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

Similar documents
WYSE Academic Challenge 2004 Sectional Physics Solution Set

WYSE Academic Challenge Regional Physics 2008 SOLUTION SET

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

element k Using FEM to Solve Truss Problems

2010 Sectional Physics Solution Set

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

Work and Energy (Work Done by a Varying Force)

2015 Regional Physics Exam Solution Set

Chapter 2 Linear Mo on

PHY2053 Summer 2012 Exam 2 Solutions N F o f k

PH2200 Practice Exam I Summer 2003

Phys101 Lecture 4,5 Dynamics: Newton s Laws of Motion

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

REVIEW OF ENGINEERING THERMODYNAMICS

GENERAL PHYSICS PH 221-1D (Dr. S. Mirov) Test 4 (Sample) ALL QUESTIONS ARE WORTH 20 POINTS. WORK OUT FIVE PROBLEMS.

Chapter 3, Solution 1C.

EE 221 Practice Problems for the Final Exam

The Spring. Consider a spring, which we apply a force F A to either stretch it or compress it

v v at 1 2 d vit at v v 2a d

Frequency Is Neither A Pole Or A Zero In Fran De Aquino's m g Equation Jerry E. Bayles April 10, 2000

Use 10 m/s 2 for the acceleration due to gravity.

Solution to HW14 Fall-2002

As we have already discussed, all the objects have the same absolute value of

Chapter Newton-Raphson Method of Solving a Nonlinear Equation

Uniform Circular Motion

Energy & Work

Electric Potential Energy

ME 236 Engineering Mechanics I Test #4 Solution

Physics 321 Solutions for Final Exam

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

Physics 102. Final Examination. Spring Semester ( ) P M. Fundamental constants. n = 10P

Solutions to Problems. Then, using the formula for the speed in a parabolic orbit (equation ), we have

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

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

ADORO TE DEVOTE (Godhead Here in Hiding) te, stus bat mas, la te. in so non mor Je nunc. la in. tis. ne, su a. tum. tas: tur: tas: or: ni, ne, o:

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

CHAPTER 5 Newton s Laws of Motion

R th is the Thevenin equivalent at the capacitor terminals.

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

CYLINDER MADE FROM BRITTLE MATERIAL AND SUBJECT TO INTERNAL PRESSURE ONLY

Chapter 7 Impulse and Momentum

Demand. Demand and Comparative Statics. Graphically. Marshallian Demand. ECON 370: Microeconomic Theory Summer 2004 Rice University Stanley Gilbert

Version 001 HW#6 - Circular & Rotational Motion arts (00223) 1

Faculty of Engineering

2015 Sectional Physics Exam Solution Set

WYSE Academic Challenge Sectional Physics 2007 Solution Set

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

Solutions to Physics: Principles with Applications, 5/E, Giancoli Chapter 16 CHAPTER 16

ANALOG ELECTRONICS 1 DR NORLAILI MOHD NOH

Physic 231 Lecture 33

Physics for Scientists and Engineers. Chapter 9 Impulse and Momentum

Haddow s Experiment:

CHAPTER 10 ROTATIONAL MOTION

Conduction Heat Transfer

Work, Energy, and Power

Chapter Newton-Raphson Method of Solving a Nonlinear Equation

OVERVIEW Using Similarity and Proving Triangle Theorems G.SRT.4

Sample Test 3. STUDENT NAME: STUDENT id #:

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

Solutions for Homework #9

Quiz: Experimental Physics Lab-I

Name: SID: Discussion Session:

VECTORS VECTORS VECTORS VECTORS. 2. Vector Representation. 1. Definition. 3. Types of Vectors. 5. Vector Operations I. 4. Equal and Opposite Vectors

CHAPTER 6 WORK AND ENERGY

Solubilities and Thermodynamic Properties of SO 2 in Ionic

Phys101 First Major-131 Zero Version Coordinator: Dr. A. A. Naqvi Wednesday, September 25, 2013 Page: 1

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

PHYSICS ASSIGNMENT-9

5.1 Properties of Inverse Trigonometric Functions.

_J _J J J J J J J J _. 7 particles in the blue state; 3 particles in the red state: 720 configurations _J J J _J J J J J J J J _

Chapter 2 Introduction to Algebra. Dr. Chih-Peng Li ( 李 )

Review of linear algebra. Nuno Vasconcelos UCSD

How does the momentum before an elastic and an inelastic collision compare to the momentum after the collision?

