MODUL Fizik TINGKATAN 5 MODUL FIZIK TINGKATAN 5 JAWAPAN / ANSWER JAWAPAN / ANSWERS. Nilam Publication Sdn. Bhd.

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MODUL • FIZIK TINGKATAN 5

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DAYA DAN GERAKAN II

FORCE AND MOTION II Unit

1

1.1

Contoh / Example

2 12 N – 5 N = 7 N; Arah ke kanan / To the right 3 Daya paduan, F / Resultant force, F

= 500 N + 200 N

= 700 N ke kanan / to the right 4 Daya paduan, F / Resultant force, F = 500 N – 200 N

= 300 N ke kanan / to the right

Latihan / Exercises 1 1 cm : 1 N (a) 8 N 6 N 37° Daya pa duan = 10.0 N Resulta nt forc e

10 N pada sudut 37° dengan daya 8 N

10 N at angle of 37° with the 8 N force

(b) 8 N 10 N Daya pad uan = 1 5.6 N Resultan t force

15.6 N pada sudut 27° dengan daya 10 N

15.6 N at an angle of 27° with the 10 N force

(c) 5 N 5 N Daya pad uan = 5. 0 N Resu ltant forc e

5.1 N pada sudut 60° dengan daya 5 N

5.1 N at an angle of 60° with 5 N force

2 2 cm : 1 N

2 N

5 N

120° 16°

Daya paduan / Resu

ltant force = 6.2 N

6.25 N pada sudut 16° dengan daya 5 N

6.25 N at an angle of 16° with 5 N force

B Takal / Pulley F = 40 N – 30 N = 10 N F = 30 N – 2 N = 28 N m = 4 kg + 3 kg = 7 kg m = 4 kg + 3 kg = 7 kg F = ma a = 10 N7 kg = 1.43 m s–2 F = ma, ∴28 N = (7 kg)(a) a = 28 N7 kg = 4 m s–2 Kaedah I / Method I

Pertimbangkan jisim 4 kg sahaja (gerak ke bawah)

Consider only the 4 kg mass (moving downwards)

40 N – T = m1a

40 N – T = (4 kg)(1.43 m s–2) ∴ T = 34.28 N Kaedah II / Method II

Pertimbangkan jisim 3 kg sahaja (gerak ke atas)

Consider only the 3 kg mass (moving upwards)

T – 30 N = m2a

T = 30 N + (3 kg)(1.43 m s–2)

T = 34.29 N

Kaedah I / Method I

Pertimbangkan jisim 3 kg sahaja (gerak ke bawah)

Consider only the 3 kg mass (moving downwards) 30 N – T = ma 30 N – T = (3 kg)(4 m s–2) ∴ T = 30 N – 12 N = 18 N Kaedah II / Method II

Pertimbangkan jisim 4 kg sahaja (gerak ke atas)

Consider only the 4 kg mass (moving upwards) T – 2 N = ma ∴ T = 2 N + (4 kg)(4 m s–2) = 2 N + 16 N = 18 N Latihan / Exercises

1 (a) Berat budak / Mass of the boy = W = mg = 50 kg × 10 m s–2 = 500 N (b) (i) R = W = 500 N (ii) R – mg = ma R = 500 N + (50 kg)(2 m s–2) = 600 N (iii) mg – R = ma R = 500 N – (50 kg)(2 m s–2) = 400 N (iv) R – mg = ma

Tetapi a = 0 (kerana halaju malar)

but a = 0 (because constant velocity)

∴ R = mg = 500 N 2 (a) (i) F = ma

3 0 N = [(2 + 3) kg][a] a = 6 m s–2 (ii) dari gerakan troli

from the motion of the trolley

T = ma

= (2 kg)(6 m s–2)

= 12 N

atau dari gerakan jisim 3 kg

or from the motion of the 3 kg-mass

30 N – T = ma 30 N – T = (3 kg)(6 m s–2) T = 30 N – 18 N = 12 N (b) (i) 30 N – 10 N = (3 + 2) kg × a 20 N = 5 kg × a a = 4 m s–2

(ii) dari gerakan troli

from the motion of the trolley

T – 10 N = ma

T – 10 N = (2 kg)(4 m s–2)

