Physics Final Exam Three Hours, Closed Book

Transcription

1 Physics Final Exam Three Hours, Closed Book The instructions for this exam are identical to those of the midterm exams, except that this exam is three hours long. This is a closed-book exam: do not use any references (text, notes, etc.). You will need a calculator capable of basic functions (arithmetic, logarithms, trigonometric functions, etc.), but you are not allowed to store formulas, etc., in the calculator memory. This booklet contains: This cover page. Three pages of formulas. The exam questions. Five blank pages for extra work. Remove the cover page and formula pages from your exam. Discard them when the exam is over. Write your name on the first page of questions. Write your answers directly on the exam booklet. Show your work and box your final answers. If you write any final answers on blank pages, indicate that you have done so on the page on which the problem was posed. I don t want to miss any of your work when grading your exam. In multi-part questions, if you get stuck on an early part, try to answer the later parts anyway either make up some answer for the early parts to use in the later parts, or describe what approach you would use to solve the later parts. Sometimes later parts won t depend on earlier parts at all. It has been an honor to have you in class this semester. I truly appreciate the opportunity to share my love of physics with you and to help you learn about the way we physicists view the world. Good Luck!!!

2 v = x t a = v t v f = v i + at x f = x i + v i t at2 v 2 f = v 2 i + 2a(x f x i ) a = v2 r = ω2 r F = µ k n F µ s n F = mg F = Gm 1m 2 r 2 F = k x F = 1 4 ρav2 (air) F = 6πηrv (liquid) F = m a F = p t p = m v p = J J = area under Force time curve Physics Formulas Mech Energy = K + U W = F d = F d cos θ K = 1 2 mv2 K = 1 2 Iω2 U = 1 2 kx2 U = mgy P = W E or t t τ = r F = rf = rf sin θ τ = Iα I = i L = Iω m i r 2 i x = θ R v = ωr a = αr ω = θ t α = ω t ω f = ω i + αt θ f = θ i + ω i t αt2 ω 2 f = ω 2 i + 2α(θ f θ i ) ρ = m V p = F A F = ρvg Q = V t = va v 1 A 1 ρ 1 = v 2 A 2 ρ 2 p 2 = p 1 + ρgd p ρv2 1 + ρgy 1 = p ρv2 2 + ρgy 2 sin 2 θ + cos 2 θ = 1 sin θ cos θ = tan θ sin(90 ± θ) = cos θ cos(90 ± θ) = sin θ The solution to ax 2 + bx + c = 0 is x = b ± b 2 4ac. 2a Volume of a sphere: V = 4 3 πr3 Surface area of a sphere: A = 4πr 2 Area of a circle: A = πr 2 Circumference of a circle: l = 2πr

3 Physics Formulas x(t) = A cos(ωt) v(t) = Aω sin(ωt) a(t) = Aω 2 cos(ωt) k ω = m g ω = l ω = 2πf T = 1 f = 2π ω y(x, t) = A cos ( 2π x ωt) λ v = fλ F t v = µ, µ = m l f o = (1 ± vo v ) (1 vs v )f s f = 2 v reflector f s v ( ) I β = 10 log I 0 I = P A I = P 4πr 2 f n = nf 1 v = c n n 1 sin θ 1 = n 2 sin θ 2 1 s + 1 s = 1 f m = h h = s s θ λ D λ peak = nm K T Q = Mc T P = eσat 4 F = K q 1q 2 r 2 E = K q r 2 E = Q ɛ 0 A F = qe String or open-open tube: f 1 = v 2L Open-closed tube: f 1 = v 4L Bragg scattering: 2d cos θ = mλ { n λ constructive Interference in general: L = (n + 1 )λ destructive 2 { L 1 L 2 n constructive = λ 1 λ 2 n + 1 destructive 2 0 constructive Single-slit interference: y = n L λ a destructive n λ d constructive Two-slit or Multi-slit interference: sin θ = ( ) n + 1 λ 2 d destructive n L λ d constructive y = ( ) n + 1 λ 2 L d destructive

