nhung bai tap vat ly hay lop 8

... k  ng ca qu cu là: A. 1 ,87 8s;14,4cm B. 1 ,88 7s; 7,2cm C. 1 ,88 3s; 7,2cm D. 1 ,88 1s;14,4cm Câu 44: Trong giao thoa Y-âng có a = 0,8mm, D = 1,2m. Ching thi hai bc x c ... & kho khó 2.0 | 48 tan 3 = 2 L3 C3 L3 C3 2 L2 C2 1 180 πL Z Z Z Z 4 ( 180 π) LC 1 180 πC = = = . 1 R Z Z 3 (240π) LC 1 240πL 240πC      tan 3 = 2 2 2 2 ( 180 π) 1 4 4 5 5 (120π) == (240π) 3 ... 06,0.10. 08, 0.1,0 2 04,0.2 2 1,0.2 2 08, 0 22 2 max  v = 0,0036  2 max v = 0,09  v max = 0,3(m/s) = 30cm/s. Cách 2: Độ giảm biên độ sau nửa chu kỳ 12 2μmg 2.0.1.0, 08. 10 A A = = = 0,08m = 8cm k2  ...

... S 1 = 3S 2 hay a + x = 2 2 h x+ 2 2 2 2 2 2 ( ) 9( ) 8 2ax+9h 0a x h x x a⇔ + = + ⇔ − − = thay s vào ta có:ố 8x 2 -100 15 -6.50 2 = 0 ph ng trình có 2 nghi m: xươ ệ 1 = 73 ,8( m), và x 2 ... v i tớ 2 = 1 2( ) 2(300 1, 225) 7 ,84 9 ,8 h h s g + + = = - V y th i gian t ng c ng đ v t r i ch m đ t làậ ờ ổ ộ ể ậ ơ ạ ấ : t = t 1 + t 2 = 0,5 + 7 ,84 = 8, 34(s) Bài 6.10 : M t ca nô chuy n ... ớ α v i:ớ 0 0 1 73 ,8 tan 1, 476 55,9 55 54' 50 x h α α = = = ⇒ = = + H ng th 2, ch y xi t vào trong đo n AH, h p v i MH góc ướ ứ ạ ế ạ ợ ớ β v i:ớ 0 0 2 25, 4 tan 0,5 08 26,93 26 56 ' 50 x h β...

... nơi tiền tuyến . “Giếng nước gốc đa” hay chính là đôi mắt hẹn ngày về của người bạn gái, làm ấm lòng người lính phương xa ?Tất cả đều có thể ,bởi một chút nhung nhớ ấy cùng với ngôi nhà ,ruộng ... hồn nhiên như cây cỏ ngỡ không bao giờ quên cái vầng trăng tình nghĩa . Vậy mà, hoàn cảnh sống thay đổi ,hết chiến tranh , con người trở về thành phố, quen với cửa gương và ánh điện của cuộc ... nhưng con người giờ đã khác xưa, quen với ánh sáng nhân tạo nên coi trăng hoàn toàn xa lạ. Một sự thay đổi đến phũ phàng,tê tái… Người lính đã quên những tình cảm chân thành ,những tháng năm gian...

... 06,0.10. 08, 0.1,0 2 04,0.2 2 1,0.2 2 08, 0 22 2 max  v = 0,0036  2 max v = 0,09  v max = 0,3(m/s) = 30cm/s. Cách 2: Độ giảm biên độ sau nửa chu kỳ 12 2μmg 2.0.1.0, 08. 10 A A = = = 0,08m = 8cm k2  ...  2 max v = 0 ,88  v max = 2204, 088 ,0  = 0,2. 22 (m/s) = 20 22 cm/s. Câu 14: Vật xuất phát từ M đến N thì đi được quãng đường S = 2 + 2 2 . Thời gian: T T 5 Δt = + = (s) 12 8 96 Câu ... 22 C U 170 150 80 V R U 2 .80 . 170 80 120V Hệ số công suất của mạch là 2 120 cos cos 0 ,8 150 Công suất tiêu thụ của mạch khi đó là: 2 2 2 2 max U P .cos P .cos 150.0, 8 96W R Câu 34:...

