nhung bai tap vat ly hay lop 6

... 80 6 cos(ωt + π 6 ) V, uC = 402cos(ωt – 2π3)V, điện áp hiệu dụng ở hai đầu điện trở là UR = 60 3V. Hệ số công suất của đoạn mạch trên là A. 0, 862 . B. 0,908. C. 0,753. D. 0 ,66 4. ... 1πu=3cos(40πt + )cm 6 ; 25πu=3cos(40πt - )cm 6 . Vn tc truyn sóng 60 cm/s. Tìm s ng v n O1O2? A. 16 B. 8 C. 9 D. 18 Câu 48: AB ... C. 126W D. 96W Câu 34. Cho mạch điện RLC mắc nối tiếp, trong đó RC2 < 2L. Đặt vào hai đầu đoạn mạch điện áp xoay chiều u = U2cos 2ft, trong đóng U có giá trị không đổi, f có thể thay...

... 0173,8tan 1, 4 76 55,9 55 54'50xhα 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 ớβ v i:ớ0 0225, 4tan 0,508 26, 93 26 56 '50xhβ ... =Vì v1 = 3v2 => S1 = 3S2 hay a + x = 2 2h x+2 2 2 2 2 2( ) 9( ) 8 2ax+9h 0a x h x x a⇔ + = + ⇔ − − = thay s vào ta có:ố8x2-10015 -6. 502 = 0 ph ng trình có 2 nghi m: xươ ... n t c ta có:ứ ộ ậ ố13 12 23v v v= +r r rV i: vớ12 = 10000,9( / )15 .60 ABm st= =v23 = 500 5( / )15 .60 9BCm st= =Vì phà luôn chuy n đ ng vuông góc v i b sôngể ộ ớ ờnên 12vr...

... 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...

... A. 6 10 6 s B. 6 103s C. 6 102s D. 6 1012s Câu 21: Mắc vào hai đầu đoạn mạch RLC nối tiếp gồm một nguồn điện xoay chiều có tần số f thay đổi đượC. Khi tần số f1 = 60 Hz, hệ ... 80 6 cos(ωt + π 6 ) V, uC = 402cos(ωt – 2π3)V, điện áp hiệu dụng ở hai đầu điện trở là UR = 60 3V. Hệ số công suất của đoạn mạch trên là A. 0, 862 . B. 0,908. C. 0,753. D. 0 ,66 4. ... maxTt = 0,1= T = 0,6s S = 2A +A = 60 cm 6  Câu 6: Vật có tốc độ cực đại khi gia tốc bằng 0; tức là lúchl dh msF = F +F = 0 lần đầu tiên tại N Là lạ & kho khó 1.0 | 16 Xét đoạn chứa...

... 139:MộtnhàmáyđiệnnguyêntửcócôngsuấtP= 6. 105kW,hiệusuất20%.NhiênliệulàUđượclàmgiàu25%.Muốnchonhàmáyhoạtđộngliêntụctrong1nămcầnphảicungcấpchonókhốilượngquạngcầnlàbaonhiêu,biếtnănglượngtrungbìnhtỏaracủamộtphânhạchlà200MeV.1nămcó 365 ngày.A.1154kg. B. 461 6kg. C. 461 6tấn. D.185kg.ChọnB. 27AP.t P.tH N 2,9 565 .10E.N H. ENm M 1153540 ,6 gN     *Khốilượngquặngcầndùng: ... 32:NguồnsóngởOdaođộngvớitầnsố10Hz,daođộngtruyềnđivớivậntốc0,4m/strêndâydài.Trênphươngnàycó2điểmPvàQtheothứtựđóPQ=15cm.Chobiênđộa=10mmvàbiênđộkhôngđổitrongquátrìnhtruyềnsóng.NếutạithờiđiểmnàođóPcóliđộ0,5cmdichuyểntheochiềudươngthìliđộtạiQlà:A.-1cm. B.-0,5cm. C.8 ,66 cm. D 8 ,66 mm.ChọnC.*λ=4cm.2 .157,54   *QtrễphahơnPmộtlượng=π+π/2*DựavàohìnhvẽthìuQ=10cos30=8 ,66 mm.Câu 33:M,NlàhaiđiểmtrêncùngmộtphươngtruyềnsóngcủasóngmặtnướcMN=5,75λ.TạimộtthờiđiểmnàođóMvàNđangcóliđộlàuM=3mm,uN=-4mm,mặtthoángởNđangđilêntheochiềudương.Coibiênđộlàkhôngđổi.BiênđộsóngtạiMvàchiềutruyềnsónglà:M ... ,nướctrongtựnhiêncólẫn0,015%D2O,vớikhốilượngriêngcủanướclà1000kg/m3,1u=931,5MeV/c2,NA =6, 022.1023mol-1.Nếutoànbộ21Dđượctáchratừ1m3nướclàmnhiênliệuchophảnứngtrênthìnănglượngtỏaralà:A. 1, 863 .10 26 MeV. B. 1,0812.10 26 MeV. C. 1, 061 4.10 26 MeV....

