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PHYS 2306  Foundations Of Physics I  Virginia Tech Study Resources
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PHYS 2306 Solution Quiz 8
School: Virginia Tech
Course: Foundations Of Physics II
22) A bird is flying south with a positive charge in its beak. It flies through a homogeneous magnetic field that points towards the center of the earth. What is the direction of the magnetic force on the bird? A. B. C. D. E. F. north east south west away

Q10_c29_sol
School: Virginia Tech
Course: Physcs 1
Quiz 10 Phys. 2306 F09 You may use your text book, notes, and a calculator. A long straight wire is carrying constant current I in the +x direction. In (A) a metal rod of very small diameter is moving with constant velocity v in the +x diection. The rods

Q9_sol_c28
School: Virginia Tech
Course: Physcs 1
Quiz 9 Phys. 2306 F09 You may use your text book, notes, and a calculator. Three innitely long wires run from z and carry currents of magnitude I. The current in wire 1 that passes through the point (x,y)=(0,a) ows in the +z direction, the current in the

Q8_c27_sol
School: Virginia Tech
Course: Physcs 1
Solutions to Quiz 8 Phys. 2306 F09 You may use your text book, notes, and a calculator. A particle with negative charge q and mass m = 2.00 1015 kg is traveling at t = 0 with (vx , vy , vz ) = 1.00106 (3, 4, 5) m/s , at position (x, y, z ) = (a, 0, 0) m.

Q7_sol_25
School: Virginia Tech
Course: Physcs 1
Quiz 725 Use lower 2 loops: a 4.00 E1 b E2 c 3.00 I1 I2 3.00 A 6.00 5.00 A d I in 3.00 resistor is I1+I2=8.00 A (e) This I is positive so sense stays the way it is drawn up (a) 0=(V(a)V(b)+(V(d)V(a)+(V(b)V(d) = E1 12.0 24.0 E1=36.0 V (c) 0=(V(c)

Q7_sol_1
School: Virginia Tech
Course: Physcs 1
Quiz 71 Use outer loop: R a 2.00 A c 4.00 6.00 3.00 A 5.00 A d (V(a)V(c)+(V(d)V(a)+V(c)V(d)=0 2.00R 12.00 +30.00 = 2.00R = 18.00 R = 9.00 (c) R 2.00 A a c b 4.00 E1 E2 6.00 Original Circuit 3.00 5.00 A 3.00 A d

Q6_sol_35
School: Virginia Tech
Course: Physcs 1
Quiz 6 3,4,5 C CP CP C = C C CS = V V V CP = C+C = 3C 1/CS = 1/3C+1/3C = 2/3C CS = (3/2)C U = CSV2/2 = (3/4)CV2 UU = (3/4 2/3)CV2 = (1/12)CV2 C=2C Switch closed Q (QS QP) = (Cs  CP) V = (3/24/3)CV = (1/6)CV >0 Potential difference across battery is co

Q6_sol_12
School: Virginia Tech
Course: Physcs 1
Quiz 6 1,2 C C d a c C C V CS CS = V CP = V C=2C Switch open 1/CS = 1/C+1/C = (1/C)(1+1/2) = 3/(2C) CS = (2/3)C CP = 2CS = (4/3)C U = CPV2/2 = (2/3)CV2 QP = CPV = (4/3)CV QS = QP/2 = (2/3)CV = QC = QC VcVd = (VcVa) + (VaVd) = QC/C + QC/C = (2/3)(CV)(

Q5_sol
School: Virginia Tech
Course: Physcs 1
Quiz 5 Phys. 2306 F09 Crn94823 You may use your text book, notes, and a calculator. Two very large area parallel, conducting plates A,B are kept at a potential dierence V by a battery. The plates are separated by a distance d. Small holes are drilled in t

