WYSE Academic Challenge 2004 State Finals Physics Solution Set

Transcription

1 WYSE Acaemc Challenge 00 State nals Physcs Soluton Set. Answer: c. Ths s the enton o the quantty acceleraton.. Answer: b. Pressure s orce per area. J/m N m/m N/m, unts o orce per area.. Answer: e. Aerage elocty s splacement e by the tme oer whch the splacement occurre. The splacement s zero because the runner returne to the startng pont, so the aerage elocty s zero.. Answer: c. Usng the relatonshp between elocty an acceleraton or constant acceleraton an a coornate system wth east as the poste recton: + at.00 m/s + (.00 m/s )( 8.00 s).0 m/s The negate sgn on the result means that the nal elocty s.0 m/s towar the west. 5. Answer: b. Usng the relatonshp between splacement, elocty, an acceleraton or constant accelerate moton an agan usng a coornate system wth east as the poste recton: x x t + (.00 m/s)( 8.00 s) + (.00 m/s )( 8.00 s).0 m at The negate sgn on the result means that the splacement s.0 m to the west. 6. Answer: c. The object traels to the east, comes to a stop, an then traels west. Use the knematcal equaton relatng ntal elocty,, nal elocty,, acceleraton, a, an splacement, x - x, to etermne how ar east the object raels beore comng to rest: a 0 ( ) (.00 m/s) x x x x a (.00 m/s ).00 m Next, etermne how ar the objects nal poston usng the relatonshp between, a, x, x,, an tme, t: x (.00 m/s)( 8.00 s) + (.00 m/s )( 8.00 s).0 m x + t + at 0 + Ths poston s 6.0 m west o the arthest east poston. The object moe a stance.0 m east an then moe 6.0 m back to the west or a total o stance o 0.0 m. 00 State nals Physcs Soluton Set

2 7. Answer: c. Usng a ertcal coornate system wth the groun at y 0 an upwar poste, the ntal ertcal poston o the projectle s y +0 m an the nal poston s y 0. The ntal ertcal elocty s y ( 0.0 m/s) sn( 0.0 ) 0.0 m/s an the ertcal acceleraton s a constant 9.80 m/s. The tme that the project traels beore reachng the groun can be etermne rom the knematcal relatonshp y ( 0.0 m/s) t ( 9.80 ) y + yt + a t 0 m y m/s t Ths can be sole or t usng the quaratc ormula: 0.0 m / s ± t ( 0.0 m/s) ( 9.80 m/s )( 0.0 m) ( 9.80 m/s ).9s or 0.8s Obously, the poste tme soluton s the one we are seekng. 8. Answer: a. The horzontal component o the ntal elocty, x, s gen by ( 0.0 m/s) cos m/s cos θ x The horzontal component o the acceleraton s zero, so the horzontal elocty s constant. The horzontal component o elocty just beore the projectle reaches the groun s.6 m/s. 9. Answer: b. The mass o an object s a property o that object. It oes not epen on where the object s locate. 0. Answer: a. The acceleraton o the car s ts centrpetal acceleraton gen by a r ( 0.0 m/s) 00 m.00 m/s The roa s lat (horzontal) so the normal orce, N, s equal to the weght o the car, mg. The only horzontal orce s the statc rcton orce, r, between the tres an the roa. Rememberng that the upper lmt o the statc rcton orce s µ s N an applyng Newton's Secon Law mples r ma µ N ma µ ma N ma mg a g.0 m/s 9.80 m/s 0.0. Answer:. Impulse s equal to change n momentum an mpulse s the aerage orce multple by the tme the aerage orce s apple: ae t p p p t p p ae ( 00 kg m/s) ( 00 kg m/s) 0.0 N 0.0 s 00 State nals Physcs Soluton Set