T = 8 N + 10 N

= 18 N

atau dari gerakan jisim 3 kg

or from the motion of the 3 kg-mass

30 N – T = m1 a 30 N – T = (3 kg)(4 m s–2)

T = 30 N – 12 N

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1.2 Contoh / Example 1 (b) Fx 75 N = sin 70° ∴ Fx = 75 N sin 70° = 70.48 N Fy 75 N = kos / cos 70° ∴ Fy = 75 N kos / cos 70° = 25.65 N (c) Fx 5 N = sin 40° ∴ Fx = 5 N sin 40° = 3.21 N Fy 5 N = kos / cos 40° ∴ Fy = 5 N kos / cos 40° = 3.83 N (d) Fx 6 N = kos / cos 60° ∴ Fx = 6 N kos / cos 60° = 3.0 N Fy 6 N = sin 60° ∴ Fy = 6 N sin 60° = 5.20 N Latihan / Exercises 1 (a) Fx 6 N = kos / cos 60° ∴ Fx = 6 N kos / cos 60° = 3.0 N (b) F = ma a = 3 N2 kg = 1.5 m s–2 2 (a) Fy Fx 60° F = 5 000 N (b) Fx 5 000 N = kos / cos 60° ∴ Fx = 5 000 N kos / cos 60° = 2 500 N Fy 5 000 N = sin 60° ∴ Fy = 5 000 N sin 60° = 4 330 N 3 (a) Fy Fx 55° F = 100 N (b) Fx 100 N = kos / cos 55° ∴ Fx = 100 N kos / cos 55° = 57.36 N Fy 100 N = sin 55° ∴ Fy = 100 N sin 55° = 81.92 N

4 (a) Berat budak, W / Weight of boy, W = 400 N Wc = mg sin θ

Daya,Wc / Force, Wc

= 400 N sin 30º = 200 N

(b) Daya paduan / Resultant force = 200 + (–120) = 80 N (c) F = ma F = (40 kg)a (40 kg)a = 80 N a = 80 N40 kg = 2 m s–2 1.3

Contoh keadaan yang melibatkan daya keseimbangan daya Example of conditions that involve forces in equilibrium 2 5 cm (20 N) T2 T1 30º 30º 3 T T R θ θ Latihan / Exercises

1 Daya geseran / Friction force, FR

= mg sin θ = 5 × 10 × sin 15° = 50 × 0.2588 = 12.9 N

F(normal) = mg kos / cos θ = 5 × 10 × kos / cos 15° = 50 × 0.9659 = 48.3 N 2 (a) T F Berat Weight 3.0 N 40° Perhatian / Note:

Arah bagi tiga daya itu adalah berkitar.

The directions of the three forces are cyclic.

(b) Dari segi tiga di atas, / From the triangle above, T3.0 N = tan 40°

F = 3.0 N tan 40°

= 2.52 N

(c) Dari segi tiga di atas, / From the triangle above, 3.0 NT = kos / cos 40°

T = kos / cos 40° 3.0 N

= 3.92 N

3 Pecahkan daya kepada komponen Split the force into component (a) Bagi komponen mendatar: For horizontal component: T sin θ = 25 sin 45° T sin θ = 17.68...①

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Bagi komponen mengufuk: / For vertical component:

T kos / cos θ + 25 kos / cos 45° = 35 T kos / cos θ = 35 – 25 kos / cos 45° T kos / cos θ = 17.32...② ① ÷ ②: T sin θ T kos / cos θ = 17.6817.32 tan θ = 1.021 θ = 45.6° 35 N 45° 35 N 25 N 25 sin 45° 25 kos / cos 45° (b) Gantikan / Substitute θ

= 45.6° ke dalam persamaan ① / into equation ① T sin 45.6° = 17.68

T = 17.68sin 45.6°

T = 24.7 N

Ketegangan pada tali B / Tension in rope B = 24.7 N

1.4

kembali ke panjang dan bentuk asal

return to its original length and shape

kedudukan asalnya / original positions

Daya tolakan antara molekul / Repulsive intermolecular forces Daya tarikan antara molekul / Attractive intermolecular forces

Eksperimen / Experiment Pemboleh ubah dimanipulasi Manipulated variable

Daya / Berat / Jisim

Force / Weight / Mass

Pemboleh ubah bergerak balas Responding variable Pemanjangan spring Extension of a spring Pemboleh ubah dimalarkan Fixed variable

Diameter spring / Ketebalan spring

Diameter of the spring / Thickness of the spring

Perbincangan Discussion

1 Pemanjangan spring, x berkadar langsung dengan daya, F

Extension of the spring, x is directly proportional to the force, F

2 Pemanjangan spring / Extension of springDaya / Force 3 Jika spring diregangkan dengan berat yang

berlebihan, ia mungkin tidak akan kembali ke panjang asal kerana telah melebihi had kenyal.

