‬$ ‫ `د' ذون آ?(رر ا*( (م‬NK Cl − = 3.C M ( AlCl 3 ) = 7,48mol.L−1

[

]

[

[

]

]

[ ]

:#$&‫ا"! ا‬ -1 :‫ ر(ط‬Z( ‫(ن‬$ .PV=Cte :"9 B!;K ‫ز و‬f 5Op ‫ )اء‬Y-. '"9 ‫ارة‬K ')‫ در‬ -2 P0V0 = P1V1

 ‫ ا*="ذ ح ا‬:‫ إاد‬

V1 =

P0V0 = 0,9 L P1

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