lý thuyết tin học a bài tập tin học a

Ngày tải lên: 03/04/2013, 11:10

... Email: Jackie9x.spb@gmail.com  3 a = - 2 x 1 A a A v 2 2 2                 Hay 1 v a v v 2 max 2 2 2 max 2    hay 2 2 2 2 max a (v v )   hay 1 a a v v 2 max 2 2 max 2  ...  A) , gia tc có đ ln cc đi : a max =  2 A. Giá tr đi s: a max = 2 A khi x= -A; a min =- 2 A khi x =A; . -  v trí cơn bng (x = 0), gia tc bng 0. + Lc tác dng lên vt dao ... LC 1   A 3 A 1 B 1 A 2 B 2 B 3 A 3 A 1 B 1 A 2 Mcăsaoăăăăăăă Mcătamăgiácă tamăgiác UYN TI I C MễN VT í Email: Jackie9x.spb@gmail.com 30 in t. Trong thc t, cỏc mch dao...

Ngày tải lên: 05/01/2014, 08:10

... A 1 để A 2max : A 2max = 2 1 sin( ) A ϕ ϕ − và A 1 = 2 1 tan( ) A ϕ ϕ − Chú ý: Nếu cho A 2 thì từ 2 công thức trên ta tìm được A = A min A min = A 2 sin(ϕ 2 - ϕ 1 ) = A 1 tan(ϕ 2 ... 0986338189 34 Tóm Tắt Lý Thuyết Và PP Giải Bài Tập Vật Lí 12 - Năm Học 2013 – 2014 – Tập 1 BẢNG CHỦ CÁI HILAP Kí hiệu in hoa Kí hiệu in thường Đọc Kí số A α alpha 1 B β bêta 2 Γ γ gamma 3 ∆ δ denta 4 E ε epxilon ... c a lò xo ở VTCB (đơn vị là mét) + Đề cho x, v, a, A: ω  2 2 v A x −  a x  max a A  max v A 2. Cách xác định A: + A = x max : vật ở VT biên (kéo vật khỏi VTCB 1 đoạn rồi buông x = A) . Gv...

Ngày tải lên: 14/03/2014, 09:21

... ngược pha: π+=ϕ∆ )1k2( thì: A = A 1 – A 2 . D. Nếu hai dao động lệch pha nhau bất kì: 21 AA − ≤ A ≤ A 1 + A 2 Gv soạn: ThS. Nguyễn Vũ Bình - Tel: 0986338189 23 Bài Tập Ôn Thi Đại Học Môn ... qua ma sát gi a m 1 và mặt sàn nằm ngang, ma sát gi a m 1 và m 2 là µ 12 = 0,8. Biên độ dao động c a vật m 1 bằng bao nhiêu để hai vật không trượt lên nhau: A. A ≤ 0,8 cm. B. A ≤ 2 cm C. A ... đúng: A. Độ lệch pha c a các dao động thành phần đóng vai trò quyết định tới biên độ dao động tổng hợp. B. Nếu hai dao động thành phần cùng pha: π=ϕ∆ 2k thì: A = A 1 + A 2 C. Nếu hai dao động...

Ngày tải lên: 14/03/2014, 09:21

... đang ở vị trí x =A/ 2, người ta thả nhẹ nhàng lên m một vật có cùng khối lượng và hai vật dính chặt vào nhau. Biên độ dao động mới c a con lắc? A. ' 7 2 A A = B. ' 2 2 A A = C. ' 7A ... pha có cùng biên độ A gây ra tại M sự giao thoa với biên độ 2A. Nếu tăng tần số dao động c a hai nguồn lên 2 lần thì biên độ dao động tại M khi này là A. 0 . B. A C. A 2 . D. 2A Câu 13: Hai ... khoảng gi a A và B có giao thoa sóng do hai nguồn trên gây ra. Phần tử vật chất tại trung điểm O c a đoạn AB dao động với biên độ bằng: A. 2 a B. 2a C. 0 D .a Câu 2: Để khảo sát giao thoa sóng...

Ngày tải lên: 14/03/2014, 09:22

... 4>$)9B$'7&'&=%$'7  A. )& A 4<3 B.   A 4&'53 C. -4 A 4$3 D. -4 A 4 &'53 Câu ... HỎI LÝ THUYẾT ÔN THI ĐẠI HỌC MÔN VẬT LÍ - NĂM HỌC 2013 - 2014 – TẬP 3 Câu 49: >$)9B&'&=%2-4%   A >$45&'H P3I*#&'53 Q3I*#9)53 I3M ... )"#- 43 Câu 15: ~4")>$ A. )9>&'3 B. &')9>&'3 C. )9&'&=3 Gv...

