Final on December Physics 106 R. Schad. 3e 4e 5c 6d 7c 8d 9b 10e 11d 12e 13d 14d 15b 16d 17b 18b 19c 20a

Similar documents
Turn in scantron You keep these question sheets

Physics 208, Spring 2016 Exam #3

Louisiana State University Physics 2102, Exam 3 April 2nd, 2009.

YOUR NAME. Exam 3 NOV Physics 106 -R. Schad

Turn in scantron You keep these question sheets

PHY 131 Review Session Fall 2015 PART 1:

Exam 2 Solutions. Note that there are several variations of some problems, indicated by choices in parentheses.

Where k = 1. The electric field produced by a point charge is given by

8. (6) Consider the circuit here with resistors R A, R B and R C. Rank the

Exam 4 Solutions. a. 1,2,and 3 b. 1 and 2, not 3 c. 1 and 3, not 2 d. 2 and 3, not 1 e. only 2

Gen. Phys. II Exam 2 - Chs. 21,22,23 - Circuits, Magnetism, EM Induction Mar. 5, 2018

we can said that matter can be regarded as composed of three kinds of elementary particles; proton, neutron (no charge), and electron.

Louisiana State University Physics 2102, Exam 2, March 5th, 2009.

A) I B) II C) III D) IV E) V

Final Exam: Physics Spring, 2017 May 8, 2017 Version 01

Physics 2B Winter 2012 Final Exam Practice

Describe the forces and torques exerted on an electric dipole in a field.

PHY2049 Fall11. Final Exam Solutions (1) 700 N (2) 350 N (3) 810 N (4) 405 N (5) 0 N

PHYS 241 EXAM #2 November 9, 2006

b) (4) How large is the current through the 2.00 Ω resistor, and in which direction?

Name (Print): 4 Digit ID: Section:

SUMMARY Phys 2523 (University Physics II) Compiled by Prof. Erickson. F e (r )=q E(r ) dq r 2 ˆr = k e E = V. V (r )=k e r = k q i. r i r.

Physics 420 Fall 2004 Quiz 1 Wednesday This quiz is worth 6 points. Be sure to show your work and label your final answers.

Physics 202 Final Exam Dec 20nd, 2011

P202 Practice Exam 2 Spring 2004 Instructor: Prof. Sinova

Physics Final. Last Name First Name Student Number Signature

1. In Young s double slit experiment, when the illumination is white light, the higherorder fringes are in color.

Yell if you have any questions

A) n 1 > n 2 > n 3 B) n 1 > n 3 > n 2 C) n 2 > n 1 > n 3 D) n 2 > n 3 > n 1 E) n 3 > n 1 > n 2

Questions A hair dryer is rated as 1200 W, 120 V. Its effective internal resistance is (A) 0.1 Ω (B) 10 Ω (C) 12Ω (D) 120 Ω (E) 1440 Ω

Profs. P. Avery, A. Rinzler, S. Hershfield. Final Exam Solution

n Higher Physics 1B (Special) (PHYS1241) (6UOC) n Advanced Science n Double Degree (Science/Engineering) n Credit or higher in Physics 1A

Fundamental Constants

2. Waves with higher frequencies travel faster than waves with lower frequencies (True/False)

Exam 3 Solutions. Answer: 1830 Solution: Because of equal and opposite electrical forces, we have conservation of momentum, m e

Physics 227 Final Exam December 18, 2007 Prof. Coleman and Prof. Rabe. Useful Information. Your name sticker. with exam code

FINAL EXAM - Physics Patel SPRING 1998 FORM CODE - A

For more sample papers visit :

Chapter 1 The Electric Force

PHY2054 Exam II, Fall, Solutions

PHYS 212 Final Exam (Old Material) Solutions - Practice Test

Exam 2, Phy 2049, Spring Solutions:

Circuits Capacitance of a parallel-plate capacitor : C = κ ε o A / d. (ρ = resistivity, L = length, A = cross-sectional area) Resistance : R = ρ L / A

