SAT II Physics (Gary Graff) Episode 1 Part 7 pdf

... )(. ) (. ) 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 ... problem is the volume (V 2 ). ()() ()() : ()()( PV T PV T V V PVT 11 1 22 2 2 11 = = Rearranging to find 2 22 21 2 1 4 5 325 1 75 273 306 ) ()() ()(.)( ) (. )( )...

Ngày tải lên: 22/07/2014, 10:22

... 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 =++ + =+++ = = ΩΩ Ω Ω ΩΩΩ Ω Ω . . The value of ... switch. CHAPTER 5 Peterson’s: www.petersons.com 17 3 Next we find the equivalent resistance for the R 6 – R 7 pair. 11 1 11 10 1 15 1 16 67 6 67 RRR R R R t t t t =+...

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 . 00 27 1 304 3 29 cm 1 cm cm = =− = − q q CHAPTER 3 Peterson’s: www.petersons.com 12 9 Concave lenses ... length of 3.6 cm. Describe the image. Solution 11 1 11 1 36 1 37 1 277 0 27 fpq fpq q =+ −= −= − rearranges to 1 cm cm cm cm . 1 cm 4cm = == q q 1 025. The...

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 = ×       × () ×+× =× − 6 67 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

... (deceleration). v vas v a s s s f o o 2 2 2 2 2 0 02 2 17 5 212 5 1 = =+− − − = − − = = () () () (. ) (. m/s m/s) 2 27 6.m KINEMATICS Peterson’s SAT II Success: Physics 74 Solution Σ Σ Σ Σ ΣΣ F F F F FF x y z xz = = = = == 0 0 0 0 00 ... 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

... (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 13 1 . . SCALARS ... AND VECTORS Peterson’s SAT II Success: Physics 64 Rope c extends 75 ° into the first quadrant. Side y = (side r) (sin 75 ° ) = (20N) (. 97) = 19 .4N Side x = (side r) (cos...

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 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 Success: Ph...

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

... ANSWERS 1. E 2. D 3. A 4. A 5. D 6. C 7. C 8. E 9. A 10 . C 11 . B 12 . E 13 . C 14 . E 15 . D 16 . C 17 . D 18 . C 19 . D 20. B 21. B 22. A 23. B 24. A 25. D 26. C 27. C 28. E 29. E 30. B 31. C 32. ... the following? I. Negative work II. An isothermal process III. Increasing energy (A) I only (B) II only (C) I and III only (D) II and III only (E) I, II, and III...

Ngày tải lên: 22/07/2014, 10:22

... 235.04392 31 The mass of is 13 9.9 235 14 0 Uu Ba 11 05995 The mass of is 91. 92 615 28 92 u Kr u 1 0 235 92 14 0 56 92 36 4 1 0 1 008665 235 04392 31 139 9 nU BaKrn uu +→ ++ +→ .11 0599 91 92 615 28 4 03466 236 ... the carbon atom we have: ()(. ). ()(. ). 6 1 672 6 10 1 0036 10 6 1 0 072 76 6 0 27 26 p p u + −− + ×=× = kg kg or 443656 6 1 674 9 10 1 0049 10 6...

Ngày tải lên: 22/07/2014, 10:22