WYSE Academic Challenge 2014 Sectional Physics Exam SOLUTION SET. [ F][ d] [ t] [ E]

Similar documents
WYSE Academic Challenge 2004 Sectional Physics Solution Set

PHY2053 Summer 2012 Exam 2 Solutions N F o f k

Energy & Work

Spring 2002 Lecture #17

element k Using FEM to Solve Truss Problems

WYSE Academic Challenge Sectional Physics 2007 Solution Set

Phys101 Second Major-061 Zero Version Coordinator: AbdelMonem Saturday, December 09, 2006 Page: 1

CHAPTER 6 WORK AND ENERGY

10/24/2013. PHY 113 C General Physics I 11 AM 12:15 PM TR Olin 101. Plan for Lecture 17: Review of Chapters 9-13, 15-16

( ) WYSE ACADEMIC CHALLENGE Regional Physics Exam 2009 Solution Set. 1. Correct answer: D. m t s. 2. Correct answer: A. 3.

Kinetics of Particles. Chapter 3

ME2142/ME2142E Feedback Control Systems. Modelling of Physical Systems The Transfer Function

SAFE HANDS & IIT-ian's PACE EDT-04 (JEE) Solutions

Conservation of Energy

10/23/2003 PHY Lecture 14R 1

CIRCUIT ANALYSIS II Chapter 1 Sinusoidal Alternating Waveforms and Phasor Concept. Sinusoidal Alternating Waveforms and

EE 204 Lecture 25 More Examples on Power Factor and the Reactive Power

f = µ mg = kg 9.8m/s = 15.7N. Since this is more than the applied

Final Exam Spring 2014 SOLUTION

Chapter 7. Systems 7.1 INTRODUCTION 7.2 MATHEMATICAL MODELING OF LIQUID LEVEL SYSTEMS. Steady State Flow. A. Bazoune

Q1. In figure 1, Q = 60 µc, q = 20 µc, a = 3.0 m, and b = 4.0 m. Calculate the total electric force on q due to the other 2 charges.

Q1. A string of length L is fixed at both ends. Which one of the following is NOT a possible wavelength for standing waves on this string?

Physics for Scientists and Engineers. Chapter 9 Impulse and Momentum

Important Dates: Post Test: Dec during recitations. If you have taken the post test, don t come to recitation!

Diodes Waveform shaping Circuits. Sedra & Smith (6 th Ed): Sec. 4.5 & 4.6 Sedra & Smith (5 th Ed): Sec. 3.5 & 3.6

Solution to HW14 Fall-2002

Circuits Op-Amp. Interaction of Circuit Elements. Quick Check How does closing the switch affect V o and I o?

Diodes Waveform shaping Circuits

kg C 10 C = J J = J kg C 20 C = J J = J J

ME306 Dynamics, Spring HW1 Solution Key. AB, where θ is the angle between the vectors A and B, the proof

Phys101 Final Code: 1 Term: 132 Wednesday, May 21, 2014 Page: 1

Chapter 11: Angular Momentum

2/4/2012. τ = Reasoning Strategy 1. Select the object to which the equations for equilibrium are to be applied. Ch 9. Rotational Dynamics

Final Exam Spring 2014 May 05, 2014

Analysis The characteristic length of the junction and the Biot number are

Sample Test 3. STUDENT NAME: STUDENT id #:

Study Guide For Exam Two

Electric potential energy Electrostatic force does work on a particle : Potential energy (: i initial state f : final state):

Section 3: Detailed Solutions of Word Problems Unit 1: Solving Word Problems by Modeling with Formulas

τ rf = Iα I point = mr 2 L35 F 11/14/14 a*er lecture 1

Chapter 5: Force and Motion I-a

Spring 2002 Lecture #13

K = 100 J. [kg (m/s) ] K = mv = (0.15)(36.5) !!! Lethal energies. m [kg ] J s (Joule) Kinetic Energy (energy of motion) E or KE.

