FE Other Disciplines Practice Exam 9781947801042


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NCEES advancing licensure (or engineers and surveyors

~

other disciplines practice exam

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Copyright© 2020 by NCEES®. All rights reserved .

All NCEES sample questions and solutions are copyrighted under the laws of the United States. No part of this publication may be reproduced , stored in a retrieval system, or transmitted In any form or by any means without the prior written permission of NCEES. Requests for permissions should be addressed in writing to [email protected]. ISBN 978-1-947801-04-2

Printed in the United States of America 1st printing January 2020

CONTENTS Introduction to NCEES Exams

1

About NCEES Exam Format Examinee Guide Scoring and reporting Updates on exam content and procedures Exam Specifications ..... .......... ... ....................... ....... .... ...... . ..... .. .. ...... 3 Practice Exam ............. ... ........ ..... ..... ..... ............................. ............ .... 7 Solutions ................. ..... ..... ..... .... .. .... ,... ........... .... .............. .. ... ....... ... 57

iii

About NCEES NCEES is a nonprofit organization made up of the U.S . engineering and surveyi ng licensing boards in all 50 states, US . territories, and the District of Columbia. We develop and score the exams used for engineering and surveying licensure in the United States. NCEES also promotes professional mobility through its services for licensees and its member boards. Engineering licensure in the United States is regulated by licensing boards in each state and territory. These boards set and maintain the standards that protect the public they serve; As a result, licensing requirements and procedures vaty by jurisdiction, so stay in touch with your board (ncees.org/licensingboards).

Exam Format The FE exam contains 110 questions and is administered year-round via computer at approved Pearson VUE test centers. A 6-hour appointment time inc.ludes a tutorial, the exam, and a break. You'll have 5 hours and 20 minutes to complete the actual exam.

In addition to traditional multiple-choice questions with one correct answer, the FE exam uses common alternative item types such as • • •



Multiple correct options- allows multiple choices to be correct Point and click- requires examinees to click on part of a graphic to answer Drag and drop-requires examinees to click on and drag items to match, sort, ra~ or label Fill in the blank-provides a space for examinees to enter a response to the question

To familiarize yourself with the format, style, and navigation of a computer-based exam, view the demo on ncees.org/ExamPrep.

Examinee Guide The NCEES Examinee Guide is the official guide to policies and procedures for all NCEES exams. During exam registration and again on exam day, examinees must agree to abide by the conditions in the Examinee Guide, which includes the CBT Examinee Rules and Agreement. You can download the Examinee Guide at ncees.org/exams. It is your responsibility to make sure you have the current version. Scoring and reporting Exam results for computer-based exams are typically available 7- 10 days after you take the exam. You will receive an email notification from NCEES with instructions to view your results in your MyNCEES account. All results are reported as pass or fail. Updates on exam content and procedures Visit us at ncees.org/exams for updates on everything exam-related, including specifications, exam-day policies, scoring, and corrections to published exam preparation materials. This is also where you will register for the exam and find additional steps you should follow in your state to be approved for the exam.

DrJBl!lfffl

EXAM SPECIFICATIONS

3

-•..!&-~:.~l?J,t,Zi/1~.ZW,··9~-iQr.

Fundamentals of Engineering (FE} OTHER DISCIPLINES CBT Exam Specifications Effective Beginning with the July 2020 Examinations



n,e FE exam is a computer-based test (CBT). It is closed book with an electronic reference.



Examinees have 6 hours to complete the exam, which contains 110 questions. The 6-hour time also includes a tutorial and an optional scheduled break.

The FE exam uses both the International System of Units (SI) and the U.S. Customary System (USCS).



Number of Questions Knowledge

8-12

Mathematics

1.

A. Analytic geometty and trigonometty

B. Differential equations , . C. Numerical methods (e.g., algebraic equations, roots of equat10ns, approximations, precision limits, convergence) o. Linear algebra (e.g., matrix operations) E. Single-variable calculus

Probability and Statistics A. Estimation (e.g., point, confidence intervals) B. Expected value and expected error in decision making . . c. Sample disb-ibutions and sizes (e.g., significance, hypothesis testing, non-nonnal distributions) 2 D. Goodness of fit (e.g., correlation coefficient, standard enors, R )

2.

6-9

5-8

3.

