SAT II Physics (Gary Graff) Episode 1 Part 4 ppsx

... 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

... 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. ... 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 49 . E 50...

Ngày tải lên: 22/07/2014, 10:22

... ANSWERS 1. A 2. E 3. C 4. E 5. A 6. C 7. E 8. A 9. B 10 . B 11 . D 12 . E 13 . A 14 . C 15 . A 16 . B 17 . E 18 . D 19 . A 20. E 21. C 22. D 23. E 24. D 25. E 26. A 27. A 28. D 29. C 30. C 31. A 32. ... A 33. D 34. D 35. D 36. E 37. C 38. C 39. C 40 . C 41 . E 42 . C 43 . A 44 . E 45 . C 46 . D 47 . B 48 . B 49 . A 50. D 51. B 52. C 53. B 54. A 55. A 56....

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 ... resistance for the R 2 – R 3 – R 4 parallel set of resistors: 11 11 11 10 1 8 1 4 1 25 4 12 3 RRRR R R R t t t t =++ =++ = = ΩΩΩ Ω Ω . CHAPT...

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 3 04 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 ... t J Joules gC gt fo f = =− = ° −° () , . ()( ) ()( 12 500 4 18 4 300 20 CC J Joules gC t f ) , . ()( ) () (,       = °             12 500 12 55 2...

Ngày tải lên: 22/07/2014, 10:22

... 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 3 04 3 29 cm 1 cm cm = =− = − q q CHAPTER 3 Peterson’s: www.p...

Ngày tải lên: 22/07/2014, 10:22

... other and rearranging yields: v v = ×       × () ×+× =× − 667 10 610 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 the t...

Ngày tải lên: 22/07/2014, 10:22

... 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 583 m @ 211 ° from ... (sin 35°) = (16 N) (.57) = 9.1N Side x = (side r) (cos 35°) = (16 N) (.82) = 13 .1N Both components of rope b are located in the third quadrant. y x negative N negative N =∴− =∴− 91...

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

... 10 .(.nm × −−− ×× ×× 7 14 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 7 14 .(. ).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 10...

Ngày tải lên: 22/07/2014, 10:22