3 . Answer: e. Work, W, s orce,, multple by the component o the splacement n the recton o the orce, r : W r W 0.0 J 5.0 N r.00 m. Answer:. The work-energy theorem states that the work s equal to the change n knetc energy: W KE m m Solng ths equaton or : + W / m (.00 m/s) + ( 0.0 J) /( 0.0 kg).5 m/s. Answer: a. The total change n gratatonal potental energy s PE gra mg h mg snθ 5 ( 80.0 kg)( 9.80 m / s )( m) sn J The bcyclst s 0% ecent at conertng chemcal potental energy nto gratatonal potental energy: PE t gra PE PE chem gra ( 0.00) ( 0.00)( 000 W) 00 W t.7 0 s T 00 W 00 W 5 J 5. Answer:. The translatonal spee,, o a pont n crcular moton s relate to the angular spee, ω, an the raus o the crcular moton, r, by ra re ωr π.00 re s ( 5.0 cm) 8cm/s 6. Answer:. I the angular elocty o the grnng wheel s constant, the net torque on the wheel must be zero. There are two torques actng on the wheel, the motor an the rctonal torque o the metal aganst the wheel These must be equal n magntue to a to zero.. (Note that the normal orce o the metal aganst the wheel causes no torque.) The rctonal orce, r acts a shown n the agram N m 5.00 N m 5.00 N m τ r R r sn 90 Rµ N µ 0.8 R N ( 0.00 m)( 60.0 N) 00 State nals Physcs Soluton Set

4 7. Answer:. Balancng requres the total torque to be zero. I the 0.0 kg s place a stance x to the let o the ulcrum, the total torque about an axs through the ulcrum s τ τ (.0 kg) gx + ( 0) m g( 0.00 m) ( 0.0 kg) g(.80 m) 0 0 kg + τ ulcrum + τ beam + τ 0kg 0 ulcrum beam Solng ths or x. m x beam g ( 0.00 m) + (.00 kg) g(.80 m) ( 0.0 kg) g ( 0.0 kg)( 0.00 m) + (.00 kg)(.80 m) ( 0.0 kg) 8. Answer: b. Drawng ree-boy agrams or each o the masses: m.00 kg 5.00 kg 0.0 N T s the tenson n the rope connectng the two masses. Newton's Secon Law can be wrtten or each mass. Here we only conser components n the horzontal recton wth a coornate system that has towar the rght as the poste recton. or the.00 kg mass: m a T (.00 kg) a or the 5.00 kg mass: 5 m 5 a N T ( 5.00 kg) a 5 Solng each equaton or the acceleraton an settng the two acceleratons equal: T 0.0 N T.00 kg 5.00 kg 0.0 N T 5.00 kg +.00 kg 5.00 kg. N 9. Answer: b. Uner these contons, the olume low rate o the water s unorm. Ths mples A A where A s the cross-sectonal area o the ppe an s the spee o the low. The ppes hae crcular cross-sectons so ths can be rewrtten as π π Applyng Bernoull's equaton to ths system ( h h ) + ( ) + ρ gh + ρ P + ρgh + ρ P P + ρg ρ P Makng the substtuton or ere aboe, P P P + ρg.8 0 ( ) 6 ( ) ( ) ( 0.0 cm) h h + ρ.00 0 N/m kg/m 6 N/m ( 0.0 cm) ( 5.00 m/s) 00 State nals Physcs Soluton Set

