Exam 2 -- Fall 2003 solutions

Exam 2 -- Fall 2003 solutions - Physics 240 Fall 2003: Exam...

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Physics 240 Fall 2003: Exam #2 Solutions Please print your name:___________________________________________________ Please list your discussion section number:___________________________________ Please list your discussion instructor:________________________________________ Form #1 Instructions 1. Fill in your name above 2. This will be a 1.5 hour, closed book exam. The exam includes 20 questions. 3. You may use a calculator, please do not share calculators 4. You may use two 3”x5” note cards (one from the first exam, plus one new one) with notes and equations you think may be useful. You can write on both sides of the card if you like. 5. You will be asked to show your University student ID card when you turn in your exam. ID checked by:___________________________________ Table of constants: ε 0 = 8.85 x 10 -12 C 2 /Nm 2 k = 1/(4 πε 0 ) q electron =-1.6 x 10 -19 C q proton =1.6 x 10 -19 C m electron =9.1 x 10 -31 kg m proton =1.67 x 10 -27 kg µ 0 = 4 π x 10 -7 Tm/A G = 6.67 x 10 -11 Nm 2 /kg 2
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1: Which of the following is equivalent to the quantity 1000 Ohms * 20 microFarads? A) 20000 Joules/second B) 0.02 kilogram*meters/second C) 0.0002 volts D) *0.02 seconds E) 20000 Newtons/Coulomb 2: What is the equivalent resistance of the resistor network seen by the battery shown in the figure? A) 5R/2 B) *5R/3 C) 2R D) 11R/5 E) R/2 This is an RC time: 1000 * 20 µ F = 0.02 seconds V R 2R R 2R This problem has one feature you have to be careful about. The resistor coming down on the diagonal is actually shorted out. That is, it has a wire connected from one end to the other. This means the potential difference across this resistor will be zero. No current will flow through it and there will be no voltage drop across it. It has no effect on the circuit. You can see this most easily in the redrawn circuit below. Working on this redrawn circuit, we have resistors of R and 2R in parallel, which has an equivalent resistance of 2R/3. This combination is in series with the resistor R, so the total resistance is 5R/3. V R 2R R 2R
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3: Three identical light bulbs are connected in a circuit as shown. Each of the bulbs has a resistance of 30 , and the battery supplies a voltage of 30V. How much power does light B consume? A) *30W B) 60W C) 10W D) 20W E) 50W 4: A mass spectrometer accelerates a singly charged ion to a velocity of 4.5x10 5 m/s before injecting it into region with a constant 0.400 Tesla field. A ‘singly charged ion’ has a charge equal in magnitude to the electron charge. If it bends to the left in a circular path with a radius of 25 cm, what is the mass of the ion? A) 2.6x10 -27 kg B) 9.1x10 -31 kg C) *3.6x10 -26 kg D) 8.3x10 -25 kg E) 3.3x10 -24 kg A B C Each bulb sees 30V across it, so the power through each is just: P = V 2 /R = 30 2 / 30 = 30W The centripetal force for circular motion in a mass spectrometer is provided by the magnetic force: mv 2 / r = qvB or m = rqB / v m = (0.25m*1.6x10 -19 C*0.4T) / 4.5x10 5 m/s m = 3.6x10 -26 kg
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5: Five ions, labeled A-E, all enter a region of uniform magnetic field with the same velocity. Which ion lands at position 3? 6: In Oersted’s laboratory a compass needle is aligned with a constant magnetic field.
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This note was uploaded on 04/01/2008 for the course PHYSICS 240 taught by Professor Davewinn during the Fall '08 term at University of Michigan.

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Exam 2 -- Fall 2003 solutions - Physics 240 Fall 2003: Exam...

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