= 1.23 m/s 2 [W] Required: t. Solution:!t = = 17 m/s [W]! m/s [W] (two extra digits carried) = 2.1 m/s [W]

Exponents and Powers

Objective of curve fitting is to represent a set of discrete data by a function (curve). Consider a set of discrete data as given in table.

16 Reflection and transmission, TE mode

DCDM BUSINESS SCHOOL NUMERICAL METHODS (COS 233-8) Solutions to Assignment 3. x f(x)

Microelectronics Circuit Analysis and Design. NMOS Common-Source Circuit. NMOS Common-Source Circuit 10/15/2013. In this chapter, we will:

6 Roots of Equations: Open Methods

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

g r mg sin HKPhO 香港物理奧林匹克 2014 Multiple Choices:

Chapter 6 : Gibbs Free Energy

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

Physics Dynamics: Atwood Machine

Dennis Bricker, 2001 Dept of Industrial Engineering The University of Iowa. MDP: Taxi page 1

Torsion, Thermal Effects and Indeterminacy

, where. This is a highpass filter. The frequency response is the same as that for P.P.14.1 RC. Thus, the sketches of H and φ are shown below.

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.

The Atwood Machine OBJECTIVE INTRODUCTION APPARATUS THEORY

Q x = cos 1 30 = 53.1 South

ALGEBRA 2/TRIGONMETRY TOPIC REVIEW QUARTER 3 LOGS

Types of forces. Types of Forces

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

Math 497C Sep 17, Curves and Surfaces Fall 2004, PSU

SOLUTIONS TO CONCEPTS CHAPTER

SOLUTIONS TO CONCEPTS CHAPTER 6

Center of Mass and Momentum. See animation An Object Tossed Along a Parabolic Path.

Homework Notes Week 7

PHYSICS Unit 3 Trial Examination

Transcription:

YSE CDEMIC CHLLENGE Regnl hyscs E 009 Slutn Set. Crrect nswer: D d hrzntl v hrzntl 3 345 t s t 0.3565s t d d d ll ll ll gt 9.80 s 0.63 ( 0.3565s). Crrect nswer: (-70. 0 ) ( 3 /s) t ( 4. 0 /s ) ( 4. 0 /s ) 8. 0 /s 3. Crrect nswer: v t (-70. 0 ) ( 3 /s)( t) ( 4. 0 /s )( t) ( t 0.50s) (-70. 0 ) ( 3 /s)( 0.50s) ( 4. 0 /s )( 0.50s) ( t 0.00s) (-70. 0 ) ( 3 /s)( 0.00s) ( 4. 0 /s )( 0.00s) ( t 0.50 s) ( t 0.00 s) ( 55) ( 70 ) ve 30. / s 0.50 s 0.00 s 0.50 s 4. Crrect nswer: Let nrth, suth - Δt v v t ( ) ( ) ( 40.0 N ) Δt ( 00. kg / s) ( Δt 0.0 s 5. Crrect nswer: C t 300. kg / s KE knetc energy, GE grvttnl ptentl energy ) 55 70 ΔEnergy ΔKE ΔGE drver drver. 0 ( ) g( y y ) ( 58.0 kg)( 0 4 ) / s g( y y ) 4 J drver drver drver gy drver drver gy

