RED. Physics 105 Final Exam Hess Colton CID NO TIME LIMIT. CLOSED BOOK/NOTES. 1 m 3 = 1000 L stress = F/A; strain = L/L. g=9.

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1 Fall 007 Pysics 105 Final Exam Hess -108 Colton NO IME IMI. COSED BOOK/NOES =9.80 m/s b b 4ac If ax bx c 0, x a 1 x xo vot at v vo ax xo w = m, PE = my f = N (or f N, for static friction) F = -kx, PE s = ½ kx P = F // v = Fvcos arc lent: s = r v = r a tan = r a c = v /r GMm GMm F, PE r r G = N. m /k R eart = m Meart = k Msun = k Ispere = (/5) MR Ioop = MR Idisk = ½ MR Irod (center) = (1/1) M I rod (end) = (1/3) M water = 1000 k/m 3 k B = J/K N A = R = k B N A = J/mol K R = liter atm/mol K = W/m K 4 1 atm = Pa = 14.7 psi 9 F C 3 5 K C cal = J RED Barcode Here CID 1 m 3 = 1000 stress = F/A; strain = / P P 0 A 1 v 1 = A v 1 1 P1 v1 y1 P v y 0 V 3 V0 1 3 KEave mv ave k B Q ka 1 t 4 P e A W on as = PV (constant pressure) 3 3 U nr Nk B (monatomic ideal as) Wnet Qc e 1 Q Q e max ec 1 k, m, c m k B Y v, = m/ v v v air = 340 m/s, unless oterwise specified A spere = 4r I 10 lo I 0 = 10-1 W/m I 0 v v0 f ' f v v S v v f n n n 1,,3,... f n n n 1,3,5,... 4

2 Keep four sinificant diits trouout your calculations; do not round up to less tan four. Wen data is iven, assume it as at least four sinificant diits. For example 15 meters means meters. You are encouraed to write your work and full answers on te exam (but record your final answers on te bubble seet as explained above). [? ] wit coices means simple multiple coice, marked on te bubble seet (scantron). Fall 008 students: ast year we did numerical answers a different way on te bubble seets. (Problems wit numerical answers are marked wit a [ S].) As a result, I opted not to include answer ranes on tis practice exam, but don t worry I will on your actual final exam. Since you re not allowed calculators, te problems on te final will eiter use easier numbers or no numbers at all. Also: your first pae will be similar, but not identical to te first pae of tis exam. And I will list te problems wit te problem numbers in front, and te answer coices oin vertically down te pae (te way I ve done te last couple of midterms). --Dr Colton Make sure your calculator is in DEGREES, not radians. On te surface of a pond, waves travel at a speed of 1. m/s. Kermit te Fro splases te surface 7 times eac second as e listens to music. e period of te water waves e makes is [1S] sec. e wavelent of te waves is [S] m. 1. = 1/ f=1/7 sec = = v/f =1. m/s/7 =0.171 m A 0. k mass vibrates on a sprin wit sprin constant 5 N/m. e frequency of te oscillation is [3S] Hz (not rad/s). = (k/m) = sqrt(5/0.) =5.0 rad/sec f= = Hz If you sake te end of a rope up and down, te resultin wave will be [4?]. 1) lonitudinal ) trianular 3) nodal 4) circular 5) transverse 4. coice 5 A boy is listenin to two speakers tat oscillate identically. He is 10 m from te first, and 11.5 m from te second. e lowest frequency of sound tat will ive im an intensity minimum at tis location because of interference is [5S] {74.7, 10} Hz. (Speed of sound is 340 m/s). 5. difference of 1.5 m in distance=/. en = 3m for destructive interference. f = v/ = 340/3 = 113 Hz A car drives very fast at 11 m/s toward a ill and onks its orn wit frequency of 480 Hz. (Speed of sound is 340 m/s). A man on te ill ears a frequency of [6S] Hz. 6. f' = fo V ± Vobs V ± Vsource. f =fo= 480*340/( ). v = f /(f -f o ) v s = 600 /(10) =716 Hz A sound meter measures te noise intensity level of a macine to be 58 db. e intensity arrivin at te meter is [7S] W/m. If tere were two identical macines te same distance from te meter, te noise level would be [8S] db. 7. = 10 lo(i/10-1 ) 5.8 = lo(i/10-1 ) I/10-1 = I = 6.31E-7 W/m 8. Eac factor of ives 3.01 db more: 61 db Wen a small part of an artery ets narrowed wit plaque, te blood in tis narrow portion of te artery as a blood pressure tat is [9?] 1) larer ) smaller 3) te same compared to te blood pressure in te wide parts just outside it in te same artery tube. 9. Coice

