WYSE caemic Challenge Sectinal Physics 7 Slutin Set. Crrect answer: E. Energy has imensins f frce times istance. Since respnse e. has imensins f frce ivie by istance, it clearly es nt represent energy. Dimensins fr each respnse are wrke ut belw using the fllwing efinitins: [P] pwer, [E] energy, [t] time, [F] frce, [M] mass, [a] acceleratin, an [] istance. a. watt ay [P]x[t] [E]/[t]x[t] [E] b. ft puns []x[f] [E] c. kg m /s [M] x [] x [] / [t] / [t] [M] x { [] / [t] } x [] [M] x [a] x [] [F] x [] [E]. N m [F] x [] [E] e. yne / cm [F] / [] nt energy. Crrect answer: B. ΔGPE m g h f - m g h i m g ( h f - h i ) Weight m g ΔGPE / ( h f - h i ) -3.4 kj / (-4. m) 557 N 3. Crrect answer: B. Frictin is an example f a nn-cnservative frce since the wrk ne by frictin is nt etermine by the starting an ening psitin, but by the path taken between the starting an ening psitin. 4. Crrect answer: E. Since n external impulse acts upn the system, the ttal mmentum f the system must be cnstant. Hwever the bjects in the system may exchange mmentum as they cllie. Thus if ne f the bjects gains mmentum thrugh the cllisin, the ther bject in the system must lse the same amunt f mmentum. Since the cllisin is perfectly elastic, the ttal kinetic energy f the system must be cnstant. Hwever the bjects in the system may exchange kinetic energy as they cllie. Thus if ne f the bjects gains kinetic energy thrugh the cllisin, the ther bject in the system must lse the same amunt f kinetic energy. 5. Crrect answer: D. ω v / r (6.8 m / s) / (.33 m) 8. ra/s 6. Crrect answer: C. t any instant, the tire appears t be rtating abut the pint in cntact with the highway. Since the tp f the tire is a istance r frm the cntact pint, the linear spee f the tp f the tire is v (r) ω (.66 m) (8. ra/s) 53.6 m/s 7 Sectinal Slutin Set
7. Crrect answer:. F m a m ( Δv / Δt ) Δv F Δt / m (. N-East) (3. s) / (5. kg). m/s East (answer a.) 8. Crrect answer: D. The wrk ne by the frce equals the area uner the curve frm x 3. m t x 5. m. Frce (N) 5.. 3. 5. x (m) W ½ (5 N + N) (5. m 3. m) 5 N m 5 J 9. Crrect answer: B. Cnsiering the chil+earth system, frictin is an external frce acting n the system which will change the energy f the system. et W the wrk ne by frictin, ΔKE the change in kinetic energy, an ΔGPE the change in gravitatinal ptential energy. W ΔKE + ΔGPE ½ (9. kg) (.3 m/s) + (9. kg) (9.8 m/s ) (-.5 m) -449 J. Crrect answer: D. The sum f frces in the vertical irectin equals zer. F sin(33. ) W + N F sin(33. ) (9. kg)(9.8 m/s ) + (5 N) F 6.8 N 33. F W N. Crrect answer: C. ccring t Newtn s n aw, the irectin f the acceleratin must be in the irectin f the frce, which is in a irectin twar the center f the earth. Even thugh the satellite is mving at cnstant spee, since it is mving in a circular rbit, an thus changing irectin, it experiences a change in velcity an is therefre accelerating. 7 Sectinal Slutin Set