009 Regnl hyscs Slutn Set 6. Crrect nswer: E ( ) 0 ( ) ( 5) 7. Crrect nswer: D 4 / s.54 / s The chnge n entu syste s equl t the ttl pulse ctng n the syste. Ipulse Σ Δt 8. Crrect nswer:, sys, sys The unversl lw grvttn s credted t Newtn. 9. Crrect nswer: ccrdng t the unversl lw grvttn, the gntude the grvttnl rce between tw pnt sses s prprtnl t the prduct ther sses, nd nversely prprtnl t the squre ther seprtn dstnce. 0. Crrect nswer: D n pprprte unt esure r pressure s scl.. Crrect nswer: E ccrdng t the specl thery reltvty, n bserver wuld bserve tht clck vng reltve t the bserver runs t slwly. Let Δt prper the te ntervl seen by n bserver vng wth the clck. Let Δt prper the te ntervl seen by n bserver wh sees the clck vng wth speed v. c dentes the speed lght. Then: Δt prper Δtprper v c. Crrect nswer: C ecuse the net eternl rce ctng n the syste s zer, the entu the syste wll be cnserved. The knetc energy the syste wll be less ter the cllsn snce se energy s used t dent the pucks.

009 Regnl hyscs Slutn Set 3. Crrect nswer: KE KE 4 8 ( 4 ) 4. Crrect nswer: E Sung trques wth respect t the center the ruler: Στ Στ C CC c c (. g 5.0 g) 980 ( 50.0c 0.0c) sn( 90 ) ( M 5.0 g) 980 ( 75.0c 50.0c) sn( 90 ) 00 s s M 75 g 5. Crrect nswer: up ulcru ulcru ulcru dwn stck g hlder g ( 0.50 kg) g ( 0.00 kg) g ( 0.50 kg) g ( 0.050kg) g ( 0.50 kg) g ( 0.00 kg) ( 0.550 kg)( 9.8 / s ) 5.39 N 6. Crrect nswer: C net kg 000 g s 000 889 7. Crrect nswer: D y b rse y ntercept run 50 s y 3000 T ( 54 g) 3500 0. N 3 s 0.05 3 50 s R 3 50 s g

009 Regnl hyscs Slutn Set 8. Crrect nswer: E The rble s vng wth the crt bere t s prjected nd thus hs cnstnt hrzntl speed t the rght s seen by n bserver n the lb. hen the rble s prjected upwrd, t stll hs the se hrzntl speed t the rght, but nw t ls hs vertcl cpnent tn. The cbntn these tw tns yelds prblc pth, E, s shwn. C D E 9. Crrect nswer: E Usng cnservtn energy: X k v N 70 3.3J k gy N 3.3J 70 v.43 / s gy k ( 0. ) (.5 kg) 9.8 ( 0.3) (.5 kg) v gy s v s ( 0.3) (.5 kg) 9.8 ( 0.4 ) (.5 kg).4 s v k gy v 0. Crrect nswer: Gven : 80N @55 ˆ80N cs( 55 ) ˆ80 j N sn( 55 ) ˆ7.5N ˆ33.8 j N nd such tht : 33.8N @90 ˆ33.8 j N ˆ33.8 j N ˆ7.5N ˆ33.8 j N ˆ33.8 j N ˆ7.5N 7.5N @80 3 r. Crrect nswer: D τ τ w. r. t. w. r. t. r r r r ( )[ C ] 3r( 3 )[ CC ] ( 9 )( r )[ CC ] 7r[ CC ]

009 Regnl hyscs Slutn Set. Crrect nswer: C Snce n eternl trques ct n the syste 3 pnt sses nd ssless rd, the ngulr entu the syste s cnserved. th respect t pnt, the ntl ngulr entu the vng pnt ss s vr(c). The ntl ngulr entu the rd nd ttched sses s zer. Therere the ttl ntl ngulr entu s vr(c) 0 vr(c). terwrd, the rd nd ss syste spns wth cnstnt ngulr speed. In ters ths ngulr speed, ω, the trnsltnl speed the sses t dstnce r r s uter (r)ω, whle the trnsltnl speed the ss clsest t s nner rω. S the ttl nl ngulr entu the syste s: ( uter )(r)(c) ( uter )(r)(c) ( nner )(r)(c). v r r r Replcng nner nd uter nd splyng ne btns: (rω)(r)(c) (rω)(r)(c) (rω)(r)(c) 9r ω (C). Settng the ttl ntl ngulr entu equl t the ttl nl ngulr entu, ne hs: vr(c) 9r ω(c). Slvng: ω v / (9r). 3. Crrect nswer: E In the precedng stutn, whch physcl qunttes re cnserved n the syste 3 pnt sses nd rd? Snce n eternl trques ct n the syste 3 pnt sses nd ssless rd, the ngulr entu the syste s cnserved. Knetc energy s nt cnserved snce se energy s lst t het durng the nelstc cllsn. Lner entu s nt cnserved becuse the pn t eerts n eternl pulse n the rd t keep t r trnsltng. 4. Crrect nswer: π π π ω π T 48s 4s 5 π ωt cs 0 3 π π 4s t 8.00s 3ω 3 π 0 c ωt g h j 5 c 5. Crrect nswer: C T π k Snce perd depends upn ss nd sprng cnstnt, nd nt pltude, the perd des nt chnge. T 48.0 s.