Q3_sol
School: Virginia Tech
Course: Physcs 1
Quiz 3 Phys. 2306 F09 CRN 94850 You may use your text book, notes, and a calculator. In order to illustrate that matter is electrically neutral to a very high degree, consider two Cu spheres of radii 0.1 cm that are 1 m apart. Thus they look like point ch

Q2_sol
School: Virginia Tech
Course: Physcs 1
Quiz 2 Phys. 2306 F09 CRN 94850 You may use your text book, notes, and a calculator. Two loud speakers A,B are a distance L= 2.00 m apart. They each emit pure tones of frequency f= 206 Hz. The speed of sound in air is 344 m/s. 1) Along the line joining th

Q1_sol
School: Virginia Tech
Course: Physcs 1
Quiz 1 Phys. 230694850 F09 You may use your text book, notes, and a calculator. A transverse wave on a string of mass/length is generated by xing one end of the string and attaching the other end to a motor that has an amplitude A, maintains a tension T

Vectors
School: Virginia Tech
Course: Physcs 1
Physics 2306 M. Blecher August 18, 2009 Chapter 1 Vectors 1.1 Representation of Vectors Vectors in three dimensional (3D) space are quantities possessing magnitude and direction. They are written in bold font in this note. In component form,vector A is wr

Q3_sol
School: Virginia Tech
Course: Physcs 1
Quiz 3 Phys. 2306 F09 CRN 94850 You may use your text book, notes, and a calculator. In order to illustrate that matter is electrically neutral to a very high degree, consider two Cu spheres of radii 0.1 cm that are 1 m apart. Thus they look like point ch

PHYS 2306 Solution Quiz 7
School: Virginia Tech
Course: Foundations Of Physics II
a R=3 = 12 volts R=4 R=5 b C=6 F S1 c 19, 20 and 21) Consider the circuit shown above. The switch is closed at time t =0. 19) Just after the switch is closed, what is the current (in Amps) through the 4.0 resistor? A. B. C. D. E. F. G. H. 60/47 12/7 2

PHYS 2306 Solution Quiz 6
School: Virginia Tech
Course: Foundations Of Physics II
16, 17 and 18) Consider the circuit shown below. The current through the 4 resistor is 3.0 A. 16) What is the equivalent resistance (in ohms) of the three resistors? A. B. C. D. E. F. G. H. 26/7 9 20/9 16/3 22/3 7 5 3 R=2 R=3 R=4 3 and 4 are in parallel.

PHYS 2306 Solution Quiz 5
School: Virginia Tech
Course: Foundations Of Physics II
13) You move a positive charge q = 3.0 C from the negative to the positive terminal of a 6.0 V battery. How much work (in Joules) did you do? A. B. C. D. E. F. G. H. 3.0 3.0 6.0 6.0 12 12 18 18 Change in electric potential energy of the charge q is q

PHYS 2306 Solution Quiz 4
School: Virginia Tech
Course: Foundations Of Physics II
10) A charge q = 4.0E06 is inside a spherical surface, but not at the center of the sphere? What is the electric flux (in Nm2/C) through the surface of the sphere A. Cant determine the flux because the charge is not at the center of the sphere B. 2.3E+0

PHYS 2306 Solution Quiz 3
School: Virginia Tech
Course: Foundations Of Physics II
7) A positive charge q is placed at (x = a, y = 0, z = 0). A negative charge 2q is placed at (x = 2a, y = 0, z = 0). What is xcomponent of the electric field at the origin? A. B. C. D. E. F. G. H. 3kq/a2 3kq/2a2 3kq/a2 kq/2a2 kq/2a2 3kq/4a2 3kq/4a2 3

PHYS 2306 Solution Quiz 2
School: Virginia Tech
Course: Foundations Of Physics II
4.) A pipe is 0.40 meters long. It is open at one end and closed at the other end? If the frequency produced by the pipe is 1075 Hz, what is the harmonic of that tone? The speed of sound in air is 344 m/s. A. 1 B. 2 Pipe open at one end and closed at the

PHYS 2306 Solution Quiz 1
School: Virginia Tech
Course: Foundations Of Physics II
1. and 2.) The displacement of a transverse traveling wave is given by y(x,t) = (0.30 m)cos[2(x/0.50 m + t/0.0025 s)] 1.) The velocity (in meters/sec) of the wave is (positive velocity is in the +xdirection, negative velocity is in the xdirection) A. B.

