SAT II Physics (Gary Graff) Episode 1 Part 6 docx

... 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 located 13 .25 cm from the lens on the ... convex mirror. 11 1 11 1 36 1 37 1 277 − =+ − −= − −= −− fpq fpq q rearranges to 1 cm cm cm . 0027 1 304 3 29 cm 1 cm cm = =− = − q q CHAPTER 3 Peterson’s: www.pe...

Ngày tải lên: 22/07/2014, 10:22

... other and rearranging yields: v v = ×       × () ×+× =× − 66 7 10 61 0 64 10 3 81 10 59 11 24 65 . . N m kg kg mm 2 2 11 0 6 m/s WEIGHTLESSNESS The strength of the earth’s gravitational field ... 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 th...

Ngày tải lên: 22/07/2014, 10:22

... series). 11 111 1 11 2 1 3 1 4 1 5 1 6 1 5 12 345 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 t t...

Ngày tải lên: 22/07/2014, 10:22

... )(. ) (. ) 12 2 9 2 2 19 19 2 910 16 10 16 10 01 N     = ו × =× − − − F m m F 2 304 10 23 10 28 2 2 24 . . N 11 0 4 That’s the force operating on the particles at a distance of 10 mm. Now ... ) −       − ° ++ = + ° − 26 13 1 25 19 5 19 50 8 36 8 ((, . ) ,. . . , 26 13 1 25 26 13 1 25 19 50 8 36 8 19 5 28 0 81 J JJ J C t J C t J ff +=+ ° + ° ==...

Ngày tải lên: 22/07/2014, 10:22

... the already known F 1 . The calculation of the torques is: TF F T wt wt =+()( ) 11  There are two torque-producing entities. ==•+• =•+• =• ()(. )10 1 2 5 10 11 Nm N m N m 1N m Nm T T Should ... operation svt at o =+ 1 2 2 is suitable for part one. Remember that the train started from rest, so the (v o t ) term is zero. The working equation is: sat s s = = = 1 2 1 2 12...

Ngày tải lên: 22/07/2014, 10:22

... components ROPE Rope a Rope b Rope c Rope d Y axis + 7.8N 9.1N +19 .4N 26. 1N X axis +29.1N 13 .1N + 5.2N +15 .0N –8N in x direction + 36. 2N in y direction After combining the x components and the ... 1 unit. The slope of the graph is calculated as rise over run. GRAPHS Peterson’s SAT II Success: Physics 62 Rope a extends 15 ° into the first quadrant. Side y = Side r (sin...

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. ... C 62 . B 63 . C 64 . E 65 . B 66 . D 67 . B 68 . A 69 . D 70. B 71. A 72. E 73. D 74. B 75. D Peterson’s SAT II Su...

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 decim...

Ngày tải lên: 22/07/2014, 10:22

... ANSWERS 1. B 2. E 3. D 4. D 5. B 6. A 7. C 8. E 9. A 10 . D 11 . C 12 . A 13 . A 14 . B 15 . D 16 . E 17 . D 18 . E 19 . A 20. D 21. B 22. E 23. A 24. B 25. A 26. D 27. B 28. C 29. A 30. C 31. C 32. ... faster. II. The voltage decreases. III. The current increases. (A) I only (B) II only (C) I and III only (D) II and III only (E) I, II, and III PRACTICE TEST 3 P...

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 ... in the red range is 3.9 × 10 14 Hz. Calculate the work function: Φ= ×       =× • × × − − hf eV J Js Hz 1 16 10 66 10 39 10 62 5 10 19 34 14 1 . (. )...

Ngày tải lên: 22/07/2014, 10:22