1 / 36

Review for Final Exam

Review for Final Exam. Schedule. Review Session on Monday at 4:00 PM in 101 UPL if it extends beyond 6:00 PM it will move to 707 Keen. Final exam – Tuesday, April 23 at 7:30 AM in 101 Carroway building.

belva
Download Presentation

Review for Final Exam

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript


  1. Review for Final Exam

  2. Schedule • Review Session on Monday at 4:00 PM in 101 UPL if it extends beyond 6:00 PM it will move to 707 Keen. • Final exam – Tuesday, April 23 at 7:30 AM in 101 Carroway building. • This Power Point presentation contains 36 slides – most likely I won’t get to more than about 1/3 of them. Make sure you understand the concepts and problems presented here.

  3. B B I I B-Field Due to Currents • Electric currents produce magnetism. • B = oI/(2r) (due to a long straight wire) • Direction from the right-hand rule. Curl your fingers as if following B. Your thumb is in the direction of the current.

  4. q,v R B is out of the page. Example: Mass Spectrometer. F = qvB F = ma ma = qvB a = qvB/m qvB/m = v2/R  R = mv/(qB)

  5. q,v R B is out of the page. Example: Mass Spectrometer. F = qvB F = ma = mv2/R = qvB mv = qBR or p = qBR ½mv2 = (mv)2/(2m) = p2/(2m) Energy = ½mv2 = ½(qBR)2/m q < 0 !!

  6. B2 I1 B2 I2 F12 B2 I1 F21 F12 Forces Between Currents F12/L = I1B2 = I1oI2/(2r) = o I1I2/(2r)

  7. B1 B1 F12 F21 F21 B1 I2 Forces Between Currents I1 I2

  8. Major Concepts • Battery (voltages) • Ohm’s Law: V = IR • Resistance: R = V/I (in ohms “”) • Resistivity: R = L/A ( in /m) • Power: P = I2R (resistors) • Power: P = VI (batteries or resitors)

  9. Major Concepts • Series Circuit • Current must go through all resistors • RT = R1 + R2 + R3 + … • Parallel Circuit • Current is divided between the resistors • 1/RT = 1/R1 + 1/R2 + 1/R3 + … • Terminal Voltage • Accounts for the resistance within the battery. • Vterminal = V – Ir(internal)

  10. A B C Arrow shows the direction that positive charges move. + R = 24  V = 8 V - Current in a Simple Circuit V = IR  I = V/R I = 8 V/24  I = 0.33 Amp Current is the same everywhere in the circuit! Current at A = 0.33 Amps Current at B = 0.33 Amps Current at C = 0.33 Amps

  11. B C D A + R = 24  V = 8 V - Energy in a Simple Circuit Recall: I = 1/3 Amp Consider the energy of a proton moving through the circuit. (recall: q = +1 e) Energy(A) = qVA = 8 eV Energy(B) = qVB = 8 eV Energy(C) = qVC – q(IR) = 8eV – 1e * ( (1/3)*24 V ) = 0 eV Energy(D) = qVD = 0 eV Proton loses energy moving from B to C. It gains energy moving from D to A.

  12. B C D A + R = 24  V = 8 V - Energy in a Simple Circuit Recall: I = 1/3 Amp Consider the energy of a proton moving through the circuit. (recall: q = -1 e) Energy(D) = qVA = -8 eV Energy(C) = qVB = -8 eV Energy(B) = qVC – q(IR) = -8eV – -1e*((1/3)*24 V ) = 0 eV Energy(A) = qVD = 0 eV An electron loses energy moving from C to B. It gains energy moving from A to D.

  13. Circuit Analysis • Kirchhoff’s Rules: • Loop Rule: • The sum of the voltage drops around any closed loop is zero. • Conservation of Energy • V = 0 • Junction Rule: • The net current into and out of any point in a circuit is zero. • Charge conservation. • I = 0

  14. A B C I2 I3 H I1 E D Complex Circuits R1 • I1 goes from A to B and from E to H. • I2 goes from B to F to G to E. • I3 goes from B to C to D to E. F V R2 R3 G R1 = 10 , R2 = 3 , R3 = 6  and V = 24 Volts

  15. Complex Circuits R1 A B C • Junction Rule at B: • I1 - I2 - I3 = 0 • Loop Rule: ABFGEHA • -I1R1 – I2R2 + V = 0 • Loop Rule: ABCDEHA • -I1R1 – I3R3 + V = 0 F V I2 I3 R2 R3 H I1 G E D

  16. A B C F I2 I3 H I1 G E D Complex Circuits R1 • I1 - I2 - I3 = 0 • -I1R1 – I2R2 + V = 0 • So: -10I1 – 3I2 + 24 = 0 • -I1R1 – I3R3 + V = 0 • So: -10I1 – 6I3 + 24 = 0 V R2 R3

  17. Solving the equations. I1 - I2 - I3 = 0 10I1 + 3I2 - 24 = 0 10I1 + 6I3 - 24 = 0 10I1 + 3I2 - 24 = 0 - ( 10I1 + 6I3 - 24 = 0 ) 3I2 – 6I3 = 0 3I2 = 6I3 I2 = 2I3

  18. Solving the equations. I1 - I2 - I3 = 0 I2 = 2I3 I1 - 2I3 - I3 = 0 I1 - 3I3 = 0 I1 = 3I3 10I1 + 6I3 - 24 = 0  10*3I3 + 6I3 - 24 = 0 OR: 36I3 - 24 = 0  I3 = 24/36 = 0.667 A I1 = 3I3  I1 = 3* 0.667 A = 2.0 A

