SAT II Physics (Gary Graff) Episode 1 Part 3 doc

... exactly 18 0° opposite the direction the bird must fly. Take the resultant vector and add (or subtract) 18 0° to or from the vector’s direction. 18 0° + 31 = 211 ° The bird must fly 5 83 m @ 211 ° from ... 2as Peterson’s: www.petersons.com 63 Rope b extends 35 ° into the third quadrant . Side y = (side r) (sin 35 °) = (16 N) (.57) = 9.1N Side x = (side r) (cos 35 °) = (16 N) (...

Ngày tải lên: 22/07/2014, 10:22

... convex mirror. 11 1 11 1 36 1 37 1 277 − =+ − −= − −= −− fpq fpq q rearranges to 1 cm cm cm . 0027 1 30 4 3 29 cm 1 cm cm = =− = − q q CHAPTER 3 Peterson’s: www.petersons.com 12 9 Concave lenses ... describe the image. Solution 11 1 fpq =+ Rearrange and substitute: 11 1 1 8 11 12 5 05 1 1 075 fpq q q q −= −= −= == cm 20cm cm cm cm 13 .25cm . The image is...

Ngày tải lên: 22/07/2014, 10:22

... 10 24 kg). Solution F mv r FG mm r v Gm r c == = 2 12 2 and ()() Setting the equations equal to each other and rearranging yields: v v = ×       × () ×+× =× − 667 10 610 64 10 3 81 ... sin cos sin cos tan θ θ θ θ θ NEWTON’S LAWS OF MOTION Peterson’s SAT II Success: Physics 11 8 • Diagram D is the same as Diagram B except that the trailing parts of the two waves...

Ngày tải lên: 22/07/2014, 10:22

... correctly completes the reaction. 1 1 7 3 7 4 HLi Be+→ +? (A) (B) 1 0 (C) (D) (E) 1 1 0 1 H n e − α Υ PRACTICE TEST 1 Peterson’s SAT II Success: Physics 264 33 .33 . 33 .33 . 33 . The corThe cor The corThe ... ANSWERS 1. B 2. E 3. B 4. D 5. E 6. C 7. A 8. A 9. E 10 . C 11 . E 12 . B 13 . B 14 . B 15 . D 16 . D 17 . E 18 . D 19 . B 20. E 21. E 22....

Ngày tải lên: 22/07/2014, 10:22

... series). 11 111 1 11 2 1 3 1 4 1 5 1 6 1 5 12 34 5 CCCCCC Cfffff C t t t =++++ =++++ = µµµµµ µ. fffff f Cf t ++++ = . 33 25 2 16 67 14 5 µµµ µ µ CHAPTER 5 Peterson’s SAT II Success: Physics 17 0 SERIES ... of a set of parallel resistors is found in the following manner: 11 11 12 3 RRRR t =++ 11 2 1 5 1 10 1 40 1 5 2 1 025 1 825 12 1 R R R R t t...

Ngày tải lên: 22/07/2014, 10:22

... ) −       − ° ++ = + ° − 26 13 1 25 19 5 19 50 836 8 ((, . ) ,. . . , 26 13 1 25 26 13 1 25 19 50 836 8 19 5 28 0 81 J JJ J C t J C t J ff +=+ ° + ° == ° = °= 856 3 28 0 81 32 8 . , . J C t J J t Ct f f f It ... )(. ) (. ) 12 2 9 2 2 19 19 2 910 16 10 16 10 01 N     = ו × =× − − − F m m F 2 30 4 10 23 10 28 2 2 24 . . N 11 0 4 That’s the force oper...

Ngày tải lên: 22/07/2014, 10:22

... equation below. ΣFTT TT T T T yy y y =− = = = = 12 12 1 2 2 0 10 0N 10 0 N is found by substitut iing and solving 10 0N N 200N 2 =° = = T T T 2 2 30 10 0 5 (sin ) . T 3 could have been found before T 2 , ... the (v o t ) term is zero. The working equation is: sat s s = = = 1 2 1 2 12 5 14 12 2 2 22 ( . )( sec)m/s m CHAPTER 2 Peterson’s SAT II Success: Physics 80 K...

Ngày tải lên: 22/07/2014, 10:22

... AnsAns AnsAns Ans ww ww w erer erer er ss ss s 1. C 2. B 3. E 4. B 5. B 6. E 7. B 8. E 9. A 10 . A 11 . C 12 . D 13 . C 14 . B 15 . C 16 . D 17 . B 18 . C 19 . D 20. E 21. E 22. D 23. D 24. C 25. B 26. C 27. C 28. C 29. C 30 . A 31 . B 32 . ... B 32 . A 33 . D 34 . D 35 . A 36 . A 37 . D 38 . C 39 . C 40. C 41. D 42. C 43. B 44. D 45. D 46. A 47. E 48. B...

Ngày tải lên: 22/07/2014, 10:22

... Prefixes T tera 1 × 10 12 10 12 G giga 1 × 10 9 10 9 M mega 1 × 10 6 10 6 hK hectokilo 1 × 10 5 10 5 ma myria 1 × 10 4 10 4 K kilo 1 × 10 3 10 3 h hecto 1 × 10 2 10 2 d deka 1 × 10 1 10 1 Basic UnitBasic ... Unit 1 meter – 1 gram – 1 liter d deci 1 × 10 1 10 1 c centi 1 × 10 –2 10 –2 m milli 1 × 10 3 10 3 dm dec...

Ngày tải lên: 22/07/2014, 10:22

... 10 .(.nm × −−− ×× ×× 714 19 14 39 10 26 10 63 8 10 32 1 mHzJ nm Hz ). . / photon Orange 4 00 58 52 10 34 10 5 19 14 19 − − ×× J nm Hz J / / . photon Yellow photon Green 33 5 710 3 810 56 65 10 ... Wavelength Frequency Red 77 7 710 3 910 714 .(. ).nm m H×× − zz nm Hz nm Hz nm Orange Yellow Green 63 48 10 58 52 10 53 57 14 14 × × 11 0 56 65...

Ngày tải lên: 22/07/2014, 10:22