Form #425 Page 1 of 6

Physics 2020 Exam 2 Constants and Formulae

Physics GRE: Electromagnetism. G. J. Loges 1. University of Rochester Dept. of Physics & Astronomy. xkcd.com/567/

Profs. Y. Takano, P. Avery, S. Hershfield. Final Exam Solution

Exam 2 Fall 2014

Solution for Fq. A. up B. down C. east D. west E. south

The next two questions pertain to the situation described below. Consider a parallel plate capacitor with separation d:

Select the response that best answers the given statement. Be sure to write all final multiple choice answers on your Scantron answer sheet.

PHYSICS : CLASS XII ALL SUBJECTIVE ASSESSMENT TEST ASAT

University of the Philippines College of Science PHYSICS 72. Summer Second Long Problem Set

Physics 208 Final Exam December 15, 2008

Physics 208 Final Exam December 15, 2008

Last Homework. Reading: Chap. 33 and Chap. 33. Suggested exercises: 33.1, 33.3, 33.5, 33.7, 33.9, 33.11, 33.13, 33.15,

Name (Last, First): You may use only scientific or graphing calculators. In particular you may not use the calculator app on your phone or tablet!

YOUR NAME Sample Final Physics 1404 (Dr. Huang)), Correct answers are underlined.

Mansfield Independent School District AP Physics C: Electricity and Magnetism Year at a Glance

MASSACHUSETTS INSTITUTE OF TECHNOLOGY Department of Physics Spring 2014 Final Exam Equation Sheet. B( r) = µ o 4π

1 cm b. 4.4 mm c. 2.2 cm d. 4.4 cm v

r r 1 r r 1 2 = q 1 p = qd and it points from the negative charge to the positive charge.

Exam 2 Solutions. ε 3. ε 1. Problem 1

= 8.89x10 9 N m 2 /C 2

Physics 240 Fall 2005: Exam #3 Solutions. Please print your name: Please list your discussion section number: Please list your discussion instructor:

Physics 2B Spring 2010: Final Version A 1 COMMENTS AND REMINDERS:

Figure 1 A) 2.3 V B) +2.3 V C) +3.6 V D) 1.1 V E) +1.1 V Q2. The current in the 12- Ω resistor shown in the circuit of Figure 2 is:

Yell if you have any questions

Capacitance, Resistance, DC Circuits

Magnets. Domain = small magnetized region of a magnetic material. all the atoms are grouped together and aligned

Good Luck! Mlanie LaRoche-Boisvert - Electromagnetism Electromagnetism and Optics - Winter PH. Electromagnetism and Optics - Winter PH

2) A linear charge distribution extends along the x axis from 0 to A (where A > 0). In that region, the charge density λ is given by λ = cx where c

Magnetic Fields; Sources of Magnetic Field

Phys102 Final-132 Zero Version Coordinator: A.A.Naqvi Wednesday, May 21, 2014 Page: 1

Physics 6B Summer 2007 Final

= e = e 3 = = 4.98%

Two point charges, A and B, lie along a line separated by a distance L. The point x is the midpoint of their separation.

Physics 2020 Exam 1 Constants and Formulae

21 MAGNETIC FORCES AND MAGNETIC FIELDS

Exam 2 Solutions. PHY2054 Spring Prof. Paul Avery Prof. Pradeep Kumar Mar. 18, 2014

Physics 212 Midterm 2 Form A

On my honor, I have neither given nor received unauthorized aid on this examination.

cancel each other out. Thus, we only need to consider magnetic field produced by wire carrying current 2.

