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

Size: px
Start display at page:

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

Transcription

1 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 / tp launh

2 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 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 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

3 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 N +. Crret answer: C atnal rgnal atnal atnal.00 kg (.00 m/s ma ( j N 0.0 N.00 kg ( j m/s ( j N a + a 000 m m 600 s 600 s 9.0 m.8m/s

4 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 ( m/s ( ( u m / s ( 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 ( 0 m ( 9 C an ε up + up + let let / C.0 0 /( N m..00 mm N, where.00 mm θ.00 mm

5 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 π, s s π 8. v λ m/s. v T ρ T v 8. Crret answer: B mm h h 6. mm h h 9. Crret answer: D 8. x + s 8. ρ m/s. 000 mm 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 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

6 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 N. 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 Ns 0 6.N rtn Cnserng the vertal mpnents re: nrmal mg + 60Nsn nrmal mg 60Nsn 0.0 ( 0.0 kg( 9.8 m/s 60Nsn 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 kg ( 8.00 kg( 9.8 m/s. m/s

7 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

8 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 ( ( (.00 W 6.0 W

WYSE Academic Challenge 2004 Sectional Physics Solution Set

WYSE Academic Challenge 2004 Sectional Physics Solution Set WYSE Acadec Challenge 004 Sectnal Physcs Slutn Set. Answer: e. The axu pssble statc rctn r ths stuatn wuld be: ax µ sn µ sg (0.600)(40.0N) 4.0N. Snce yur pushng rce s less than the axu pssble rctnal rce,

More information

PHY2053 Summer 2012 Exam 2 Solutions N F o f k

PHY2053 Summer 2012 Exam 2 Solutions N F o f k HY0 Suer 0 Ea Slutns. he ree-bdy dagra r the blck s N F 7 k F g Usng Newtn s secnd law r the -cnents F a F F cs7 k 0 k F F cs7 (0 N ( Ncs7 N he wrk dne by knetc rctn k r csθ ( N(6 cs80 0 N. Mechancal energy

More information

Energy & Work

Energy & Work rk Dne by a Cntant Frce 6.-6.4 Energy & rk F N m jule () J rk Dne by a Cntant Frce Example Pullng a Sutcae-n-heel Fnd the wrk dne the rce 45.0-N, the angle 50.0 degree, and the dplacement 75.0 m. 3 ( F

More information

Spring 2002 Lecture #17

Spring 2002 Lecture #17 1443-51 Sprng 22 Lecture #17 r. Jaehn Yu 1. Cndtns fr Equlbrum 2. Center f Gravty 3. Elastc Prpertes f Slds Yung s dulus Shear dulus ulk dulus Tday s Hmewrk Assgnment s the Hmewrk #8!!! 2 nd term eam n

More information

element k Using FEM to Solve Truss Problems

element k Using FEM to Solve Truss Problems sng EM t Slve Truss Prblems A truss s an engneerng structure cmpsed straght members, a certan materal, that are tpcall pn-ned at ther ends. Such members are als called tw-rce members snce the can nl transmt

More information

WYSE Academic Challenge Sectional Physics 2007 Solution Set

WYSE Academic Challenge Sectional Physics 2007 Solution Set 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.

More information

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

Phys101 Second Major-061 Zero Version Coordinator: AbdelMonem Saturday, December 09, 2006 Page: 1 Crdinatr: AbdelMnem Saturday, December 09, 006 Page: Q. A 6 kg crate falls frm rest frm a height f.0 m nt a spring scale with a spring cnstant f.74 0 3 N/m. Find the maximum distance the spring is cmpressed.

More information

CHAPTER 6 WORK AND ENERGY

CHAPTER 6 WORK AND ENERGY CHAPTER 6 WORK AND ENERGY CONCEPTUAL QUESTIONS 16. REASONING AND SOLUTION A trapeze artist, starting rm rest, swings dwnward n the bar, lets g at the bttm the swing, and alls reely t the net. An assistant,

More information

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

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 0/4/03 PHY 3 C General Physcs I AM :5 PM T Oln 0 Plan or Lecture 7: evew o Chapters 9-3, 5-6. Comment on exam and advce or preparaton. evew 3. Example problems 0/4/03 PHY 3 C Fall 03 -- Lecture 7 0/4/03

More information

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

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

More information

Kinetics of Particles. Chapter 3

Kinetics of Particles. Chapter 3 Kinetics f Particles Chapter 3 1 Kinetics f Particles It is the study f the relatins existing between the frces acting n bdy, the mass f the bdy, and the mtin f the bdy. It is the study f the relatin between