4 More Physics Formulas V = q 4πɛ 0 r U e = qv V = E x E = V x Q = CV C = κɛ 0A d U = 1 Q 2 2 C = 1 2 QV = 1 2 CV 2 I = q t V = IR R = ρl A P = IV = I 2 R = V 2 1 R equiv = 1 R R 2 R equiv = R 1 + R 2 ΣI in = ΣI out Σ V = 0 R τ = RC V = V 0 e t τ V = V f (1 e t τ F = q v B F = l I B Φ = AB cos θ E = N Φ t ) Wire: B = µ 0I 2πr Solenoid: B = µ 0IN L Circular loops: B = µ 0IN 2R Fundamental Constants and Commonly Used Data gravitational constant 11 N m2 G = kg 2 electrostatic constant K = N m2 C 2 permittivity of free space 12 C2 ɛ 0 = N m 2 permeability of free space µ 0 = 4π 10 7 T m A Stefan-Boltzmann constant σ = W m 2 K 4 grav. acceleration at Earth surface g = 9.80 m s threshold of hearing I 0 = W m 2 speed of light in vacuum c = m s fundamental charge e = C electron mass m e = kg proton mass m p = kg speed of sound in air v = 343 m s air at 20 atmospheric pressure 1 atm = N m 2 specific heat capacity of water c water = 4186 J kg K density of water ρ water = 1000 kg m 3 power output of the sun W radius of Sun m radius of Earth m radius of Moon m mass of Sun kg mass of Earth kg mass of Moon kg distance from Sun to Earth m

5 Physics Spring 2009 Final Exam Name Grading note: There are 11 problems on 11 pages. Point values are given with each problem. They add up to 100 points. In multi-part problems, points are not necessarily evenly divided between the parts. 10 cm 10 cm laser 5 m 1. [5 points] Light from a green laser pointer, λ = 532 nm, shines onto a pair of slits, producing a pattern of bright spots on a wall 5 m away. A bright spot is seem immediately across from the pair of slits, and the next nearest bright spots are 10 cm away from this central point. What is the distance between the two slits? 1

6 2. [5 points] For an MRI imaging apparatus, it is desired to create a magnetic field of 1 T within a solenoid of length 2.00 m and radius 0.20 m, with 1000 coils per centimeter over the length of the solenoid. What current should run through the solenoid? 3. [5 points] Some airport immigration stations use thermal imaging cameras to detect incoming passengers who have fevers. The idea is to test and isolate anyone who carrying a highly contagious disease. In practice, such detection may be of limited use in containing an epidemic, such as the present-day H1N1 scare, because carriers of the disease are highly contagious well before they show any symptoms of the disease. A typical healthy face is 34 C=307 K. Thermal imaging systems are capable of distinguishing feverish people whose faces are 3 degrees warmer than this. What is the ratio of the power emitted by a feverish face to the power emitted by a healthy face? 2

7 4. [10 points] (a) The Earth s distance from the Sun is m. Imagine that the Earth is suddenly removed from our own solar system and put in orbit around a star which has twice the power output of our own Sun (but with the same color spectrum). At what distance from that star should the Earth be placed in order to be at the same temperature that it has in its present orbit around the Sun? (b) If that star has twice the power output of our own Sun but the same color spectrum, what would have to be different about that star (compared to our Sun), and what would have to be the same? 3

8 5. [10 points] A 1 µf capacitor was charged up using a 5 V power source. The power source was then disconnected. Later on, at time t = 0, a 10 kω resistor was connected to the capacitor, and the capacitor began to discharge through the resistor. What was the capacitor voltage at time t = 5 ms? What was the current through the resistor at this time? What was the charge on the capacitor at this time? 4

