Technical Sciences Grade 12 Paper 1 NSC November 2020 Memo

Downloaded from hlayiso.com SENIOR CERTIFICATE/SENIOR SERTIFIKAAT NATIONAL SENIOR CERTIFICATE/ NASIONALE SENIOR SERTIFIKAAT GRADE/GRAAD 12 TECHNICAL SCIENCES P1/ TEGNIESE WETENSKAPPE V1 NOVEMBER 2020 MARKING GUIDELINES/NASIENRIGLYNE MARKS/PUNTE: 150 These marking guidelines consist of 16 pages./ Hierdie nasienriglyne bestaan uit 16 bladsye. Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Technical Sciences/P1/Tegniese Wetenskappe/V1 3 DBE/November 2020 SC/NSC/SS/NSS – Marking Guidelines/Nasienriglyne QUESTION 1/VRAAG 1 1.1 D  (2) 1.2 D  (2) 1.3 A  (2) 1.4 B  (2) 1.5 B  (2) 1.6 A  (2) 1.7 B  (2) 1.8 B  (2) 1.9 A  (2) 1.10 C  (2) [20] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Technical Sciences/P1/Tegniese Wetenskappe/V1 4 DBE/November 2020 SC/NSC/SS/NSS – Marking Guidelines/Nasienriglyne QUESTION 2/VRAAG 2 2.1 An object continues in a state of rest or uniform (moving with constant) velocity  unless it is acted upon by a net (resultant) force . OR An object will remain at rest or continue moving at a constant velocity  unless a non-zero resultant /net force acts on it.  ’n Voorwerp sal volhard in sy toestand van rus (of uniforme snelheid) tensy ’n net (resulterende) krag daarop inwerk. OF 'n Liggaam sal in sy toestand van rus of uniforme snelheid (beweeg teen konstante snelheid) volhard tensy 'n ongebalanseerde krag/(netto of resulterende krag) daarop inwerk. (2) 2.2 OPTION/OPSIE 1 OPTION/OPSIE 2 Fnet = ma Fnet = ma Fnet = 0 N Fnet = 0 N Fnet = F Cos 40° + fk Fnet = F Cos 40° + fk F Cos 40° + fk = 0 N Any one  F Cos 40° + fk = 0 N Any one  Fnet = FH + fk Fnet = FH + fk FH + fk = 0 N FH + fk = 0 N (Choose right to be positive) (Choose left to be positive) (Kies regs as positief) (Kies links as positief) F Cos 40° + fk = 0 F Cos 40° + fk = 0 80 Cos 40° + fk = 0  - 80 Cos 40° + fk = 0  61,28 + fk = 0 - 61,28 + fk = 0 fk = - 61,28 fk = 61,28 N (to the left) fk = 61,28 N (to the left) (na links) (na links) (3) 2.3.1 Inertia is a property/tendency of an object/body to resist a change  in its state of rest or motion (in a straight line). Traagheid is die eienskap van 'n voorwerp om die verandering in sy toestand van rus of beweging teen te staan. (2) 2.3.2 Apply Negative marking/Pas negatiewe nasien toe Increase.  Inertia of an object is directly proportional to its mass.  When the mass of an object increases, its inertia also increases.  Verhoog Traagheid van ’n voorwerp is direk eweredig aan sy massa. Indien die massa van 'n voorwerp verhoog, sal die traagheid ook verhoog. (3) [10] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Technical Sciences/P1/Tegniese Wetenskappe/V1 5 DBE/November 2020 SC/NSC/SS/NSS – Marking Guidelines/Nasienriglyne QUESTION 3/VRAAG 3 3.1.1  Tension is a (pulling) force acting in a string or rope.   Force applied by Zane (FZane)/F160/160 N   Spanning is 'n (trek)krag wat in 'n ketting of tou werk.  