More information

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

ME2142/ME2142E Feedback Control Systems. Modelling of Physical Systems The Transfer Function Mdellng Physcal Systems The Transer Functn Derental Equatns U Plant Y In the plant shwn, the nput u aects the respnse the utput y. In general, the dynamcs ths respnse can be descrbed by a derental equatn

More information

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

SAFE HANDS & IIT-ian's PACE EDT-04 (JEE) Solutions ED- (JEE) Slutins Answer : Optin () ass f the remved part will be / I Answer : Optin () r L m (u csθ) (H) Answer : Optin () P 5 rad/s ms - because f translatin ωr ms - because f rtatin Cnsider a thin shell

More information

Conservation of Energy

Conservation of Energy Cnservatn f Energy Equpment DataStud, ruler 2 meters lng, 6 n ruler, heavy duty bench clamp at crner f lab bench, 90 cm rd clamped vertcally t bench clamp, 2 duble clamps, 40 cm rd clamped hrzntally t

More information

10/23/2003 PHY Lecture 14R 1

10/23/2003 PHY Lecture 14R 1 Announcements. Remember -- Tuesday, Oct. 8 th, 9:30 AM Second exam (coverng Chapters 9-4 of HRW) Brng the followng: a) equaton sheet b) Calculator c) Pencl d) Clear head e) Note: If you have kept up wth

More information

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

CIRCUIT ANALYSIS II Chapter 1 Sinusoidal Alternating Waveforms and Phasor Concept. Sinusoidal Alternating Waveforms and U ANAYSS hapter Snusdal Alternatng Wavefrs and Phasr ncept Snusdal Alternatng Wavefrs and Phasr ncept ONNS. Snusdal Alternatng Wavefrs.. General Frat fr the Snusdal ltage & urrent.. Average alue..3 ffectve

More information

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

EE 204 Lecture 25 More Examples on Power Factor and the Reactive Power EE 204 Lecture 25 Mre Examples n Pwer Factr and the Reactve Pwer The pwer factr has been defned n the prevus lecture wth an example n pwer factr calculatn. We present tw mre examples n ths lecture. Example

More information

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

f = µ mg = kg 9.8m/s = 15.7N. Since this is more than the applied Phsics 141H lutins r Hmewrk et #5 Chapter 5: Multiple chice: 8) (a) he maimum rce eerted b static rictin is µ N. ince the blck is resting n a level surace, N = mg. the maimum rictinal rce is ( ) ( ) (

More information

Final Exam Spring 2014 SOLUTION

Final Exam Spring 2014 SOLUTION Appled Opts H-464/564 C 594 rtland State nverst A. La Rsa Fnal am Sprng 14 SOLTION Name There are tw questns 1%) plus an ptnal bnus questn 1%) 1. Quarter wave plates and half wave plates The fgures belw

More information

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

Chapter 7. Systems 7.1 INTRODUCTION 7.2 MATHEMATICAL MODELING OF LIQUID LEVEL SYSTEMS. Steady State Flow. A. Bazoune Chapter 7 Flud Systems and Thermal Systems 7.1 INTODUCTION A. Bazune A flud system uses ne r mre fluds t acheve ts purpse. Dampers and shck absrbers are eamples f flud systems because they depend n the

More information

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. 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. Phys10 Secnd Majr-08 Zer Versin Crdinatr: Dr. I. M. Nasser Saturday, May 3, 009 Page: 1 Q1. In figure 1, Q = 60 µc, q = 0 µc, a = 3.0 m, and b = 4.0 m. Calculate the ttal electric frce n q due t the ther

More information

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?

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? Term: 111 Thursday, January 05, 2012 Page: 1 Q1. A string f length L is fixed at bth ends. Which ne f the fllwing is NOT a pssible wavelength fr standing waves n this string? Q2. λ n = 2L n = A) 4L B)

More information

Physics for Scientists and Engineers. Chapter 9 Impulse and Momentum

Physics for Scientists and Engineers. Chapter 9 Impulse and Momentum Physcs or Scentsts and Engneers Chapter 9 Impulse and Momentum Sprng, 008 Ho Jung Pak Lnear Momentum Lnear momentum o an object o mass m movng wth a velocty v s dened to be p mv Momentum and lnear momentum