  19. Solving the equations. I1 - I2 - I3 = 0 I2 = 2I3 I3 = 24/36 = 2/3 A I1 = 3I3  I1 = 3* 2/3 A = 2 A I2 = 2I3 = 2 * 2/3 = 4/3 A

  20. Check the Equations • I1 = 2 A, I2 = 4/3 A, I3 = 2/3 A • I1 - I2 - I3 = 0 • 2 – 4/3 – 2/3 = 0  • 10I1 + 3I2 - 24 = 0 • 10*2 + 3*4/3 – 24 = 0  • 10I1 + 6I3 - 24 = 0 • 10*2 + 6*2/3 – 24 = 0 

  21. Balancing the Energy • P(in) = VI1 = 24 * 2 = 48 W • P1 = I12R1 = 22*10 = 40 W • P2 = I22R2 = (4/3)2*3 = 16/3 W • P3 = I32R3 = (2/3)2*6 = 8/3 W • PT = 40 + 16/3 + 8/3 = 48 W 

  22. Reflected Light The angle of incidence equals the angle of reflection. in out

  23. in out Refracted Light The direction changes when the light moves from material to another. The change depends on the material. Snell’s Law determines the angles: N1 sin 1 = N2 sin 2 The index of refraction, N, is the ratio of the speed of light in vacuum to the speed of light in the material. N = c/v c = speed of light in vacuum. v = speed of light in material. glass air

  24. 1 2 image Optic Axis object 3 Lenses 1 – Parallel Ray 2 – Central Ray 3 – Focal Ray

  25. image Optic Axis object Lens Equation 1 1 1 — = — + — F O I F O I

  26. Optic Axis Diverging Lens 1 3 object image 1 – Parallel Ray 2 – Central Ray 3 – Focal Ray 2

  27. Optic Axis Diverging Lens 1 3 object image F = - 20 cm O = 50 cm Use the Lens Equation 2

  28. Diverging Lens w/ Lens Equation 1 1 1 — = — + — F O I F = - 20 cm O = 50 cm 1/(-20) = 1/50 + 1/I  1/I = -1/20 – 1/50 = -5/100 – 2/100 1/I = -7/100  I = -100/7 = - 14.28 cm m = -I/O = - (-14.28 cm)/( 50 cm ) = + 2/7 m > 0  Image is upright. I < 0  image is virtual.

  29. Atomic Model • Electrons moved around nucleus only in certain stable orbits. • Stable orbits are those in which an integral number of wavelengths fit into the diameter of the orbit (2rn = n) • They emitted (absorbed) light only when they changed from one orbital to another. • Orbits have quanta of angular momenta. L = nh/2 • Orbit radius increases with energy rn = n2 r1 (r1 = .529 x 10-10 m)

  30. Ionized atom n = 3 n = 2 - 3.4 eV n = 1 E = - 13.6 eV Atomic Energy Levels • En = Z2/n2 E1 • Hydrogen • En = - 13.6 eV/n2

  31. Emission & Absorption • Energy is conserved. • E = Eatom = Ef - Ei • Photon energy = hf = hc/ • Absorption  photon disappears a electron in the atom changes from a lower energy level to a higher energy level. • Emission  an electron in atom goes from higher energy level to a lower energy level. This change in energy is the energy of the photon.

  32. Heisenberg Uncertainty Principle • Impossible to know both the position and the momentum of a particle precisely. • A restriction (or measurement) of one, affects the other. • x p  h/(2) • Similar constraints apply to energy and time. • E t  h/(2) EXAMPLE: If an electron's position can be measured to an accuracy of 1.96×10-8 m, how accurately can its momentum be known? x p  h/(2)  p = h/(2x) p = 6.63x10-34 Js /(2 1.96x10-8 m) = 5.38 x 10-27 N s

  33. Nuclear Decay Rates • N = -No t • Number of decaying nuclei (in a given time t), depends on: • Number of remaining nuclei, No • Nuclear decay constant for that type of nuclei,  • Number of nuclei remaining after a time t: • N = No e-t

  34. Nuclear Decay Rates At t =  N is 1/e (0.368) of the original amount

  35. ( , ) +  + ( , ) Example • Start with 5 protons  end up with 5 protons. • Start wit 11 baryons  end up with 11 baryons. • Q-value: (Mass Energy of Final State – Mass Energy of Initial State) Mn = 1.008665 u + M(B) = 10.012936 u = 11.021601 u M = 4.002602 u + M(L) = 7.016003 u = 11.018605 u Q = M (in MeV) = Mf – Mi = 11.018605 u - 11.021601 u Q = -0.002996 u = -0.002996 u * 931.5 MeV/u = -2.79 MeV Negative Q-value means we get extra energy out of the reaction.

  36. Example: Nuclear Power • Suppose that the average power consumption, day and night, in a typical house is 340 W. What initial mass of 235U would have to undergo fission to supply the electrical needs of such a house for a year? (Assume 200 MeV is released per fission.) Energy used = rate * time = 340 W * 3.15 x 107 s/yr = 1.07 x 1010 J Energy from each nucleus is: 200 MeV/nuclei = 200 MeV /nuclei * 1.60 x 10-13J/MeV = 3.2 x 10-11 J/nuclei Number of nuclei required = Energy/(Energy per nucleus) Number of nuclei = 1.07 x 1010 J/3.2 x 10-11 J/nuclei = 3.34 x 1020 nuclei Total mass of 235U = number of nuclei * mass/nucleus Mass = 3.34 x 1020 nuclei * 235 amu * 1.66 x 10-27 kg/amu = 1.31 10-4 kg

More Related