Chemistry A. Oxidation and reduction (e.g., reactions, corrosion control) B. Acids and bases (e.g., pH, buffers) C. Chemical reactions (e.g., stoichiometry, equilibrium, bioconversion)

4.

Instrumentation and Controls A. Sensors (e.g., temperature, pressure, motion, pH, chemical constituents) B. Data acquisition (e.g., logging~ sampling rate, sampling range, filtering, amplification, signal interface, signa1 processing, analog/digital [AID], digital/analog [D/A], digital) C. Logic diagrams

4-6

Engineering Ethics and Societal Impacts A. Cod~s of ethic_s (e;g., identifying and solving ethical dilemmas) B. Pubhc protection issues (e.g., licensing boards) C. Soc~etal impacts (e.g., economic, sustainability, life~cycle analysis, environmental, public safety)

5-8

5.

4

6.

6-9

Safety, Health, and Environment A. Industrial hygiene ( e.g., carcinogens, toxicology, exposure limits, radiation exposure, biohazards, half-life) B. Basic safety equipment (e.g ., pressure-relief valves, emergency shutoffs, fire prevention and control, personal protective equipment) C. Gas detection and monitoring (e.g., 0 2, CO, CO2, CH4, I-hS , radon) D. Electrical safety E. Confined space entJ.y and ventilation rates F. Hazard co1mn unications (e.g., SDS, proper labeling, concenfrations, fire ratings, safety equipm ent)

7.

Engineering Economics A. Time value of money (e.g., present w orth, ann ual worth, future wmth,

B. C. D. E.

8.

6- 9

rate of return) Cost analysis ( e.g., incremental, average, sunk, estimating) Economic analyses (e.g ., break-even, benefit-cost, optimal economic life) Uncertainty (e.g ., expected value and risk) Project selection (e.g., compruison of projects with unequal lives, lease/buy/make, depreciation, discounted cash flow, decision trees)

9-14

Statics A. Vector analysis B . Force systems (e.g., resultants, concurrent, distributed) C. Force couple systems D. Equilibrium of rigid bodies (e.g., suppo1t reactions) E. Internal forces in rigid bodies (e.g., trusses, frames, machines) F. Area properties (e.g. , centroids, moments of ine.rtia, radius of gyration, parallel axis theorem) G. Static friction H. Free-body diagrams I. Weight and mass computations (e.g ., slug, lbm, lbr, kg, N, ton, dyne, g, gc)

9.

9-14

Dynamics A. Particle and rigid-body kinematics B. Linear motion (e.g ., force, mass, acceleration) C. Angular motion (e.g ., torque, inertia, acceleration) D. Mass moment of inertia E . Impulse and momentum ( e.g., linear, angular) F . Work, energy, and power G. Dynamic friction H. Vibrations (e.g., natural frequency)

10.

Strength of Materials A. Stress types (e.g. , normal, shear)

9-14

B. Combined loading-principle of superposition C. Stress and strain caused by axial loads, bending loads, torsion, or transverse shear forces D. Shear and moment diagrams E. Analysis of beams, trusses, frames, and columns F . Loads and defonnations (e.g., axial-extension, torque-angle of twist, moment-rotation)

5

. d

stress-based inclu mg al . d rincipal stresses,. aximutn nonn G S·u·ess transfonnatton ru~ ~ (e g Mohr's clfcle, m · t •e cntetta · ·• Yiel ding and frac mM' ) . , , brittle fractuJe, . . fatigue, n 1ses stress,_Tres~a,.~o . Eul er bucklmg, cteep, d all owable stress) fl ctor of safety, an H Material fa1lu1 e (e,g' . t1·ation factors, a d' g1·ams stress concen all o phase ia ' ' Y . . f terials ( e.g., 11 Matenals d1 s) properltes o ma · . (pl agram and phase change) A. Physical . .; s.e phase equ1h mun,_ . of materials t 'als B Mechanical prope1t1es . hemical properties of ma e~t ~ - ~hennal properties of maten~ls Electrical prop~rties of matertal s F. Mateiial select10n

6-9

E:

12

12-18

Fluid Mechanics . Newtonian liquids and gases) ., b . . ( Newtoman, nonA. Fluid properties e.g.., nolds number, Froude nurn er, B. Dimensionless numbers (e.g., Rey Mach number) C. Laminar and tmbulent flow . D. Fluid statics (e.g., hydrostatic head) . (e g Bernoulli equation) . d entwn equations • ·, . . E. Energy, nnpulse, an mdfi oi:ntI. . losses ( e.g., pipes, valves, fittmgs, lammar, F. Pipe and duct flow an ,.c on transitional and turbulent flow) . ) G Open-channel flow (e.g., M'anning's equation, drag . ) , • . erations H. Fluid transport systems (e.g., senes and pa~allel op . I Flow measurement (e.g., pitot tube, ventun meter, weu) J. Turbomachinery (e.g,, pumps, turbines, ~ans, compressors) K. Ideal gas law (e.g., mixtures of nonreactive gases) L. Real gas law (e.g. , z factor)

13. Basic Electrical Engineering . A. Electrical fundamentals (e.g., charge, cunent, voltage, resistance, power, energy) B. Current and voltage laws (e.g., Kirchhoff: Ohm) C. AC and DC circuits (e.g., real and imaginary components, complex numbers, power factor, reactance and impedance, series, parallel, capacitance and inductance, RLC circuits) D. Measuring devices (e.g., voltmeter, ammeter, wattmeter) E. Three-phase power (e.g., motor efficiency, balanced loads, power equation) 14. Thermodynamics and Heat Transfer A. Thermodynamic laws (e.g., first law, second law) B. TileIU1odynamic equilibrium

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FE·, O.liHERiDISCIPLINES-PiRACTICE EXAM 46.

The needle of a pressure gauge with a mass of 2 g consists of a uniform bar that is 150 mm long and pivoted at its center. To remove the vibration of the needle in use, the moment of inertia of the needle about its pivot point is tripled by adding equal weights at each end of the needle. To accomplish this and keep the needle balanced about its pivot, the mass (g) of each weight must be most nearly:

o A. o B. o C. o D.

47.

0.333 0.667

1.0 2.67

In the figure below, Block A weighs 50 N, Block B weighs 80 N, and Block C weighs 100 N. The coefficient of friction at all surfaces is 0.30. The maximum force F (N) that can be applied to Block B without disturbing equilibriwn is most nearly: CABLE

F

0 A. o B. o C. o D.

15 54 69 84

Copyright © 2020 by NCEES

27 NEXT•

FE'OTHER DISCIPl::INES PRACTICE'EXAM A box in the figure below has a mass of 6 kg and is supported as shown, Match a free-body diagram to the box, the cord CE, the cord AB, and the knot at/ pill(/(11/////////iju C,

48.

Free-body diagrams FCB (FORCE OF CORD CE)

J

[

1.

58.9 N (WEIGHT OR GRAVITY)

I FCBA (FORCE OF CORD CBA)

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60Q?\.

2.

Box Knot

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J

CordAB I

j

CordCE I

Fen (FORCE OF SPRING)

FcE (FORCE OF CORD CE)

FEc (FORCE OF KNOT)

t\

3,

*

F CE (FORCE OF BOX)

FORCEOFPULLEYATB

L;

B

4.

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FE,©THERDISCll?LINES PRACTICE EXAM A flat plate moving at a speed of 0.15 m/s is separated from a fixed plane surface by an oil film 3 0.25 mm thick. If the viscosity of the oil is 40 x 10- N·s/m2, the shear stress in the oil film is

74.

most nearly : A.

0

O B.

o C. o D.

0.024 Pa 2.4 Pa 24 Pa 2.4 kPa

A Newtonian fluid flows through 2-in. commercial steel pipe (I.D. = 0.0525 m).

75.

Liquid Properties: Density = 1.00 g/mL Viscosity = 1.05 cP 3 Flow = 0.500 m /min The Reynolds number is most nearly: 0

A.

o B. o C. o D.

1.93 7.94 1.93 1.15

X X X

X

3

10 3 10 5 10 7 10

Which of the following statements best describes conditions in turbulent flow?

76.

o o o o

A.

Fluid particles move along a smooth, straight path.

B. C.

Energy loss varies directly with the square of the velocity. Energy loss varies directly with the velocity.

D.

Reynolds number is always less than 2,000.

Copyright © 2020 by NCEES

42

NEX'f--t

FE o:rHER.01sc1 PILI NES·PRAC.:lil.CE EXAM 1

77.