More information

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

Important Dates: Post Test: Dec during recitations. If you have taken the post test, don t come to recitation! Important Dates: Post Test: Dec. 8 0 durng rectatons. If you have taken the post test, don t come to rectaton! Post Test Make-Up Sessons n ARC 03: Sat Dec. 6, 0 AM noon, and Sun Dec. 7, 8 PM 0 PM. Post

More information

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

Diodes Waveform shaping Circuits. Sedra & Smith (6 th Ed): Sec. 4.5 & 4.6 Sedra & Smith (5 th Ed): Sec. 3.5 & 3.6 des Waefrm shapng Cruts Sedra & Smth (6 th Ed): Se. 4.5 & 4.6 Sedra & Smth (5 th Ed): Se. 3.5 & 3.6 Tw-prt netwrks as buldng blks Reall: Transfer funtn f a tw-prt netwrk an be fund by slng ths rut ne.

More information

Solution to HW14 Fall-2002

Solution to HW14 Fall-2002 Slutin t HW14 Fall-2002 CJ5 10.CQ.003. REASONING AND SOLUTION Figures 10.11 and 10.14 shw the velcity and the acceleratin, respectively, the shadw a ball that underges unirm circular mtin. The shadw underges

More information

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

Circuits Op-Amp. Interaction of Circuit Elements. Quick Check How does closing the switch affect V o and I o? Crcuts Op-Amp ENGG1015 1 st Semester, 01 Interactn f Crcut Elements Crcut desgn s cmplcated by nteractns amng the elements. Addng an element changes vltages & currents thrughut crcut. Example: clsng a

More information

Diodes Waveform shaping Circuits

Diodes Waveform shaping Circuits des Waefrm shapng Cruts Leture ntes: page 2-2 t 2-31 Sedra & Smth (6 th Ed): Se. 4.5 & 4.6 Sedra & Smth (5 th Ed): Se. 3.5 & 3.6 F. Najmabad, ECE65, Wnter 212 Tw-prt netwrks as buldng blks Reall: Transfer

More information

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

kg C 10 C = J J = J kg C 20 C = J J = J J Seat: PHYS 1500 (Spring 2007) Exam #3, V1 Name: 5 pts 1. A pendulum is made with a length of string of negligible mass with a 0.25 kg mass at the end. A 2nd pendulum is identical except the mass is 0.50

More information

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

ME306 Dynamics, Spring HW1 Solution Key. AB, where θ is the angle between the vectors A and B, the proof ME6 Dnms, Spng HW Slutn Ke - Pve, gemetll.e. usng wngs sethes n nltll.e. usng equtns n nequltes, tht V then V. Nte: qunttes n l tpee e vets n n egul tpee e sls. Slutn: Let, Then V V V We wnt t pve tht:

More information

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

Phys101 Final Code: 1 Term: 132 Wednesday, May 21, 2014 Page: 1 Phys101 Final Cde: 1 Term: 1 Wednesday, May 1, 014 Page: 1 Q1. A car accelerates at.0 m/s alng a straight rad. It passes tw marks that are 0 m apart at times t = 4.0 s and t = 5.0 s. Find the car s velcity

More information

Chapter 11: Angular Momentum

Chapter 11: Angular Momentum Chapter 11: ngular Momentum Statc Equlbrum In Chap. 4 we studed the equlbrum of pontobjects (mass m) wth the applcaton of Newton s aws F 0 F x y, 0 Therefore, no lnear (translatonal) acceleraton, a0 For

More information

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

2/4/2012. τ = Reasoning Strategy 1. Select the object to which the equations for equilibrium are to be applied. Ch 9. Rotational Dynamics /4/ Ch 9. Rtatna Dynamcs In pue tansatna mtn, a pnts n an bject tae n paae paths. ces an Tques Net ce acceeatn. What causes an bject t hae an angua acceeatn? TORQUE 9. The ctn ces an Tques n Rg Objects

More information

Final Exam Spring 2014 May 05, 2014

Final Exam Spring 2014 May 05, 2014 95.141 Final Exam Spring 2014 May 05, 2014 Section number Section instructor Last/First name Last 3 Digits of Student ID Number: Answer all questions, beginning each new question in the space provided.

