De thi dai hoc nam 2016 lan 1 mon vat li cua Nhat Ban

De thi dai hoc nam 2016 lan 1 mon vat li cua Nhat Ban De thi dai hoc nam 2016 lan 1 mon vat li cua Nhat Ban De thi dai hoc nam 2016 lan 1 mon vat li cua Nhat Ban De thi dai hoc nam 2016 lan 1 mon vat li cua Nhat Ban De thi dai hoc nam 2016 lan 1 mon vat li cua Nhat Ban

A B C D (acceleration due to gravity) A E F g 1 h t t1 t4 t1 t2 t3 t4 B M B A A A B A B B µ F F m F0 F B A F0 µmg µ(M + m)g µ(M + m)mg M µ(M + m)M g m C M AB CD AB v CD m AB m AB AB v H CD h h H M M +m √ m M +m M M +m √ m M +m D 3.0m/s 4.0kg 4.0s F [N] t[s] 4 12 16 28 32 44 kg·m/s E 1 T1 T2 T2 T1 √ 2 √ 2 F O r A v0 B ∠AOB = θ v0 √ √ √ (3 sin θ − 2)gr √ (2 sin θ + cos θ − 2)gr (3 cos θ − 2)gr (sin θ + cos θ − 2)gr A( 1) B( −20◦ C A 2) C( 3) −20◦ C 1.0kg 0◦ C 1.0 × 101 K 1.0 × 103 J/K 2.0 × 103 J/kg·K 3.3 × 105 J/kg 0◦ C 1.1 2.2 3.3 11 22 33 B V A 4V B A 8p T nA [mol] B p T A A n′A [mol] n′A nA 10 12 C A B A B C A B C TA TB TC TA TB TC TA = TB = TC TA = TB < TC TA < TC < TB TB < TC < TA TC < TA = TB A( A 1) B( x 2) C( v x=L 3) t= t=0 x=0 A B 2L v 10 10 B f0 P Q P Q (a) (d) 11 11 P Q (a) (c) (b) (d) (c) (a) (d) (b) C L D λ ∆x L = 10 cm λ = 5.0 × 10−7 m ∆x = 0.50 mm D mm 12 mm 0.010 0.050 0.10 12 0.50 A( A 1) B( 2) c( xy 3) D( O 4) E( Q 5) F( 6) (a, 0) −Q k ( P a , √ 3a ) 13 kQ 2a2 √ 3kQ 2a2 kQ a2 13 √ 3kQ a2 2kQ a2 B 2 14 14 C X Y R Y X X X R Y Y PX X Y PY PY PX 15 9 2 15 9 D V CA S1 CB S2 ,2 S1 S2 S1 S1 S1 S2 S2 16 1C V2 B (C + C )V B A CA V 2 CB CA (CA + CB ) V 2 CC CA V2 CA + CB CA CB V 2 CA + CB CB2 V2 CA + CB 16 E O ◦ 30 O P # ‌ OP OP r H0 17 rH0 √ 3rH0 πrH0 √ 3πrH0 rH0 √ 3rH0 πrH0 √ 3πrH0 17 F B q V O P OP d m′ q P′ m OP′ 4d m′ m 18 16 2 18 A( 1) A X X 19 19 ... C A 2) C( 3) −20◦ C 1. 0kg 0◦ C 1. 0 × 10 1 K 1. 0 × 10 3 J/K 2.0 × 10 3 J/kg·K 3.3 × 10 5 J/kg 0◦ C 1. 1 2.2 3.3 11 22 33 B V A 4V B A 8p T nA [mol] B p T A A n′A [mol] n′A nA 10 12 C A B A B C A B C... A 1) B( x 2) C( v x=L 3) t= t=0 x=0 A B 2L v 10 10 B f0 P Q P Q (a) (d) 11 11 P Q (a) (c) (b) (d) (c) (a) (d) (b) C L D λ ∆x L = 10 cm λ = 5.0 × 10 −7 m ∆x = 0.50 mm D mm 12 mm 0. 010 0.050 0 .10 ... 15 9 D V CA S1 CB S2 ,2 S1 S2 S1 S1 S1 S2 S2 16 1C V2 B (C + C )V B A CA V 2 CB CA (CA + CB ) V 2 CC CA V2 CA + CB CA CB V 2 CA + CB CB2 V2 CA + CB 16 E O ◦ 30 O P # ‌ OP OP r H0 17 rH0 √ 3rH0