If the spring is stretched by too large weight, it might not return to its original length due to its exceeding its elastic limit.

Faktor-faktor yang mempengaruhi kekenyalan Factors that affect elasticity

Kurang kenyal / Less elastic Lebih kenyal / More elastic Kurang kenyal / Less elastic Lebih kenyal / More elastic Kurang kenyal / Less elastic Lebih kenyal / More elastic

Latihan / Exercises 1 (a) 20 g → 7 cm – 5 cm = 2 cm 40 g → 4 cm (b) 20 g menghasilkan pemanjangan 2 cm 20 g gives an extension of 2 cm ∴ 60 g → pemanjangan / extension 6 cm

∴ panjang spring dengan beban 60 g = 5 cm + 6 cm = 11 cm

length of spring with 60 g load = 5 cm + 6 cm = 11 cm

2 (a) x = 13 cm – 10 cm = 3 cm k = Fx = 3 cm = 2 N cm6 N –1 (b) (i) 6 N → 3 cm 12 N → 6 cm

Jumlah panjang / Total length

= 10 cm + 10 cm + 6 cm + 6 cm = 32 cm (ii) k = Fx = 12 cm = 1 N cm12 N –1

(c) (i) 4 N → 1 cm 12 N → 3 cm

Jumlah panjang / Total length = 10 cm + 3 cm = 13 cm (ii) k = 12 N3 cm = 4 N cm–1

3 (a) A: 10 g → 2 cm B: 10 g → 4 cm

20 g → 4 cm 20 g → 8 cm Jumlah pemanjangan / Total extension = 4 cm + 8 cm = 12 cm

(b) 10 g → 1 cm 50 g → 5 cm

Pemanjangan sistem / Extension in the system = 5 cm (c) Sistem B / System B : 10 g → 2 cm

∴ 40 g → 8 cm

A : 10 g → 2 cm

∴ 40 g → 8 cm

∴ Pemanjangan sistem / Extension in the system = 8 cm + 8 cm = 16 cm

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4 E = 12 Fx = 12 × (0.02 kg × 10 m s–2) × 0.03 m = 0.003 J 5 E = 12 Fx = 12 (20 N) (0.4 m) = 4.0 J 6 (a) E = 12 kx2 = 12 (200 N m–1) (0.04 m)2 = 0.16 J (b) 12 mv2 = 0.16 J v2 = 2 × 0.16 J 0.01 kg = 32 m2 s–2 v = 5.66 m s–1 Perhatian / Note: kg = J Nmkg = (kg m skg–2)(m) = m2 s–2

LATIHAN PENGUKUHAN / ENRICHMENT EXERCISE

Soalan Objektif / Objective Questions

1 A 2 C 3 A 4 B 5 B 6 B

7 D 8 D 9 A 10 A

Soalan Struktur / Structure Question

1 (a) (i) x berkadar langsung dengan W, asalkan had kenyal tidak dilebihi.

x is directly proportional to W, provided the elastic

limit is not exceeded.

(ii) Hukum Hooke / Hooke’s Law (b) 0 1 2 3 4 5 0 2 4 6 8 14 6 7 12 10 Spring Spring Pemanjangan, x / cm Extension, x / cm Pemberat berslot Slotted weight Berat, W / N Weight, W / N (c) k = 10 N0.04 m = 250 N m–1 (d) E = 12 Fx = 12 (10 N) (0.04 m) = 0.2 J

2 (a) Kuantiti jirim / Amount of matter

(b) (i) W 30º T T 30º

(ii) 250T = kos / cos 30° T = 289 N

(c) 1. Sudut antara tali dan palang yang lebih besar untuk mengurangkan tegangan tali.

Bigger angle between the rope and the bar to reduce

the tension on the rope.

2. Tali yang kurang kenyal supaya tidak berayun.

Lower elasticity rope so that no swinging.