. Crrect answer: C. View lking wn n table tp One shul push at the center f mass f the system in rer that the system have n tenency t rtate. 3. cm frm center f mass t center f mass 7. kg 3. kg Push (7. kg)(. cm) + (3. kg)(3. cm) x C.M. 7.5 cm 7. kg + 3. kg 3. Crrect answer:. Using Newtn s n aw: a F / M sys. N / (7. kg + 3. kg).5 m/s 4. Crrect answer: E. et wn be in the irectin f the negative y axis, with y at the surface f the table. et Viy the initial y cmpnent f velcity, V fy the final y cmpnent f velcity, a y the y cmpnent f acceleratin, y the final y psitin, an y the initial y psitin. f Then, V fy Viy + ay ( y f yi ) + ( 9.8m / s )(.5m ). Thus V fy 5.4m/ s. Since the marble still has an x cmpnent f velcity equal t 4. m/s, use the Pythagrean Therem t fin the final spee. S, V f fx fy / V + V 4 + 5.4 m / s 6.74m s. 5. Crrect answer: C. The current in each element f the series circuit will be I V / tt (. V) / (. Ω + 47. Ω).5. The pwer issipate in the 47. Ω resistr will be P I (.5 ) (47. Ω).938 W. The energy issipate in the 47. Ω resistr is then E P t (.938 W)(. s) 58.8 J. 6. Crrect answer: B. i F qvbsin( θ) 5. sin ( θ).83 θ 54. 3 6 r N (4.8 ( 8 54. ) C)(64. ms)(. T) sin( θ ) 7 Sectinal Slutin Set
7. Crrect answer: E. + +.769 cm f i. cm +.5 cm f. 3 cm 8. Crrect answer:. M i.cm 8.7.5cm 9. Crrect answer: The angular velcity is cnstant, therefre the angular acceleratin is zer.. Crrect answer: D. The tranlatinal spee, v, f a pint a istance r frm the axis f rtatin f a by rtating with angular velcity ω is given by. Crrect answer: C. (./s)(.m).m / s v ω r Using cnservatin f energy: 35. ( sin 35. ) + mv mg Δh + mv mv mg mv ( 5.m / s) (.m / s) mv mv v v mg sin 35. g sin 35. 9.8m / s ( ) sin 35..87 m. Crrect answer: B. The ifference B is equal t the sum + ( B). Graphical aitin f the vectrs: B + ( B) 7 Sectinal Slutin Set
3. Crrect answer:. Using a crinate system with x east an y nrth: x x. km x x + + x y y x 4.km sin 6. 34.64 km. km + 34.64 km 4.64 km y y 4. km cs6.. km +. km. km. km 4.64 km y ( 4.64 km) + (. km) 4.8km θ tan tan 53.8 ( Nf E) + y y x 4. Crrect answer: D. verage speee path length time. km + 4. km. minutes + 4. minutes.km / minute 5. Crrect answer: B. The wrk ne by a frce is the area between the frce versus psitin curve an the psitin axis. W ( 6N)( m) + ( 3 N)( m) 45J 6. Crrect answer: B. s in 5: ( 6N)( m) 6J W an ΔK W 6 J 7. Crrect answer: C. t an instant when the acceleratin f an bject is perpenicular t its velcity, the bject is neither speeing up nr slwing wn. It is nly changing irectin. Travelling suth with acceleratin east is turning t the river s left. 8. Crrect answer: D. n bject mving with cnstant spee has centripetal acceleratin. When the bject is at pint D, the centripetal irectin is twar the right f the page. D C B 9. Crrect answer: C. The mass f the astrnaut es nt epen n the astrnaut's lcatin s the astrnaut's mass is kg n the istance planet. 7 Sectinal Slutin Set
3. Crrect answer: E. v v a ( x x ) v v a( x x ) ( 9.8m / s )(.m) 9.8m / s 3. Crrect answer: E. F qvbsinθ 6 3 (. C)( m /s)( 3.T) sin 6..4 N The right han rule inicates that the magnetic frce is vertically upwar. 3. Crrect answer:. ( m / s) ( 4.Hz) c c λ f λ.5 m f 33. Crrect answer:. The fcal length f a mirrr is ne-half its raius f curvature, f / (3. cm)/ 5. cm. sf + s f s s s f (.cm)( 5.cm) (.cm) ( 5.cm) 3.cm The negative image istance means the image is 3. cm behin the mirrr. 34. Crrect answer: D. 3 K 4 K In the steay state, the rate f heat flw thrugh each piece will be the same: Q Q K ΔT K ΔT K ( T 3 K) 3K ( 4 K T ) J J Slving fr T J : T J ( 3K) + 3 ( 4K) + 3 36K 35. Crrect answer: E. ln ( 4. ).38 ecays per secn ln Nλ N t.s / 7 Sectinal Slutin Set