... 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 2 . 024 5m m/s 2 298 2 800 2 •• •− =− . 26 .26 . 26 .26 . 26 . The corThe cor The corThe cor The cor rr rr r ect ... 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...
Ngày tải lên: 22/07/2014, 10:22
... the denominator. FK qq r F m m F = = ו × =× − − − ()() . (. ) . 12 2 28 2 52 24 23 10 25 10 92 10 N N Comparing the two forces: Ratio of forces N N = × × = − − 92 10 23 10 4 1 24 24 . . The force on the electrons is four ... 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...
Ngày tải lên: 22/07/2014, 10:22
SAT II Physics (Gary Graff) Episode 1 Part 8 pps
... equivalent values to the values of R 1 , R 5, and R 8, yielding: R R t t =++++ = 24 369 24 ΩΩΩΩΩ Ω 2. 2. 2. 2. 2. Use Ohm’s La Use Ohm’s La Use Ohm’s La Use Ohm’s La Use Ohm’s La w to fw to ... 5 Peterson’s SAT II Success: Physics 190 Solution V V V V V V V s p s p s ps p s s = = = = N N N N turns turns ()( ) ()( ) , 8000 900 30 000 24 00V CHAPTER 5 Peterson’s SAT II S...
Ngày tải lên: 22/07/2014, 10:22
SAT II Physics (Gary Graff) Episode 1 Part 6 docx
... PROPERTIES Peterson’s SAT II Success: Physics 126 The mathematical relationship between the velocity of light in one material (usually air) compared to another material was determined to be nn 1 12 2 sin sinθθ= ... index of refraction (n) for water is 1.33 and for air is 1. Stating Snell’s Law nn n n 1 12 2 2 11 2 2 2 133 5 1 655 sin sin sin sin sin ( . )(. ) . sin θθ θ θ θ θ =...
Ngày tải lên: 22/07/2014, 10:22
SAT II Physics (Gary Graff) Episode 1 Part 5 docx
... PROPERTIES Peterson’s: www.petersons.com 97 Work Nm PE kg m/s m Nm KE 1 2 kg m 2 =• =• = = =• • =• = = =• Fs Joule mgh Joule mv 2 22 2 s Nm=• = Joule It should be pointed out that the Work and PE equations ... SUMMARY Peterson’s SAT II Success: Physics 98 The 25 0N boulder shown in the diagram above possesses PE, which can be stated in equation form below. PE = mgh = wt h = (25...
Ngày tải lên: 22/07/2014, 10:22
SAT II Physics (Gary Graff) Episode 1 Part 4 ppsx
... (deceleration). v vas v a s s s f o o 2 2 2 2 2 0 02 2 17 5 21 25 1 = =+− − − = − − = = () () () (. ) (. m/s m/s) 22 7 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 ... becomes: 0 1 2 2 98 26 22 6 12 74 2 2 2 =+ −= −= −= vt gt v gt t v v o o o o (. )(. ) (. ) . m/s s s m/ 2 ss 2 Note: The negati...
Ngày tải lên: 22/07/2014, 10:22
SAT II Physics (Gary Graff) Episode 1 Part 3 doc
... 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 ... begin by isolating a in the equation below: V f 2 = V 0 2 + 2as We then subtract V o 2 from both sides: V f 2 –V o 2 = V o 2 –V o 2 +...
Ngày tải lên: 22/07/2014, 10:22
SAT II Physics (Gary Graff) Episode 1 Part 1 pot
... q 1 = n 2 sin q 2 (Snell’s Law) nλ = d sinq PV = n r t P 1 V 1 = P 2 V 2 V T V T 1 1 2 2 = P T P T 1 1 2 2 = ()() ()()PV T PV T 11 1 22 2 = ∆Q = ∆U + ∆W Q = cm∆T FK qq r = ()() 12 2 F = EQ V ... Modern Physics 193 Particulate Theory of Light 195 Photoelectric Effect 198 Relativity 20 1 Chapter Summary 21 1 Chapter 7: The Atom 21 3 The Atom 21 5 Radioactivity 22 4 Pa...
Ngày tải lên: 22/07/2014, 10:22
SAT II Physics (Gary Graff) Episode 2 Part 8 pptx
... 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 C O D O E 18 O A...
Ngày tải lên: 22/07/2014, 10:22
SAT II Physics (Gary Graff) Episode 2 Part 7 pot
... 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. A 33. ... together. III. The cooler gas decreases in temperature when placed into the second container. (A) I only (B) II only (C) I and III only (D) II and III only (E) I, II, a...
Ngày tải lên: 22/07/2014, 10:22