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UVM PHYS 012 - Final Exam Study Guide
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Physics 012 1st EditionFinal Exam Study Guide: Chapters 18-22, 24-30Chapter 18Electric Forces and Electric FieldsProblem: In the rectangle in the drawing, a charge is to be placed at the empty corner to make the net force on the charge at corner A point along the vertical direction. What charge (magnitude and algebraic sign) must be placed at the empty corner?Answer:q = +3.0 μCQ = unknown chargeθ = angle between A and empty corner kq2/(4d)2 = kqQ/(√(17) d)2cosθQ = -17√(17)/64 = -3.3μCChapter 19Electric Potential Energy and Electric PotentialProblem: The potential at location A is 452 V. A positively charged particle is released there fromrest and arrives at location B with a speed vB. The potential at location C is 791 V, and when released from rest from this spot, the particle arrives at B with twice the speed it previously had, or 2vB. Find the potential at B.Answer:qVA = qVB + ½ mvB2qVC = qVB + ½ m(2vB)2q(VC – VA) = 4(½ mvB2) – ½ mvB2[q(VC – VA)]/3 = ½ mvB2qVA = [q(VC – VA)]/3 + qVBVB = VA (VC-VA)/3 = +339 VChapter 20Electric CircuitsThe drawing shows a circuit that contains a battery, two resistors, and a switch. What is the equivalent resistance of the circuit when the switch is(a) open and (b) closed? What is the total power delivered to the resistors when the switch is(c) open and (d) closed?Answer:(a) Req = R1 = 65 Ω(b) Req = (1/R1 + 1/R2)-1 = (1/65 + 1/96)-1 = 38.76 Ω(c) P = V2/Req = 92/65 = 1.25 W(d) P = V2/Req = 92/38.76 = 2.09 WChapter 21Magnetic Forces and Magnetic FieldsProblem: A copper rod of length 0.85 m is lying on a frictionless table (see the drawing below). Each end of the rod is attached to a fixed wire by an unstretched spring that has a spring constant of k = 75 N/m. A magnetic field with a strength of 0.16 T is oriented perpendicular to the surface of the table.(a) What must be the direction of the current in the copper rod that causes the springs to stretch?(b) If the current is 12 A, by how much does each spring stretch?Answer: (a) Counter-clockwise(b) FB = ILBsinθ = ILBFs = 2kx2kx = ILBx = ILB/2k = (12 A)(0.85m)(0.16 T)/2(75/m) = 0.0109 mChapter 22Electromagnetic InductionProblem: A solenoid has a cross-sectional area of 6.0 x 10-4 m2, consists of 400 turns per meter, and carries a current of 0.40 A. A 10-turn coil is wrapped tightly around the circumference of the solenoid. The ends of the coil are connected to a 1.5-Ω resistor. Suddenly, a switch is opened, and the current in the solenoid dies to zero in a time of 0.050 s. Find the average current induced in the coil.Answer: Bsol = μ0ni = (4π x 10-7)(400m-1)(0.4A) = 2.01 x 10-4 Tε = -N(πr2)ΔB/Δt = -NAΔB/Δt = -(10)(6 x 10-4)((2.01x10-4)/0.05) = -2.4 x 10-5 VIavg. = ε/R = -2.4 x 10-5 / 1.5 = -16.1 μAChapter 24Electromagnetic WavesProblem: A mirror faces a cliff located some distance away. Mounted on the cliff is a second mirror, directly opposite the first mirror and facing toward it. A gun is fired very close to the first mirror. The speed of sound is 343 m/s. How many times does the flash of the gunshot travel the round-trip distance between the mirrors before the echo of the gunshot is heard?Answer: tsound = 2L/vsoundtlight = 2L/c# = tsound/tlight = c/vsound= (3 x 108)/343 = 874,635 timesChapter 25The Reflection of Light: MirrorsProblem: A concave mirror has a focal length of 12 cm. This mirror forms an image located 36 cm in front of the mirror. What is the magnification of the mirror?Answer: 1/do = 1/f – 1/di = 1/12 – 1/36 = 1/18do = 18 cmm = -di/do = -36/18 = -2 x Chapter 26The Refraction of Light: Lenses and Optical InstrumentsProblem: The drawing below shows a ray of light traveling through three materials whose surfaces are parallel to each other. The refracted rays (but not the reflected rays) are shown as the light passes through each material. A ray of light strikes the a–b interface at a 50 degree angle of incidence. The index of refraction of material a is 1.20. The angles of refraction in materials b and c are, respectively, 45 degrees and 56.7 degrees. Find the indices of refraction inthese two media.Answer: nasinθa = nbsinθb(1.20)sin(50) = nbsin(45)nb = 1.30nbsinθb = ncsinθc(1.30)sin(45) = ncsin(56.7)nc = 1.10Chapter 27Interference and Wave Nature of LightProblem: Light of wavelength 469 nm enters a layer of gasoline (ng= 1.4) on top of a layer of water (nw = 1.33) at an angle. What is the minimum non-zero value of t so the light coming from gasoline is bright?Answer: 2t + λg/2 = λg/2, 3λg/2, 5λg/2, etc.t = m λg/2 = [(1)λair]/[2ng] = 168 nmChapter 28Special RelativityProblem: In the frame of reference of a stationary person on Earth, the distance between Earth and Alpha Centauri is 4.3 light years. A rocket leaves Alpha Centauri at 0.95 c heading toward Earth. How much will passengers age by the time they reach Earth?Answer: Passengers on ship measure proper time, Earth observers measure dilated time.Δt = d/v = 4.3 lyr/ 0.95c = 4.3(1yr)(c)/0.95c = 4.53 yrsΔt0 = Δt/γ = Δt√(1-[v2/c2]) = 4.53 √(1-[0.952]) = 1.4yrsChapter 29Particles and WavesProblem: From a cliff that is 9.5 m above a lake, a 41 kg woman jumps from rest, straight down into the water. At the instant she strikes the water, what is her de Broglie wavelength?Answer: λB = h/mvv = √(2gh) = 13.65 m/sλB= (6.626 x 10-34 m2kg/s)/(41 kg)(13.65 m/s) = 3.71 x 10-31 mChapter 30The Nature of the AtomProblem: A hydrogen atom is in the ground state. It absorbs energy and makes a transition to the n = 3 excited state. The atom returns to the ground state by emitting two photons. What aretheir wavelengths?Answer: 1/λ = R(1/nf2 – 1/ni2) = (1.097 x 107)(1/[1]2 – 1/[3]2) = (1.097 x 107)(1 – 1/9) = 9751111.1λ = 102.55 nm1/λ = R(1/nf2 – 1/ni2) = (1.097 x 107)(1/[2]2 – 1/[3]2) =(1.097 x 107)(1/4 – 1/9) = 1523611.1λ = 656.34


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