Krag toegepas deur Zane (FZane)/F160/160 N (3) 3.1.2 Decrease /Verlaag (2) 3.1.3 OPTION 1/OPSIE 1 OPTION 2/OPSIE 2 FN/ N  FN/ N  FZane FY  for both fk  fk  FTom  FTom  Fx Fg/w Fg/w ACCEPTABLE LABELS: NOTE: AANVAARBARE BYSKRIFTE: LET OP:  Fg/w/weight/Gewig Penalise once if  fk/f/friction/Wrywing  Force diagram used  FTom/ F200/ 200 N/Force by Tom/Krag van Tom  Arrows are not shown  FZane/ Force by Zane/F160/160 N/Tension  If force does not touch /T/Krag van Zane the dot  FN/N/Normal force/Normaalkrag  An additional force  FY/FV/Vertical component of force by  Using broken lines Zane/Vertikale komponent van krag deur Zane Penaliseer eenmalig  FX/FH/ Horizontal component of force by indien: Zane/Horisontale komponent van krag deur  Pyle nie aangedui nie Zane  Kragtediagram  FA /Fzane/FTom geteken  Die krag raak nie die kol nie  Voeg ’n addisionele krag by  Gebruik stippellyne (5) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Technical Sciences/P1/Tegniese Wetenskappe/V1 6 DBE/November 2020 SC/NSC/SS/NSS – Marking Guidelines/Nasienriglyne 3.2 OPTION 1/OPSIE 1 OPTION 2/OPSIE 2 Choose east to be positive Kies oos as positief Fnet = Fx + FTom + fk  Fnet = FZane Cos 65° + FTom + fk Fnet = Fx + FTom + fk  Fnet = FZane Cos 65° + FTom + fk 205 = 200 + 160 Cos 65° + fk fk = - 62,62 N 205 = 200 + 160 Cos 65° + fk fk = - 62,62 N N = Fg - FZane Sin 65) = mg - FZane Sin 65°) fk = μk N = 350 x 9,8 – 160 Sin 65°  fk = μk (Fg - FZane Sin 65°) Any one = 3284,99 N fk = μk (mg - FZane Sin 65°)  fk = μk N  62,62 = μk (350 x 9,8 – 160 Sin 65°)  62,62 = μk (3284,99) 62,62 = μk (3284,99) μk = 0,019/ 0,02  μk = 0,019/ 0,02  OPTION 3/OPSIE 3 Choose east to be positive Kies oos as positief fk = μk N  Fnet – (FZane Cos 65° + FTom) = μk N Fnet – (FZane Cos 65° + FTom) = μk (mg - FZane Sin 65°) 205 – 200  – 160 Cos 65°  = μk (350 x 9,8 – 160 Sin 65°)  - 62,62 = μk (3284,99) (Ignoring direction/Ignoreer rigting) 62,62 = μk (3284,99) μk = 0,019 /0,02  (6) [16] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Technical Sciences/P1/Tegniese Wetenskappe/V1 7 DBE/November 2020 SC/NSC/SS/NSS – Marking Guidelines/Nasienriglyne QUESTION 4/VRAAG 4 4.1.1 Momentum (of an object) is the product of the object's mass  and its velocity (in a straight line).  Momentum (van 'n voorwerp) is die produk van die voorwerp se massa en sy snelheid (in 'n reguitlyn.) (2) 4.1.2 OPTION/OPSIE 1 OPTION/OPSIE 2 1000 vi truck/bakkie = 120 x vi truck/bakkie = 120 x 1 3600 3,6 -1 -1 = 33,33 m.s , east  = 33,33 m.s , east  (2) 4.1.3 p = mcar vi car  = 1 050 x 16,67  = 17 503,5 kg.m.s-1, west/wes  (3) 4.2.1 The total linear momentum of an isolated system remains constant (is conserved) in magnitude and direction. Die totale liniëre momentum in ’n geïsoleerde sisteem is konstant. OR/OF The total linear momentum of an isolated system before collision/explosion is equal to total linear momentum after