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### ass10-4

Course: ASSIGN 312, Fall 2009
School: UVA
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10 Problem Assignment 4 4. Lifetime of the sun. The flux of radiant energy from the sun at the surface of the earth is approximately 1:4 kW=m2 . (a) From this estimate the total power produced by nuclear reactions in the sun; from your knowledge of the proton-proton cycle, estimate the rate of hydrogen burning in the sun; finally, estimate the amount of time before the sun ``burns out&quot;. (b) Now assume...

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10 Problem Assignment 4 4. Lifetime of the sun. The flux of radiant energy from the sun at the surface of the earth is approximately 1:4 kW=m2 . (a) From this estimate the total power produced by nuclear reactions in the sun; from your knowledge of the proton-proton cycle, estimate the rate of hydrogen burning in the sun; finally, estimate the amount of time before the sun ``burns out". (b) Now assume that the sun's energy comes from chemical reactions. Estimate the lifetime of the ``chemical'' sun. [Note: this very calculation was carried out by Lord Kelvin at the end of the 19th century. His estimate was substantially larger than ``biblical'' estimates, but much smaller than geological estimates based on sedimentation rates.] Given that the flux of radiant energy form the sun is 1:4 kW=m2 (which is energy/areatime) we can compute the total power (energy/time) of the sun by multiplying by the surface area of a sphere with a radius equal to the distance between the sun and the earth. or 2:5 1045 eV=sec. The energy produced by the sun comes from fusion. In a series of steps known as the proton-proton cycle or chain, protons (Hydrogen nuclei) are combined to form an alpha particle (Helium nuclei). The steps are (2) 1 H + 1 H ! 2H + e + 1 2 3 (2) H + H ! He + 3 3 4 He + He ! He + 1 H + 1 H (1:19 MeV) (5:49 MeV) (12:85 MeV) Psun = 1:4 kW=m2 4(1:5 1011 m)2 = 4:0 1026 Joules=sec The net result is 4 1 H ! 4 He + 2 2 + + 2 e i.e. that four hydrogens are transmuted into one helium with an energy of 26:21 MeV given off. Alternatively we can say that to get a helium atom we need to add four protons with a rest mass of 938:280 MeV each and two electrons with a rest mass of 0:5110 MeV each to get an particle of rest mass 3727:409 MeV. The difference is 26:733 MeV. If we assume that this is the only source of energy, then we can ask the rate at which protons are turned into helium. 1 proton proton cycle 4 protons 2:51045 eV=sec = 3:741038 protons=sec 7 eV 2:6733 10 1 proton proton cycle Now if we assume that the mass of the sun (1:99 1030 kg) started off as entirely 1 1:99 1030 kg = 1:24 1057 protons: 1:67 1027 kg The lifetime of the sun is then sec = 3:32 1018 sec sun = 1:24 1057 protons 3:74 1038 protons or 1:051011 years. This assumes among other things t...