... 18: Mộtconlắcđơngồmvậtnặngcóm=250gmangđiệntíchq=10 –7 Cđượctreobằngmộtsợidây khôngdãn,cáchđiện,khốilượngkhôngđángkể,chiềudài90cmtrongđiệntrườngđềucóE=2.10 6 V/m(Ecó phươngnằmngang).Banđầuvậtđứngyênởvịtrícânbằng.Ngườitađộtngộtđổichiềuđườngsứcđiệntrường nhữngvẫngiữnguyênđộlớncủaE,lấyg=10m/s 2 .Chukìvàbiênđộdaođộngcủaquảcầulà: A.1 ,87 8s;14,4cm. B.1 ,88 7s;7,2cm.C.1 ,88 3s;7,2cm. D.1 ,88 1s;14,4cm. ChọnD. *Chukìdaođộngcủaconlắc:   2 2 l l T 2 2 1 ,88 1 s g' qE g m    ... 120: TrongthínghiệmgiaothoaánhsángvớihaikheY-âng,khoảngcáchtừmặtphẳngchứahaikheS 1 vàS 2 đến mànMlàD 1 =250cm.KheSlànguồncungcấpánhsángđơnsắcchohaikheS 1 vàS 2 .LúcđầukhesángScáchđềuhai kheS 1 vàS 2 vàcáchmặtphẳngchứahaikhenàymộtđoạnD 2 =49cm,sauđóchokheSchuyểnđộngtịnhtiến20mm theophươngtạomộtgóc30 0 vớimànMvàhướngraxamànM.Khiđóvântrungtâmtrênmàndờimộtđoạnbằngbao nhiêu?Khoảngvântrênmàncóthayđổikhông? A. Dời 8, 66cm,khoảngvântrênmànkhôngthayđổi. B. Dời7,07cm:khoảngvântrênmàncóthayđổi. C. Dời 8, 84mm,khoảngvântrênmànkhôngthayđổi. D. Dời 8, 84mm,khoảngvântrênmàncóthayđổi. ChọnA  Câu ... 16:Haiconlắcđơncùngchiềudàivàcùngkhốilượng,cácvậtnặngcoilàchấtđiểm,chúngđượcđặtở cùngmộtnơivàtrongđiệntrườngđều E  cóphươngthẳngđứnghướngxuống,gọiT 0 làchukỳchưatíchđiện củamỗiconlắc,cácvậtnặngđượctíchđiệnlàq 1 vàq 2 thìchukỳtrongđiệntrườngtươngứnglàT 1 vàT 2 ,biết T 1 =0,8T 0 vàT 2 =1,2T 0 .Tỉsốq 1 /q 2 là: A. 44 /81 . B. -81 /44. C. -44 /81 . D. 81 /44. ChọnB.         1 1 0 1 q E l l 9 T 2 0,8T 0 ,8. 2 g q E g m 16 g m  ...

... 1 2 2 d d   mt khc nhìn hình v ta có 2 2 2 1 2 2 1 2 1 2 2 2 2 2 1 2 1 (8 ) 8 32 2( ) 32 (8 ) 8 ( ) 16 8 2 d x d d x d d x d x d d x d x                        ... 1) d k         Hay : 1 1,6 (2 1) (2 1) (2 1).0 ,8 2 2 d k k k        (1) . Theo hình v ta thy 1 AO d AC   (2). Thay (1) vào (2) ta có : 2 2 (2 1)0 ,8 2 2 AB AB k OC    ... B B 2 2 8 8 C C 1 1 8 8 ...

... độ ánh Sáng trong chân không 3.1 08 (m/S). Tốc độ của một êlectron tăng tốc qua hiệu điện thế 10^5 V là: A. 0.4.10^8m/S B. 0 .8. 10^8m/S C. 1,2.10^8m/S D. 1,6.10^8m/S Tương tự như câu của thầy Điền ... điểm ban đầu có n2 = 1,8n1 nguyên tử bị phân rã. Xác định chu kì bán rã của chất phóng xạ này: A: 8, 7h B: 9,7h C: 15h D: 18h Ta có: (1) (2) Lấy (2) chia (1) => = 1 ,8 => Giải phương ... 10:(2 38, 92)U sau một chuỗi các phóng xạ và biến thành hạt nhân bền (206 ,82 )Pb . Tính thể tích He tạo thành ở điều kiện chuẩn sau 2 chu kì bán rã biết lúc đầu có 119g urani: A. 8, 4lít B. 2 ,8 lít...

