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200 Pages

### Chapter 32

Course: PHYS 221, Spring 2012
School: Clemson
Rating:

Word Count: 10012

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Solve: From 32.1. the circuit in Figure EX32.1, we see that 50 and 100 resistors are connected in series across the battery. Another resistor of 75 is also connected across the battery. 32.2. Solve: In Figure EX32.2, the positive terminal of the battery is connected to a resistor. The other end of that resistor is connected to resistor and a capacitor in parallel. 32.3. Model: Assume that the connecting...

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Clemson - PHYS - 221
33.1. Model: A magnetic field is caused by an electric current.Visualize: Please refer to Figure EX33.1. Solve: The magnitude of the magnetic field at point 1 is 2.0 mT and its direction can be determined by using the right-hand rule. Grab the current ca
Clemson - PHYS - 221
34.1.Visualize:To develop a motional emf the magnetic field needs to be perpendicular to both, so lets say its direction is into the page. Solve: This is a straightforward use of Equation 34.3. We havev=Assess:E 1.0 V = = 2.0 104 m/s lB (1.0 m ) ( 5.
Clemson - PHYS - 221
35.1. Model: Apply the Galilean transformation of velocity.Solve: (a) In the laboratory frame S, the speed of the proton isv=(1.41106m/s ) + (1.41 106 m/s ) = 2.0 106 m/s2 2The angle the velocity vector makes with the positive y-axis is = tan 1
Clemson - PHYS - 221
36.1. Model: A phasor is a vector that rotates counterclockwise around the origin at angular frequency .Solve: (a) Referring to the phasor in Figure EX36.1, the phase angle ist = 180 + 30 = 210 (b) The instantaneous value of the emf is rad180= 3.665
Clemson - PHYS - 221
37.1. Model: S and S are inertial frames that overlap at t = 0. Frame S moves with a speed v = 5.0 m/salong x-direction relative to frame S. Visualize: theThe figure shows a pictorial representation of the S and S frames at t = 1.0 s and 5.0 s. Solve: F
Clemson - PHYS - 221
38.1. Model: Current is defined as the rate at which charge flows across an area of cross section.Solve: Since the current is Q / t and Q = N / e , the number of electrons per second isN 10 nA 1.0 108 C/s = = = 6.25 1010 s 1 6.3 1010 s 1 t e 1.60 1019 C
Clemson - PHYS - 221
39.1. Solve: A steady photoelectric current of 10 A is indicated in the graph. The number of electrons persecond is10 A = 10Cs= 1.0 10 5C 1 electron = 6.25 1013 electrons/s s 1.6 10 19 C39.2. Model: Light of frequency f consists of discrete quanta,
Clemson - PHYS - 221
40.1. Model: The sum of the probabilities of all possible outcomes must equal 1 (100%).Solve: The sum of the probabilities is PA + PB + PC + PD = 1. Hence, 0.40 + 0.30 + PC + PD = 1 PC + PD = 0.30 Because PC = 2PD, 2PD + PD = 0.30. This means PD = 0.10 a
Clemson - PHYS - 221
41.1. Model: Model the electron as a particle in a rigid one-dimensional box of length L.Solve: Absorption occurs from the ground state n = 1. Its reasonable to assume that the transition is from n = 1 to n = 2. The energy levels of an electron in a rigi
Clemson - PHYS - 221
42.1.Solve: (a) A 4p state corresponds to n = 4 and l = 1. From Equation 42.3, the orbital angularmomentum is L = 1(1 + 1) = 2 . (b) In the case of a 5f state, n = 5 and l = 3. So, L = 3 ( 3 + 1) = 12 .42.2. Solve: (a) Excluding spin, a state is descri
Clemson - PHYS - 221
43.1. Model: The nucleus is composed of Z protons and neutrons. Solve: (a) 3H has Z = 1 proton and 3 1 = 2 neutrons. (b) 40Ar has Z = 18 protons and 40 18 = 22 neutrons. (c) 40Ca has Z = 20 protons and 40 20 = 20 neutrons. (d) 239Pu has Z = 94 protons and
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ENG 106Homework #1Winter 2012Question 1. Youve decided to buy a Toyota Prius because of the (recently) high price of gasoline. Like every car, aPrius wears as it ages, providing less in psychological benefits and costing more for maintenance and repai
Laney College - ENGIN - 106
ENG106Homework #1 SolutionWinter 2012Question 1. Youve decided to buy a Toyota Prius because of the (recently) high price of gasoline. Likeevery car, a Prius wears as it ages, providing less in psychological benefits and costing more formaintenance a