1. An incident ray from the object to the mirror, parallel to the principal axis and then reflected through the focal point F.

Phys102 First Major-122 Zero Version Coordinator: Sunaidi Wednesday, March 06, 2013 Page: 1

Important: This test consists of 15 multiple choice problems, each worth points.

CHAPTER 10 ROTATIONAL MOTION

MTH 263 Practice Test #1 Spring 1999

ω = 0 a = 0 = α P = constant L = constant dt = 0 = d Equilibrium when: τ i = 0 τ net τ i Static Equilibrium when: F z = 0 F net = F i = ma = d P

Hooke s Law (Springs) DAVISSON. F A Deformed. F S is the spring force, in newtons (N) k is the spring constant, in N/m

b) (6) What is the volume of the iron cube, in m 3?

PHYS 219 Spring semester Lecture 02: Coulomb s Law how point charges interact. Ron Reifenberger Birck Nanotechnology Center Purdue University

Chapter 11 Torque and Angular Momentum

EN40: Dynamics and Vibrations. Homework 4: Work, Energy and Linear Momentum Due Friday March 1 st

Grade 12 Physics Exam Review

Conservation of Energy

Harold s AP Physics Cheat Sheet 23 February Electricity / Magnetism

Literal Equations Manipulating Variables and Constants

a) No books or notes are permitted. b) You may use a calculator.

PHYS 1443 Section 004 Lecture #12 Thursday, Oct. 2, 2014

Introduction to Electronic circuits.

Physics 207: Lecture 27. Announcements

WYSE Academic Challenge Regional Physics 2008 SOLUTION SET

PHYSICS 231 Review problems for midterm 2

Kinematics (special case) Dynamics gravity, tension, elastic, normal, friction. Energy: kinetic, potential gravity, spring + work (friction)

Grade 11/12 Math Circles Conics & Applications The Mathematics of Orbits Dr. Shahla Aliakbari November 18, 2015

EXAMPLE 2: CLASSICAL MECHANICS: Worked examples. b) Position and velocity as integrals. Michaelmas Term Lectures Prof M.

Flipping Physics Lecture Notes: Simple Harmonic Motion Introduction via a Horizontal Mass-Spring System

"NEET / AIIMS " SOLUTION (6) Avail Video Lectures of Experienced Faculty.

10/9/2003 PHY Lecture 11 1

Honors Physics Final Review Summary

43. A person sits on a freely spinning lab stool that has no friction in its axle. When this person extends her arms,

Chapter 07: Kinetic Energy and Work

Review questions. Before the collision, 70 kg ball is stationary. Afterward, the 30 kg ball is stationary and 70 kg ball is moving to the right.

Q1. A) 48 m/s B) 17 m/s C) 22 m/s D) 66 m/s E) 53 m/s. Ans: = 84.0 Q2.

Question Mark Max

BME 5742 Biosystems Modeling and Control

Physics 231 Ch 9 Day

LECTURE 2 1. THE SPACE RELATED PROPRIETIES OF PHYSICAL QUANTITIES

PHYSICS 231 Lecture 18: equilibrium & revision

Physics 220: Classical Mechanics

CHAPTER 12 OSCILLATORY MOTION

Quiz 3 July 31, 2007 Chapters 16, 17, 18, 19, 20 Phys 631 Instructor R. A. Lindgren 9:00 am 12:00 am

CHAPTER 8 Potential Energy and Conservation of Energy

Motion in Space. MATH 311, Calculus III. J. Robert Buchanan. Fall Department of Mathematics. J. Robert Buchanan Motion in Space

Problem While being compressed, A) What is the work done on it by gravity? B) What is the work done on it by the spring force?

