SAT II Physics (Gary Graff) Episode 1 Part 8 pps

... 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 82 5 12 1 R R R R t t t t =++ + =+++ = = ΩΩ Ω Ω ΩΩΩ Ω Ω . . The value of ... 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 . CHAPTER 5 Peterson’s SAT II Success: Physics 1...

Ngày tải lên: 22/07/2014, 10:22

... one another. STATICS Peterson’s SAT II Success: Physics 86 Notice the g instead of a. The object starts from rest, which allows us to eliminate the (v o t) term, leaving the working equation: sgt s s = = = 1 2 1 2 98 88 5 2 (. . m/s ... the already known F 1 . The calculation of the torques is: TF F T wt wt =+()( ) 11  There are two torque-producing entities. ==•+• =•+•...

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. ... same throughout. Both teams must pull with a force of 12 00N to maintain static equilibrium, so the total tension is 2400N....

Ngày tải lên: 22/07/2014, 10:22

... ) −       − ° ++ = + ° − 26 13 1 25 19 5 19 50 83 6 8 ((, . ) ,. . . , 26 13 1 25 26 13 1 25 19 50 83 6 8 19 5 28 0 81 J JJ J C t J C t J ff +=+ ° + ° == ° = °= 85 6 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 2 304 10 23 10 28 2 2 24 . . N 11 0 4 That’s t...

Ngày tải lên: 22/07/2014, 10:22

... convex lens of 8 cm focal length. Locate and 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 ... 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.pet...

Ngày tải lên: 22/07/2014, 10:22

... 2 Peterson’s SAT II Success: Physics 10 8 Example A bicyclist coasts a distance of 18 0 m in 30 seconds. How many revolutions do the wheels make in that time? The bicycle wheels are 80 cm in diameter. Solution The ... 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...

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 =∴− =∴−...

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 decimil...

Ngày tải lên: 22/07/2014, 10:22

... WS 1 O A O B O C O D O E 2 O A O B O C O D O E 3 O A O B O C O D O E 4 O A O B O C O D O E 5 O A O B O C O D O E 6 O A O B O C O D O E 7 O A O B O C O D O E 8 O A O B O C O D O E 9 O A O B O C O D O E 10 O A O B O C O D O E 11 O A O B O C O D O E 12 O A O B O C O D O E 13 O A O B O C O D O E 14 O A O B O C O D O E 15 O A O B O C O D O E 16 O A O B O C O D O E 17 O A O B O...

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. ... immediately because: I. Heat is melting the ice. II. Condensation is occurring. III. Phase change occurs. (A) I only (B) II only (C) I and III only (D) II and III...

Ngày tải lên: 22/07/2014, 10:22