collision/explosion. In 'n geïsoleerde sisteem, is die totale liniêre momentum voor die botsing gelyk aan die totale liniêre momentum na die botsing. (2) 4.2.2 POSITIVE MARKING FROM 4.1.2/POSITIEWE NASIEN VAN 4.1.2 1 1 ∑Eki = m truck/bakk ie v i2 truck/bakk ie + m car v i2car  2 2 1 1 = (1 350)(33,33)2 + (1 050)(-16,67)2  2 2 = 895 741,68 J 1 1 ∑Ekf = m truck/bakk ie v 2f truck/bakk ie + mcar v 2f car 2 2 1 1 = (1 350)(20,3)2 + (1 050)(5,32)2  2 2 = 293 019,51 J ∑Eki  ∑Ekf /(Kinetic energy is not conserved/Kinetiese energie nie behoue nie)  Therefore, collision was inelastic  / Die botsing was dus onelasties. NOTE: If a learner starts: ∑Eki = ∑Ekf take 1 mark/ Indien leerder met ∑Eki = ∑Ekf begin gee 1 punt (5) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Technical Sciences/P1/Tegniese Wetenskappe/V1 8 DBE/November 2020 SC/NSC/SS/NSS – Marking Guidelines/Nasienriglyne 4.3.1 Inversely proportional.  /Omgekeerd eweredig OR 1 Fnet  Δt (2) OR When the contact time increases/decrease, the net force decreases/increase. Wanneer die kontaktyd verhoog, sal die netto krag verlaag. NOTE/NB: GIVE full mark for mathematical expression/Gee volpunte vir wiskundige uitdrukking 4.3.2 Equal to  Gelyk aan (1) 4.3.3  Impulse remains constant.   Airbags increase the contact time during the crash.   The longer the contact time, the smaller the force  exerted by the driver on the car and the lesser is the extent of injuries.  Impuls bly konstant.  Lugsakke verleng die kontaktyd tydens die botsing.  Hoe langer die kontaktyd, hoe kleiner die krag wat deur die drywer op die motor uitgeoefen word en hoe minder die beserings. (3) 4.3.4 OPTION/OPSIE 1 OPTION/OPSIE 2 Let the direction towards the tree Let the direction towards the tree be positive be negative Neem die rigting na die boom as Neem die rigting na die boom as positief negatief Fnet∆t = ∆p Fnet∆t = ∆p  for any Fnet∆t = m(vf – vi) Fnet∆t = m(vf – vi) - 57 500∆t  = 1 150(0 – 15)  57 500∆t  = 1 150{0 – (-15)}  ∆t = 0,30 s  ∆t = 0,30 s  (4) [24] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Technical Sciences/P1/Tegniese Wetenskappe/V1 9 DBE/November 2020 SC/NSC/SS/NSS – Marking Guidelines/Nasienriglyne QUESTION 5/VRAAG 5 5.1.1 Wlearner/leerder = Fapp ∆y Cos θ  = (25)(0,9)(Cos 0°) = (25)(0,9)(1)  for any = 22,5 J  (3) 5.1.2 OPTION 1/OPSIE 1 OPTION 2/OPSIE 2 Choose up to be positive Positive marking from 5.1.1 Kies op as positief Merk positief vanaf 5.1.1 Fnet = Fa + Fg Wg = Fg ∆y Cos α Fnet = Fa + mg = mg ∆y Cos α = 25 + 2(- 9,8)  = 5,4 N = (2)(9.8)(0,9)(Cos 180°)  for any = (2)(9.8)(0,9)(-1) Wnet = Fnet ∆y Cos θ  = - 17,64 J = (5,4) (0,9)(Cos 0°) Wnet = W learner/leerder + Wg   for any = (5,4)(0,9)(1) = 22,5 + (- 17,64)  = 4,86 J  = 4,86 J OPTION 3/OPSIE 3 OPTION 4/ OPSIE 3 Wnet = Fnet ∆y Cos θ Wnet = W learner/leerder + Wg  for any = W learner/leerder + mg ∆y Cos α  for any = (Fa + mg) ∆y Cos θ = {25 + 2(- 9,8) } (0,9)(Cos 0°)  = 22,5 + (2)(9.8)(0,9)(Cos 180°)  = (5,4)(0,9)(1) = 22,5 + (2)(9.8)(0,9)(-1) = 4,86 J  = 4,86 J  (4) 5.2.1 The total mechanical energy of an isolated system is constant.  