ON-LINE PHYSICS 122 EXAM #2 (all online sections)

Chapter 10: Rotation

Translational Motion Rotational Motion Equations Sheet

in state i at t i, Initial State E = E i

Physics 106 Lecture 6 Conservation of Angular Momentum SJ 7 th Ed.: Chap 11.4

Chapter 11 Angular Momentum

Chapter 6 Work and Energy

Physics 102. Second Midterm Examination. Summer Term ( ) (Fundamental constants) (Coulomb constant)

Module 7: Solved Problems

Part C Dynamics and Statics of Rigid Body. Chapter 5 Rotation of a Rigid Body About a Fixed Axis

Physics 107 HOMEWORK ASSIGNMENT #20

AP Physics. Harmonic Motion. Multiple Choice. Test E

Flipping Physics Lecture Notes: Simple Harmonic Motion Introduction via a Horizontal Mass-Spring System

Homework for Diffraction-MSE 603: Solutions May 2002

Transcription:

WYSE Aaem Challenge 0 Setnal hss Exam SOLUTION SET. Crret answer: E Unts Trque / unts pwer: [ r ][ ] [ E] [ t] [ r ][ ][ t] [ E] [ r ][ ][ t] [ ][ ] [ r ][ t] [ ] m s m s. Crret answer: D The net external re (an therere the mpulse atng n the tw blk sstem s zer. r example, the upwar nrmal re that the surae exerts n eah blk s equal an ppste t the gravtatnal re n eah blk. The re that the sprng re exerts n the.00 kg blk s equal an ppste t the re that the sprng exerts n the.00 kg blk. Therere the mmentum the tw blk sstem must be nserve. ( m + ( m ( m + ( m (.00 kg. m/s + (.00 kg 0.00 m/s (.00 kg kg kg, nt 0.70 m/s kg kg, nt kg kg, kg kg,. m/s.00 kg.00 kg.00 kg.00 kg. Crret answer: E There are n external res that wrk n the sstem nsstng the tw blks an sprng. Therere the energ that sstem must be nstant. Thus the ntal knet energ the.00 kg blk must be equal t the knet energ the tw blks n the state plus the elast ptental energ n ths state. In equatn rm: m m + m + kx kg kg, nt kg, kg,. Crret answer: C At the tp ts trajetr, the prjetle has nl a hrzntal mpnent velt. Ths hrzntal mpnent velt s nstant sne there are n res atng n the prjetle n the hrzntal retn. Thus the hrzntal mpnent velt at the tp s the same as at launh: ( 6.0 m/s s( 6..6 m/s. hrz S /.6 / 6.0 0.. tp launh

0 Setnal hss Exam Slutn Set. Crret answer: D Usng nservatn energ, the sum gravtatnal ptental energ plus knet energ s nstant. GE + KE GE + KE 6. Crret answer: A ntal ntal 6.0 J + 7.0 J 7.0 J + KE GE h 8.0 J KE m.66 m/s GE ntal 6.0 J (.00 kg( 9.80 m/s 7.0 J - 6.0 J 0.88 m (.00 kg mg h 7. Crret answer: A A hrzntal B ur res at upn the lner, the tensn (T n the tensn, T gravt R strng pullng t the let at B, the rtnal re ( at D C atng alng the plane as shwn, the nrmal re (N N mg the plane at D atng alng a lne rm D t C, an the gravtatnal re (mg atng wnwar at C. Wth D hrzntal β respet t a lne perpenular t the page thrugh pnt C, N an mg nt prue trques sne ther lnes atn pass thrugh pnt C. s the nl res prung trques wth respet t C are the rtnal re an the tensn. Sne the lner s n stat equlbrum, the sum these tw trques must be zer. w + R sn( 90.0 ΣTrqueswrt, C RT sn 90.0 w 0 RT w R w RT w R w RT R 8. Crret answer: B Reslvng all res alng the nrmal (t the plane retn: Nrmal: N s( 0 N β β T C β mg gravt rtn: s( 90 0 Tensn: T sn(β mg s β Weght: hrzntal D N β The sum res n the nrmal retn must be zer. N mg s ( β T sn( β 0