More information

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

Analysis The characteristic length of the junction and the Biot number are -4 4 The temerature f a gas stream s t be measured by a thermule. The tme t taes t regster 99 erent f the ntal ΔT s t be determned. Assumtns The juntn s sheral n shae wth a dameter f D 0.00 m. The thermal

More information

Sample Test 3. STUDENT NAME: STUDENT id #:

Sample Test 3. STUDENT NAME: STUDENT id #: GENERAL PHYSICS PH -3A (Dr. S. Mirv) Test 3 (/7/07) ke Sample Test 3 STUDENT NAME: STUDENT id #: -------------------------------------------------------------------------------------------------------------------------------------------

More information

Study Guide For Exam Two

Study Guide For Exam Two Study Gude For Exam Two Physcs 2210 Albretsen Updated: 08/02/2018 All Other Prevous Study Gudes Modules 01-06 Module 07 Work Work done by a constant force F over a dstance s : Work done by varyng force

More information

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

Electric potential energy Electrostatic force does work on a particle : Potential energy (: i initial state f : final state): Electc ptental enegy Electstatc fce des wk n a patcle : v v v v W = F s = E s. Ptental enegy (: ntal state f : fnal state): Δ U = U U = W. f ΔU Electc ptental : Δ : ptental enegy pe unt chag e. J ( Jule)

More information

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

Section 3: Detailed Solutions of Word Problems Unit 1: Solving Word Problems by Modeling with Formulas Sectn : Detaled Slutns f Wrd Prblems Unt : Slvng Wrd Prblems by Mdelng wth Frmulas Example : The factry nvce fr a mnvan shws that the dealer pad $,5 fr the vehcle. If the stcker prce f the van s $5,, hw

More information

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

τ rf = Iα I point = mr 2 L35 F 11/14/14 a*er lecture 1 A mass s attached to a long, massless rod. The mass s close to one end of the rod. Is t easer to balance the rod on end wth the mass near the top or near the bottom? Hnt: Small α means sluggsh behavor

More information

Chapter 5: Force and Motion I-a

Chapter 5: Force and Motion I-a Chapter 5: rce and Mtin I-a rce is the interactin between bjects is a vectr causes acceleratin Net frce: vectr sum f all the frces n an bject. v v N v v v v v ttal net = i = + + 3 + 4 i= Envirnment respnse

More information

Spring 2002 Lecture #13

Spring 2002 Lecture #13 44-50 Sprng 00 ecture # Dr. Jaehoon Yu. Rotatonal Energy. Computaton of oments of nerta. Parallel-as Theorem 4. Torque & Angular Acceleraton 5. Work, Power, & Energy of Rotatonal otons Remember the md-term

More information

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.

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. Knetc Energy (energy of moton) E or KE K = m v = m(v + v y + v z ) eample baseball m=0.5 kg ptche at v = 69 mph = 36.5 m/s K = mv = (0.5)(36.5) [kg (m/s) ] Unts m [kg ] J s (Joule) v = 69 mph K = 00 J

More information

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

1. An incident ray from the object to the mirror, parallel to the principal axis and then reflected through the focal point F. Hmewrk- Capter 25 4. REASONING Te bject stance ( = cm) s srter tan te cal lengt ( = 8 cm) te mrrr, s we expect te mage t be vrtual, appearng ben te mrrr. Takng Fgure 25.8a as ur mel, we wll trace ut: tree

More information

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

Phys102 First Major-122 Zero Version Coordinator: Sunaidi Wednesday, March 06, 2013 Page: 1 Crdinatr: Sunaidi Wednesday, March 06, 2013 Page: 1 Q1. An 8.00 m lng wire with a mass f 10.0 g is under a tensin f 25.0 N. A transverse wave fr which the wavelength is 0.100 m, and the amplitude is 3.70

More information

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

Important: This test consists of 15 multiple choice problems, each worth points. Physics 214 Practice Exam 1 C Fill in on the OPSCAN sheet: 1) Name 2) Student identification number 3) Exam number as 01 4) Sign the OPSCAN sheet Important: This test consists of 15 multiple choice problems,

More information

CHAPTER 10 ROTATIONAL MOTION

CHAPTER 10 ROTATIONAL MOTION CHAPTER 0 ROTATONAL MOTON 0. ANGULAR VELOCTY Consder argd body rotates about a fxed axs through pont O n x-y plane as shown. Any partcle at pont P n ths rgd body rotates n a crcle of radus r about O. The

More information

MTH 263 Practice Test #1 Spring 1999

MTH 263 Practice Test #1 Spring 1999 Pat Ross MTH 6 Practce Test # Sprng 999 Name. Fnd the area of the regon bounded by the graph r =acos (θ). Observe: Ths s a crcle of radus a, for r =acos (θ) r =a ³ x r r =ax x + y =ax x ax + y =0 x ax