Laney College - ENGIN - 106
ENG 106Homework #2Winter 2012Time Value of Money (continued):2.32d What is the present worth of \$9000 at the end of each year for 30 years at 8.75% interest compoundedannually?2.36 Five annual deposits in the amounts of \$1,200, \$1,000, \$800, \$600 an
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ENG 106Homework #2 solutionWinter 20 12Time Value of Money (continued):2.32d What is the present worth of \$9000 at the end of each year for 30 years at 8.75% interest compoundedannually? (use equation 2.10, because appendix B doesn't have i= r = 8.75
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ENG 106Homework #3Winter 2012Due: Wed, Feb 1, 3 pm, homework box3.3 A California bank, Berkeley Savings and Loan, advertised the following information: interest 7.55% andeffective annual yield 7.842%. No mention is made of the interest period in the
Laney College - ENGIN - 106
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Laney College - ENGIN - 106
ENG 106Homework #4Winter 2012Due: Wed, Feb 8, 3 pm, homework box4.7 An annuity provides for 10 consecutive end-of-year payments of \$8,500. The average general inflation rate isestimated to be 5% annually, and the market interest rate is 12% annually.
Laney College - ENGIN - 106
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Laney College - ENGIN - 106
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Laney College - ENGIN - 106
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Laney College - ENGIN - 106
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Laney College - ENGIN - 106
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Laney College - ENGIN - 106
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Laney College - ENGIN - 106
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Laney College - ENGIN - 106
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Laney College - ENGIN - 106
ENG 106Homework #9 SolutionsDue: Wed, Mar 14, 3 pm, HW boxWinter 20129.X The J.F. Manning Metal Company is considering the purchase of a new milling machineduring year zero. The machines base price is \$135,000, and it will cost another \$15,000 tomod
Laney College - ENGIN - 106
Laney College - ENGIN - 106
IRR10yrcorporatebondPays\$40interestevery6monthsBondlife:10yrsSoldat9yearsfor\$950FindIRR1) Initial screening Benefits = \$40 x 18 + 950 = \$1,670 &gt; \$1000 continue2) Set up equationsPWcosts = -\$1,000PWbenefits = A (P/F, i, n1) + A (P/F, i, n2) + A (P
Laney College - ENGIN - 106
A123456789101112131415161718192021222324BCDEFGHIIRRofanoilwellusingERR=MARR(\$M)MARR=IRR=Yr01234567CashFlow\$4.0\$3.5\$2.5\$1.5\$0.5\$0.5\$1.5\$2.5InterestRate23.1%23.1%15.0%15.0%15.0%15.0%15.0%=IF(E14&gt;
Laney College - ENGIN - 106
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Laney College - ENGIN - 106
I. Types of Cash Flows (in year n)A. From Operating ActivitiesVarDescriptionTaxInflRnGross Revenue or Income (or savings)from sales of goods and services (or savings attributed to the project)Operating Expensescosts of items &quot;used up&quot; within the
Laney College - ENGIN - 106
ReplacementAnalysisForkliftDefenderMARR'=EOY0123410%P75000POps6000450030001500Yearskept,NEOYSG*8250990011700132007500110217500S GOps8250PW012375006000ATCF321423CR30009900475004500825008250990015
Laney College - ENGIN - 106
4500ReplacementAnalysisForkliftDefenderMarginalAnalysisMARR'=10%EOY01234P75000PNOps6000450030001500Yearskept,NEOYSG*825099001170013200750012131260004234500343000S G6000450030001500Ops825099001170013200A
Laney College - ENGIN - 106
ReplacementAnalysisPumpwithexpensesonlyMARR'=EOY01234Yearskept,NEOYP10%P2500350700105014001012500021322500012325000142342500CR1500800350OpsPWOps1500800400200S GATCFSG*3504007003507002001050350
Laney College - ENGIN - 106
-\(\- \;viI V\
Laney College - ENGIN - 106
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Laney College - ENGIN - 106
Laney College - ENGIN - 106
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Laney College - ENGIN - 106
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Laney College - ENGIN - 106
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Laney College - ENGIN - 106
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