1

The rectangular homogeneous gate sho,w below is 3.00 m high x 1.00 m wide and has a frictionless hinge at the bottom. If the fluid on the left side of the gate has a density of 3 1,600 kg/m , the magnitude of the force F (kN) required to keep the gate closed is most nearly:

__ ____ _____________ F~~- .. ------~

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------------------------------------------------------------·---------------------------------~- ---------------------------

__ ________________________________ ~=~~-:-":-:~~-:: FLU1D ~-:-:-:~~-~~~-::__ ____________________________~:::

IT · 3.00m

--------------------------------------FRICTIONLESS HINGE 0

o o o

A. B. C. D.

0

22 24 220

~

Copyright © 2020 by NCEES

43

NEXT•

FE O:JiHER DISCIPLINES PRACTICE EXAM A jet of water is shot vertically upward and hits a flat circular plate of mass M, as shown in th figure below. The diameter of the exhaust is 6 cm. If friction losses are neglected, the flow

rat

78.

(ro3/s) of the water exiting the jet is most nearly:

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J.

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--=-

r

M

CD

WATER

RESERVOIR

10m I 0

e

I I MASS DENSITY= p

+~D~6cm 3m

1 0

o o

0,015 0.022 0.033 0.040

A. B.

C. O D.

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MOMENT DIAGRAM

Maximum magnitude of the bending moment is 26 kN ·m.

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F.E- ,Qffilf-lER!IDISCIPtlNESiSOLUTI.ONS 63.

/l

lO kN/m

I I Q I

25kN~

Desired q

25 kN

Q1 =(0.150m)(0.025m) x [0.175m-0.11875m - 0.0125m] = 1.6406 X 10-4

6

3 m

4

I= 21.58 x 10 mm x (

= 0.00002158 VQ

q=-=

J

4 m

25, 000

I

4

lm 1,000mm

4 N[1.6406 x l0-

3 m ]

4 0.00002158 m

= 190,060 NI m =190kN / m

THE CORRECT ANSWER IS: B

Copyright@ 2020 by NCEES

y'

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Vo,ax~25kN

T

81

f ;E C)ifi,flER 101s·e1Pt.lNE-S", SOLUTIONS 1

64.

Rotating the plane 45° physically creates twice the angle on the Mohr's circle, resulting in a 90° Mohr's circle rotation. Thus, the correct answer is the center. ls

''

Ga\

500 psi

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THE CORRECT ANSWER IS SHADED ABOVE. 65.

Refer to Mohr's Circle in the Mechanics of Material chapter of the FE Reference Handbook. From a constructed Mohr's Circle, the maximum in-plane shear stress is

R ==

(j X - (j' y (

2

j2 -'- 2 I

'Cxy

R = •{ 40;20J +l02

R ==

Jwo

R == 14.1 ksi

THE CORRECT ANSWER IS: B

Copyright © 2020 by NCEES

82

'tmax =

R.

FE' OTHER DIS.CIPILINESISOLl!JTIONS

66.

Refer to the Columns section in the Mechanics of Materials chapter of the FE Rejerence Handbook to determine the critical buckling stress. Pinned on both ends would be pinnedpinned, so K = 0.7. \11·t f~ Examinees are expected to know that Eis the elastic modulus in this question. From the Typical Material Properties table, determine that steel has E = 200 GPa, and GPa = 109 N/m2. 2

cr r =

c

1t

2

n (2oox10

E

(K'r )

9

~)

m

2

2

12 1.974x10 _li._

m2

=-------~ = 226.6 MPa

(0.7x 20 mJ 0.15m

8 711 ,

THE CORRECT ANSWER IS: B 67.

The beam loading resembles one of the loadings enumerated in the simply supported beam slopes and deflection table in the Mechanics of Materials chapter of the FE Reference Handbook. This avoids solving differential equations to derive the appropriate fo1mula. 3

The table indicates that the maximum deflection under this loading is vmax = -PL 48EI. The maximum deflection occurs where the load is applied in the middle of the beam, as stated in 3

the problem. This equation can be reananged to find the elastic modulus E = 48- 1PL

vmax

.

Take the deflection as Vmax= -0.004 m (0.4 cm), using negative for downward deflection. Convert everything from cm to meter. The second moment of the area of the cross section is found in the Statics chapter of the FE Reference Handbook. This gives I xc = b;; . In this type of problem~ the second moment of

the area about a horizontal (x) axis passing through the centroid of the cross section is used, hence the Ix value is used. Also in this problem, the width b and height h are the same; both are C

5 cm, or 0.05 m.