More information

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

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 Equilibrium when: F net = F i τ net = τ i a = 0 = α dp = 0 = d L = ma = d P = 0 = I α = d L = 0 P = constant L = constant F x = 0 τ i = 0 F y = 0 F z = 0 Static Equilibrium when: P = 0 L = 0 v com = 0

More information

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

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 HYIC 534 XRCI-4 ANWR Hke s Law (prings) DAVION Clintn Davissn was awarded the Nbel prize fr physics in 1937 fr his wrk n the diffractin f electrns. A spring is a device that stres ptential energy. When

More information

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

b) (6) What is the volume of the iron cube, in m 3? General Physics I Exam 4 - Chs. 10,11,12 - Fluids, Waves, Sound Nov. 14, 2012 Name Rec. Instr. Rec. Time For full credit, make your work clear to the grader. Show formulas used, essential steps, and results

More information

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

PHYS 219 Spring semester Lecture 02: Coulomb s Law how point charges interact. Ron Reifenberger Birck Nanotechnology Center Purdue University PHYS 19 Spring semester 016 Lecture 0: Culmb s Law hw pint charges interact Rn Reifenberger Birck Nantechnlg Center Purdue Universit Lecture 0 1 Earl Develpments in Electrstatics Tw f the ur rces in Nature:

More information

Chapter 11 Torque and Angular Momentum

Chapter 11 Torque and Angular Momentum Chapter Torque and Angular Momentum I. Torque II. Angular momentum - Defnton III. Newton s second law n angular form IV. Angular momentum - System of partcles - Rgd body - Conservaton I. Torque - Vector

More information

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

EN40: Dynamics and Vibrations. Homework 4: Work, Energy and Linear Momentum Due Friday March 1 st EN40: Dynamcs and bratons Homework 4: Work, Energy and Lnear Momentum Due Frday March 1 st School of Engneerng Brown Unversty 1. The fgure (from ths publcaton) shows the energy per unt area requred to

More information

Grade 12 Physics Exam Review

Grade 12 Physics Exam Review Grade 12 Physcs Exam Revew 1. A 40 kg wagn s pulled wth an appled frce f 50 N [E 37 degrees abve the hrzntal. The wagn mves 8 m [E] hrzntally whle 5 N f frctn act. Fnd the wrk dne n the wagn by the...

More information

Conservation of Energy

Conservation of Energy Lecture 3 Chapter 8 Physcs I 0.3.03 Conservaton o Energy Course webste: http://aculty.uml.edu/andry_danylov/teachng/physcsi Lecture Capture: http://echo360.uml.edu/danylov03/physcsall.html 95.4, Fall 03,

More information

Harold s AP Physics Cheat Sheet 23 February Electricity / Magnetism

Harold s AP Physics Cheat Sheet 23 February Electricity / Magnetism Harold s AP Physics Cheat Sheet 23 February 206 Kinematics Position (m) (rad) Translation Horizontal: x = x 0 + v x0 t + 2 at2 Vertical: y = y 0 + v y0 t 2 gt2 x = x 0 + vt s = rθ x = v / Rotational Motion

More information

Literal Equations Manipulating Variables and Constants

Literal Equations Manipulating Variables and Constants Literal Equations Manipulating Variables and Constants A literal equation is one which is expressed in terms of variable symbols (such as d, v, and a) and constants (such as R, g, and π). Often in science

More information

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

a) No books or notes are permitted. b) You may use a calculator. PHYS 050 Sprng 06 Name: Test 3 Aprl 7, 06 INSTRUCTIONS: a) No books or notes are permtted. b) You may use a calculator. c) You must solve all problems begnnng wth the equatons on the Inormaton Sheet provded

More information

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

PHYS 1443 Section 004 Lecture #12 Thursday, Oct. 2, 2014 PHYS 1443 Secton 004 Lecture #1 Thursday, Oct., 014 Work-Knetc Energy Theorem Work under rcton Potental Energy and the Conservatve Force Gravtatonal Potental Energy Elastc Potental Energy Conservaton o

More information

Introduction to Electronic circuits.

Introduction to Electronic circuits. Intrductn t Electrnc crcuts. Passve and Actve crcut elements. Capactrs, esstrs and Inductrs n AC crcuts. Vltage and current dvders. Vltage and current surces. Amplfers, and ther transfer characterstc.