TOPPER SAMPLE PAPER 4 Class XII- Physics Solutions. Time: Three hours Max. Marks: 70

General Physics II Summer Session 2013 Review Ch - 16, 17, 18

Physics 102 Spring 2006: Final Exam Multiple-Choice Questions

EXAM 3: SOLUTIONS. B = B. A 2 = BA 2 cos 0 o = BA 2. =Φ(2) B A 2 = A 1 cos 60 o = A 1 2 =0.5m2

Final Exam Solutions

AP Physics C. Magnetism - Term 4

/20 /20 /20 /60. Dr. Galeazzi PHY207 Test #3 November 20, I.D. number:

Physics 55 Final Exam Fall 2012 Dr. Alward Page 1

Review. Spring Semester /21/14. Physics for Scientists & Engineers 2 1

Exam 4 (Final) Solutions

Physics 208 Final Exam May 12, 2008

PHYS 1102 EXAM - II. SECTION: (Circle one) 001 (TH 9:30 AM to 10:45AM) 002 (TH 3:30 PM to 4:45 PM) You have 1 hr 45 minutes to complete the test

Calculus Relationships in AP Physics C: Electricity and Magnetism

ELECTROMAGNETIC INDUCTION AND FARADAY S LAW

Maxwell s equations and EM waves. From previous Lecture Time dependent fields and Faraday s Law

Problem Score 1 /30 2 /15 3 /15 4 /20 5 /20 6 /20 7 /15 8 /25 9 /20 10 /20 Total /200

a. Clockwise. b. Counterclockwise. c. Out of the board. d. Into the board. e. There will be no current induced in the wire

Transcription:

Final on December11. 2007 - Physics 106 R. Schad YOUR NAME STUDENT NUMBER 3e 4e 5c 6d 7c 8d 9b 10e 11d 12e 13d 14d 15b 16d 17b 18b 19c 20a

1. 2. 3. 4. This is to identify the exam version you have IMPORTANT Mark the A This is to identify the exam version you have IMPORTANT Mark the B A 120 volt outlet is protected by a 10 amp circuit breaker. What is the maximum number of 60 watt light bulbs that can be connected in parallel to the outlet before setting off the circuit breaker? a. 2 b. 8 c. 10 d. 14 e. 20 Radio waves differ from light waves in empty space because they a. Travel slower than light waves. b. Travel faster than light waves. c. Have a higher frequency and longer wavelength than light. d. Have a shorter wavelength and lower frequency than light. e. Have a longer wavelength and lower frequency than light.

5. At what incident angle θ does a fish in the ocean see the sunset? (The index of refraction for water is n = 1.333.) 6. a. 90º b. 0.75 radians c. 49º d. 41º e. The sunset is not visible to the fish. The figure shows a circular loop of wire being dropped toward a wire carrying a constant current to the left. The direction of the induced current in the loop of wire is which of the following? a. clockwise b. zero c. impossible to determine d. counterclockwise

7. You observe an interference pattern like the one shown in the figure for three slits. If you then cover up the center slit so that no light gets through, the interference pattern that you will observe will be most like which of the following? a. the original three slit pattern but with every other maximum missing b. the original three slit pattern but with every third maximum missing c. the two slit pattern shown in the figure but with half the distance between maxima d. the two slit pattern shown in the figure but with every other maximum missing e. the two slit pattern shown in the figure 8. A positively charged particle is moving in the +y-direction when it enters a region with a uniform electric field pointing in the +x-direction. Which of the diagrams below shows its path while it is in the region where the electric field exists. The region with the field is the region between the plates bounding each figure. The field lines always point to the right. The x-direction is to the right; the y-direction is up. q q q q q (a) (b) (c) (d) (e)