009 Regnl hyscs Slutn Set 6. Crrect nswer: τ I α I R T R TR I R T T I R 0.0600 / s (.500 kg ) 4.69 N ( ) 0 0.0800 T T.50 kg 7. Crrect nswer: eternl 4kg &6kg syste ( 6.00kg) g ( 4.00kg) g ( 0.00kg)(.60 / s ) g 8.00 / s T T T Sttnry supprt T OR 4.00 kg T n 4 kg ss ( 4. 00kg) 4 kg ss T ( 4.00kg) g ( 4.00kg)(.60 / s ) (4 kg) g (6 kg) g 6.00 kg n 6 kg ss ( 6. 00kg) 6 kg ss ( 6.00 kg) g T ( 6.00kg)(.60 / s ) ddng these tw equtns yelds the equtn bve, resultng n the se nswer. 8. Crrect nswer: D Cnsderng the pulley: T T n pulley T 8.00 kg g 9. Crrect nswer: C bt reltvet erth / E θ sn T T ( 4.00 kg g 4.00kg.60 / s ).8 N pulley pulley 0 [ Sˆ ] ( 8.50k / h)[]?ˆ ( 5.00k / h)[ Eˆ ] bt reltvet rver rver reltve t erth 5 36.0 ( west suth) 8.5 ( 8.50k / h) θ ( 5.00k / h) Ŝ / E Ê

009 Regnl hyscs Slutn Set 30. Crrect nswer: C T blnce the dwnwrd rce grvty, the gnetc rce ust be drected upwrd. qv qv suth est qv up gnetc Due suth s the crrect respnse. ( ) ( ) ( ) 3. Crrect nswer: The ptentl derence crss the 4.00 kω resstr s gven by Oh s Lw. 4k Ω (5)(4kΩ) 0..00 kω.00 kω Thus the vltge crss the seres cbntn cnsstng the.00 kω nd 3.00 kω resstrs s ls 0. The current thrugh the.00 kω nd 3.00 kω cbntn s: 0 I k Ω r 3kΩ 4. kω 3kΩ Thus the ptentl derence crss the 3.00 kω resstr s: k (4)(3kΩ) 3 Ω 4.00 kω 3.00 kω 3. Crrect nswer: E 8 3 slpe 8 3 (equtn the prcess lne) rk re under 'curve' rk (verge pressure)(chnge n vlue) 8 3 3 8 8 3 rk ( 8 3 ) 7.5 33. Crrect nswer: ( 5 )( 5 ) ( 3 )( 3 ) T nrt nrt nrt nrt 5 9 5 T.78T 9 nrt 5 nrt 9

009 Regnl hyscs Slutn Set 34. Crrect nswer: D hen the ptentl derence crss the cpctr s 4.00, the ptentl derence crss the resstr s 9.00 4.00 5.00. Usng Oh s Lw t deterne the current n the resstr t tht nstnt (whch s ls the current n ech the seres eleents), ne hs 5.00 I 3. 33. Snce ths s ls the current n the cpctr t ths nstnt, ne hs.50kω q cp C Δcp Δq Δt C Δt 35. Crrect nswer: I C 3.33.00.67.67 s r t N s 0.5 wvelength. S there re 5 tes 0.5 wvelengths n the ppe. 5 L λ 4 4L λ 5 N N N.5 / L λ 4L / 5