5 0. Answer: a. The work ae to the gas urng ths process can be etermne rom the rst Law o Thermoynamcs U Q W ( 60 J) 80 J Q U + W 0 J + The change n entropy s gen by S Q T 80 J 0. J/K.5 K Note that the temperature must be on an absolute scale. The numercal alue o the temperature n Kelns s obtane rom the numercal alue o the Celsus temperature by ang Answer: a. Coulomb's law relates the electrostatc orce to the charges an the separaton stance. Takng the rato o the rst orce to the secon orce: kq Q R kq Q R q ( q) R q( q) ( R) 8. Answer: c. The buoyant orce on an object epens on the ensty o the lu, ρ, the olume submerge, V, an the acceleraton o graty, g, an s gen by buoyant ρ Vg (.50 0 kg/cm )( 00 cm )( 9.80 m/s ) 5.88 N. Answer: c. Usng the rght-han rule, the magnetc el n the plane o the crcular loop ponts nto the page n the regon nse the loop. The magnetc el create by a long straght wre has a magntue nersely proportonal to the stance rom the wre. As the crcular loop moes away rom the wre, the magntue o the magnetc el, an hence the lux, s ecreasng. The magnetc el create by the nuce current opposes ths ecrease by ang to the magnetc el pontng nto the page. Agan applyng the rghthan rule, the current nuce n the crcular loop must low clockwse.. Answer: c. Ths may be remembere as a act that currents n opposte rectons are repelle by the magnetc nteracton between them. Howeer, the rght han rules or the recton o magnetc el aroun a straght wre an recton o magnetc orce on current can be use. Conser the two parallel wres shown below wth currents lowng n opposte rectons: The upper wre creates a magnetc el pontng nto the page at the poston o the lower. Usng the rght-han rule or the magnetc orce on the lower current ue to a magnetc el pontng nto the page, the magnetc orce on the lower wre s towar the bottom o the page. Ths s a repulse orce. 00 State nals Physcs Soluton Set

6 5. Answer: c. The.00 Ω an the let.00 Ω resstors are n parallel. They can be replace by an equalent resstance R A (.00 Ω)(.00 Ω) (.00 Ω) + (.00 Ω). Ω The equalent resstance R A wll be n seres wth the rght.00 Ω resstor. The equalent resstance wll be R +.00 Ω.Ω +.00 Ω. Ω R A 6. Answer:. rom the knetc theory o gases, the rms spee o gas molecules s proportonal to the square root o the absolute temperature. Ths mples rms rms T rms rms T T T 00 K 00 K ( 00 m/s) 00 m/s 7. Answer: c. In two-slt ntererence, constructe ntererence rnges occur uner the conton y n λ L where n s the orer o the rnge, λ s the waelength o the lght llumnatng the slts, s the separaton o the slts, y s the poston on the screen, an L s the stance rom the slts to the screen. The erence between the poston o the seenth orer an thr orer brght rnges on the screen, D, wll be gen by (.00 0 m)(.00 0 m) 7λL λl λl D 7 D y7 y λ m 500 nm L (.00 m) 8. Answer: b. In a seres crcut, whch contans an nuctor an no capactor, the current lags the oltage n phase. 9. Answer: b. The relatonshp between waelength an photon energy s hc E λ hc λ E 8 (.66 0 J s)( m/s) 6 7 (.00 ev)(.60 0 J/eV). 0 m nm 0. Answer: a. Applyng the thn lens equaton whch relates object poston, s, mage poston, s', an ocal length, : + s s s s s ( 0 cm)( 0 cm) ( 0 cm) ( 0 cm) cm The mage stance s negate because the mage s rtual. 00 State nals Physcs Soluton Set

7 . Answer: e. The ncent angle s the complementary angle to the 5.0 n the agram. Applyng Snell's Law to ths stuaton: (.00)( sn 6.0 ) n snθ n snθ n snθ θ arcsn 8. 5 n.600 Angle θ s the complementary angle to the reracte angle θ, so θ Answer: e. The two orces actng on the person are the orce o the scale an the orce o graty. Newton's Secon Law apple to the person s scale + graty ma 980 N mg m where we hae use a coornate system wth upwar poste. Solng ths or the mass m: (.00 m/s ) 980 N m.00 m/s + g 980 N.00 m/s m/s 8.kg. Answer: b. Ths s a act rom partcle physcs.. Answer: a. The Bohr moel o the atom prects energy leels or one-electron atoms to be E n.6 ev Z n where n s the prncpal quantum number an Z s the atomc number o the atom. Regarless o the atomc number, the lowest energy leel occurs when n. The next hgher energy leel occurs or n. 5. Answer: e. The mean letme, τ, o the sotope s relate to the hal-le, t / by /.00 yr τ t 5.77 yr ln ln The actty o a sample o raoacte sotope s relate to the number o nucle, N, an the mean letme, τ, o the sample: N A τ yr.7 0 / yr 00 State nals Physcs Soluton Set