3 In a toy un, a compressed sprin wit sprin constant 300 N/m puses on a ball, speedin it up. If te ball as mass 0.00 k and leaves te un wit speed 4 m/s, te sprin must initially ave been compressed [10S] m. If te boy soots te ball into te air wit initial velocity 4 m/s, it will rise a distance of [11S] m before it stops. It will take [1S] sec to come to a stop at te top /kx = 1/mv. x = v*sqrt(m/k) = 4 *sqrt(0.0/300) = 3.6 cm 11. 1/mv = m. =v / = 4^//9.8; = m. t= v/a = 4 m/s/9.8 =0.408 sec A car is slowin down wile travelin in te neative-x direction. e direction of te acceleration of te car [13?] 1) is neative ) is positive 3) is zero 4) depends on speed. 13. direction of force: positive wo steamboats use ideal Carnot enines. e steam in one is otter tan te oter, and te temperature of te outside air is te same for bot. e efficiency of te enine will be reatest [14?] 1) in te cooler enine ) in te otter enine 3) in neiter enine. 14. Coice A cyclist is travelin at 15 m/s and is 83 m from te finis line of a strait track. o complete te race in te next 5 seconds, is constant acceleration must be [15S] m/s. 15. x = x o + v o t + ½ a t a =(x-v o t)/t = *(83-15*5)/5^;= 0.64 m/s A 100 k defensive end runnin nort at 4 m/s tackles a 75 k quarterback runnin sout at 9 m/s. eir velocity rit after te tackle is [16S] m/s to te [17] 1) nort ) sout. If instead te defensive end is runnin east before te tackle, teir velocity rit after te tackle is [18S] {3.15, 4.94} m/s *9 = 175*v v= (400-75*9 )/175=-1.57 m/s 17. Coice 18. pf = sqrt(400^+675^) Vf = pf/(m+m) = sqrt(400^+675^)/175=4.48 m/s A man weiin 480 N stands m from te left end of a plank weiin 350 N tat is 6 m lon. It pivots on its left end, were it is ancored to te wall. On te rit it is supported by a vertical cable. e torque on te plank about te left end applied by te man s weit is [19S] Nm. e tension in te cable at te rit end is [0S] {49, 349} N. = 480* = 960 Nm 0. e plank: = 350*3. For s to cancel, rope = *3=010 Nm. = = 010/6 m = 335 N

4 A brass rod of radius cm is too bi to fit in te cm radius ole tat is drilled for it in a plate. o fit te rod in te ole, you must lower its temperature by a minimum number of derees: [1S] C. e termal expansion coefficient of brass is / C. Assume tat te plate temperature does not cane /1.9e-5/1.5016; = 56.1 C A 500 k satellite orbitin te eart as an altitude of m above te surface of te eart. e potential enery of te satellite is [S] J lower tan if it were infinitely far from te eart.. PE = - GMm/r = 6.67e-11*500*5.98e4/(5e6+6.38e6);=8.76e10 Satellites in ier orbits ave a speed tat is [3?] 1) slower ) faster 3) same speed a satellite in lower orbit. 3. Coice 1 compared to A bar as lent 3 m and as a cm cm square cross section. If one end is placed in ice at -10 C, and te oter in water at 90 C, te eat flow is measured to be 500 W. e termal conductivity of te bar material is [4S] J/s m C. 4. Q/t = ka (-1)/ = Power. k=p/(a) = 500 * 3/(0.0*0.0)/100 = 3.75e4 W/m/ C A cannon is bein carried by a boat in a lake. e boat displaces a volume of water tat as te same weit as te [5?] 1) boat alone ) cannon and boat 3) cannon alone. e boat tips and te cannon falls into te water. e cannon is lifted slowly from te water by a cable. e tension in te cable is [6?] 1) equal to te weit of te water displaced by te cannon ) equal to te weit of te cannon 3) equal to te sum of #1 and # 4) equal to te difference of #1 and #. 5. Coice 6. Coice 4 A as is contained in a metal cylinder of volume 1 m3, at a pressure of 4 atmosperes at room temperature, 0 C. e number of moles contained in te tank is [7S] moles. If te as in te cylinder is cooled and compressed at constant pressure to a volume of 0.3 m 3, te work done was [8S] J. 7. PV=nR. n=pv/r = 4*1.01e5*1/8.314/(0+73) = moles 8. W = PV = 4*1.01e5*0.7=.836e5 J

5 A boy pulls orizontally a 15 k sled wit force 50 N alon a orizontal pat were te coefficient of friction is 0.5. e frictional force between te sled and te pat is [9S] N. e acceleration of te sled is [30S] m/s. If te boy pulls wit te same force at an anle of 30 above te orizontal te frictional force will be [31?] 1) less ) same 3) more 9. f = m = 0.5*15*9.8 = N F= ma= P-f = P-m. a = (P-f)/m= ( )/15 = m/s. 31. Normal force drops; f drops. Coice 1. A cannonball is sot orizontally off a cliff tat is 30 m i. e cannonball its te round 80 m from te base of te cliff. e cannonball was sot wit initial speed [3S] {.9, 3.6} m/s. Wen a cannonball sot at an anle above te orizontal is at its iest part of its arc, te acceleration of te cannonball is in te direction [33?] Nelect any friction. 1) forward ) backward 3) up 4) down 5) forward and up 6) forward and down 7) backward and up 8) backward and down 3. ime to fall 30 m: x = x o + v o t + ½ a t t= sqrt( y/) = sqrt(*30/9.8) =.47 sec. en v x = x/t = 80 m/.47 sec = 3.3 m/s 33. a is always down. Coice 4 A piano of mass 300 k is pulled at constant speed up a ramp inclined at 30 to te orizontal, by a rope parallel to te ramp. e acceleration of te piano is [34S] m/s. ere is no friction. e tension in te rope is [35] {140, 000} N. 34. V is constant, terefore a=0 F=0 = -msinmsinsin(30) = A 300 k orse is tied to a 55 k waon wit a rope. It pulls so te tension in te rope is 00 N, and tey bot accelerate. Assume no friction on te waon. e acceleration of te waon and te orse toeter is [36] m/s. e orizontal force tat te orse exerts on te round to accelerate tem bot is (and te round puses back te same) is [37S] {95, 390} N. 36. is te net force on te waon, so a = P/M = 00 N/ 55 k = m/s 37. F= (m+m)a = 85 K (0.381m) = 314.N. Or F- = ma, F = +ma = * = 314. N