WYSE Academic Challenge Sectional Physics 2006 Solution Set

Transcription

1 WYSE Acadeic Challenge Sectional Physics 6 Solution Set. Correct answer: d. Using Newton s nd Law: r r F 6.N a.kg 6./s.. Correct answer: c. 6. sin θ Correct answer: b. o N 6. N Using Newton s nd Law: r r F ( )(g) (3M)(g) a ( + ) (M + 3M) A B B 3g/5 θ θ 4. Correct answer: c. W Fdcos θ ( ) (9.N)(35.)cos(8 o ) -665 J 5. Correct answer: b. Using Hook s Law twice and diiding: F F k k 34 N 48 N k k 9.4 c 6.4 c 6. Correct answer: b. 3.8 /s 3 (.5 s).5 (.5 s) s 3 (. s).5 (. s) s 3 (.5 s) (. s) V 3.8 /s.5 s. s.5 s ( ) 6 Sectional Solution Set

2 7. Correct answer: e. [ ] [ ][ ] ) [] τ [][ r F] ()(kg /s ) a. a force. F a (kg)(/s No b. a torque. No c. an electrical current. (C/s) No d. power. [ ] [ W] [ F][ d] (kg)(/s )() P [] t [] t s No e. none of the aboe. Correct answer 8. Correct answer: b. y oyt gt ot Soling the equation for t and substituting into the y equation, one obtains: y oy g o V. Thus y has a quadratic dependence upon. o 9. Correct answer: a. Joule Joule second. Watt Joule second. Correct answer: a. Using the right hand rule, position your right hand with fingers wrapped and pointing in the sae sense as the wheel is spinning. The thub then points to the right. R Ale. Correct answer: d. Using the right hand rule, one finds that the two forces produce a torque which is into the of the paper. Since torque equals the rate of change of angular oentu, the change in angular oentu is into the of the paper. Since the direction of the torque is perpendicular to the direction of the angular oentu, the agnitude of the angular oentu reains the sae. Only the direction of the angular oentu changes. The wheel is said to precess. 6 Sectional Solution Set

3 . Correct answer: c. Since the object is said to be in free flight, friction is assued to be negligible. Therefore, the only force acting on the object is graity, which is a conseratie force. Therefore, one ay use conseration of energy to sole the proble. The intial energy of the object and earth syste is 36. J J 9. J. A little later, the systes total energy is still 9. J, but now its potential energy is 54. J, so the kinetic energy of the object is 9. J 54. J 36. J. Using the forula for kinetic energy: 36. J (. kg) V. Soling for the speed, one obtains: V 8.49 /s. 3. Correct answer: e. The change in potential energy between the two eents is 54. J 36. J 8. J. Using the forula for potential energy: 8. J g Δy (. kg)(9.8 /s ) Δy. Soling, Δy Correct answer: e. Since the wires M and N are oing apart, the length after tie is related to the initial length by: (Lf Li ) L f L i + V + V. Thus, V + V. The ef is then found using: Δ(BA) BA f BA i B(A f A i ) B(WL f WL i ) BW(L f Li ) BW( V + V ε ). 5. Correct answer: d. Series cobination of 5. Ω and 5. Ω gies: (5. Ω + 5. Ω 67. Ω) 67. Ω ιn parallel wih 4. Ω gies: ( 67. )( 4. ) ( ) 6. Correct answer: a. Ω 5.Ω Using a for of Kirchhoff s Current Law, (8 A) + (-3A) + I 3 I 3-5 A ΣI ΣI, one has: entering leaing 6 Sectional Solution Set

4 7. Correct answer: d. Here one akes use of Archiede s Principle. The weight of the displaced fluid equals the buoyant force on the object. Since the object floats, the buoyant force ust equal the weight of the cylinder. The weight of the lower fluid that is displaced can be calculated as follows: (olue displaced) X (density) X (acceleration due to graity) (.5)( 6. c 3 )(. g/c 3 )(. kg / g)(9.8 /s ).94 N The weight of the top fluid that is displaced can be calculated as follows: (olue displaced) X (density) X (acceleration due to graity) (.5)( 6. c 3 )(. g/c 3 )(. kg / g)(9.8 /s ).94 N Total weight of fluid displaced weight of floating object.94 N +.94 N.588 N. 8. Correct answer: e. The nth haronic has n lobes ( waelength ( λ/ ). ). A lobe represents one half of a If L represents the length of the string, then for n lobes one has nλ L. The speed of waes on the string, the frequency of of the waes, and the waelength are related by the equation: f n λ n, where n indicates the nth haronic. The speed,, of waes on the string is independent of the haronic. That is, is constant for a gien string and string tension. Thus, using the relationship aboe to replace the waelength, L The net higher haronic would thus gie, f + n + Equating s and substituting in nubers, one has: n. L fn. n L L L L fn + fn. Or n + n n + n Cancelling L and soling for n, one obtains n Correct answer: b. Rolling without slipping can be epressed as c ω R ω R c. c/s. rad/s. c 6 Sectional Solution Set