0 Setnal hss Exam Slutn Set 9. Crret answer: D Applng Newtn s n Law t the allng mass an lettng wn be pstve: mg T ma T m g T ( a (.0 kg( 9.80 m/s.00 m/s T.N, where T s the tensn n the rpe..0 kg Applng the rtatnal rm Newtn s n Law t the wheel: τ Iα, (where τ trque, I mment nerta, anα angular aeleratn. RT Iα Ia / R I R T / a 0. Crret answer: A ( 0. m (.N /(.00 m/s 0.96kg m 0. m A B A ρ ρ Dg D D (.00 atm + ρntghnt + ρmantg( D + H (.00 atm + ρ g( H + H + D + ρ g( H ghnt + ρmantg( D + H ρntg( Hmtn + Hnt + D + ρmantg( H ghnt + ρmantg( D ρntg( Hmtn + Hnt + D ( ρmant ρnt ρntg( Hmtn + Hnt Hnt ρnt ( Hmtn + Hnt Hnt ρnt ( Hmtn ρnt ( Hmtn ( ρmant ρnt ( ρmant ρnt ( ρmant ρnt.87( 6. km ( 0.8 km (..87 nt nt B nt mtn nt mant Cntnent.87 g/m Mantle. g/m A Rt B D H Muntan 6. km. km Depth Cmpensatn. Crret answer: E Usng Newtn s n Law: ma rgnal rgnal + 0.0 N +. Crret answer: C atnal rgnal atnal atnal.00 kg (.00 m/s ma (.0 + 9.0 j N 0.0 N.00 kg (.0 +.8 j m/s (7.0 + 9.0 j N a + a 000 m 78000 m 600 s 600 s 9.0 m.8m/s

0 Setnal hss Exam Slutn Set. Crret answer: B t vert hrz gt g vert t hrz 7.0 m g vert 7.0 m.0 m 9.80 m/s 8.87 m/s.0 m 7.0 m. Crret answer: B Let { } ente mass energ nt energ { H} + { H} + { He} + { 0n} + energ { H} + { H} { He} + { 0n} + energ release. Crret answer: E energ nt energ release ({ H} + { H} { He} { 0n} 8 (.9979 0 m/s (.0 +.0609.0060.008669 7 (.678 0 u m / s (.66089 0 kg/u.76 0 J 9 (.76 0 J / (.60 0 J/e 7.Me u E average kt 6 (.8 0 J/K( + 7. K.77 0 J 0 6. Crret answer: C + ( 0 m ( 0 m ( 0 m + + ( s( θ let + sn( θ up let + up 6 ( 0 m up 6 ( 0 m ( πε e e.60 0 up + 9 9 9 6 ( 0 m ( 9 C an ε up + up + let let + 8.8 0 / C.0 0 /( N m..00 mm N, where.00 mm θ.00 mm

0 Setnal hss Exam Slutn Set 7. Crret answer: E. 0.0 m s m π A s x + ( π t λ π. π, λ m λ m. 8. 8. - π, s s π 8. v λ m/s. v T ρ T v 8. Crret answer: B + + 6.0 mm h h 6. mm h h 9. Crret answer: D 8. x + s 8. ρ m/s. 000 mm 6. 000 t ( 70 mm 9. mm (.67 0 kg/m 6.mN The mtn s nstant aelerate mtn wth an aeleratn 9.8 m/s vertall wnwar. Usng upwar as the pstve retn an the at that the greatest heght s reahe when the vertal velt s zer: v v 0 0 ( 0.0 m/s v v 0 a 0 + 0 +.0 m 7.9 m a 9.8 m/s 0. Crret answer: B The mtn s nstant aelerate mtn wth an aeleratn 9.8 m/s vertall wnwar. Usng upwar as the pstve retn, v v 0 + at 0.0 m/s 9.8 m/s (.0 s 0. m/s