Ngày tải lên: 14/03/2014, 09:22

Ngày tải lên: 08/07/2014, 16:22

... )) SD ABCD = · · · 0 tan 3 60 SA SDA SDA SDA AD = = =Þ Þ 4) SA ⊥ (ABCD) (gt) ⇒ AO là hình chiếu c a SO trên ( ABCD) ⇒ · ( ,( )) SO ABCD = · · tan 6 SA SO A SO A a AO = =Þ 5) BC ⊥ ( SAB) ⇒ ... (SAD))⇒ AK ⊥ ( SCD) (1) • SA⊥ ( ABCD) ⇒ SA ⊥ AD, AD⊥ AB ⇒ AD ⊥ ( SAB)(2) Từ (1) và (2) ta có · · · (( ),( )) ( , )SCD SAB AD AK DAK= = và do · · · 0 0 tan 3 60 30SDA SDA DAK= ⇒ = ⇒ = 7) Ta ... chiếu c a SA trên ( SCD) ⇒ · · · ( ,( )) ( , ) SA SCD SAAK ASK = = · tan AK ASK SK = , SK= 3 2 a ,AK = 3 2 a · · 0 1 tan 30 3 AK ASK ASK SK = = =Þ Þ 10) AH ⊥ ( SBC) ⇒ SH là hình chiếu c a SA trên...

Ngày tải lên: 07/01/2013, 22:47

Ngày tải lên: 10/11/2013, 10:28

... 22. A 23. C 24. B 25. B 26. D 27. D 28. D 29. B 30. A 31. B 32. D 33. B 34. A 35. C 36. D 37. C 38. D 39. A 40. A 41. A 42. A 43. C 44. C 45. B 46. D 47. D 48. A 49. A ... trung trc c a AB cách AB 100 m và M là đim nm trên đng thng qua O song song vi AB, gn O nht mà ti đó nhn đc âm to nht. Cho rng AB << OI (vi I là trung đim c a AB ). Khong ... mt khong NA = 1m, có mc cng đ âm là L A = 90 dB. Bit ngng nghe c a âm đó I o = 0,1 nW/m 2 . Cng đ âm đó ti A là A. I A = 0,1 nW/m 2 . B. I A = 0,1 mW/m 2 . C. I A = 0,1 W/m 2 ....

Ngày tải lên: 17/02/2014, 23:16

... dây. C. khong cáh gi a hai nút sóng hay hai bng sóng liên tip. D. hai ln khong cách gi a hai nút sóng hay hai bng liên tip. LÝ THUYT TNG HP V SÓNG DNG (BÀI TP T LUYN) Giáo ... mt si dây mà hai đu đc gi c đnh, bc sóng bng A. đ dài c a dây. B. mt n a đ dài c a dây. C. khong cách gi a hai bng sóng liên tip. D. hai ln khong cách gi a hai bng sóng liên ... Mô t s i dây AB co chiê u da i 1 m cng ngang , đâ u A cô đi nh , đâ u B g n v i mô t nha nh cu a âm thoa dao đô ng điê u hoa v i tâ n sô 20 Hz. Trên dây AB co mô t so...

Ngày tải lên: 17/02/2014, 23:16

... có u AB và u AM lệch pha nhau  Ở đây 2 đoạn mạch AB và AM có cùng i và u AB chậm pha hơn u AM   AM –  AB =   tan tan tan 1 tan tan          AM AB AM AB Nếu u AB ...  1 2 1 2 tan tan tan 1 tan tan          Trường hợp đặc biệt  = /2 (vuông pha nhau) thì tan 1 tan 2 = -1. LUYỆN THI ĐẠI HỌC MÔN VẬT LÝ Email: Jackie9x.spb@gmail.com  ... 2 kA A S mg g * gim biờn sau mi chu k l: 2 4 4 mg g A k gim biờn sau N chu k: N 0 N 4NF ms A = A - A = K * S dao ng thc hin c: 2 4 4 A Ak A N A mg g * Thi gian...

Ngày tải lên: 27/02/2014, 19:41

... H + c a axit dễ bị khử hơn các ion H + c a nước b) Khả năng phóng điện c a các anion ở anot: Ở anot xảy ra quá trình oxi h a các anion gốc axit như Cl - , S 2- …hoặc ion OH- c a bazơ kiềm ... Ngược lại trên anot sẽ diễn ra sự oxi h a dạng khử c a cặp có thế oxi h a – khử nhỏ nhất trước. a) Khả năng phóng điện c a các cation ở catot: Ở catot có thể xảy ra các quá trình khử sau đây: - ... Na, K, Ba, Ca, Mg, Al Ví dụ 1: Điện phân NaCl nóng chảy có thể biểu diễn bằng sơ đồ: Catot ( – ) NaCl Anot ( + ) 2| Na + + e → Na 2Cl - → Cl 2 + 2e Phương trình điện phân là: 2NaCl 2Na...