More information

Physics 207: Lecture 27. Announcements

Physics 207: Lecture 27. Announcements Physcs 07: ecture 7 Announcements ake-up labs are ths week Fnal hwk assgned ths week, fnal quz next week Revew sesson on Thursday ay 9, :30 4:00pm, Here Today s Agenda Statcs recap Beam & Strngs» What

More information

WYSE Academic Challenge Regional Physics 2008 SOLUTION SET

WYSE Academic Challenge Regional Physics 2008 SOLUTION SET WYSE cdemic Chllenge eginl 008 SOLUTION SET. Crrect nswer: E. Since the blck is mving lng circulr rc when it is t pint Y, it hs centripetl ccelertin which is in the directin lbeled c. Hwever, the blck

More information

PHYSICS 231 Review problems for midterm 2

PHYSICS 231 Review problems for midterm 2 PHYSICS 31 Revew problems for mdterm Topc 5: Energy and Work and Power Topc 6: Momentum and Collsons Topc 7: Oscllatons (sprng and pendulum) Topc 8: Rotatonal Moton The nd exam wll be Wednesday October

More information

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

Kinematics (special case) Dynamics gravity, tension, elastic, normal, friction. Energy: kinetic, potential gravity, spring + work (friction) Kinematics (special case) a = constant 1D motion 2D projectile Uniform circular Dynamics gravity, tension, elastic, normal, friction Motion with a = constant Newton s Laws F = m a F 12 = F 21 Time & Position

More information

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

Grade 11/12 Math Circles Conics & Applications The Mathematics of Orbits Dr. Shahla Aliakbari November 18, 2015 Faculty of Mathematics Waterloo, Ontario N2L 3G1 Centre for Education in Mathematics and Computing Grade 11/12 Math Circles Conics & Applications The Mathematics of Orbits Dr. Shahla Aliakbari November

More information

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

EXAMPLE 2: CLASSICAL MECHANICS: Worked examples. b) Position and velocity as integrals. Michaelmas Term Lectures Prof M. CLASSICAL MECHANICS: Worked examples Michaelmas Term 2006 4 Lectures Prof M. Brouard EXAMPLE 2: b) Position and velocity as integrals Calculate the position of a particle given its time dependent acceleration:

More information

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

Flipping Physics Lecture Notes: Simple Harmonic Motion Introduction via a Horizontal Mass-Spring System Flipping Physics Lecture Ntes: Simple Harmnic Mtin Intrductin via a Hrizntal Mass-Spring System A Hrizntal Mass-Spring System is where a mass is attached t a spring, riented hrizntally, and then placed

More information

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

NEET / AIIMS  SOLUTION (6) Avail Video Lectures of Experienced Faculty. 07 NEET EXAMINATION SOLUTION (6) Avail Vide Lectures f Exerienced Faculty Page Sl. The lean exressin which satisfies the utut f this lgic gate is C = A., Whichindicates fr AND gate. We can see, utut C

More information

10/9/2003 PHY Lecture 11 1

10/9/2003 PHY Lecture 11 1 Announcements 1. Physc Colloquum today --The Physcs and Analyss of Non-nvasve Optcal Imagng. Today s lecture Bref revew of momentum & collsons Example HW problems Introducton to rotatons Defnton of angular

More information

Honors Physics Final Review Summary

Honors Physics Final Review Summary Hnrs Physics Final Review Summary Wrk Dne By A Cnstant Frce: Wrk describes a frce s tendency t change the speed f an bject. Wrk is dne nly when an bject mves in respnse t a frce, and a cmpnent f the frce

More information

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

43. A person sits on a freely spinning lab stool that has no friction in its axle. When this person extends her arms, 43. A person sits on a freely spinning lab stool that has no friction in its axle. When this person extends her arms, A) her moment of inertia increases and her rotational kinetic energy remains the same.

More information

Chapter 07: Kinetic Energy and Work

Chapter 07: Kinetic Energy and Work Chapter 07: Knetc Energy and Work Conservaton o Energy s one o Nature s undamental laws that s not volated. Energy can take on derent orms n a gven system. Ths chapter we wll dscuss work and knetc energy.

More information

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.

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. 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. 30 kg 70 kg v (a) Is this collision elastic? (b) Find the

More information

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.

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. Phys10 Final-133 Zer Versin Crdinatr: A.A.Naqvi Wednesday, August 13, 014 Page: 1 Q1. A string, f length 0.75 m and fixed at bth ends, is vibrating in its fundamental mde. The maximum transverse speed

More information

Question Mark Max

Question Mark Max PHYS 1021: FINAL EXAM Page 1 of 11 PHYS 1021: FINAL EXAM 12 December, 2013 Instructor: Ania Harlick Student Name: Total: / 100 ID Number: INSTRUCTIONS 1. There are nine questions each worth 12.5 marks.

