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T580(E)(M30)TAPRIL EXAMINATIONNATIONAL CERTIFICATEENGINEERING SCIENCE N3(15070413)30 March (X-Paper)09:00–12:00Candidates need drawing instruments.This question paper consists of 10 pages, 1 information sheet and 1 formula sheet.

(15070413)-2-T580(E)(M30)TDEPARTMENT OF HIGHER EDUCATION AND TRAININGREPUBLIC OF SOUTH AFRICANATIONAL CERTIFICATEENGINEERING SCIENCE N3TIME: 3 HOURSMARKS: 100INSTRUCTIONS AND INFORMATION1.Answer ALL the questions.2.All calculations should consist of at least THREE steps:2.12.22.3The formula used or manipulation thereofSubstitution of the given data in the formulaThe answer with the correct SI unit3.The constant values, as they appear on the attachment information sheet,must be used wherever possible.4.Number the answers according to the numbering system used in this questionpaper.5.Keep subsections of questions together.6.Rule off on completion of each question.7.Drawing instruments must be useddrawings/diagrams must be fully labelled.8.Use g 9,8 m/s2.9.Answers must be rounded off to THREE decimal places.10.Write neatly and legibly.Copyright reservedforalldrawings/diagrams.AllPlease turn over

(15070413)-3-T580(E)(M30)TQUESTION 1: MOVEMENT1.1A car travelling at 20 m.s-1 applies its brakes and stops after 50 m.Determine the following:1.1.1The average acceleration1.1.2The time taken to stop(2 2)1.2(4)A Bricklayer throws a brick upwards to a next level to a bricklayer who missesthe catch. The brick falls to the ground after 6 seconds.Calculate the following:1.31.2.1The height that the brick reaches(3)1.2.2The total distance covered(1)1.2.3The velocity at which the brick hits the ground(1)A motor vehicle with a mass of 2,5 tons accelerates uniformly from rest up anincline of 1 in 25 and reaches a speed of 55 km/h after 3 minutes.Calculate the following:1.3.1The acceleration of the motor vehicle(2)1.3.2The kinetic energy of the motor vehicle after 3 minutes(2)1.3.3The gain in potential energy of the motor vehicle after 3 minutesCopyright reserved(3)[16]Please turn over

(15070413)-4-T580(E)(M30)TQUESTION 2: MOMENTS2.1State the definition of a force.2.2A uniform beam is 18 m long and rests horizontally on two supports. The onesupport is at the left end and the other support is 4 m from the right end. Thebeam carries concentrated loads of 200 N and 100 N 5m and10 m from the left end respectively. The beam also carries a uniformlydistributed load of 15 N/m over the first 6 m from the right end.(2)2.2.1Calculate the reactions at the supports and test your answers.(5)2.2.2Draw the shear force diagram to scale.(4)15 N/m5m5m2m6m14 m4mRLFIGURE 12.3Calculate the turning moment about the fulcrum in FIGURE 2 below.200 N120 N1,2 mxx5m30FIGURE 2Copyright reservedOF ? kN(2)[13]Please turn over

(15070413)-5-T580(E)(M30)TQUESTION 3: FORCES3.1State the definition of equilibrant of forces.3.2Solve analytically the magnitude and direction of the resultant in FIGURE 3below.(2)300 N150 N30º45º200 N250 NFIGURE 33.3(4)Three forces shown in FIGURE 4 below are in equilibrium.Calculate the magnitude of the unknown forces analytically.P560 N35oo40QFIGURE 4Copyright reserved(5)Please turn over

(15070413)3.4-6-T580(E)(M30)TCalculate the nature and magnitude of forces with reference to thecomponents AB and BC in the roof truss in FIGURE 5 below.500 NAoB60o30C350 N150 NFIGURE 5(4)[15]QUESTION 4: FRICTION4.1Distinguish between static friction and kinetic friction.(2)4.2Name TWO applications of friction.(2)4.3The weight of an object in FIGURE 6 below is 2 500 N. The object is placedon an inclined plane that forms an angle of 20 to the horizontal. Thecoefficient of friction is 0,4 for the surface.Po30o20FIGURE 6

(15070413)-7-T580(E)(M30)TCalculate the following:4.3.14.3.2The pushing force P required to push the object upwards againstthe slopeThe angle in degrees that the plane must be raised so that thebody will slide down by itself(6)(1)[11]QUESTION 5: HEAT5.1Explain the law of conservation of energy.(2)5.245 steel shafts are cooled down in 5 litres of oil. The initial temperature of theshafts is 180 C and that of the oil 20 C. Each shaft has a mass of750 grams.Calculate the final temperature of the mixture. Use the shc of steel as500 J/kg C and that of oil 1,5 kJ/kg C.5.3(4)An overhead crane is used to carry a mould in a foundry workshop. Thebeam of the overhead crane has a length of 19 m at a temperature of 18 C.After the moulding process the temperature of the beam of the overheadcrane is expected to rise to 100 C in the foundry.Use the coefficient of linear expansion of the beam, which is 25 x 10-6/ C.Calculate the following:5.3.1The increase in temperature of the beam(1)5.3.2The expected expansion of the beam in mm(2)5.3.3The expected final length of the beam in metres(2)