PHYS 2306 Test 2 Solution
School: Virginia Tech
Course: Foundations Of Physics II
Physics 2306 Second Exam Fall 2008 FORM D 1.) An ideal 12 V EMF is in series with a 2 F capacitor. The left plate of the capacitor has 1.5E05 C charge and the right plate has +1.5E05 C charge as shown in the figure below. What is the potential difference

PHYS 2306 Test 1 Solution
School: Virginia Tech
Course: Foundations Of Physics II
Physics 2306 Fall 2008 First Exam FORM A 1) The pressure fluctuation in air due to a sound wave is given by p(x,t) = (20 Pa) sin[(5 radians/meter) x (2500 radians/second) t] Which one of the following statements is true? A. The wave is a transverse wave a

PHYS 2306 Final
School: Virginia Tech
Course: Foundations Of Physics II
Physics 2306 Spring 2008 Final FORM A 1) What is the first Maxwell equation (based on the order given in my lecture notes)? A. B. C. D. E. F. G. H. around perimeter of surfaceBdl = 0[Ienclosed 0 dE/dt] over closed surface EnonconservativedA = Qenclosed/0

Q4_sol
School: Virginia Tech
Course: Physcs 1
Quiz 4 Phys. 2306 F09 CRN 94850 You may use your text book, notes, and a calculator. Consider a hollow conductor. Its outer surface is a sphere with radius, R=0.100 m, while the inner surface is irregular. In the hollow cavity and insulated from the condu

Exam_f_ins
School: Virginia Tech
Course: Physcs 1
Phys. 2306 Final Exam F09 Instructions December 7, 2009 0 INSTRUCTIONS & FORMULAS FINAL EXAM F09 PHYS2306 This is a closed book exam. No books, notes, computers or other paper may be brought into the exam. All cell phones must be shut o. The honor system

Exam_3a_09_sol
School: Virginia Tech
Course: Physcs 1
Phys. 2306 Exam 3 F09 Form A November 25, 2009 0 Phys. 2306 Exam 3 F09 Form A Solutions In cosmic ray experiments, particles travel downward (j) with speed v= 3.00 106 m/s. They enter a region of constant magnetic eld, directed into the paper (k) at (x,y,

Exam_3_ins
School: Virginia Tech
Course: Physcs 1
Phys. 2306 Exam 3 F09 Instructions November 3, 2009 0 INSTRUCTIONS AND FORMULAS EXAM 3 F09 PHYS. 2306 This is a closed book exam. No books, notes, computers or other paper may be brought into the exam. All cell phones must be shut o. The honor system is i

30_InstructorSolutions
School: Virginia Tech
Course: Foundations Of Physics II
INDUCTANCE 30 Apply Eq.(30.4). di (a) E2 = M 1 = (3.25 104 H)(830 A/s) = 0.270 V; yes, it is constant. dt 30.1. IDENTIFY and SET UP: EXECUTE: (b) E1 = M di2 ; M is a property of the pair of coils so is the same as in part (a). Thus E1 = 0.270 V. dt EVALU

29_InstructorSolutions
School: Virginia Tech
Course: Foundations Of Physics II
ELECTROMAGNETIC INDUCTION 29 29.1. 29.2. IDENTIFY: Altering the orientation of a coil relative to a magnetic field changes the magnetic flux through the coil. This change then induces an emf in the coil. SET UP: The flux through a coil of N turns is = NBA