TEKANAN PRESSURE Unit

2

1 P = FA = 160 N0.2 m2 = 800 N m–2 2 (a) PMaksimum = FA Minimum = mgA Minimum = 0.5 kg × 10 m s(0.05 × 0.10) m–22 = 1 000 N m–2 (b) PMinimum = A F Maksimum = mgA Maksimum = 0.5 kg × 10 m s(0.2 × 0.10) m2–2 = 250 N m–2 3 P = mgA 2 × 104 Pa = 60 kg × 10 m s–2 A A = (60 × 10) N2 × 104 Pa (Perhatian / Note: 1 Pa = 1 N m–2) = 3.0 × 10–2 m2 2.1 Eksperimen A / Experiment A Inferens / Inference

Tekanan dalam cecair bergantung kepada kedalamannya.

The pressure of the liquid depends on its depth.

Hipotesis / Hypothesis

Apabila kedalaman cecair bertambah, tekanannya juga bertambah.

When the depth of the liquid increases, its pressure also increases.

Tujuan / Aim

Untuk menyiasat hubungan antara tekanan sesuatu cecair dengan kedalamannya.

To investigate the relationship between the pressure of a liquid and its depth.

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Pemboleh ubah / Variables

Kedalaman cecair / Depth of liquid Tekanan cecair / Pressure of liquid Ketumpatan cecair / Density of liquid

Susunan radas / Arrangement of the apparatus

Kaki retort Retort stand Pembaris meter Metre rule Manometer Manometer Bekas Container Corong tisel Thistle funnel Kepingan getah Rubber sheet Cecair Liquid h y Tiub getah Rubber tube Prosedur / Procedure

1 Radas disusun seperti dalam rajah.

The apparatus is set up as shown in the diagram.

2 Bekas diisi dengan cecair.

A container is filled with a liquid.

3 Satu corong tisel dipasang pada manometer.

A thistle funnel is attached to the manometer.

4 Mulut corong tisel diturunkan secara mencancang ke dalam cecair sehingga kedalaman, h = 10 cm.

The mouth of thistle funnel is lowered vertically into the liquid until the depth, h = 10 cm.

5 Perbezaan paras air di dalam manometer, y direkodkan.

The different of water level in manometer, y is recorded.

6 Langkah 4 dan 5 diulang untuk kedalaman, h = 20 cm, 30 cm, 40 cm dan 50 cm.

Steps 4 and 5 are repeated for the depths of h = 20 cm, 30 cm, 40 cm and 50 cm.

Menganalisis data / Analysing data

y (cm)

h (cm) 0

Eksperimen B / Experiment B Inferens / Inference

Tekanan dalam cecair bergantung kepada ketumpatan cecair.

The pressure of liquid depends on the density of liquid.

Hipotesis / Hypothesis

Apabila ketumpatan cecair bertambah, tekanannya juga bertambah.

When the density of liquid increases, its pressure also increases.

Tujuan / Aim

Untuk menyiasat hubungan di antara tekanan sesuatu cecair dengan ketumpatannnya.

To investigate the relationship between the pressure of liquid and its density.

Pemboleh ubah / Variables

Ketumpatan cecair / Density of liquid Tekanan cecair / Pressure of liquid Kedalaman / Depth Latihan / Exercises 1 P = ρgh = (1 150 kg m–3) (10 m s–2) (40 m) = 460 000 Pa / 460 kPa 2 P = ρgh = (1.36 × 104 kg m–3)(10 m s–2)(0.8 – 0.2) m = 8.16 × 104 Pa / 8.16 kPa 3 P = ρgh = (1 000 kg m–3) (10 m s–2) (5 m) = 50 000 Pa / 50 kPa

Aplikasi tekanan dalam cecair Applications of pressure in liquids (d) • tekanan / pressure

• tekanan atmosfera + ρgH / atmospheric pressure + ρgH • tinggi / higher

2.2

Aktiviti untuk menunjukkan kewujudan tekanan atmosfera Activities to show the existence of atmospheric pressure • lebih besar / greater • berkurang / decreases

Latihan / Exercises

1 P = hρg

= (0.76 m) × (1.36 × 104 kg m–3) × (10 m s–2) = 1.03 × 105 Pa

2 Tekanan gas X / Pressure of gas X = (76 cm Hg) – (40 cm Hg) = 36 cm Hg 3 (a) 75 cm (b) (i) 75 cm (ii) 75 cm (c) (i) P = hρg = (0.75 m) × (1.36 × 104 kg m–3) × (10 m s–2) = 1.02 × 105 Pa

(ii) Tekanan atmosfera merkuri = Tekanan atmosfera air

Atmospheric pressure of mercury

= Atmospheric pressure of water

Pm = Pa

hmρmg = haρag

∴ 1.02 × 105 Pa = (h

a) × (1 × 103 kg m–3) × (10 m s–2)