0 Setnal hss Exam Slutn Set. Crret answer: D There are tw perpenular mpnents t the ar s aeleratn, the entrpetal aeleratn, a, an the tangental aeleratn, a t : a v a + at + at R. Crret answer: E re ( 9.00 m/s 70. m + ( 0.00 m/s 0.00 m/s Wrk s the area uner the re versus pstn graph between the lmts the mtn. The re at pstn x.00 m s (.00 N/m(.00 m6.00 N. The re at pstn x.00 m s (.00 N/m(.00 m.00 N. The wrk s the area the trapez wth base.00 m an the rst heght 6.00 N an sen heght.00 N. The wrk s stn W (.00m 8.0J (.00N + 6.00N. Crret answer: C Applng mpulse equals hange n mmentum an usng a rnate sstem wth pstve x twar the nrth: Impulse mv mv. Crret answer: B Impulse m v v 80.0N s ( 0.0 m/s ( 0.0 m/s.60 kg As the velt the blk s nstant, the net re n the bjet s zer. Cnserng the hrzntal mpnents re: rtn 60 Ns 0 0 60Ns 0 6.N rtn Cnserng the vertal mpnents re: nrmal mg + 60Nsn 0.0 0 nrmal mg 60Nsn 0.0 ( 0.0 kg( 9.8 m/s 60Nsn 0.0 69.N µ µ rtn k nrmal k rtn nrmal 6.N 69. N 0.0. Crret answer: B Applng Newtn s Sen Law an usng vertall upwar as the pstve retn: net ma a net m 00N 0N sn 0.0 8.00 kg ( 8.00 kg( 9.8 m/s. m/s

0 Setnal hss Exam Slutn Set 6. Crret answer: C I τ α ( 9.00N m (.00 kg m.80 ra/s ( θ θ ( π / 0 0 θ θ 0 + ω0t + αt an ω0 0 t. s α.80 ra/s 7. Crret answer: E N s/kg J 8. Crret answer: E Applng nservatn energ: ( 6.00 kg (.00 m/s (.00 m/s 0 ( 0.00 m mv mv mv + kx mv + kx k x x 9. Crret answer: A Applng the parallel axs therem: 88.N/m I I m + m I I m m ( 8.0 kg m ( 0.0 kg m.00 m.00 kg 0. Crret answer: C Applng nservatn angular mmentum: L rtr + L b I ω L rtr rtr ω I + L rtr rtr. Crret answer: B Reatane b + I + I b b ω I b rtr ω + I b ω I rtr ω rtr + I b b (.00 0 kg m (.00 0 ra/s + (.00 0 kg m ( 0 (.00 0 kg m + (.00 0 kg m ω 9. ra/ s. Crret answer: A R.00 Ω The equvalent resstane the 6.00 Ω an the.00 Ω restrs s R eq +.00 Ω 6.00 Ω.00 Ω Ths equvalent resstane s n seres wth the.00 Ω, resultng n a net resstane 6.00 R eq.00 Ω R eq.00 Ω R eq.00 Ω +.00 Ω. 00 Ω 6.00

0 Setnal hss Exam Slutn Set Usng Ohm s Law, the urrent lwng thrugh ths net equvalent resstane s 6.00 I.00 Ω.00 A Ths wll als be the urrent thrugh equvalent resstane R eq, resultng n a vltage arss R eq, I R.00 A.00Ω.00 eq eq Ths wll als be the vltage arss the rgnal.00 Ω resstr, resultng n a urrent thrugh the.00 Ω resstr I (.00 (.00 Ω.00 Ω. Crret answer: E. A (.00 sn(.0 sn(.0 n snθ n snθ n snθ n.9 snθ. Crret answer: C σ AT T σ AT σ AT T. Crret answer: C prtn ( 7. + 7. ( 0 + 7. (.00 W 6.0 W