_ m) = x 107m4 . This is the value of I that will be used in the (0.05 m)x(0.05 5 208 1 Xe 12 12 other equation. 3 = bh =

~

Solvin E- -PL3 -(3,000 N)x(0.7 m)3 = 10.29 x 109 = 10.29 GPa 4 7 g, - 481vmax - 48x (5.208 x 10- m )x(-0.004m) . m . . Note that the deflection Vrnox is taken as negative, which takes care of the mmus s1.gn m the numerator. Deflection down is taken as negative.

TRE CORRECT ANSWER IS: B CoPYrlght@ 2020 by NCEES

83

~ ~ ~~;l;t)-"~ t -•~ ~ .~ ~ /

FE:O.THER DISCIPC.INES:SOLUTIONS·· The solidification path of an Ag-70 w/o Cu alloy is shown as the vertical line. The temperature at which solidification begins is shown as the horizontal dashed line and is 920°C.

68.

1,100

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I

I

I~ + L

I

I

962°

l? 9 0 0 ~ a+L

1 800 ~

100

~

600

~

CL

L/

/

I

8.8

28.J

I

I

I

I

I ~

92

'i-780°

~

~

500

I

a. + ~

I I

400

0 Ag

20 80

40

60

80

60

40

20

Cu 0

COMPOSITION, % BY WEIGHT

THE CORRECT ANSWER IS: B

Refer to the Thennal and Mechanical Processing section in the Materials Science chapter of the

69.

FE Ref erence Handbook.

Cold-working metal decreases the recrystallization temperature, ductility, and slipping or twining takes place. During cold work, grains become elongated instead of equiaxed. THE CORRECT ANSWER IS: C

-

Refer to the Electrochemistry section in the Chemistry and Biology chapter of the FE Reference Handbook.

70.

_.:-'..._'ii·:~~ \,·

.,,,.•t,,-,.,·f.

- . ~ l. . .

'•t'

Aluminum is anodic relative to copper and, therefore, will corrode to protect the copper.

•-.·1~t .. ,. . •-fl. __ -...~ -~.

t.·•~ "-:"·~-' 'it/4-. •'• -,~ ,, .. ~-.

_ ..,-

··1

THE CORRECT ANSWER IS: B

-=-1 ·,

\

. ~--· • • •_-:,

,;