More information

BME 5742 Biosystems Modeling and Control

BME 5742 Biosystems Modeling and Control BME 5742 Bsystems Mdeln and Cntrl Cell Electrcal Actvty: In Mvement acrss Cell Membrane and Membrane Ptental Dr. Zv Rth (FAU) 1 References Hppensteadt-Peskn, Ch. 3 Dr. Rbert Farley s lecture ntes Inc Equlbra

More information

Physics 231 Ch 9 Day

Physics 231 Ch 9 Day Physs Ch 9 Day 0 0 Wed., /6 ab F, /8 n., / Tues. / 9. Rtatnal Enegy Quz 8 8 Enegy Quantzatn Reew Exam (Ch 5-8) Exam (Ch 5-8) RE 9.b bng laptp, smatphne, pad, Pate Exam (due begnnng lass) 9.-.5 (.9) The

More information

LECTURE 2 1. THE SPACE RELATED PROPRIETIES OF PHYSICAL QUANTITIES

LECTURE 2 1. THE SPACE RELATED PROPRIETIES OF PHYSICAL QUANTITIES LECTURE. THE SPCE RELTED PROPRIETIES OF PHYSICL QUNTITIES Phss uses phsl prmeters. In ths urse ne wll del nl wth slr nd vetr prmeters. Slr prmeters d nt depend n the spe dretn. Vetr prmeters depend n spe

More information

PHYSICS 231 Lecture 18: equilibrium & revision

PHYSICS 231 Lecture 18: equilibrium & revision PHYSICS 231 Lecture 18: equlbrum & revson Remco Zegers Walk-n hour: Thursday 11:30-13:30 am Helproom 1 gravtaton Only f an object s near the surface of earth one can use: F gravty =mg wth g=9.81 m/s 2

More information

Physics 220: Classical Mechanics

Physics 220: Classical Mechanics Lecture 10 1/34 Phys 220 Physics 220: Classical Mechanics Lecture: MWF 8:40 am 9:40 am (Phys 114) Michael Meier mdmeier@purdue.edu Office: Phys Room 381 Help Room: Phys Room 11 schedule on course webpage

More information

CHAPTER 12 OSCILLATORY MOTION

CHAPTER 12 OSCILLATORY MOTION CHAPTER 1 OSCILLATORY MOTION Before starting the discussion of the chapter s concepts it is worth to define some terms we will use frequently in this chapter: 1. The period of the motion, T, is the time

More information

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

Quiz 3 July 31, 2007 Chapters 16, 17, 18, 19, 20 Phys 631 Instructor R. A. Lindgren 9:00 am 12:00 am Quiz 3 July 31, 2007 Chapters 16, 17, 18, 19, 20 Phys 631 Instructor R. A. Lindgren 9:00 am 12:00 am No Books or Notes allowed Calculator without access to formulas allowed. The quiz has two parts. The

More information

CHAPTER 8 Potential Energy and Conservation of Energy

CHAPTER 8 Potential Energy and Conservation of Energy CHAPTER 8 Potental Energy and Conservaton o Energy One orm o energy can be converted nto another orm o energy. Conservatve and non-conservatve orces Physcs 1 Knetc energy: Potental energy: Energy assocated

More information

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

Motion in Space. MATH 311, Calculus III. J. Robert Buchanan. Fall Department of Mathematics. J. Robert Buchanan Motion in Space Motion in Space MATH 311, Calculus III J. Robert Buchanan Department of Mathematics Fall 2011 Background Suppose the position vector of a moving object is given by r(t) = f (t), g(t), h(t), Background

More information

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?

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? Problem 07-50 A 0.25 kg block s dropped on a relaed sprng that has a sprng constant o k 250.0 N/m (2.5 N/cm). The block becomes attached to the sprng and compresses t 0.12 m beore momentarl stoppng. Whle

More information

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

ON-LINE PHYSICS 122 EXAM #2 (all online sections) ON-LINE PHYSIS EXAM # (all nline setins) ) Bubble in the ID number setin f the santrn. ) This Exam is hurs lng - 34 multiple-hie questins. hse the ne BEST answer fr eah questin. Yu are nt penalized fr

More information

Chapter 10: Rotation

Chapter 10: Rotation Chapter 10: Rotation Review of translational motion (motion along a straight line) Position x Displacement x Velocity v = dx/dt Acceleration a = dv/dt Mass m Newton s second law F = ma Work W = Fdcosφ