(15070413)-8-T580(E)(M30)T5.4WaterWater &SteamSteamTemperature; in ºC43215.4.15.4.25.4.35.4.4FIGURE 7FIGURE 7 above shows a water-to-steam process.Complete the process in FIGURE 7 by filling in the missing descriptions andsymbols. Write only the description and the symbol next to the questionnumber (5.4.1–5.4.4) in the ANSWER BOOK.(4)[15]QUESTION 6: HYDRAULICS6.16.2Explain the following terminology in .1.3Fluidity(2)A single-action water pump with a piston diameter of 120 mm has to deliver1,4 10-3 m3 water per stroke. The average force exerted on the piston is2,4 kN per working stroke.Calculate the following:6.2.1The pressure of the fluid during the working stroke(3)6.2.2The length of the stroke(2)6.2.3The work done per stroke lengthCopyright reserved(2)[12]Please turn over

(15070413)-9-T580(E)(M30)TQUESTION 7: ELECTRICITY7.17.2Explain the concept power factor, and distinguish between a high powerfactor and a low power factor.(2)An electrical circuit as shown in FIGURE 8 below consists of 4 cellsconnecting in series, each having an EMF of 4 volts and a internal resistanceof 0,15 ohms per cell, connected in series.R1 8 ΩRbR2 5 ΩR4 4,5 ΩR3 5 ΩARa 0,5 ΩAR5 2 ΩBVVFIGURE 8Calculate the following:7.2.1The total resistance of the circuit7.2.2The voltage across AB(2 3)7.3(6)A single-phase transformer has a supply voltage of 220 V. The turn ratio ofthe primary to the secondary is 12 : 1. The secondary current at full load is 20A and the secondary coil has 50 windings.Calculate the following:7.3.1The secondary voltage7.3.2The primary current(2 2)Copyright reserved(4)[12]Please turn over

(15070413)-10-T580(E)(M30)TQUESTION 8: CHEMISTRY8.18.2Name TWO precautions to keep moisture and water away from metals whichcan corrode.(2)Name the chemical composition of the following compounds:8.2.1Limestone (marble)8.2.2Caustic soda(2 1)(2)8.3Name a non-metal that is a good conductor of electricity.(1)8.4Name ONE property of solder.(1)[6]TOTAL:Copyright reserved100

(15070413)T580(E)(M30)TFORMULA SHEETENGINEERING SCIENCE N3All the formulae needed are not necessarily included.Any applicable formula may also be used.W F.sWP tUitset / Output Inset / InputF m.aF NR tan N R FC FT sin .a 0m1 .u1 m2 .u 2 m1 .v1 m2 .v2De ( D t )hnat / wet h f xh fgP 2. .T .n.T F .rFRAM FPL D 2 . A ARAM APL4VRAM VPL nARAM .H RAM APL .LPLP FS w sin FX F cos FC w cos FY F sin FT cos F FS . a 0 FX F1 cos 1 . Fn cos nFe T1 T2T1 tension ratioT2 FX F1 sin 1 . Fn sin nP Fe .vtan v .d .n. n N60R FX FY22 FY FXQ m.c. t.t f to tW F .sm.w.w Q m.hv E P m.g . hP 1mv 22Q I 2 .R.t L L0 . . t. L f L0 L A A0 . . t. A f A0 Am I .z.tVP N P I S VS N S I P2.a.s v 2 u 21s u.t .a.t 22v u a.tM F. s F F CWM ACWM E K PABS PATM PMET111 . RPAR R2RnV1 V2 e(U 1 U 2 )Copyright reservedQtP g hRSER R1 .RnV I RPlease turn over

(15070413)T580(E)(M30)TINFORMATION SHEETPHYSICAL CONSTANTSQUANTITYAtmospheric pressureDensity of copperDensity of aluminiumDensity of goldDensity of alcohol (ethyl)Density of mercuryDensity of platinumDensity of waterDensity of mineral oilDensity of airElectrochemical equivalent ofsilverElectrochemical equivalent ofcopperGravitational accelerationHeat value of coalHeat value of anthraciteHeat value of petrolHeat value of hydrogenLinear coefficient of expansion ofcopperLinear coefficient of expansion ofaluminiumLinear coefficient of expansion ofsteelLinear coefficient of expansion ofleadSpecific heat capacity of steamSpecific heat capacity of waterSpecific heat capacity ofaluminiumSpecific heat capacity of oilSpecific heat capacity of steelSpecific heat capacity of copperCopyright reservedCONSTANTSKONSTANTE101,3 kPa8 900 kg/m32 770 kg/m319 000 kg/m3790 kg/m313 600 kg/m321 500 kg/m31 000 kg/m3920 kg/m31,05 kg/m31,118 mg/CAtmosferiese drukDigtheid van koperDigtheid van aluminiumDigtheid van goudDigtheid van alkohol (etiel)Digtheid van kwikDigtheid van platinumDigtheid van waterDigtheid van minerale olieDigtheid van lugElektrochemiese ekwivalent van silwer0,329 mg/CElektrochemiese ekwivalent van koper9,8 m/s230 MJ/kg35 MJ/kg45 MJ/kg140 MJ/kg17x 10-5/0CSwaartekragversnellingWarmtewaarde van steenkoolWarmtewaarde van antrasietWarmtewaarde van petrolWarmtewaarde van waterstofLineêre uitsettingskoëffisiënt van koper23 x 10-5/0C12 x 10-5/0CLineêre uitsettingskoëffisiënt vanaluminiumLineêre uitsettingskoëffisiënt van staal54 x 10-5/0CLineêre uitsettingskoëffisiënt van lood2 100 J/kg.0C4 187 J/kg.0C900 J/kg.0CSpesifieke warmtekapasiteit van stoomSpesifieke warmtekapasiteit van waterSpesifieke warmtekapasiteit vanaluminiumSpesifieke warmtekapasiteit van olieSpesifieke warmtekapasiteit van staalSpesifieke warmtekapasiteit van koper2 000 J/kg.0C500 J/kg.0C390 J/kg.0CHOEVEELHEID

