SAT II Physics (Gary Graff) Episode 2 Part 1 potx

... 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 ... potential difference of 1 volt. (Remember, 1 volt is equal to 1 Joule per coulomb of charge.) Work = = × () () =× = × − − QV ev .C V eV J 1 16 10 1 1 1 6...

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 II...

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. 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...

Ngày tải lên: 22/07/2014, 10:22

... reciprocals. That is to do the following 11 111 12 3 4 RRRRR t =+++ yielding the decimal solution 1 05 025 016 7 0 12 4 R t =+ + + . Which is 1 10 42 9 6 R t == Ω . 17 .17 . 17 .17 . 17 . The corThe cor The corThe ... ideal gas law equation PV T PV T T VT V T LK L 11 1 22 2 2 21 1 2 2 300 4 15 ==== converts into . ()( ) 00K . PRACTICE TEST 2 Peterson’s SAT II...

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. ... light. II. The beam is below threshold. III. The beam does not contain any photons. (A) I only (B) II only (C) I and III only (D) II and III on...

Ngày tải lên: 22/07/2014, 10:22

... 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 . C 23 . D 24 . A 25 . E 26 . E 27 . C 28 . D 29 . C 30. B 31. C 32. ... than 2. 4 m/s 2 . (B) toward him at 2. 4 m/s 2 . (C) downward at 2. 4 m/s 2 . (D) greater than 2. 4 m/s 2 . (E) none of these. PHYSICS...

Ngày tải lên: 22/07/2014, 10:22

... is 23 5.043 923 1 The mass of is 13 9.9 23 5 14 0 Uu Ba 11 05995 The mass of is 91. 926 1 528 92 u Kr u 1 0 23 5 92 14 0 56 92 36 4 1 0 1 008665 23 5 043 923 1 139 9 nU BaKrn uu +→ ++ +→ .11 0599 91 926 1 528 ... . ()(. = ×=× − −− − 6 0 519 9 6 9 10 94 10 5 4656 10 65 31 10 kg kg or 4486 10 3 29 16 10 43 ×= × −− uu). The total mass of the carbon 12 at...

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. ... of . 024 5m. Stating the equations: and but an Vgd Vad VV f f 2 0 2 2 0 2 2 2 = =− = dd m/s m...

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 3 45 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...

Ngày tải lên: 22/07/2014, 10:22