... > d = (k +21) = 1,6k + 0,8 d2 = AO2 + x2 = 6 2 + x2 > (1,6k +0,8)2 = 36 + x2 > 0  x2 = (1,6k +0,8)2 – 36  64 O C N B A NGUOITHAY.VN 104 724AB AB ... tan32t = 3 = tan 6  > 32t = 6  + k > t = 41+23k x = 6cos(32t - 6  ) = x = 6cos[32(41+23k) -3 ] x = 6cos(k - 6 ) = ± 3 3 cm = ± 5,19 ... +1,5k ≤ 20 > - 6 ≤ k ≤ 6 > Trên đng tròn có 26 đim dao đng vi biên đ cc đi im gn đng thng AB nht ng vi k = 6 im M thuc cc đi th 6 d1 – d2 = 6 = 18 cm; d2...

... 2(T/12+T/4)=2T/3=2.0.02/3=1/75s.2phut =60 00T.vậy đáp số là 60 00.1/75=80s.AS6:Chiếu sáng hai khe Yang đồng thời 2 bức xạ đơn sắc có lamda1 = 0.45 micromet và lamda2 (từ 0 .64 đến 0. 76 micromet ). O và M là ... là ?A.n1=240 (vòng/phút) và n2= 960 (vòng/phút)B.n1= 360 (vòng/phút) và n2 =64 0 (vòng/phút)C.n1=120 (vòng/phút) và n2=1920 (vòng/phút)D.n1=300(vòng/phút) và n2= 768 ( vòng/phút )Bài này giải tà ... điện năng là 60 %. Để hiệu suất truyền tải tăng đến 90% mà công suất truyền đến nơi tiêu thụ vẫn không thay đổi thì điện áp hiệu dụng đưa lên 2 đầu đường dây bằng bao nhiêu?D.1 46, 67V nhờ giúp...

... 410cos(20 ) 10 6 ix t x eπππ= + ↔ = Bấm: 10 6 3 4 3 10 6 2 6 π π π∠ − + ∠ − − + ∠, SHIFT, 2, 3 máy sẽ hiện 6 64π∠ − Kết quả: 6 6 cos(20 )( )4x t cmππ⇒ = − Ví dụ 7: Hai ... máy FX570ES : 6 1 110cos(20 ) 10 6 ix t x eπππ−= − ↔ = , 22 2 6 3 cos(20 ) 6 32ix t x eπππ−= − ↔ = 3 14 3cos(20 ) 4 3x t xπ= − ↔ = −, 6 4 410cos(20 ) 10 6 ix t x eπππ= ... - 09 068 48238 Trang 16 Email: doanvluong@yahoo.com ; doanvluong@gmail.com Trang 16 CLABRM A. ABu 15 6 cos(200 t / 6) (V)= π − π B. ()ABu 15 6 cos 200 t / 6 (V)= π + π C....

... đến Anot:12mv22= eUAK=⇒ v2= 0 .66 .10 6 m/sVậy 0, 66 .10 6 ≤ v2≤ 0, 97.10 6 19 Cho dây AB cố định cố thể thay đổi l bằng ròng rọc. f = 20 Hz, khi thay đổi l ta thấy giữa2 lần có sóng ... anot là v thõa mãn:A. 0m/s ≤ v ≤ 0, 97.10 6 m/s B. 0, 66 .10 6 m/s ≤ v ≤ 0, 97.10 6 m/sC. 0, 71.10 6 m/s ≤ v ≤ 0, 97.10 6 m/s D. 0m/s ≤ v ≤ 0, 71.10 6 m/sHướng Dẫnhttp.//boxmath.vn/ TUYỂN TẬP ... =m.v22=> v =13, 92.1, 6. 10−13.24, 0015.1, 66 .10−27= 6, 7.1014Ta có: q.v.B =m.v2r=> r =m.vq.B⇐⇒4, 0015.1, 66 .10−27. 6, 7.10142.1, 66 .10−19.0, 5= 1, 07m21 Một sợi dây căng...

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... xx++3 6, 32,4I = BDRU = xx++3 6, 32,4 6 = 6, 32,4)3 (6 ++xxU13 = I. R13 = 6, 32,4)3 (6 ++xx.1,2 = 6, 32,4)3(2,7++xxI1 = 113RU = 3 6, 32,4)3(2,7++xx = 6, 32,4)3(4,2++xxU24 ... I2 - I1 = 6, 32,4 .6 +xx - 6, 32,4)3(4,2++xx = 6, 32,42, 76, 3+−xx IA = xx 6, 32,42,7 6, 3+− (2)Biện luận :+ Khi x tăng từ 2 (Ω) trở lên thì x2,7 và x 6, 3 đều giảm ... = I1 - I2 = 6, 32,4)3(4,2++xx - 6, 32,4 .6 +xx = 6, 32,4 6, 32,7+−xx(1)Biện luận :Khi x = 0 → IA = 2 (A)Khi x tăng thì (7,2 - 3 ,6. x) giảm ; (4,2.x + 3 ,6) tăng do đó IA...