∴ ha = 10.2 m

(iii) Tekanan / Pressure = (10.2 + 0.40) m = 10.6 m

(tekanan meningkat, jadi panjang h meningkat)

(pressure increases, so the length of h increases)

4 Jumlah tekanan / Total pressure

= Kedalaman ikan dari paras air + tekanan atmosfera

Depth of fish from the water surface + atmospheric pressure

= 3 m + 10 m = 13 m

2.3

Latihan / Exercises

1 (a) Pgas = Patm + h cm Hg = 76 cm Hg + 15 cm Hg = 91 cm Hg

(b) Pgas = hρg

= (0.91 m) × (1.36 × 104 kg m–3) × (10 m s–2)

= 1.24 × 105 Pa

2 (a) Tekanan di titik B / Pressure at point B

= Tekanan atmosfera + Tekanan disebabkan jalur AB Atmospheric pressure + Pressure due to column AB = 76 cm Hg + 8 cm Hg

= 84 cm Hg

= (0.84 m)(1.32 × 104 kg m–3)(10 N kg–1) = 110 880 N m–2

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Tekanan di titik C = Tekanan di titik B (pada sama paras)

Pressure at point C = Pressure at point B (at the same level)

= 82 cm Hg = 1.11 × 105 Pa

(b) Tekanan bekalan gas / Pressure of gas supply = Tekanan pada titik C / Pressure at point C = 82 cm Hg

= 1.11 × 105 Pa

(c) Ketinggian jalur merkuri tidak berubah kerana tekanan tidak bergantung pada saiz tiub manometer.

The height of the mercury column does not change because

the pressure is independent of the size of the manometer tube. 2.4 Latihan / Exercises 1 F1 A1 = FA22 F1 50 cm2 = 2 500 N20 m2 F1 = 50 cm2 × 2 500 N20 × 104 cm2 = 0.625 N 2 F1 A1 = FA22 250 N5 cm2 = F 200 cm2 F = 10 000 N

3 (a) tekanan yang dipindahkan / the pressure transmitted, P = 20 N0.005 m2 = 4 000 N m–2 (b) P = mgA 2 4 000 N m–2 = jisim / mass × 10 m s–2 0.1 m2 ∴ jisim / mass = 40 kg 2.5

Menghubungkaitkan daya apungan dengan berat air yang disesarkan dan isi padu air yang disesarkan

Relate buoyant force to the weight of the water displaced and volume of the water displaced

(c) bertambah / increases (d) bersamaan / equal

Daya apungan dan keapungan / Buoyant force and flotation

(a) = (b) < (c) >

Sesebuah kapal akan tenggelam lebih dalam ke dalam air jika berat yang lebih diletakkan di dalamnya.

The hull of the ship will sink deeper in the water if extra weight is put into it.

• bertambah; bertambah / increases; increases • bertambah / increases

• besar / larger

Peranan simbol Plimsoll pada kapal laut The purpose of Plimsoll symbol on a ship (a) streamline / Streamlined

(b) tinggi / high (c) stabil / stable

(d) daya apung yang besar / high buoyant force

Kapal selam / Submarine

Kapal selam tenggelam / Submarine sinks • lebih besar / larger

• tenggelam / sink

Kapal selam timbul semula / Submarine rise up • lebih kecil / smaller

• timbul semula / rise up

Ciri-ciri kapal selam: / Characteristics of a submarine:

(b) menahan / withstand

(c) tenggelam; terapung / sink; float (d) memantau / observe

(e) pernafasan / respiration

Belon udara panas / Hot air balloons

• daya apungan; berat belon / buoyant force; weight of the balloon

Ciri-ciri belon udara panas: / Characteristics of hot air balloons:

(a) lebih besar / bigger (b) lebih besar / bigger (d) mengurangkan / reduce (e) rintangan udara / air resistance

Hidrometer / Hydrometer

• terapung ke atas / float upright • sensitif / sensitive

• lebih dalam / more

Ciri-ciri sebuah hidrometer: / Characteristics of a hydrometer:

(b) kecil / small (d) terkakis / corrode

Latihan / Exercises

1 (a) Daya keapungan / Buoyant force, FB

= berat di udara – berat ketara

weight in the air – apparent weight

= 65 N – 30 N = 35 N

(b) Berat air yang disesarkan / Weight of water displaced = daya apungan / buoyant force

