Download presentation

Presentation is loading. Please wait.

Published byAshlynn Bach Modified over 3 years ago

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

4
Example: Mass Spectrometer. F = qvB F = ma ma = qvB a = qvB/m qvB/m = v 2 /R R = mv/(qB) R q,v B is out of the page.

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

6
Forces Between Currents B2B2 B2B2 I1I1 F 21 F 12 I2I2 B2B2 I1I1 F 12 /L = I 1 B 2 = I 1 o I 2 /(2 r) = o I 1 I 2 /(2 r)

7
Forces Between Currents I1I1 F 21 F 12 I2I2 B1B1 B1B1 B1B1 F 21 I2I2

8
Major Concepts Battery (voltages) Ohm’s Law: V = IR Resistance:R = V/I (in ohms “ ”) Resistivity:R = L/A ( in /m) Power:P = I 2 R (resistors) Power:P = VI (batteries or resitors)

9
Major Concepts Series Circuit Current must go through all resistors R T = R 1 + R 2 + R 3 + … Parallel Circuit Current is divided between the resistors 1/R T = 1/R 1 + 1/R 2 + 1/R 3 + … Terminal Voltage Accounts for the resistance within the battery. V terminal = V – Ir (internal)

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

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

12
Energy in a Simple Circuit V = 8 V + - R = 24 Recall: I = 1 / 3 Amp Consider the energy of a proton moving through the circuit. (recall: q = -1 e) Energy(D) = qV A = -8 eV Energy(C) = qV B = -8 eV Energy(B) = qV C – q(IR) = -8eV – -1e*(( 1 / 3 )*24 V ) = 0 eV Energy(A) = qV D = 0 eV B C D A 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
Complex Circuits R1R1 R2R2 V R3R3 I1I1 I2I2 I3I3 F G A B C D E H 1.I 1 goes from A to B and from E to H. 2.I 2 goes from B to F to G to E. 3.I 3 goes from B to C to D to E. R 1 = 10 , R 2 = 3 , R 3 = 6 and V = 24 Volts

15
Complex Circuits R1R1 R2R2 V R3R3 I1I1 I2I2 I3I3 A B C D E F G H 1.Junction Rule at B: I 1 - I 2 - I 3 = 0 2.Loop Rule: ABFGEHA -I 1 R 1 – I 2 R 2 + V = 0 3.Loop Rule: ABCDEHA -I 1 R 1 – I 3 R 3 + V = 0

16
Complex Circuits R1R1 R2R2 V R3R3 I1I1 I2I2 I3I3 A B C D E F G H I 1 - I 2 - I 3 = 0 -I 1 R 1 – I 2 R 2 + V = 0 So: -10I 1 – 3I 2 + 24 = 0 -I 1 R 1 – I 3 R 3 + V = 0 So: -10I 1 – 6I 3 + 24 = 0

17
Solving the equations. I 1 - I 2 - I 3 = 0 10I 1 + 3I 2 - 24 = 0 10I 1 + 6I 3 - 24 = 0 3I 2 – 6I 3 = 0 3I 2 = 6I 3 I 2 = 2I 3 10I 1 + 3I 2 - 24 = 0 - ( 10I 1 + 6I 3 - 24 = 0 )

18
Solving the equations. I 1 - I 2 - I 3 = 0 I 2 = 2I 3 I 1 - 2I 3 - I 3 = 0 I 1 - 3I 3 = 0 I 1 = 3I 3 10I 1 + 6I 3 - 24 = 0 10*3I 3 + 6I 3 - 24 = 0 OR: 36I 3 - 24 = 0 I 3 = 24/36 = 0.667 A I 1 = 3I 3 I 1 = 3* 0.667 A = 2.0 A

19
Solving the equations. I 1 - I 2 - I 3 = 0 I 2 = 2I 3 I 3 = 24/36 = 2/3 A I 1 = 3I 3 I 1 = 3* 2/3 A = 2 A I2 = 2I3 = 2 * 2/3 = 4/3 A

20
Check the Equations I 1 = 2 A, I 2 = 4/3 A, I 3 = 2/3 A I 1 - I 2 - I 3 = 0 2 – 4/3 – 2/3 = 0 10I 1 + 3I 2 - 24 = 0 10*2 + 3*4/3 – 24 = 0 10I 1 + 6I 3 - 24 = 0 10*2 + 6*2/3 – 24 = 0

21
Balancing the Energy P (in) = VI 1 = 24 * 2 = 48 W P 1 = I 1 2 R 1 = 2 2 *10 = 40 W P 2 = I 2 2 R 2 = (4/3) 2 *3 = 16/3 W P 3 = I 3 2 R 3 = (2/3) 2 *6 = 8/3 W P T = 40 + 16/3 + 8/3 = 48 W

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

23
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: N 1 sin 1 = N 2 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. in out air glass

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

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

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

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

28
Diverging Lens w/ Lens Equation 1 1 1 — = — + — F O I 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. F = - 20 cm O = 50 cm

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 r n = 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 r n = n 2 r 1 (r 1 =.529 x 10 -10 m)

30
Atomic Energy Levels E n = Z 2 /n 2 E 1 Hydrogen E n = - 13.6 eV/n 2 Ionized atom E = - 13.6 eV - 3.4 eV n = 1 n = 2 n = 3

31
Emission & Absorption Energy is conserved. E = E atom = E f - E i 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 = -N o t Number of decaying nuclei (in a given time t), depends on: Number of remaining nuclei, N o Nuclear decay constant for that type of nuclei, Number of nuclei remaining after a time t: N = N o 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 1. Suppose that the average power consumption, day and night, in a typical house is 340 W. What initial mass of 235 U 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 10 7 s/yr = 1.07 x 10 10 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 10 10 J/3.2 x 10 -11 J/nuclei = 3.34 x 10 20 nuclei Total mass of 235U = number of nuclei * mass/nucleus Mass = 3.34 x 10 20 nuclei * 235 amu * 1.66 x 10 -27 kg/amu = 1.31 10 -4 kg

Similar presentations

Presentation is loading. Please wait....

OK

Direct-Current Circuits

Direct-Current Circuits

© 2018 SlidePlayer.com Inc.

All rights reserved.

To make this website work, we log user data and share it with processors. To use this website, you must agree to our Privacy Policy, including cookie policy.

Ads by Google

Training ppt on communication skills Ppt on area of parallelogram and triangles for class 9 Ppt on mahatma gandhi biography Ppt on environment and development Ppt on email etiquettes presentation tips Ppt on real numbers for class 9th computer Convert pdf ppt to ppt online training Ppt on water activity meter Download ppt on fibre to fabric Download ppt on third law of motion