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

... 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

... 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 5 14 12 2 2 22 ( . )( sec)m/s m CHAPTER ... (deceleration). v vas v a s s s f o o 2 2 2 2 2 0 02 2 17 5 21 2 5 1 = =+− − − = − − = = () () () (. ) (. m/s m/s) 22 7 6.m KINEMATICS Peterson’...

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...

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 ... The missing value in the 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...

Ngày tải lên: 22/07/2014, 10:22

... convex mirror. 11 1 11 1 36 1 37 1 27 7 − =+ − −= − −= −− fpq fpq q rearranges to 1 cm cm cm . 0 027 1 304 3 29 cm 1 cm cm = =− = − q q CHAPTER 3 Peterson’s: www.petersons.com 12 9 Concave lenses ... placed 20 cm from a 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...

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 = ×       × () ×+× =× − 667 10 610 64 10 3 81 ... 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 gr...

Ngày tải lên: 22/07/2014, 10:22

... quadrant. y x =∴+ =∴+ positive N positive N 19 4 52 . . CHAPTER 1 Peterson’s SAT II Success: Physics 48 This leads to: V f 2 – V o 2 = 2as Then we divide both sides by 2s: VV s as s f 2 0 2 2 2 2 − = Clearing fractions ... 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...

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 ... f f = 1 Τ h h q p i 0 = 11 1 fpq =+ n 1 sin q 1 = n 2 sin q 2 (Snell’s Law) nλ = d sinq PV = n r t P 1 V 1 = P 2 V...

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 ... 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

... 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