Mathematical Homework and Review Solutions

Homework: you must do 3 problems from each of the 11 groups of problems, you must show work, not copy answers or my calculations and you must choose real problems or essays, no one or two word answers. Its due Friday at the exam.

Also, BRING IN YOUR BOOK

Here are the solutions . Let me know if I have to post a link or you can read this well enough.

Mathematical Physics Final Exam Review

Solutions to Problems

Oscillations and Waves

1. 1/5 or .2 Hz 2a. .04 s 2b. T = 2pi sqrt(m/k) ; m= 2x10^-4kg 3. T =5.44 s

4. 2250 N 5. 5. 900 N 6. 3.11 m 7. 136 Hz 8. 670 nm 9.a because the wavespeed increased 10. no drawings here come see me if you need to 11. c = srqt ( F_T/(M/L)) = 44.7 m/s first harmoci l = 2l = 4m F 44.7/4 = 11.2 Hz 12. c 13. b 14. a 15. b 16. b

17. 1 s ; 1Hz; new period =1.41 s 18. Waves in phase amplitudes add, produces maxima in light or sound or whatever type of wave 19. b 20a 21 k is spring constant in N/m; x is distance from equilibrium point 22.more

More Waves page 8

1. 1.3m 2. 30C 3. 103,000 Hz 4.no for both 5. c =256 x( 4x.33) = 338 m/s 6. 2L = c/f , L= 340/786 = .43 m 7. 6.3 Hz 8a. f = 850 (340/340-35) = 944 Hz 8b. f = 771 Hz

9. waves reflected from bottom arrive at top in phase with waves created or reflected at the top. 10. Open tube 2L =l or f =c/2L =343 Hz

More waves page 9

10. L = Nl/2 so f= c/l = Nc/2L

putting a cap on it changes it to one end closed L=Nodd l/4 or f=Nodd x c/4L

11.B 12. A 13. D 14.B 15. hot

Optics page 15

1. try it but don’t waste any time 2. di = 30cm Mag = 2x 3. AM radio low end of frequency spectrum gamma rays high end 4. 4.37x1014 5. 1.18x10^-14 6.Light passes through real image, light only seems to come from virtual image 7. q_incident = q _refl

8. di =27.5 cm Mi = .83 9.missing in action 10.di =26.7 cm M =1.667 11. f = 30 cm 12 do =15 cm

Refraction and more lenses Page 17

1. ni index of refraction for 1st ( incident) material nr index for 2nd (refracting ) material

qi angle incidence, qr angle of refraction. refraction: bending of light at surface between two materials as speed changes total internal reflection light bends so much away from normal it doesn’t penetrate second material i.e. qr = 90 deg happens only when nr

3. c = 1.94 x10⁸. 4.sinq water /sinq lucite = n_lucite/n_water q_water = 37 deg 5. sin 90 =1 so sinqcrit = n_air/n_water.=1/1.33 qcrit = 48 deg 5. n_gem/n_air = 1/sin 41 = 1.52 so n_obj =1.52 its glass 6. nmatl = sin 37.5/sin 25.6=1.4 7.b 8. total internal reflection and diffraction of sunlight inside water droplets ( don’t waste time on this now) 9. reflection light bounces off and is not transmitted into new medium, refraction: light bends as it enters new medium 11.C its virtual and erect 12. D bends away from normal 13. q_reflect = 45 deg, q_refraction = 32 deg 13. do = 2f so image is real inverted same size di = 2f = 20 cm

15. cglass = 2x10⁸m/s 16. c 17. A 18 still 48 deg.

Electrostatics Page 23 etc.

1. gained extra electrons and now repelled by extra electrons on the dome. 2. Gives electrons somewhere to escape to and can provide electrons when needed. 3. 9.6 x10⁴ C

4. .225 N away from 5. r² = q_Tkq_S/F so r = .135 N 6. E= F/q =3x10⁶

7. q = .6/24 = .025 C 8. 9.6x10^-12 9. excess electrons flow to leaves and repel each other, finger drains excess electrons 10. conduction tough rod to scope; induction hold rod near scope and allow some electrons to leave scope and go to finger. 11 Lines go from positive to negative. Lines are twice as crowded around positive charge since it has twice the charge.12. Missing info

Capacitance and Potential Page 25 Note correction last equation for DV on upper page 25 is DV =E dot d

1. Capacitance, voltage drop, charge, potential energy, distance, 9x10⁹,electric field

2. Coulombs, joules, Farad, Newton, Volt, meter, N/c or V/m. [the other]2. .101 J

3. 7.2x10^-8J 4. 10^-6;10^-9;10^-12 5.q=CV = 1.125 nCoul. 6. Cq/V =5.8 mF.

PE =QV/2=1.05x10^-4 J 7.this one is on you do it 8. 5V 9. store charge and energy ( really they pass AC and stop DC but that’s to advanced for you,) metal plates close to each other with air, plastic, or vacuum in between.

Currents and Circuits Page 29

1. electrical potential ( voltage); current; resistance; power. 2. beginning with Watts ending with farads: 10,9,7,1,5,6,8,4,2,3. 3 17 ohms 4 5V ( again) 5.d 6 b 7. 14.7 ohms, VI = 3300Watts, 9.9kWh 8. your turn 9. P =V²/R so R = V²/P = 192 ohms 10. 5kW x 12 x 92 days = 5520 ( about $800 these days) 11 ran away 12. a 13. b 14. rubber 15. copper 16. c 17. b 18. 1 ampere 19. potential energy per coulomb, volts 22 resistance to electrons flowing when voltage is across element, ohms 20.same voltage at all points on bird, no field in bird so no current 21 no effect I = V/R =2V/2R 22.( right place this time) dimmer, share the voltage each gets fewer volts. 23. a. 12V of course, b. I=12/7 amps, c. 7 ohms d. 12/7 amp – same- everywhere in series circuit. 24. pull a bulb out if others light parallel if not series. 25. b) 30 ohms c) 4V d) 1/5 amp 26. b) 6.67 ohms c) 12 V across each parallel resistor, d) 12/20 amps = .6 amps

27. R equiv = 4.4 ohms I circuit = 48/4.4 = 10.9 A

	2 ohm	4 ohm	6 ohm
I (A)	10.9	6.55	4.37	10.9
DV	21.8	26.2	26.2	48
P ( W)	238	172	114	524

28.D 29. D 30. C 3Amps 31. C 60 V 32. D 360 W 33. 2 A Total resistance is no 30 ohms total current is 4A and half passes through meter. 34 all of a,b, and c are true 35. d

36.75 watts = .075 kW .075 x 168 hrs = 12.6 kWh at $.1 per kWh costs= $1.26

37. b 38 T 39. T 40. T 41. F 42. F 43. T

Magnetism Page 34

1. A 2. 2 x10^-5 T 3. No only component perpendicular to field yields a force

4. F = 1.6x10^-19 x 3x10⁶ x 2 =2.3 x10^-12 N 5. EMF = -200* -1.6*.02/.2 =32 V

Fluids page 35

1. 90,000 N 2. 131,000 Pa 3. FB = 2.5 m3 x 1000 x10N/kg = 25000N 4. Use Bernoullis’s P₁-P₂ = 0 v₁ =0 so v2 = -2 g(h₂-h₁) = -2x10x (-3) = 60 so v₂ = 7.75 m/s

Thermo dynamics page 35

1. isothermal 2. n = 60 moles 3. 2x106 Pa 4. PDV = 5 x10⁶ J , T intitial = PV/nR =241 K, T final = 1444 K