= 35 N

(c) Daya keapungan / Buoyant force, FB = ρVg

35 N = (1 000 kg m–3)(V)(10 m s–2) V = (1 000 kg m35 N–3 × 10 m s–2) = 0.0035 m3

Blok tenggelam sepenuhnya, maka isi padu blok = isi padu air yang disesarkan = 0.0035 m3

The block is completely submerged, so volume of the block

= volume of water displaced = 0.0035 m3

2 (a) (i) Jumlah berat yang bertindak pada bola pantai X > Jumlah berat yang bertindak pada bola pantai Y

Total weight acting on the beach ball X > Total weight

acting on the beach ball Y

(ii) Isi padu air laut yang disesarkan oleh bola pantai X > Isi padu air laut yang disesarkan oleh bola pantai Y

Volume of sea water displaced by the beach ball X >

Volume of sea water displaced by the beach ball Y

(iii) Berat air laut yang disesarkan oleh bola pantai X > Berat air yang disesarkan oleh bola pantai Y

Weight of sea water displaced by the beach ball X >

Weight of sea water displaced by the beach ball Y

(iv) Daya apungan yang bertindak ke atas bola pantai X > Daya apungan yang bertindak ke atas bola pantai Y

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Buoyant force acting on the beach ball X > Buoyant

force acting on the beach ball Y

(v) Ketumpatan air laut adalah sama atau tidak berubah

Density of sea water is same or unchanged

(b) (i) Daya apungan / Buoyant force

(ii) Prinsip Archimedes / Archimedes’ principle (c) (i) Semakin bertambah isi padu air laut yang disesarkan,

semakin bertambah daya apungan ATAU sebaliknya

As the volume of sea water displaced increases, the

buoyant force increases OR vice versa

(ii) Semakin bertambah berat air laut disesarkan, semakin bertambah daya apungan ATAU sebaliknya

As the weight of sea water displaced increases, the

buoyant force increases OR vice versa

(d) (i) Isi padu air disesarkan / Berat air disesarkan

Volume of water displaced / Weight of water displaced

(ii) Daya apungan / Buoyant force

(iii) Ketumpatan air laut / Density of sea water

2.6

Aktiviti / Activity 1 (c) bertambah / increases (d) lebih tinggi / higher (e) daya paduan / resultant force Aktiviti / Activity 2

(c) bertambah / increases (d) perbezaan / difference (e) mendekati / closer Aktiviti / Activity 3 (2) • rendah / lowest (3) • paling rendah / lowest Aktiviti / Activity 4 (c) rendah / decreases (d) tinggi / higher

(e) Daya paduan / resultant force

Penunu Bunsen / Bunsen burner

(c) atmosfera / Atmospheric (d) tekanan / pressure (e) gas bahan api / fuel gas

LATIHAN PENGUKUHAN / ENRICHMENT EXERCISE

Soalan Objektif / Objective Questions

1 A 2 D 3 C 4 D 5 A 6 D

7 D 8 A 9 C 10 C 11 C

Soalan Struktur / Structure Question

1 (a) (i) Bacaan tolok Bourdon A > bacaan tolok Bourdon B

Reading of Bourdon gauge A > Reading of Bourdon

gauge B

(ii) Tekanan udara di X > tekanan udara di Y

Air pressure at X > Air pressure at Y

(iii) Diameter tiub kaca di X > diameter tiub kaca di Y

Diameter of glass tube at X > Diameter of glass tube

at Y

(iv) Tinggi paras air di X < tinggi paras air di Y

Height of water level at X < Height of water level at Y

(v) Perbezaan tekanan di X < perbezaan tekanan di Y

Difference of pressure at X < Difference of pressure at

Y

(vi) Laju aliran udara di X < laju aliran udara di Y

Speed of air flow at X < Speed of air flow at Y

(b) Semakin berkurang laju aliran udara, semakin bertambah tekanan udara ATAU sebaliknya

As the speed of air flow decreases, the air pressure

increases OR vice versa

(c) Prinsip Bernoulli / Bernoulli’s principle

(d) • Kapal terbang bergerak dengan laju yang tinggi.

Aeroplane moves with high velocity.

• Bahagian atas sayap kapal terbang: Udara lebih laju, tekanan lebih rendah.

Upper part of the wings: air flow is faster, pressure is

lower.