9. A conducting spherical shell (below) is concentric with a solid conducting sphere. Initially, each conductor carries zero net charge. A charge of +2Q is placed on the outer sphere. After equilibrium is achieved, the charges on the surface of the solid sphere, q 1, the inner surface of the spherical shell, q 2, and the outer surface of the spherical shell, q 3, are which of the following? A) q 1 = -2Q, q 2 = +2Q, q 3 = 0 B) q 1 = 0, q 2 = 0, q 3 = +2Q C) q 1 = 0, q 2 = +2Q0, q 3 = 0 D) q 1 = +2Q, q 2 = -2Q, q 3 = +2Q E) q 1 = 0, q 2 = -2Q, q 3 = +2Q 10. R1 = 1 kω R2 = 1 kω Capacitance = 10 µf Battery voltage = 5 V We close the switch, and immediately after this the voltage drop across R2 is: a. 1.0 V b. 2.0 V c. 5.0 V d. 2.5 V e. zero

11. R1 = 1 kω R2 = 1 kω Capacitance = 10 µf Battery voltage = 5 V We close the switch, and after we waited very long the voltage drop across R2 is: a. 1.0 V b. 2.0 V c. 5.0 V d. 2.5 V e. zero 12. R1 = 1 kω R2 = 1 kω Capacitance = 10 µf Inductance L = 2H Battery voltage = 5 V We close the switch, and after we waited very long the voltage drop across R2 is: a. 1.0 V b. 2.0 V c. 5.0 V d. 2.5 V e. zero

13. The switch in the figure is closed at t = 0 when the current I is zero. When I = 15 ma, what is the potential difference across the inductor? I 240 V 12 mh 4.0 kω a. 240 V b. 60 V c. 0 d. 180 V e. 190 V 14. A bar magnet is dropped from above and falls through the loop of wire shown below. The north pole of the bar magnet points downward towards the page as it falls. Which statement is correct? S S N a. The current in the loop always flows in a clockwise direction. b. The current in the loop always flows in a counterclockwise direction. c. The current in the loop flows first in a clockwise, then in a counterclockwise direction. d. The current in the loop flows first in a counterclockwise, then in a clockwise direction. e. No current flows in the loop because both ends of the magnet move through the loop.

15. 16. A 3 cm high virtual image of a 2cm high object is seen through a lens. The object is 15 cm s away from the lens. What is the focal length of the lens? a. 22.5 cm b. 45 cm c. 25 cm d. 3.5 cm e. 33 cm The figure shows a graphical representation of the field magnitude versus time for a magnetic field that passes through a fixed loop and is oriented perpendicular to the plane of the loop. The magnitude of the magnetic field at any time is uniform over the area of the loop. Rank the magnitudes of the emf = V generated in the loop at the five instants indicated, from largest to smallest. (Use only the symbols > or =, for example a > b > c = d > e.) a. d > c > e > b > a b. a > b > c > d = e c. c = d > e = b > a d. c > d = e > b > a e. b = c = d = e > a

17. The electric potential energy of an electron is monotonously increasing in the direction of the +x axis. From this we can conclude that a) V is monotonously increasing in the +x-direction and E is constant and in the +x-direction b) V is monotonously decreasing in the +x-direction and E is constant and in the +x-direction c) V is monotonously increasing in the +x-direction and E is constant and in the -x-direction d) V is monotonously decreasing in the +x-direction and E is constant and in the -x-direction e) V is constant and E is zero 18. 19. A particle (charge = +2.0 mc) moving in a region where only electric forces act on it has a kinetic energy of 5.0 J at point A. The particle subsequently passes through point B which has an electric potential of +1.5 kv relative to point A. Determine the kinetic energy of the particle as it moves through point B. a. 3.0 J b. 2.0 J c. 5.0 J d. 8.0 J e. 10.0 J A negatively charged particle is moving in the +x-direction when it enters a region with a uniform electric field pointing in the +x-direction. Which graph gives its position as a function of time correctly? (Its initial position is x = 0 at t = 0.) x x x x x t t t t t (a) (b) (c) (d) (e)

20. A magnetic field is directed out of the page. Two charged particles enter from the top and take the paths shown in the figure. Which statement is correct? 1 2 a. Particle 1 has a positive charge and particle 2 has a negative charge. b. Both particles are positively charged. c. Both particles are negatively charged. d. Particle one has a negative charge and particle 2 has a positive charge. e. The direction of the paths depends on the magnitude of the velocity, not on the sign of the charge.