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UVA - ASSIGN - 312
Assignment 9Problem 5Applied diffraction. (10 points) One important physical limitation on the resolving power of an antenna is diffraction. Under ideal conditions: (a) From how high can an eagle see a mouse on the ground? (b) A diffraction-limite
UVA - ASSIGN - 312
Assignment 9Problem 3Phased antenna arrays and diffraction. (10 points) Obtain the radiation pattern shown in a `polar plot' by Melissinos in Fig. 4.4(b) and the corresponding `straight' plot of the time-averaged dP=d- versus angle , as shown in t
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Assignment 11 - Problem 3List three radioactive nuclides that naturally occur (in easily detectable amounts) on Earth today. What are the half-lives of these nuclides? How did they originate? At least one of them should have a half-life of less than
UVA - ASSIGN - 312
Assignment 11 - Problem 3List three radioactive nuclides that naturally occur (in easily detectable amounts) on Earth today. What are the half-lives of these nuclides? How did they originate? At least one of them should have a half-life of less than
UVA - ASSIGN - 312
Assignment 9Problem 1Fourier integrals. (8 points) Suppose that an EM pulse is described by the Gaussian function 2 2 1 f(t) = p et =2 : 2 2 (a) Calculate the Fourier transform F (!) of the function f(t). If you use MAPLE, remember to say assume(s
UVA - ASSIGN - 312
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UVA - ASSIGN - 312
&quot; Lq d ghvnwrs hohfwurvwdwlf dlu fohdqhu/ dlu lv gulyhq wkurxjk wkh fhoo e| d idq wkdw gudzv , ` +zkhq vhw dw kljk,1 Wkh furvv vhfwlrqdo glphqvlrqv ri wkh fhoo duh 2D U4 e| D U41 Qhjohfwlqj orvvhv/ |rx fdq frpsxwh wkh vshhg ri dlu iorz/ / iurp wkhvh
UVA - ASSIGN - 312
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UVA - PHYS - 111
Annual Energy Review 2007The Annual Energy Review (AER) is the Energy Information Administration's (EIA) primary report of annual historical energy statistics. For many series, data begin with the year 1949. Included are data on total energy product
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Physics 111 - Energy On This World and Elsewhere - Fall 2008 Problem Set #1 with solutionsAssigned: 15 September 2008, Due: 23:59pm, 23 September 2008 Please find below the first homework assignment. While you may begin work on the assignment right
UVA - PHYS - 111
on this world and elsewhereInstructor: Gordon D. Cates Office: Physics 106a, Phone: (434) 924-4792 email: cates@virginia.eduEnergyCourse web site available through COD and Toolkit or at http:/people.virginia.edu/~gdc4k/phys111/fall08September 2
UVA - PHYS - 111
on this world and elsewhereInstructor: Gordon D. Cates Office: Physics 106a, Phone: (434) 924-4792 email: cates@virginia.eduEnergyCourse web site available through COD and Toolkit or at http:/people.virginia.edu/~gdc4k/phys111/fall08October 7,
UVA - PHYS - 111
on this world and elsewhereInstructor: Gordon D. Cates Office: Physics 106a, Phone: (434) 924-4792 email: cates@virginia.eduEnergyCourse web site available through COD and Toolkit or at http:/people.virginia.edu/~gdc4k/phys111/fall07November 4,
UVA - PHYS - 111
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UVA - PHYS - 111
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UVA - PHYS - 111
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UVA - PHYS - 111
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UVA - PHYS - 111
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UVA - PHYS - 111
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UVA - PHYS - 111
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UVA - PHYS - 111
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UVA - PHYS - 111
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UVA - PHYS - 111
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UVA - PHYS - 111
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UVA - PHYS - 111
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UVA - PHYS - 106
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UVA - PHYS - 106
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UVA - PHYS - 106
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UVA - PHYS - 106
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UVA - PHYS - 106
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UVA - PHYS - 106
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UVA - PHYS - 106
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UVA - PHYS - 106
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UVA - PHYS - 106
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UVA - PHYS - 106
How Things Work II(Lecture #11)Instructor: Gordon D. Cates Office: Physics 106a, Phone: (434) 924-4792 email: cates@virginia.eduCourse web site available through COD and Toolkit or at http:/people.virginia.edu/~gdc4k/phys106/spring08February 1
UVA - PHYS - 106
How Things Work II(Lecture #16)Instructor: Gordon D. Cates Office: Physics 106a, Phone: (434) 924-4792 email: cates@virginia.eduCourse web site available through COD and Toolkit or at http:/people.virginia.edu/~gdc4k/phys106/spring08February 22
UVA - PHYS - 632
Lecture 13 Electromagnetic Waves Ch. 33 Cartoon Opening Demo Topics Electromagnetic waves Traveling E/M wave - Induced electric and induced magnetic amplitudes Plane waves and spherical waves Energy transport Poynting vector Pressure produced
UVA - PHYS - 632
Warm up set 10Question1. HRW6 31.TB.02. [120186] Suppose this page is perpendicular to a uniform magnetic eld and the magnetic ux through it is 5 Wb. If the page is turned by 30 around an edge the ux through it will be: (a) 4.3 Wb (b) 10 Wb (c) 5.8
UVA - PHYS - 632
41. (a) The fraction of light which is transmitted by the glasses isIf I0 = E2 f E02 = Ev2 Ev2 = 2 = 016. . Ev2 + Eh2 Ev + (2.3Ev ) 2(b) Since now the horizontal component of E will pass through the glasses, If (2.3Ev ) 2 Eh2 = = = 0.84. I 0 Ev2 +
UVA - PHYS - 632
9. (a) BP1 = 0i1/2r1 where i1 = 6.5 A and r1 = d1 + d2 = 0.75 cm + 1.5 cm = 2.25 cm, and BP2 = 0i2/2r2 where r2 = d2 = 1.5 cm. From BP1 = BP2 we get i2 = i1 r2 1.5 cm = ( 6.5 A ) = 4.3A. r1 2.25 cm(b) Using the right-hand rule, we see that the curr
UVA - PHYS - 632
16. In applying Eq. 24-27, we are assuming V 0 as r . All corner particles are equidistant from the center, and since their total charge is 2q1 3q1+ 2 q1 q1 = 0, then their contribution to Eq. 24-27 vanishes. The net potential is due, then, to the
UVA - PHYS - 632
QuestionWarm up set 71.HRW6 28.TB.05. [119859] In the context of the loop and junctions rules for electrical circuits a junction is: (a) where a wire is connected to a battery (b) where three or more wires are joined (c) where a wire is bent (d)
UVA - PHYS - 632
2 13. (a) We use I = Em /20c to calculate Em:Em = 2 0 I c = 2 4 10-7 T m / A 140 103 W / m2 2.998 108 m / s .= 103 103 V / m. .chchch(b) The magnetic field amplitude is therefore Bm = Em 103 104 V / m . = = 3.43 10-6 T. 8 c 2.99