• Bahagian bawah sayap kapal terbang: Udara kurang laju, tekanan lebih tinggi.

Lower part of the wings: air flow is slower, pressure is

higher.

• Perbezaan tekanan menolak sayap kapal terbang ke atas.

Difference of pressure pushes the wings upwards.

• Daya angkat > Berat kapal terbang.

Lift > Weight of the aeroplane.

• Luas permukaan sayap: lebih besar, menghasilkan daya angkat lebih besar

Large wing’s surface area: larger, resulting in greater

lifting power ELEKTRIK ELECTRICITY Unit

3

3.1

Medan elektrik / Electric field

3 cas positif / positive charges 4 magnitud; arah / magnitude; direction

Corak Medan Elektrik / Electric Field Pattern

Latihan / Exercises

1 (a) Kerana minyak zaitun merupakan bahan penebat yang baik

Because olive oil is a good insulator.

(b) Membentuk garisan corak yang menghubungkan antara elektrod positif dan elektrod negatif.

Form a pattern lines connecting the positive and negative

electrodes.

(c) Kekuatan medan elektrik meningkat.

The strength of the electric field increases.

Latihan / Exercises

1 ammeter; bersiri / ammeter; series 2 voltmeter; selari / voltmeter; parallel 3 (a) Cas / Charge, Q = It

= 8.0 A × (50 × 60 s)

= 8.0 C s–1 × 3 000 s

= 24 000 C

(b) Beza keupayaan / Potential difference, V = EQ = 5.76 × 1024 000 C = 240 V6 J

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4 Diberi / Given V = 6 V, Q = 40 C Kerja dilakukan / Work done, E = VQ

= 6 V × 40 C

= 6 JC × 40 C

= 240 J

3.2

Hukum ohm / Ohm’s law

• berkadar terus / directly proportional

Konduktor bukan Ohm / Non-ohmic conductors

• tidak mematuhi / not obey

Latihan / Exercises 1 V = IR 1.00 = 0.40 (R) R = 1.000.40 = 2.5 Ω Nilai y / Value of y V = IR y = (0.70)(2.5) y = 1.75 V Nilai x / Value of x V = IR 2.25 = x(2.5) x = 0.9 A 2 V = IR 12 = 2(R) R = 6 Ω

3 P mempunyai rintangan yang lebih besar dari Q.

P has a bigger resistance than Q.

∴ Kecerunan P lebih tinggi daripada Q.

Gradient of P is higher than Q.

2 sama / same 3 berkadar langsung

directly proportional

6 tidak / would not

3 sama / same

6 masih akan / still be able

Latihan / Exercises 1 (a) RPQ = 20 Ω + 10 Ω + 5 Ω = 35 Ω (b) 1R PQ = 18 Ω + 1 8 Ω + 8 Ω = 1 8 Ω3 ∴ RPQ = 8 Ω3 = 2.67 Ω (c) 1R YQ = 18 Ω + 1 8 Ω = 8 Ω2 ∴ RYQ = 4 Ω ∴ RYQ = 10 Ω + 20 Ω + RYQ = 10 Ω + 20 Ω + 4 Ω = 34 Ω (d) 1R PQ = 116 Ω + 8 Ω + 1 8 Ω 1 1R PQ = 1 + 2 + 216 Ω = 16 Ω 5 ∴ RPQ = 16 Ω5 = 3.2 Ω 2 (a) VXZ = IRXZ 1R YZ = 18 Ω + 1 8 Ω 1R YZ = 28 Ω ∴ RYZ = 4 Ω ∴ RXZ = RXY + RYZ = 8 Ω + 4 Ω = 12 Ω ∴ I = VR XZ = 20 V12 Ω = 1.67 A (b) VXZ = IXZRXZ 1R YZ = 12 Ω + 2 Ω 1 1R YZ = 22 Ω ∴ RYZ = 1 Ω ∴ RXZ = RXY + RYZ = 8 Ω + 1 Ω = 9 Ω ∴ IXZ = V XZ RXZ = 4.5 V9 Ω = 0.5 A 2 Ω I2 I I3 R2 R3 2 Ω Y Z A I = I2 + I3

Dari bahagian litar yang merentasi YZ,

From the section of the circuit across YZ,

I = IXZ = 0.5 A Tetapi / But I2 = I3 (Kerana / Because, R2 = R3) ∴ 2I2 = I 2I2 = 0.5 A I2 = 0.25 A