Ngày tải lên: 12/03/2014, 00:57

... 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...

Ngày tải lên: 12/03/2014, 09:35

... m/s. Bài 4 . Ta có: eU AK = hf max  U AK = e hf max = 26,5.10 3 V. Bài 5 . Ta có: eU AK = hf max  f max = h eU AK = 0,483.10 -19 Hz. Bài 6 . Ta có: eU AK   = hc    min = AK hc eU = ... và Cao ng Bài 2. Ta có: eU AK   = hc   U AKmax = min hc e  = 31.10 3 V. Bài 3. a) Ta có: I = U P = 0,04 A. b) Ta có: 2 1 mv 2 max = eU 0 = eU 2  v max = m eU 22 = 7.10 7 m/s. Bài ... = 26,5.10 3 V). Bài 5. Hiu n th gi a hai n cc c a ng Cu-lít-gi (ng tia X) là U AK = 2.10 4 V, b qua ng nng ban u c a êlectron khi bt ra khi catt. Tính tn s ln nht c a tia X mà...

Ngày tải lên: 14/03/2014, 09:19

... trị c a hàm số là nghiệm đơn hay nghiệm bội ba c a a thức: f ′ ( ) x = 4x 3 + 24ax 2 – 8 () x 1 + 2a = 4x ( ) 2 x + 6ax - 2 1 + 2a ⎡⎤ ⎣⎦ Tam thức g(x) = x 2 + 6ax – 2(1 + 2a) có ... –2 ( ) 2 6a + 2a + 1 x 2 + 4 ( ) 2 a + 2a x + 3 vì = = ff ′ () 1 x f ′ () 2 x ′ ( ) 3 x = 0 Vậy, phương trình Parabol đi qua 3 điểm cực trị là : y = –2 ( ) 2 6a + 2a + 1 x 2 + 4 ( ) 2 a + 2a x ... 24ax 2 – 8 () x 1 + 2a y ′′ = 12x 2 + 48ax – 8 () 1 + 2a y ′′ = 0 3x⇔ 2 + 12ax – 2 ( ) 1 + 2a = 0 (9) Vì (9) có = 3 6a ′ Δ 2 + 6 () 1 + 2a = 6 ( ) 2 6a + 2a + 1 > 0 , ∀ a...

Ngày tải lên: 19/03/2014, 15:20

... c a mắt. LUYỆN THI ĐẠI HỌC MÔN VẬT LÝ Email: Duclongtn95@gmail.com 1 HỆ THỐNG LÝ THUYẾT VÀ CÁC DẠNG BÀI TẬP VẬT LÝ 11 * Tóm tắt lý thuyết * Công thức tính nhanh * Các dạng bài ... ' 11 dd  ; k = AB BA '' = - d d' = df f  . ; ABkBA .''  Qui ước dấu d OA  : d > 0 : vật thật ; d< 0 : vật ảo. LUYỆN THI ĐẠI HỌC MÔN VẬT LÝ ... Ir, hay r I AB U   E (dòng điện chạy từ A đến B, qua nguồn từ cực âm sang cực dương) - Định luật Ôm cho đoạn mạch ch a máy thu U AB = V A – V B = Ir’ + E p , hay 'r U I pAB E-  ...

Ngày tải lên: 25/03/2014, 23:20

... 2 Hai dao động cùng pha 2 : Hai dao động ngược pha (2 1) : Hai dao động vuông pha (2 1) : 2 Hai dao động có độ lệch pha : k A A A k A A A k A A A const A A A A A 2. Khi biết một dao động ... lớn c a li độ. - Ở vị trí biên (x =  A) , gia tốc có độ lớn cực đại : a max =  2 A. Giá trị đại số: a max = 2 A khi x= -A; a min =- 2 A khi x =A; . - Ở vị trí cân bằng (x = 0), gia tốc ... y A A A A        2 2 x y A A A    và tan y x A A   với  [ Min ; Max ] 4. Dùng máy tính tìm phương trình ( dùng cho FX 570ES trở lên) LUYỆN THI ĐẠI HỌC MÔN VẬT LÝ Email:...

Ngày tải lên: 25/03/2014, 23:20

Ngày tải lên: 25/03/2014, 23:20