OR The sum of gravitational potential energy and kinetic energy in an isolated system remains constant. Die totale meganiese energie in 'n geïsoleerde sisteem bly konstant. OF Die som van die gravitasionele potensiele en kinetiese energie bly konstant in ’n geisoleerde sisteem. (2) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Technical Sciences/P1/Tegniese Wetenskappe/V1 10 DBE/November 2020 SC/NSC/SS/NSS – Marking Guidelines/Nasienriglyne 5.2.2 OPTION 1/OPSIE 1 OPTION 2/OPSIE 2 ME = Ep + Ek 1 Ek = mv2 1  for any one 2 = mgh + mv2 2 1 1 = x 6 x 02 = (6)(9,8)(5) + x 6 x 02 2 2 =0J = 294 J  Ep = mgh Accept/Aanvaar: = (6)(9,8)(5)  1 = 294 J 0 for x 6 x 02 2 ME = Ep + Ek  = 294 + 0 = 294 J  (4) 5.2.3 POSITIVE MARKING FROM 5.2.2 POSITIEWE MERK VANAF 5.2.2 OPTION 1/OPSIE 1 OPTION 2/OPSIE 2 In an isolated system/ In ’n geisoleerde ME(A) = ME(B) sisteem Ep(Top) = Ek(Bottom) ME(A) = (Ep + Ek)B 1  for any 2 NOTE/NB: If the above statement is ME(A) = (mgh + mv )B 2 omitted, learner will lose 1 mark/Indien 1 bogenoemde stelling nie ingesluit is nie 294  = 0 + (6)(v2)  2 sal leerder 1 punt verloor. 2 v = 98 1 v = 9,90 m.s-1  Ek = mv2  2 1 294  = (6)(v2)  2 v2 = 98 v = 9,90 m.s-1  (4) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Technical Sciences/P1/Tegniese Wetenskappe/V1 11 DBE/November 2020 SC/NSC/SS/NSS – Marking Guidelines/Nasienriglyne 5.2.4 POSITIVE MARKING FORM 5.2.2/POSITIEWE NASIEN VANAF 5.2.2 OPTION 1/OPSIE 1 ME = Ek + Ep (At C) Ek at C = ME – Ep at C  for any one = ME – mgh = 294 - (6x9,8x3)  = 294 - 176,4 = 117,6 J 1 mv2 = 117,6 at point C /by punt C 2 1 (6)v2 = 117,6 2 v2 = 39,2 v = 6,26 m.s-1  (4) OPTION 2/OPSIE 2 ME (At B) = ME (At C) 1 1  for any one mv2 + mgh (At B) = mv2 + mgh (At C) 2 2 1 1 (6)(9,9)2 + (6)(9,8)(0) = (6)(v)2 + (6)(9,8)(3)  2 2 294 = 3(v)2 + 176,4 v2 = 39,2 v = 6,26 m.s-1  OPTION 3/ OPSIE 3 ME (At A) = ME (At C) 1 1 mv2 + mgh (At A) = mv2 + mgh (At C)  for any 2 2 1 1 (6)(0)2 + (6)(9,8)(5) = (6)(v)2 + (6)(9,8)(3)  2 2 294 = 3(v)2 + 176,4 v2 = 39,2 v = 6,26 m.s-1  [21] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Technical Sciences/P1/Tegniese Wetenskappe/V1 12 DBE/November 2020 SC/NSC/SS/NSS – Marking Guidelines/Nasienriglyne QUESTION 6/VRAAG 6 6.1.1 Stress is the internal restoring force per unit area of body. Spanning is die interne herstelkrag per oppervlakte eenheid van 'n voorwerp. (2) 6.1.2 Strain is the ratio of change in dimension/length to the original dimension/length. Vervorming is die verhouding van verandering in dimensie tot die oorspronklike dimensie. (2) 6.2.1 σ Κ=  ε 250  106  190  10  9 ε -3 ε = 1,32 x 10 / 0,00132  (3) 6.2.2 OPTION 1/OPSIE 1 OPTION2/OPSIE 2 πd 2 Area = Area = πr 2 4 π(0,06)2 Area = π(0,03)2  Area =  = 2,827433 x 10-3 m2 4 F  = 2,827433 x 10-3m2 σ= F A σ=  F A 250 x 106   2,8274333 x 10-3 F 250 x 106   F = 706 858,35 N  2,8274333 x 10-3 F = 706 858,35 N  ACCEPT: 706 500 N OR 707 500 N AANVAAR:706 500 N OF 707 500 N ACCEPT: 706 500 N OR 707 500 N AANVAAR: 706 500 N OF 707 500 N (4) 6.3 As the temperature increases, (viscosity of a fluid) decreases.  