28_InstructorSolutions
School: Virginia Tech
Course: Foundations Of Physics II
SOURCES OF MAGNETIC FIELD 28 28.1. ! ^ EXECUTE: (a) r = ( 0.500 m ) i , r = 0.500 m ! ! ^ v r = vr^ i = vrk j ^ ! IDENTIFY and SET UP: Use Eq.(28.2) to calculate B at each point. ! ! ! ! ! qv r 0 qv r ^ r ^ B= 0 = , since r = . 4 r 2 4 r 3 r ! ! 6 ^ and

27_InstructorSolutions
School: Virginia Tech
Course: Foundations Of Physics II
MAGNETIC FIELD AND MAGNETIC FORCES 27 27.1. ! IDENTIFY and SET UP: Apply Eq.(27.2) to calculate F . Use the cross products of unit vectors from Section 1.10. ! ^ j EXECUTE: v = ( +4.19 104 m/s ) i + ( 3.85 104 m/s ) ^ ! ^ (a) B = (1.40 T ) i ! ! ! ^ ^ F

26_InstructorSolutions
School: Virginia Tech
Course: Foundations Of Physics II
DIRECTCURRENT CIRCUITS 26 26.1. 26.2. 26.3. IDENTIFY: The newlyformed wire is a combination of series and parallel resistors. SET UP: Each of the three linear segments has resistance R/3. The circle is two R/6 resistors in parallel. EXECUTE: The resista

25_InstructorSolutions
School: Virginia Tech
Course: Foundations Of Physics II
CURRENT, RESISTANCE, AND ELECTROMOTIVE FORCE 25 25.1. 25.2. IDENTIFY: I = Q / t . SET UP: 1.0 h = 3600 s EXECUTE: Q = It = (3.6 A)(3.0)(3600 s) = 3.89 104 C. EVALUATE: Compared to typical charges of objects in electrostatics, this is a huge amount of char

24_InstructorSolutions
School: Virginia Tech
Course: Foundations Of Physics II
CAPACITANCE AND DIELECTRICS 24 24.1. 24.2. 24.3. Q Vab SET UP: 1 F = 10 6 F EXECUTE: Q = CVab = (7.28 10 6 F)(25.0 V) = 1.82 10 4 C = 182 C EVALUATE: One plate has charge + Q and the other has charge Q . Q PA and V = Ed . IDENTIFY and SET UP: C = 0 ,

23_InstructorSolutions
School: Virginia Tech
Course: Foundations Of Physics II
ELECTRIC POTENTIAL 23 ra = 0.150 m rb = (0.250 m) 2 + (0.250 m) 2 rb = 0.3536 m 23.1. IDENTIFY: Apply Eq.(23.2) to calculate the work. The electric potential energy of a pair of point charges is given by Eq.(23.9). SET UP: Let the initial position of q2 b

22_InstructorSolutions
School: Virginia Tech
Course: Foundations Of Physics II
GAUSS'S LAW 22 ^ E = E cos dA, where is the angle between the normal to the sheet n and the 22.1. (a) IDENTIFY and SET UP: electric field E . EXECUTE: In this problem E and cos are constant over the surface so E = E cos dA = E cos A = (14 N/C )( cos 60 )

21_InstructorSolutions
School: Virginia Tech
Course: Foundations Of Physics II
ELECTRIC CHARGE AND ELECTRIC FIELD 21 21.1. (a) IDENTIFY and SET UP: Use the charge of one electron ( 1.602 10 19 C) to find the number of electrons required to produce the net charge. EXECUTE: The number of excess electrons needed to produce net charge

16_InstructorSolutions
School: Virginia Tech
Course: Foundations Of Physics II
SOUND AND HEARING 16 16.1. IDENTIFY and SET UP: Eq.(15.1) gives the wavelength in terms of the frequency. Use Eq.(16.5) to relate the pressure and displacement amplitudes. EXECUTE: (a) = v / f = (344 m/s)/1000 Hz = 0.344 m (b) pmax = BkA and Bk is constan

31_InstructorSolutions
School: Virginia Tech
Course: Foundations Of Physics II
ALTERNATING CURRENT 31 31.1. IDENTIFY: SET UP: EXECUTE: i = I cos t and I rms = I/ 2. The specified value is the rootmeansquare current; I rms = 0.34 A. (a) I rms = 0.34 A 31.2. (b) I = 2 I rms = 2(0.34 A) = 0.48 A. (c) Since the current is positive hal

32_InstructorSolutions
School: Virginia Tech
Course: Foundations Of Physics II
ELECTROMAGNETIC WAVES 32 32.1. IDENTIFY: Since the speed is constant, distance x = ct. SET UP: The speed of light is c = 3.00 108 m/s . 1 yr = 3.156 107 s. 32.2. x 3.84 108 m = = 1.28 s c 3.00 108 m/s (b) x = ct = (3.00 108 m/s)(8.61 yr)(3.156 107 s/yr) =

Exam_2a_09_sol
School: Virginia Tech
Course: Physcs 1
Phys. 2306 Exam 2 F09 Form A October 16, 2009 0 Solutions Phys. 2306 Exam 2 F09 Form A In Thomsons atomic model of Helium, electrons are considered point charges and are like chocolate chips in cookie dough. In this case the dough is the alpha particle, t

Exam_2_ins
School: Virginia Tech
Course: Physcs 1
Phys. 2306 Exam 2 F09 Instructions October 6, 2009 0 INSTRUCTIONS AND FORMULAS EXAM 2 F09 PHYS. 2306 This is a closed book exam. No books, notes, computers or other paper may be brought into the exam. All cell phones must be shut o. The honor system is in

Exam_1a_09_sol
School: Virginia Tech
Course: Physcs 1
Phys. 2306 Exam 1 F09 Form A Solutions Consider a horizontal (x) wire of uniform mass/length = 0.153 g/cm. The wire is attached to a wall at one end and driven up and down (y ) at frequency f= 2.00 Hz by a person applying tension T at the other end. Let y

Exam_1_ins
School: Virginia Tech
Course: Physcs 1
Phys. 2306 Exam 1 F09 Instructions September 8, 2009 0 INSTRUCTIONS AND FORMULAS EXAM 1 F09 PHYS. 2306 This is a closed book exam. No books, notes, computers or other paper may be brought into the exam. All cell phones must be shut o. The honor system is

2306fall2009
School: Virginia Tech
Course: Physcs 1
PHYSICS 2306: FOUNDATIONS OF PHYSICS I Fall 2009 Text: University Physics by Young & Freedman Week 1 Start Date 8/24 2 8/31 3 9/7 4 9/14 5 9/21 6 9/28 7 10/5 8 10/12 9 10/19 10 10/26 11 11/2 12 11/9 13 11/16 14 11/30 15 12/7 Topics Introduction, Propertie

38_InstructorSolutions
School: Virginia Tech
Course: Foundations Of Physics II
PHOTONS, ELECTRONS, AND ATOMS 38 h f  . The e e 38.1. IDENTIFY and SET UP: The stopping potential V0 is related to the frequency of the light by V0 = slope of V0 versus f is h/e. The value fth of f when V0 = 0 is related to by = hf th . EXECUTE: (a) From

37_InstructorSolutions
School: Virginia Tech
Course: Foundations Of Physics II
RELATIVITY 37 Figure 37.1 37.1. IDENTIFY and SET UP: Consider the distance A to O and B to O as observed by an observer on the ground (Figure 37.1). (b) d = vt = (0.900) (3.00 108 m s) (5.05 106 s) = 1.36 103 m = 1.36 km. 37.3. 1 IDENTIFY and SET UP: The

36_InstructorSolutions
School: Virginia Tech
Course: Foundations Of Physics II
DIFFRACTION 36 36.1. IDENTIFY: Use y = x tan to calculate the angular position of the first minimum. The minima are located by m , m = 1, 2,. First minimum means m = 1 and sin 1 = / a and = a sin 1. Use this Eq.(36.2): sin = a equation to calculate . SET

35_InstructorSolutions
School: Virginia Tech
Course: Foundations Of Physics II
INTERFERENCE 35 35.1. 35.2. IDENTIFY: Compare the path difference to the wavelength. SET UP: The separation between sources is 5.00 m, so for points between the sources the largest possible path difference is 5.00 m. EXECUTE: (a) For constructive interfer

34_InstructorSolutions
School: Virginia Tech
Course: Foundations Of Physics II
GEOMETRIC OPTICS 34 y = 4.85 cm Figure 34.1 34.1. IDENTIFY and SET UP: Plane mirror: s =  s (Eq.34.1) and m = y / y =  s / s = +1 (Eq.34.2). We are given s and y and are asked to find s and y. EXECUTE: The object and image are shown in Figure 34.1. s =

33_InstructorSolutions
School: Virginia Tech
Course: Foundations Of Physics II
THE NATURE AND PROPAGATION OF LIGHT 33 33.1. IDENTIFY: For reflection, r = a . SET UP: The desired path of the ray is sketched in Figure 33.1. 14.0 cm EXECUTE: tan = , so = 50.6 . r = 90  = 39.4 and r = a = 39.4 . 11.5 cm EVALUATE: The angle of incidence

44_InstructorSolutions
School: Virginia Tech
Course: Foundations Of Physics II
PARTICLE PHYSICS AND COSMOLOGY 44 44.1. (a) IDENTIFY and SET UP: Use Eq.(37.36) to calculate the kinetic energy K. 1 EXECUTE: K = mc 2  1 = 0.1547 mc 2 2 2 1 v / c m = 9.109 10 31 kg, so K = 1.27 1014 J (b) IDENTIFY and SET UP: The total energy of th

Test1 Solution
School: Virginia Tech
Course: FOUNDATIONS OF PHYSICS I
k = 9.0E+09 Nm2/C2 0 = 8.9E12 C2/N2m2 Physics 2306 Spring 2004 fL =[(v vL)/(v vS)]fS First Exam The displacement of a transverse wave is given by y(x,t) = (0.02 m)sin[(400 rad/sec)t (10 rad/meter)x] What is the amplitude (in meters) of the wave? A. B. C

Test2 Solution
School: Virginia Tech
Course: FOUNDATIONS OF PHYSICS I
k = 9.0E+09 Nm2/C2 0 = 8.9E12 C2/N2m2 Q(t) = Cx[1  exp(t/RC)] Physics 2306 Spring 2004 Second Exam 1.) If you move a positive charge from the negative terminal to the positive terminal of a battery, the change in electric potential energy of the charg

Test3 Solution
School: Virginia Tech
Course: FOUNDATIONS OF PHYSICS I
Physics 2306 Spring 2004 Third Exam Solution k = 9.0E+09 Nm2/C2 Q = Cx[1 exp(t/RC)] 0 = 8.9E12 C2/N2m2 I = (x/R)[1 exp(Rt/L)] m0 = 4pE07 Wb/Am Q = Q0 cos(wt + f) 1.) At t = 0 the switch is closed in an LC circuit and the capacitor has its maximum char

Chapter 34
School: Virginia Tech
Course: Foundations Of Physics
Chapter 34 Geometric Optics 34.1 Reflection and refraction at a plane surface Images Reflection 34.2 Reflection at a spherical surface Focal length Images 34.3 Refraction at a spherical surface 34.4 Thin lenses 34.7 Magnifier 34.8 Microscopes and Telescop

Chapter 32
School: Virginia Tech
Course: Foundations Of Physics
Chapter 32 Electromagnetic Waves 32.1 Maxwells Equations 32.2 Plane electromagnetic waves 32.3 Sinusoidal electromagnetic waves 32.4 Energy and momentum Momentum flow and radiation pressure: Example: NASA is giving serious consideration to the concept of

Chapter 30
School: Virginia Tech
Course: Foundations Of Physics
Chapter 30 Inductance 30.12 Mutual Inductance and SelfInductance Mutual induction and mutual inductance Selfinduction and selfinductance 30.3 Magnetic field energy The energy stored in an inductor 30.4 RL Circuits Current growth Current decay 30.5 L

Chapter 28
School: Virginia Tech
Course: Foundations Of Physics
Chapter 28 Sources of Magnetic Fields 28.1 Moving charge 28.2 Current element, RHR 28.3 Straight current carrying conductor Example: A straight wire carries a current I = 3.0 A. A particle of charge q = +6.5 x 106 C is moving parallel to the wire at a di

Cheat Sheet
School: Virginia Tech
Course: Physics
Potential Energy of a pair of charges finalinitial, Ratio of Bob to Sally = p= n= = 1kWh = 1 kW*3600s C2/n m2 = Volume of a sphere = Volume of a cylinder = m= k= m/s olu e (volume Current Density concentration charges ( () # free electrons = Potential en

Best Cheat Sheet2
School: Virginia Tech
Course: Physics
15 mechanical waves 16 sound and hearing 21 Electric charge and electric field Coulomb's law where and Electric Field wave power Electric dipoles Standing sound waves inverse square law for intensity 22 Gauss's Law Standing waves on a string node at close

Chapter+24+worksheet
School: Virginia Tech
Course: Foundations Of Physics II
Produced with a Trial Version of PDF Annotator  www.PDFAnnotator.com Chapter 24 Capacitors & Dielectrics 23.1 Capacitors, capacitance, capacitance calculations Produced with a Trial Version of PDF Annotator  www.PDFAnnotator.com Example: What is the cap

Chapter+23+worksheet
School: Virginia Tech
Course: Foundations Of Physics II
Chapter 23 Electric Potential 23.1 Electric Potential Energy Potential energy due to several point charges Example: An alpha particle and an electron are initially separated by a distance of 1010 m. An alpha particle is a helium nucleus having two proton

Chapter+22+worksheet
School: Virginia Tech
Course: Foundations Of Physics II
Chapter 22 Gausss Law 22.12 Electric Flux Calculation: Uniform Electric Field Nonuniform Electric Field 22.34 Gausss Law General Form Line Charge (like a wire) Sphere of charge 22.5 Conductors Electrostatic equilibrium. Field inside a conductor. Field

Chapter+21+worksheet
School: Virginia Tech
Course: Foundations Of Physics II
Chapter 21 Electric Charge and Electric Field 21.12 Electric Charge Types of charge. The law of conservation of charge. Elementary charge. The coulomb. Conductor. Insulator. 21.3 Coulombs Law The force that point charges exert on each other 21.46 The El

Chapter+16+worksheet
School: Virginia Tech
Course: Foundations Of Physics II
Chapter 16  Sound Waves 16.1 Sound waves Sound waves as pressure fluctuations, pressure amplitude 16.2 Speed of sound 16.3 Intensity, decibel scale 16.45 Standing sound waves Pressure and displacement nodes and antinodes Open pipe conditions Stopped pip

Chapter+15+worksheet
School: Virginia Tech
Course: Foundations Of Physics II
Chapter 15  Mechanical Waves 15.1 Properties of mechanical waves, transverse waves, longitudinal waves 15.23 Periodic waves, wave relations, wave function, sinusoidal wave, amplitude, phase, particle velocity, particle acceleration Example: A sinusoidal

PracticeTestFinal1
School: Virginia Tech
Course: Physics
Practice Test With Answers 1. Springs attached: The position as function of time of 1.5 kg mass on a spring is given by , where x is in meters and t is in seconds. a. What is the force constant of the spring? 1.5*(4.16)2 N/m b. What is the Time Period? s