More information

Translational Motion Rotational Motion Equations Sheet

Translational Motion Rotational Motion Equations Sheet PHYSICS 01 Translational Motion Rotational Motion Equations Sheet LINEAR ANGULAR Time t t Displacement x; (x = rθ) θ Velocity v = Δx/Δt; (v = rω) ω = Δθ/Δt Acceleration a = Δv/Δt; (a = rα) α = Δω/Δt (

More information

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

in state i at t i, Initial State E = E i Physcs 01, Lecture 1 Today s Topcs n More Energy and Work (chapters 7 & 8) n Conservatve Work and Potental Energy n Sprng Force and Sprng (Elastc) Potental Energy n Conservaton of Mechanc Energy n Exercse

More information

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

Physics 106 Lecture 6 Conservation of Angular Momentum SJ 7 th Ed.: Chap 11.4 Physcs 6 ecture 6 Conservaton o Angular Momentum SJ 7 th Ed.: Chap.4 Comparson: dentons o sngle partcle torque and angular momentum Angular momentum o a system o partcles o a rgd body rotatng about a xed

More information

Chapter 11 Angular Momentum

Chapter 11 Angular Momentum Chapter 11 Angular Momentum Analyss Model: Nonsolated System (Angular Momentum) Angular Momentum of a Rotatng Rgd Object Analyss Model: Isolated System (Angular Momentum) Angular Momentum of a Partcle

More information

Chapter 6 Work and Energy

Chapter 6 Work and Energy as p5309 L3 Chapter 6 rk and Energy Gals r Chapter 6 Study wrk as dened n physcs Relate wrk t knetc energy Cnsder wrk dne by a arable rce Study ptental energy Understand energy cnseratn Include tme and

More information

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

Physics 102. Second Midterm Examination. Summer Term ( ) (Fundamental constants) (Coulomb constant) ε µ0 N mp T kg Kuwait University hysics Department hysics 0 Secnd Midterm Examinatin Summer Term (00-0) July 7, 0 Time: 6:00 7:0 M Name Student N Instructrs: Drs. bdel-karim, frusheh, Farhan, Kkaj, a,

More information

Module 7: Solved Problems

Module 7: Solved Problems Mdule 7: Slved Prblems 1 A tn-walled nentr tube eat exanger f 019-m lengt s t be used t eat denzed water frm 40 t 60 at a flw rate f 5 kg/s te denzed water flws trug te nner tube f 30-mm dameter wle t

More information

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

Part C Dynamics and Statics of Rigid Body. Chapter 5 Rotation of a Rigid Body About a Fixed Axis Part C Dynamcs and Statcs of Rgd Body Chapter 5 Rotaton of a Rgd Body About a Fxed Axs 5.. Rotatonal Varables 5.. Rotaton wth Constant Angular Acceleraton 5.3. Knetc Energy of Rotaton, Rotatonal Inerta

More information

Physics 107 HOMEWORK ASSIGNMENT #20

Physics 107 HOMEWORK ASSIGNMENT #20 Physcs 107 HOMEWORK ASSIGNMENT #0 Cutnell & Jhnsn, 7 th etn Chapter 6: Prblems 5, 7, 74, 104, 114 *5 Cncept Smulatn 6.4 prves the ptn f explrng the ray agram that apples t ths prblem. The stance between

More information

AP Physics. Harmonic Motion. Multiple Choice. Test E

AP Physics. Harmonic Motion. Multiple Choice. Test E AP Physics Harmonic Motion Multiple Choice Test E A 0.10-Kg block is attached to a spring, initially unstretched, of force constant k = 40 N m as shown below. The block is released from rest at t = 0 sec.

More information

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

Flipping Physics Lecture Notes: Simple Harmonic Motion Introduction via a Horizontal Mass-Spring System Flipping Physics Lecture Ntes: Simple Harmnic Mtin Intrductin via a Hrizntal Mass-Spring System A Hrizntal Mass-Spring System is where a mass is attached t a spring, riented hrizntally, and then placed

More information

Homework for Diffraction-MSE 603: Solutions May 2002

Homework for Diffraction-MSE 603: Solutions May 2002 Hmewrk fr Diffratin-MSE 603: Slutins May 00 1. An x-ray beam f 1.5 Å impinges n a Ge single rystal sample with an inient angle θ lse t the ritial angle θ f the Ge surfae. Taking int aunt the absrptin,

More information