MARKING GUIDELINENATIONAL CERTIFICATEAPRIL EXAMINATIONENGINEERING SCIENCE N330 MARCH This marking guideline consists of 11 pages.Copyright reserved

MARKING GUIDELINE-2ENGINEERING SCIENCE N3T580(E)(M30)TNOTE: ALTERNATE CORRECT ANSWERS MUST BE CONSIDERED.QUESTION 1: MOVEMENT1.11.1.1V 2 u 2 2asv2 u22s(0) 2 (20) 2a 2(50)a Correct substitution Negative answer only 4 m/s 21.1.21.2(2)v u atv ut a.(0) (20)t 4 5 s1.2.1g -9,8 m/s2 Substitution Correct answerv u gtu v gt (0) ( 9,8)(3) 29,4 m / ss u valuev2 u2gt(0) 2 (29,4) 2 9,8(2) 44,1 m (2) Substitution Correct answeru ?1.2.21.2.3Copyright reserved(3)s 44,1 2 88,2 m Correct answeru v 29,4 m/s Correct answer(1)(1)Please turn over

MARKING GUIDELINE1.3-3ENGINEERING SCIENCE N3T580(E)(M30)T55 1 000 12,278 m/s3 600t 3 min 60 180 su 0 m/sm 2,5 1 000 2 500 kgv 55 km/h2,5 tonsh1 to 251.3.11.3.2v ut15,277 0 180 0,085 m/sa 1mv 221 (2 500)(15,277) 22 291 733,411 J( ) substitution( ) Correct answer(2)EK ( ) substitution( ) Correct answer(2)1.3.3S ut ½ at2 (0) (180) ½ (0.085) (180)2 1377m( ) s valueEp mgh( ) Substitution 2500 (9.8) (1377/25)( ) Correct answer 1349460J(3)[16]

MARKING GUIDELINE-4ENGINEERING SCIENCE N3T580(E)(M30)TQUESTION 2: MOMENTS2.12.2A force is that influence which changes or tends to change the state of rest ofa body.( ) If the body is already moving, a force is that influence which bringsthe body to rest or changes the direction of the body.( )2.21(2)Taking moments about L:Σ Anti-clockwise Moments Σ Clockwise Moments(R x 14) (200 x 5) (90 x 15) (100 x 10)( )E 239,286 N( )(2)Taking moment about R:Σ Anti-clockwise moments Σ Clockwise Moments(100 x 4) (200 x 9) (L x 14) (90 x 1)( )L 150,714 N( )(2)Test:Upward forces Downward forces239,286 150,714 kN 390 kN( )(1)2.2.2150,714N( ) 60 N( )( )00( )- 49,286N- 149,286 N-179,286 N(4)2.3Cos ɵ x F x 5 m 1,2 x 120 200 x 3,1 mCos30 x F x 5 144 620 ( )F 764Cos30 x 5F 176,443 N ( )Copyright reserved(2)[13]Please turn over

MARKING GUIDELINE-5ENGINEERING SCIENCE N3T580(E)(M30)TQUESTION 3: FORCES3.1The equilibrant of a system of forces in the same plane( ) is that single forcewhich will balance the system of forces.( )(2)HC -150 216.506 141.421 207.927 N VC 300 125 – 141.42 283.579 N R (207.927)2 (283.579)23.2 123650.686 351.64 N Tan θ 283.579207.927 1.36Θ Tan -1 1.36 53.670 E 53.670 N 3.3P(4)Psin350PCos350560 NQCos40QCopyright reservedQSin400Please turn over

MARKING GUIDELINE-6ENGINEERING SCIENCE N3T580(E)(M30)TVC F FP Sin 35 Q Sin 40Q Sin 40P Sin 35P 1,1207 Q . . (1)( ) Substitution( ) Equation (1) HP Cos 35 Q Sin 4