Kinematics v = v 0 + a t x = x 0 + v 0 t + ½ a t 2 v 2 = v 0 2 + 2a (x x 0 ) v = (v + v 0 ) / 2 Newton s Law F = m a F gravity = m g g = 9.80 m/s 2 Conservation of Energy KE 1 + U 1 + W in/out = KE 2 + U 2 Energy Kinetik (linear): KE lin = ½ mv 2 Potential (gravity): U g = m g y Work Power (electrical) Coulomb force Electric field Electric flux Gauss Law Potential energy Potential W = F d = F d cosθ P = W/t = E/t P = I V = I 2 R = ( V) 2 / R F = k e q 1 q 2 / r 2 r ke = 1/4πε o E = F/q = E = k e q r / r 2 Φ E = Φ E = dq k r = - V 2 r surface e volume E d A closedsurface U = U B - U A = V = U / q = for a point charge E d A = q inside /ε o = 4πk e q inside B q E d s = q V A B A E d s (= k e q/r point charge) Capacitance C = Q/ V [ = ε o A/d parallel plate C] C eq = C 1 + C 2 + C 3 + [parallel combination] 1/C eq = 1/C 1 + 1/C 2 + 1/C 3 + [series combination] U = Q 2 /2C = ½ Q V = ½ C ( V) 2 [energy stored in C] Charging/discharging of Capacitor q(t) = Q (1 - e -t/rc ) I(t) = ( V/R) e -t/rc q(t) = Q e -t/rc I(t) = -( V/R) e -t/rc Ohm's Law R = V/I = ρ l/a [ρ = resistivity = 1/σ] Resistivity / Resistance ρ = m e / (n q 2 τ) [τ = scattering time] ρ = ρ o [1 + α(t T o )] [temperature dependence] 1/R eq = 1/R 1 + 1/R 2 + 1/R 3 + [parallel] R eq = R 1 + R 2 + R 3 + [series]

Magnetic force Cyclotron motion [circular motion of a charge in a magnetic field] Magnetic Field by a current Ampere s Law F B = q v B F B = I L B df B = I ds B F B /L = (µ o I 1 I 2 ) / (2πa) r = (mv) / (qb) ω = (qb) / m db = µ o /4π (I ds B = (µ o I) / 2πa) r r Bds = µ 0I closedloop r ) / r 2 [force on straight conductor] [force on conductor segment] [force between 2 parallel wires] [radius] [frequency] [Biot-Savart Law] [long straight wire] Magnetic Flux Φ B = surface B d A Faraday's Law of Induction emf = ( V =) - N (dφ B / dt) Lenz Law the polarity of the induced emf is such that it wants to act against the cause. Inductor emf = ( V =) L di/dt LR circuit I = V/R (1 - e -t/τ ) switch on I = V/R e -t/τ switch off τ = L/R Speed of a wave v = f λ (light in vacuum: c = f λ) Index of refraction n = c/v = λ o /λ Snell's Law n 1 sinθ 1 = n 2 sinθ 2 Total reflection sinθ c = n 2 / n 1 Mirror and lens Equations 1/p + 1/q = 1/f m = h / h = - q / p f = R/2 (mirror) Diffraction Double slit or grating d sin θbright = m λ (m = 0, 1, 2, ) ( m + 1 ) λ d sinθdark = 2 (m = 0, 1, 2, ) Electron mass Proton mass Elementary charge Coulomb constant Permittivity of free space Permeability of free space Speed of light m e = 9 10-31 kg m p = 2 10-27 kg e = 1.6 10-19 C [electron: -e ; proton: +e] k e = 9 10 9 Nm 2 /C 2 ε o = 9 10-12 C 2 /Nm 2 k e = 1 / (4πε o ) µ o = 4π 10-7 Tm /A c = 3 10 8 m/s