Soos die temperatuur styg, verlaag die viskositeit (van 'n vloeistof). (2) 6.4 A body which does not show a tendency to regain its original shape and size when the deforming force is removed.  'n Voorwerp wat nie neig om sy oorspronklike vorm en grootte te herwin wanneer die vervormingskrag verwyder word nie. (2) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Technical Sciences/P1/Tegniese Wetenskappe/V1 13 DBE/November 2020 SC/NSC/SS/NSS – Marking Guidelines/Nasienriglyne 6.5  Clay  Any two and any other correct one  Wax  Enige twee en enige ander korrekte antwoord  Putty  Aluminium  Mild Steel  Klei  Was  Stopverf  Aluminium  Sagte staal (2) 6.6 In a continuous liquid at equilibrium, the pressure applied at any point is transmitted equally to other parts of the liquid.  In 'n kontinue vloeistof by ewewig, sal die druk wat by enige punt toegepas word eweredig na ander dele van die vloeistof versprei word. (2) 6.7 OPTION 1/OPSIE 1 OPTION 2/OPSIE 2 F1 F2 F2 =  P2  A1 A 2 A2 F1 20 000 20 000 =  P2 =  0,05 0,8 0,8 F1 = 1 250 N P2 = 25 000 Pa forboth / P2 = 25×103 Pa Vir beide NOTE: Give full marks if F2 is calculated. P2 = 25 k Pa Gee vol punte indien F2 bereken was. But P1 = P2, then F P1 = 1 A1 F 25 x 103 = 1  0,05 F1= 1 250 N  (4) 6.8 The normal force exerted by a liquid at rest on a given surface in contact with it. Die normaalkrag/stukrag wat deur 'n vloeistof in rus uitgeoefen word op 'n oppervlakte waarmee dit in kontak is. (2) [25] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Technical Sciences/P1/Tegniese Wetenskappe/V1 14 DBE/November 2020 SC/NSC/SS/NSS – Marking Guidelines/Nasienriglyne QUESTION 7/VRAAG 7 7.1.1 A is the p-type/ Positive type (semiconductor)  B is an n-type/ Negative type (semiconductor) A is die p-tipe (halfgeleier) B is die n-tipe (halfgeleier) (2) 7.1.2  ACCEPT: If the arrow is not shaded AANVAAR: Indien die pyl nie ingekleur is nie. (2) 7.1.3 Four/4 /Vier/4 (1) 7.2 A device that stores electrical charge.  'n Toestel wat elektriese lading stoor. (2) 7.3 The capacitance is directly proportional to the charge between the plates.  OR C Q Die kapasitansie is direk eweredig aan die lading tussen die plate. (2) 7.4  Decrease surface area of the plates.   Increase distance between the plates.  Use dielectric material with a low dielectric constant/ permittivity.  Verminder die plaatoppervlakte  Verhoog die afstand tussen die plate.  