... doanvluong@yahoo.com ; doanvluong@gmail.com ℡ ℡℡ ℡ Điện Thoại: 09157 181 88 – 090 684 82 38 GV: Đoàn Văn Lượng - Đ T: 09157 181 88 - 090 684 82 38 Trang 13 Email: doanvluong@yahoo.com ; doanvluong@gmail.com ... x=k f(x)=v 1 2 3 4 5 6 7 8 9 3.04 1.52 1.0133 0.76 0.6 08 0.506 0.434 0. 38 0.3377 GV: Đoàn Văn Lượng - Đ T: 09157 181 88 - 090 684 82 38 Trang 4 Email: doanvluong@yahoo.com ... tác b ấ m máy. GV: Đoàn Văn Lượng - Đ T: 09157 181 88 - 090 684 82 38 Trang 8 Email: doanvluong@yahoo.com ; doanvluong@gmail.com Trang 8 III.GIẢI NHANH TỔNG HỢP DAO ĐỘNG NHỜ MÁY TÍNH CASIO...

... 0, 8. Hãy xác định công suất định mức của quạt điện: A. 90W B. 266W C. 80 W D. 160W Hướng Dẫn  U 2 = U 2 R + U 2 q + 2U R .U q .cos(ϕ) =⇒ 380 2 = U 2 R + 220 2 + 2.U R .220.0, 8 =⇒ U R = 180 V ... người đó nghe được. Hướng Dẫn  f nghe < 180 00 =⇒ nf 1 < 180 00 =⇒ n < 180 00 420 =⇒ n < 42, 85 7; [n max ] = 42 =⇒ f max = 420.42 = 17640(Hz) 1 38 Một vật dao động điều hoà với phương trình ... hẳn là: t=N.T=12500.2= 250000s 89 Dưới tác dung của một lực có dạng F = −0, 8sin5t(N) một vật có khối lượng 400 dao động đều hòa biên độ dao đông của vật là: 8cm A. 8cm B. 20cm C. 12cm D. 32cm Hướng...

... 3 2 3 T  1 + q 2 = 7,4.10 -8  1  2  A. 1, 48. 10 -8  -8 C B. 6,4.10 -8  -8 C C. 3,7.10 -8  -8 C D. 2,4.10 -8  -8 C Câu 4:  E ... src="data:image/png;base64,iVBORw0KGgoAAAANSUhEUgAABB4AAATUCAIAAABAt1O2AAAACXBIWXMAABYlAAAWJQFJUiTwAAAgAElEQVR42uzdf2zUd/3A8ffpFUzuXAzz8CYlLbVx40cPBUaTlmqQTg10QeMwtkSMS0iG21jMYpYli2F/mBiTZYpsLBHFRGl1/oWpxLkaYLOXVJewXF0lhNgjFjl6qM3SmyZcct8/7vtlyGC0d+2+7d3j8de8q4y+PveBe+7zefUipVIpMHuRSMQQFrs5f/Ev5FeFMx0AmPd3yN5wAAAA1XufEQAAANICAABYEKJG8B4oFouNjY2XL 182 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... (A). - I ABO = ABO AO R U = 2. 8 7 14 = 8 (A) - U BO = I ABO .R BO = 8. 8 3 .2 = 6 (V) - I BO = 2 R U BO = 2 2 6 = 6 (A) - I BDO = I ABO - I BO = 8 - 6 = 2 (A) * Căn cứ vào mạch điện ... Nguyễn Xuân Ôn Giáo án bồi dưỡng học sinh giỏi - AB R 1 = R 3 8 1 + R 3 2 1 = RR 2 3 8 3 + - AB R 1 = R8 15 ⇒ R AB = R 15 8 . c. Tính điện trở tương đương giữa A và 0 : - Do đối xứng ... 3 47,2 = 0 ,82 3 (A) U 24 = U 2 = U 4 = I. R 24 = 2,10 21 . 7 12 = 3,53 (V) I 2 = 2 2 R U = 3 53,3 = 1, 18 (A) Ta có : I 2 > I 1 ⇒ I A = I 2 - I 1 = 1, 18 - 0 ,82 3 = 0,357 (A) Vậy...