-!. '. .·

~~~...

.

·j:•· '',;lk••·; -~

;.. -~I'{:

,,,J • ',

"f~-- ·,::.

t

f •.

~

~l'

~-:~·::.:[ /:t·.. -f-·r·s ..,

'

.•

~:_ !; .'",··~

J

~

1~ ~ ~

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.

·-;-.~-l•" !-·•, .... ,

,..,, . .. :#~ •.., " ,, . - •

.,.. ••

~

~ v• · ·, :•

·.·, •. '·•1,;J~ ,' .. , • •.• • II .



..>,

~

..

84

'

f, --~,.,"' ,, ·..: •e ~.,. . • !; ..... ,~, :'i1 •!-. J.'' .•~:t , . ·,' l"ll;-• ,-~t ._, --~' ·~~~---: •-' · · ~f. ' . :: • ·•. '., . • ·•.-,,~ ·, ··"..,:._, , ~ ,., .. .......... ~ •·t .·:.,'-'· ,' i• ~... . ~-u. . . ~ ~ tC ~, , "·"' · : · · ·· ,; · •· t ; ·, ~ ',..~. il-.:"•tt-• .j,~f,-~ ,.__ ·Ill.; _,ii ·~"'•1· ~!iW,-y· ~ '"a... '~,, - '-·-··, . ·, • •·,, • " ••...., . ' .• ~"-1> ... """' ~ ~ _,.'1,•:~r- ..,,· , -.....~ ""'' • .0 i· , '

~•,.

• ,· ••

,C

..,

r_

'.

.

. •

('. F

·- . • '

t .





•.

''

~•-

•'.,.



.

•••





,.,

•,•

"

·,,c

't

'•·• • . , , '

.... _.

•·

·' •

'
".,.- •

.......

- .....

.· ' • • ! • / . '·' •• . .,. ,., ' '•,.. """ •... ., .... ' ~,- .... . •,.;· ,. ·• ' • . ;... , • •

' :· ;;" : : ·• :.......... :

/· '''.:,

,., .,,)......., , :,p ........

-'ft• -·. •

,-

•.

FiE:-0 1iJiHER 1D1s.c11?1LINEs:, S0LU1ilONS·· 11

Refer to the Thennal Properties section i11 the Materfals Science chapter of the FE Reference Handbook.

71.

P

cr = A

PL 8 = AE

Constrained,

~

or = aL (T - To)

= al (T - T0 )



cr = a ( 11T) E

6

= (12.ox10- 1°c)(50°C)(200 GP~)

= 0.12 GPa = 120 MPa THE CORRECT ANSWER IS: B

Refer to the Electrostatics section in the Electrical and Computer Engineering chapter of the FE Reference Handbook.

72.

Field strength=

~

=

g~

W=-Q

= 500 V/m acting vertically between the plates. P2 -

f

P1

-

E •dl

Since P1 • P2 is in same direction as the field,

W= -QEM = -2(500)(0.05)=-501 Magnitude W

= I-SOI J = 50 J

THE CORRECT ANSWER IS: C 73

,

Refer to the Thermal Properties section in the Materials Science chapter of the FE Reference Handbook Stress = cr = Eat1T

THE CORRECT ANSWER IS: A

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85

1 FE OJrl+IER!DISCIPLINES1SOLU1i IONS The equation can be found in the Fluid Mechanics chapter of the FE Reference Handbook.

74.

1

The shear stress 't is given by

~;,7,«,&,W'ffffff,WV,#ff,w,f----tl,-

0 = 0.25 mm OIL FILM r / / / / / / / / JI I I I I

dv 't = µ dy

0.15 m/s

The velocity gradient is uniform, so that dv fl;op -=-

o

dy

µv;

y~ v (

3

40x 10- N·s/nl)(0.15 mis)

=__r =--------"'-----

't

0

0.00025 m

o

24 Pa

THE CORRECT ANSWER IS: C Refer to the Fluid Flow Characterization section in the Fluid Mechanics chapter of the FE Reference Handbook.

75.

Viscosity =µ= dynamic viscosity

J( N~: tg·:/s J(l,o:; gJ 2

2

(1.05 cP{ 0.001 P:/

so 1.05 cP = 1.05 g/(m · s) 6 g g 1,000 mL 1,000 L p= 1.00 - x - - - x---=lxlO 11 m3 m3 mL

V=QIA 2

0.0525') Area= 1t ( - - ) 2

f:

1 ~

3

v = (o.soo mm

= 0.00216 m

](min] (

2

J = J. 86 mis 0.00216 m

60 s

1

2

;:,-

\.. · ~-.~:

Re= VDplµ

-~.....

··•-;~~-·.:::, . ._'il/!,.. ~-i~~-l-'••. ·;r,1,

,-1,_.

;;,_ •

;~l~f '\ . ~ ~

0

/i-:... -_

.•

.p

·'

_,1,;_;._ .•:.

.

. • !.

...... •

-_-_ '-_•·

f · ·1'- - .~.: .'•. •\•• -· -~_• • •·· " ' ,-;.- 't' ,,.. :!'.t •.: -.~ I..:, _... . ,• *"•. .. . . -·; , /.. . . .. .. . ,,. ··

,.

_•'._.

_ ~ __ ,"; ,. . ,~·~m~-M

,,

;,

> •, ,... .~ • ••• a

~

~. h ' · W ,t \ L -.. • Iv -'i: • • • • • ., ....... • .:.Jtt • .ill ' • -~. ,,,. -~ .• 't .... ... ""' ' . :" .. S< , ·. :,· "\;,.. " . • , J ~ _.._'(t'.4°1;' 1.y..._ .. • ... ,...., . • ;., ., .'o,-~ .............. ,,,. •••-,,.... ... • ...

~~ ~

.. •

• C •

>U,

~

~

~c.'-''--.,.~ ...

"'!i.."-~

~

~··1~~·--1"" "· ::., : r>; ~; . , ·/~.:•~:'.{ •:~t~~-~j; ~ff»n.•· · · · .. '

.

.• ··: ' ·.:.,.' ,,.

'n. .

.

86

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• ·< nJ

(D

(J'Q

)

--

0

(./)

(D (D

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