9 6. [10 points] Rear-view mirrors on the passenger sides of cars are convex (diverging) mirrors. They have the words objects in mirror are closer than they appear written on them. You are riding in a car. You are 1.0 m from the passenger-side rear-view mirror. The mirror has focal length 0.50 m. (a) There is a truck 10 m behind the car. (More specifically, the distance from the truck to the mirror is 10 m.) The truck is 3 m high. You see an image of the truck in the mirror. Is the object (the truck) actually closer than the image of the truck, as the words on the mirror seem to imply, or is the image closer than the object? (b) What is the height of the image? (c) Consider what you see as you view the image of the truck, and as you view the truck itself. Is the angular size of the image of the truck smaller, larger, or the same size as the angular size of the truck itself? (Another way to ask this: is the relative angular magnification less than one, greater than one, or equal to one?) Note: as you might conclude from this problem, it is angular size, not image distance, which is responsible for the warning message on the mirror. 5

10 7. [10 points] A light bulb blew out in the dining room of my house a while back. Normally, a bulb just goes dead, but this time something funny happened. About half of the light bulb filament broke off and fell to the bottom of the bulb. One end of the remaining half of the filament remained attached to one of the leads through which the bulb gets power. The other end of the remaining half fell, by chance, on the other lead. The light bulb continued to make a complete circuit, but the filament was, in effect, only half its usual length. Assume the filament is the only resistive part of the light bulb. The bulb was plugged in to a standard 120 V socket. Before: After: (a) How does the resistance of the bulb with the half-length filament compare to the resistance of the bulb when it had a full-length filament? (Be quantitative, e.g., an answer might be its resistance was 3 times higher.) (b) How does the brightness of the bulb with the half-length filament compare to the brightness of the bulb when it had a full-length filament? (Again, be quantitative) 6

11 8. [15 points] An infinitely long, straight wire carries an upward current, I 1, which steadily increases, going from 0 to 6 A in 0.2 seconds. A coil is 0.1 m away from the wire. The coil consists of 1000 turns with radius 0.01 m. It is oriented as shown, with the coils parallel to the plane of the paper. The ends of the coil are attached to a resistor of resistance R = 1Ω. I 1 R 0.1 m (a) A current flows through the coil. What direction does it flow, clockwise or counterclockwise? (b) How much current flows through the coil? Note: Because the size of the coil is much smaller than its distance from the wire, you may assume that, at any moment, the magnetic field is the same across the entire area of the coil, and that it is equal to the magnetic field at the center. 7

12 9. [10 points] For each of the three resistors in the circuit shown below, give the voltage difference across the resistor and the current through the resistor. 45V 100Ω 300Ω 150Ω 8

13 10. [10 points] A cathode ray tube (CRT) is an old-style television set. A beam of electrons is emitted in the back of the television set and it hits the front of the set. The electrons are absorbed by phosphor compounds which glow when struck by electrons; this light forms the television picture. You have a CRT in a laboratory, oriented horizontally. The controls of the CRT have been adjusted so that, if it were working correctly, the electron beam should make a single spot of light exactly in the center of the screen (the dot in the diagram below). You observe, however, that the spot is deflected to the left when viewed from the front of the screen (the int the diagram below). You are told that there are three possible reasons for this: (a) The CRT is broken (e.g., the beam of electrons might be mis-aimed). (b) There is a uniform electric field throughout the laboratory. (c) There is a uniform magnetic field throughout the laboratory. Assuming that you do not have compasses, magnets, charged objects, etc., how can use the CRT itself to determine which of these three possibilities is the cause of the beam being off position? You can move the CRT around (changing its position or orientation), but you cannot remove it from the room. Describe how you would figure out which cause is the correct one. You should ignore gravity in this problem. beam is supposed follow the dashed path and hit the front of the CRT here......but it is observed to hit here instead Answer this question on the following page (which is otherwise blank). 9

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15 11. [10 points] Two glass microscope slides sit atop each other as shown in the diagram. On the left side, the slides are touching each other. The top slide is at a slight angle from the horizontal, θ = (see figure). The index of refraction of the glass is If the slides are illuminated form above with light of wavelength 500 nm, a series of dark and bright bands are seen (when viewed from above). Let x be the distance from the left end. How far from the left end is the first bright band, that is, what is the smallest x for which there is a bright reflection? Note: Assume that only the inner two surfaces are reflective, i.e., you can ignore the top surface of the upper slide and the bottom surface of the lower slide. 500 nm light θ=0.001 o x 11

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