5 . Correct answer: e. Using conseration of energy, ΔU + Δ d K gδh + ( ) f i [( 5. / s) (. / s) ] 9.9 ( 9.8 / s ) sin 4. and Δh d sin 4. ( ) ( ) f i f i d g sin 4. g sin 4.. Correct answer: c. The total east-west displaceent coponent will be d and the total north-south displaceent will be d y (. i) sin 6. + ( 3. i) sin i (. i) cos6. + ( 3. i) cos i The agnitude of the net dispalceent can be deterined by the Pythagorean theore: d (.3893 i) + ( 3. i) 3.5 i d + d y The direction is found using the inerse tangent function: d y 3. i θ tan tan 8.9 d.3893 i. Correct answer: b. total path length tie to traerse path. iles + 3. iles 3. inutes + 6. inutes.556 iles / inute 3. Correct answer: e. K 4. Correct answer: b. The acceleration is a d dt Δ Δ ( 6 /s) ( 6 /s) 4. s. s 3 /s Note that the instantaneous acceleration is equal to the aerage acceleration as this is a case of constant acceleration. 5. Correct answer: e. Ipulse Ft 6 Sectional Solution Set

6 6. Correct answer: c. y y t 3.8 s Projectile otion is constant acceleration otion. + y t + sinθ ± t or a yt y ( g) t.4 s + sin θ 4 sinθ t + ( g)( y y) gt y y + sinθ t gt (. /s) sin 3. ± (. /s) sin 3. 4( 9.8 /s )(.. ) 9.8 /s Obiously, the positie tie solution is the one that answers the question. 7. Correct answer: b. The difference (-y) has its least significant figure in the tenths place because y has its least significant figure in the tenths place. Calculating (-y)/z results in.64 which should be rounded to the tenth place, giing. as the correct answer. 8. Correct answer: d. c + 9. Correct answer: b. + (. kg)(. ) + (. kg)(. ) + ( 3. kg)( 3. ) (. kg) + (. kg) + ( 3. kg) Applying conseration of oentu, In a copletely inelastic collision, the final elocities are the sae: ( + ) g + g + g g ( 8 lb)(. /s) + ( 9 lb)( /s) ( 8 lb) + ( 9 lb) 9. /s 3. Correct answer: a. The equipotential lines are perpendicular to the electric field lines as shown. The electric field points fro higher electric potential to lower electric potential. D C B A 6 Sectional Solution Set

7 3. Correct answer: c. By the appropriate right-hand rule, the direction of the agnetic field produced by the currents can be deterined. The current toward the top of the page produces agnetic field out of the page in the quadrants containing points A and C and into the page in the quadrants containing points B and D. The current toward the left produces agnetic field out of the page in the quadrants containing points C and D and into the page in the quadrants containing points A and B. Because points A and D are all the sae distance fro the wires, the total agnetic field at points A and D is zero. The the total agnetic field at point C is greater than the agnetic field at point B because the agnetic field drops in agnitude as the distance fro the wires increases. 3. Correct answer: d. I r ( 3. kg)(.3 ) + (. kg)(.7 ).5 kg 33. Correct answer: c. τ rf sinθ 34. Correct answer: c. ( 3. kg)( 9.8 / s )(.3 ) sin 9 (. kg)( 9.8 / s )(.7 ) sin N There are one-fourth the original nuclei reaining after 6. inutes hae passed. This iplies that two half-lies hae passed in 6. inutes. Therefore, the half-life is 3. inutes. N Alternatiely, t ln t ln N ep t / t / ln N / N ( ) ln ( 6. in) ln [(. 9 )/( 8. 9 )] 3. in 35. Correct answer: c. I V + πfc ( 3. Ω) + π. V rs rs rs Z R V 6 ( Hz)( 4. F).78 A 6 Sectional Solution Set