Gebruik diëlektriese stof met n lae diëlektriese konstante/lae permitiwiteit (2) 7.5 OPTION 1/OPSIE 1 OPTION 2/OPSIE 2 OPTION 3/OPSIE 3 V 2 P = VI  P =I2R  P =  P =120 × 2  R P = 22 × 60  P = 240 W 1202 P = 240 W P =  P = 0,24 kW 60 P = 0,24 kW P = 240 W Energy used: P = 0,24 kW E = Pt  Energy used E = Pt  E = 0,24 × 2  Energy used: E = 0,24 × 2  E = 0,48 kWh E = Pt  E = 0,48 kWh E = 0,24 × 2  Cost of energy used: E = 0,48 kWh Cost = E used x tariff Cost of energy used: Cost = 0,48x1,75  Cost = E used x tariff Cost of energy used: Cost = R0,84  Cost = 0,48x1,75  Cost = E used x tariff Cost = R0,84  Cost = 0,48x1,75  Cost = R0,84  (7) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Technical Sciences/P1/Tegniese Wetenskappe/V1 15 DBE/November 2020 SC/NSC/SS/NSS – Marking Guidelines/Nasienriglyne OPTION 4 OPTION 5 OPTION 6 V Δt 2 W = VI∆t  W = I2R∆t  W=  R = (120)(2) (2)  = (22)(60)  (2)  (1202 ) (2) = 480 W = 480 W =  = 0,48 kWh  60 = 0,48 kWh  = 480 W = 0,48 kWh  Cost = Eused x tariff  Cost = Eused x tariff  = (0,48)(1,75)  = (0,48)(1,75)  Cost = Eused x tariff  = R0,84  = R0,84  = (0,48)(1,75)  = R0,84  [18] QUESTION 8/VRAAG 8 8.1 This is the process of generating electricity from motion.  OR The production of an emf or voltage across an electrical conductor due to relative motion between the conductor and magnetic field. Dit is die proses om elektrisiteit op te wek deur beweging. OF Die opwekking van ’n emk of spanning oor n geleier deur relatiewe beweging tussen die geleier en magneetveld. (2) 8.2  The strength of the magnetic field.   The number of turns on the coil.   The speed at which the magnet and coil are moved relative to each other. (ANY TWO)  Die sterkte van die magneetveld.  Die aantal windings op die spoel.  Die spoed waarteen die magneet en die spoel relatief tot mekaar beweeg word. (ENIGE TWEE) (2) 8.3 (Lenz's law states that) the direction of the induced emf (in the coil) opposes the effect that produces it.  (Lenz se wet sê) dat die rigting van die geïnduseerde emk (in die spoel) die effek teenwerk wat dit opgewek het. (2) 8.4  Electromagnetic braking in trains/rotating machinery.   Electric motors  Electric generators.   Induction cooking pots where the pot is heated by magnetic induction. (ACCEPT ANY OTHER CORRECT APPLICATIONS)  Elektromagnetiese remme in treine/roterende masjiene.  Elektriese motors  Generators.  Induksiepotte waar die potte deur magnetiese induksie verhit word. (AANVAAR ENIGE ANDER KORREKTE TOEPASSINGS) (3) [9] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Technical Sciences/P1/Tegniese Wetenskappe/V1 16 DBE/November 2020 SC/NSC/SS/NSS – Marking Guidelines/Nasienriglyne QUESTION 9/VRAAG 9 9.1.1 1 – (carbon) brushes 2 – commutator/ split ring  3 – magnet  1 – borsels 2 – kommutator/ spleetring 3 – magneet (3) 9.1.2 DC motor  GS motor (1) 9.2 OPTION 1/OPSIE 1 OPTION 2/OPSIE 2 Vs Ns VP NP     Vp N p VS NS 20 110 VP 1 200     VP 1 200 20 110 1 200  20 1 200  20 VP  VP  110 110 VP = 218,18 V  VP = 218,18 V  (3) [7] TOTAL/TOTAAL: 150 Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief