Physics 110. Spring Exam #1. April 23, 2008

Similar documents
Physics 120 Spring 2007 Exam #1 April 20, Name

Physics 15 Second Hour Exam

PHY2053 Summer C 2013 Exam 1 Solutions

( ) ( ) ( ) ( ) ( ) ( ) j ( ) A. b) Theorem

Calculus 241, section 12.2 Limits/Continuity & 12.3 Derivatives/Integrals notes by Tim Pilachowski r r r =, with a domain of real ( )

Go over vector and vector algebra Displacement and position in 2-D Average and instantaneous velocity in 2-D Average and instantaneous acceleration

Chapter 6 Plane Motion of Rigid Bodies

ESS 265 Spring Quarter 2005 Kinetic Simulations

2010 Sectional Physics Solution Set

X-Ray Notes, Part III

Chapter 4: Motion in Two Dimensions Part-1

Physics 207, Lecture 3

when t = 2 s. Sketch the path for the first 2 seconds of motion and show the velocity and acceleration vectors for t = 2 s.(2/63)

Introduction to Inertial Dynamics

Science Advertisement Intergovernmental Panel on Climate Change: The Physical Science Basis 2/3/2007 Physics 253

s = rθ Chapter 10: Rotation 10.1: What is physics?

5-1. We apply Newton s second law (specifically, Eq. 5-2). F = ma = ma sin 20.0 = 1.0 kg 2.00 m/s sin 20.0 = 0.684N. ( ) ( )

_ J.. C C A 551NED. - n R ' ' t i :. t ; . b c c : : I I .., I AS IEC. r '2 5? 9

() t. () t r () t or v. ( t) () () ( ) = ( ) or ( ) () () () t or dv () () Section 10.4 Motion in Space: Velocity and Acceleration

Rotations.

t s (half of the total time in the air) d?

Chapter 3: Vectors and Two-Dimensional Motion

Chebyshev Polynomial Solution of Nonlinear Fredholm-Volterra Integro- Differential Equations

Physics 201, Lecture 5

Ch.4 Motion in 2D. Ch.4 Motion in 2D

E-Companion: Mathematical Proofs

EE 410/510: Electromechanical Systems Chapter 3

Lecture 3 summary. C4 Lecture 3 - Jim Libby 1

Physics 201 Lecture 2

Physics 232 Exam II Mar. 28, 2005

Chapters 2 Kinematics. Position, Distance, Displacement

EGN 3321 Final Exam Review Spring 2017

SYMMETRICAL COMPONENTS

African Journal of Science and Technology (AJST) Science and Engineering Series Vol. 4, No. 2, pp GENERALISED DELETION DESIGNS

_ =- 314 TH / 3 RD 60M AR M NT GROUP C L) _. 5 TH AIR F0 RCE ` Pl R?N ]9. ia UNIT, - _ : --.

Motion on a Curve and Curvature

Field due to a collection of N discrete point charges: r is in the direction from

THIS PAGE DECLASSIFIED IAW EO IRIS u blic Record. Key I fo mation. Ma n: AIR MATERIEL COMM ND. Adm ni trative Mar ings.

Satellite Orbits. Orbital Mechanics. Circular Satellite Orbits

Integral Solutions of Non-Homogeneous Biquadratic Equation With Four Unknowns

-HYBRID LAPLACE TRANSFORM AND APPLICATIONS TO MULTIDIMENSIONAL HYBRID SYSTEMS. PART II: DETERMINING THE ORIGINAL

ME 141. Engineering Mechanics

Addition & Subtraction of Polynomials

Uniform Circular Motion

PHUN. Phy 521 2/10/2011. What is physics. Kinematics. Physics is. Section 2 1: Picturing Motion

graph of unit step function t

Example: Two Stochastic Process u~u[0,1]

Maximum likelihood estimate of phylogeny. BIOL 495S/ CS 490B/ MATH 490B/ STAT 490B Introduction to Bioinformatics April 24, 2002

Physic 231 Lecture 4. Mi it ftd l t. Main points of today s lecture: Example: addition of velocities Trajectories of objects in 2 = =

1. Kinematics of Particles

Chapter 2 Linear Mo on

Physics 201 Lecture 15

Mass-Spring Systems Surface Reconstruction

c- : r - C ' ',. A a \ V

ON THE EXTENSION OF WEAK ARMENDARIZ RINGS RELATIVE TO A MONOID

The Boltzmann transport equation and the diffusion equation

( ) ( ) ( ) 0. Conservation of Energy & Poynting Theorem. From Maxwell s equations we have. M t. From above it can be shown (HW)

(,,, ) (,,, ). In addition, there are three other consumers, -2, -1, and 0. Consumer -2 has the utility function

Numerical Study of Large-area Anti-Resonant Reflecting Optical Waveguide (ARROW) Vertical-Cavity Semiconductor Optical Amplifiers (VCSOAs)

A. Inventory model. Why are we interested in it? What do we really study in such cases.

Physics 101 Lecture 4 Motion in 2D and 3D

Elastic and Inelastic Collisions

TWO INTERFACIAL COLLINEAR GRIFFITH CRACKS IN THERMO- ELASTIC COMPOSITE MEDIA

T h e C S E T I P r o j e c t

P a g e 5 1 of R e p o r t P B 4 / 0 9

EECE 301 Signals & Systems Prof. Mark Fowler

STATEMENT OF ALL VOTES CAST AT THE DIRECT PRIMARY ELECTION HELD JUNE 7,1966. Twenty - ninth. Assembly District KERN COUNTY STATE OF CALIFORNIA

2D Motion WS. A horizontally launched projectile s initial vertical velocity is zero. Solve the following problems with this information.

Review: Transformations. Transformations - Viewing. Transformations - Modeling. world CAMERA OBJECT WORLD CSE 681 CSE 681 CSE 681 CSE 681

OH BOY! Story. N a r r a t iv e a n d o bj e c t s th ea t e r Fo r a l l a g e s, fr o m th e a ge of 9

Forms of Energy. Mass = Energy. Page 1. SPH4U: Introduction to Work. Work & Energy. Particle Physics:

Boyce/DiPrima 9 th ed, Ch 7.6: Complex Eigenvalues

Reinforcement learning

L4:4. motion from the accelerometer. to recover the simple flutter. Later, we will work out how. readings L4:3

PHYSICS 211 MIDTERM I 22 October 2003

Assistant Professor: Zhou Yufeng. N , ,

and ALiTO SOLO LOWELL, MICHIGAN, THURSDAY, AUGUST 9, 1928 First Results of the 1928 Nationwide Presidential Poll

THIS PAGE DECLASSIFIED IAW E

G OUP S 5 TH TE 5 DN 5. / E/ ' l / DECE 'I E THIS PAGE DECLASSIFIED IAW EO ', - , --,. . ` : - =.. r .

3 Motion with constant acceleration: Linear and projectile motion

Course Outline. 1. MATLAB tutorial 2. Motion of systems that can be idealized as particles

Version 001 test-1 swinney (57010) 1. is constant at m/s.

, _ _. = - . _ 314 TH COMPOSITE I G..., 3 RD BOM6ARDMENT GROUP ( L 5 TH AIR FORCE THIS PAGE DECLASSIFIED IAW EO z g ; ' ' Y ' ` ' ; t= `= o

Motion. ( (3 dim) ( (1 dim) dt. Equations of Motion (Constant Acceleration) Newton s Law and Weight. Magnitude of the Frictional Force

drawing issue sheet Former Royal High School - Hotel Development

TH IS PAG E D E CLA SSIFIED IAW E O 12958

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

TELEMATICS LINK LEADS

1.B Appendix to Chapter 1

On Fractional Operational Calculus pertaining to the product of H- functions

Handout on. Crystal Symmetries and Energy Bands

Answers to test yourself questions

Nonlocal Boundary Value Problem for Nonlinear Impulsive q k Symmetric Integrodifference Equation

a = Acceleration Linear Motion Acceleration Changing Velocity All these Velocities? Acceleration and Freefall Physics 114

COMP 465: Data Mining More on PageRank

Average & instantaneous velocity and acceleration Motion with constant acceleration

Week 10: DTMC Applications Ranking Web Pages & Slotted ALOHA. Network Performance 10-1

A.dr.rwarded to foreiirti count rie will be f 7 SOperann.. rsri--.-j- -.?- .JULY. 12, lsiii).,11,111. yc:tl crst.iif. lit. J. lor Sale... Kb l.

Wave Phenomena Physics 15c

ation and that his estimates of at the close of the present fis-' 1, He believed that $50.-

Transcription:

hyc Spng 8 E # pl 3, 8 Ne Soluon Mulple Choce / oble # / 8 oble # / oble #3 / 8 ol / In keepng wh he Unon College polcy on cdec honey, ued h you wll nehe ccep no pode unuhozed nce n he copleon o h wok.

I: ee Repone oble lee how ll wok n ode o ecee pl ced. I you oluon e llegble no ced wll be gen. lee ue he bck o he pge necey, bu nube he poble you e wokng on.. he Le Show ho, Dd Leen, lke o he people peo oe un o ge on V. Suppoe h Leen h peon nd on op o he Ed Sulln hee (whee he Le Show ped) nd how bll buldng ne doo. I he peon how he bll. / (gnong ny con) n ngle o o wh epec o he ecl, nd h he neghbong buldng wy.. h e he - nd y-coponen o he nl elocy? y coθ nθ co5 5. n 5 8. b. How boe o below he peon poon (on he ooop) doe he bll h? y y o oy y o 5. 8..6. (.6).9 (.6).7 c. h he nl elocy o he bll ju beoe ke he buldng? y y 5. g 8. y 9.8 @ θ n y (.6) 7. 7 @ θ.8 d. h he cceleon o he bll ju beoe ke he buldng? he cceleon due o gy nd h lue o 9.8 / n he eclly downwd decon.

. Suppoe h een cc Boeng 737 (hown below) on he unwy o lbny Inenonl po wng o keo clence. hen gen clence, he plo pple ull powe o he plne engne nd ccelee conn e o / down he unwy.. I he plne ke o when elocy eche 65 /h (whch kno o nucl le pe hou) nd no beoe, wh he nu e beoe he plne cn ke o? (Hn: k.6 nd h 36.) ; o 73.9 wh 37.. nd 65 h 73.9, b. How down he unwy doe he plne el beoe ke o? ( )( 37) 366.37. k c. Suppoe h e.5 k he plo decde h keo no e nd decde o op he plne. I he hu eee (on he engne) e engged nd he bke e ppled, he plne epeence oce h bng o e n k. h he gnude o h oce he plne h o 87, kg nd he plne elng 65 /h when he bke nd eee e ppled? he oce: ( ) ( 73.9 ) ( ).73 5 ( 87,kg)(.73 ) 7.8 N.

3. he hun body cpble o unng he lge oce whou dc njuy. Conde, o eple, n uooble ccden n whch n nlble ey dece (n bg) uddenly nd quckly nled.. Dw dg o h uon lbelng only he oce h c on he bg (once ully nled) nd on you body. h cn you conclude bou he oce o you hed on he bg nd he oce o he bg on you hed? body,bg bg,body he oce e equl n gnude nd oppoe n decon by Newon 3 d lw. b. Dee n epeon o he oce o he bg on you body ung h you wee nlly elng n nl peed o nd wee bough o e oe dnce. c. I you body (o 7 kg) coe o e oe dnce o 3 c nd you nl peed w 9.6 / (~ ph), wh he gnude o he oce h he bg eeed on you body? ( 9.6 ) 7kg.3.8 N.5 N c. hee pobbly h he bg could gge nd nle whle he c n oon nd hee w no ccden ll. h could p he on o he de nd hence he bg degned o epnd nd collpe oe ey ho e nel. Gen he noon boe, how long would ke he bg o dele he nlon w lo nnneouly nd ung h he bg ully deled oon he uooble bough o e? ( ) 7kg 9.6.8 N.3.8 O,.3 9.6. 3 7 nong h when olng he qudc equon, he que oo e zeo!

II: Mulple-Choce Ccle you be nwe o ech queon. ny ohe k wll no be gen ced. Ech ulple-choce queon woh pon o ol o pon.. o he dg below, he block o pulled o he gh conn peed by n ppled oce oened n ngle θ boe he hozonl, he gnude o he conl oce gen by. nθ b. coθ c. g d. N. Suppoe h oquo le hed-on no n oncong c nd dung he collon he oquo ee oce on he c. he oce h he c ee on he oquo. zeo. b.. c. le hn. d. gee hn. 3. bebll dond que wh de 9 ee n lengh. kng hoe ple o be he poe - nd he y- o hoe ple o 3 d be, he dplceen o unne how h double. 9 b. 8 c. 9.5 d. 7. You nd end nd on now coeed oo. You boh how nowbll wh he e nl peed, bu n deen decon. You how you nowbll downwd, o below he hozonl; you end how he nowbll, o boe he hozonl. hen he nowbll lnd on he gound, he peed (he gnude o he elocy) o you nowbll coped o you end

. he e. b. gee. c. le. d. unble o be deened. 5. lon o weghed by hngng o cle ched o he celng o n eleo whle he eleo e nd he wegh ound o be g. I he eleo cceleng downwd e o, he edng on he cle would be. N g. b. N g. c. N g. d. N g. 6. You dop ock o bdge no oe we below. hen he ock h llen, you dop nd ock. he ock connue he ee ll, he epon. ncee. b. decee. c. y he e. d. cnno be deened nce no enough noon gen o nwe. 7. I n objec el n he poe -decon 75 oe 36 econd nd he un ound nd el n he nege -decon 75 o 7 econd, he ege elocy. / b. 33.3 / c. 5 / d. 5 / 8. I he -coponen o he elocy o n objec pled eey 5 econd nd ploed, ound h he eul ncee lnely n e. hch gph below be epeen he ne oce on he pcle uncon o e?

9. Soee nge quoe e bued o ou people. Hee one eple o one uppoedly ueed by lbe Enen: Gon cnno be held eponble o people llng n loe. Could h be ue? h he gonl con beween wo people ech wh 7 kg h e eped by.5?..9 5 N b. 5. 6 N c..9-7 N d. 5. -6 N. Suppoe h you he cng nd h he blood n he o cceleed by he con o he he beng h pce nd hu chnge he elocy o / o.5 / oe dnce o.. h he cceleon o he blood?..5 / b. 6.3 / c. 9.8 / d. 87.5 /

Ueul oul: Moon n he, y o z-decon Uno Ccul Moon Geoey /lgeb G G c b b by gen e oluon whoe c b equon Qudc V bh C Sphee ngle Ccle :, : 3 3 ± Veco Ueul Conn ound K K J B ole o kg N k N G g 33 5.67.38 6. 6.67 9.8 8 3 3 σ y y eco decono eco o gnude n φ Lne Moenu/oce ok/enegy He R R ne R R N S g E E E E dco k U k gh U I K p p K p τθ θ µ [ ] ( ) ( ) ( ) Q U Q L k Q c Q k Nk V V V L L R C B B old new old new old new C C 3 5 9 9 5 3 : 3 3 εσ β β Roonl Moon lud Sple Honc Moon/e : : θ τ θ θ θ I L I gh gh gv gd V M B d L n n g l k n S ± µ λ