∴ Bacaan ammeter / Ammeter reading = 0.25 A Dari bahagian litar yang merentasi XY,

From the section of the circuit across XY,

I2 = VXY

6 Ω = 3.0 V6 Ω = 0.5 A

∴ Bacaan ammeter / Ammeter reading = 0.25 A

Teknik Menjawab [Format Kertas 2 : Perbandingan] Answering Technique [Paper 2 Format : Comparison] (a)

Rajah (a) / Diagram (a) Rajah (b) / Diagram (b)

Selari / Parallel Bersiri / Series Malap / Dim

Kurang / Less Lebih / More

Lebih / More

(b) (i) Mentol pada sambungan selari menghasilkan rintangan berkesan yang lebih kecil ATAU Mentol pada sambungan bersiri menghasilkan rintangan berkesan yang lebih besar.

Bulbs in parallel connection produces a lower effective

resistance OR Bulbs in series connection produces a greater effective resistance.

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(ii) Semakin berkurang rintangan berkesan, semakin bertambah jumlah arus di dalam litar.

As the effective resistance decreases, amount of current in

the circuit increases.

Faktor yang mempengaruhi rintangan dawai Factors that affect the resistance of wire

bertambah

increases berkurangdecreases rendahlow

Eksperimen / Experiment

Inferens Inference

Beza keupayaan yang merentasi konduktor logam bergantung kepada arus yang mengalir melalui konduktor logam.

The potential difference across a metal conductor depends on the current flowing through the metal conductor. Rintangan pada konduktor logam bergantung kepada panjang wayar.

The resistance of metal conductor depends on the length of the wire.

Hipotesis Hypothesis

Beza keupayaan yang merentasi konduktor logam meningkat apabila arus yang mengalir melalui konduktor logam meningkat. The potential difference across a metal conductor increases as the current flowing through the metal conductor increases. Rintangan pada konduktor logam meningkat apabila panjang dawai meningkat.

The resistance of metal conductor increases as the length of wire increases.

Tujuan Aim

Untuk mengkaji hubungan antara beza keupayaan, V, dan arus, I, dalam konduktor logam. To investigate the relationship between the potential difference, V, and current, I, in a metal conductor. Untuk mengkaji hubungan antara rintangan dan panjang konduktor logam.

To investigate the relationship between the resistance and the length of a metal conductor. Pemboleh ubah Variables Arus / Current Beza keupayaan Potential difference Suhu wayar Temperature of wire Panjang konduktor logam

Length of the metal conductor

Beza keupayaan

Potential difference

Suhu wayar, luas keratan rentas Temperature of wire, cross-sectional area Prosedur Procedure 1 Litar dihidupkan. The circuit is switched on. 2 Reostat dilaraskan sehingga ammeter memberikan bacaan 0.2 A. Bacaan voltmeter dicatatkan. The rheostat is adjusted until the ammeter gives a reading of 0.2 A. The reading of the voltmeter is recorded. 3 Eksperimen diulangi dengan nilai arus yang berbeza, I = 0.3 A, 0.4 A, 0.5 A, 0.6 A dan 0.7 A dengan melaraskan reostat. The experiment is repeated with different values of current, I = 0.3 A, 0.4 A, 0.5 A, 0.6 A and 0.7 A by adjusting the rheostat. 1 Dengan menggunakan pembaris meter, ukur panjang wayar konduktor, ℓ = 20.0 cm.

By using a meter ruler, measure the length of the conductor, = 20.0 cm. 2 Litar dihidupkan. The circuit is switched on. 3 Bacaan voltmeter dan bacaan ammeter dicatatkan.

The readings of the voltmeter and ammeter are recorded. 4 Rintangan dikira dengan menggunakan rumus, R = V / I The resistance is calculated using the formula, R = V / I

5 Eksperimen diulangi dengan menggunakan panjang wayar yang berbeza, ℓ = 40.0 cm, 60.0 cm, 80.0 cm dan 100.0 cm. The experiment is repeated with different lengths of the wire, = 40.0 cm, 60.0 cm, 80.0 cm and 100.0 cm. Menjadualkan data Tabulation of data I / A 0.2 0.3 0.4 0.5 0.6 0.7 V / V ℓ / cm 20.0 40.0 60.0 80.0 100.0 I / A V / V R / Ω Menganalisis data Analysis of the data 0 Beza keupayaan, Potential difference, V / V Arus, Current, I / A 0 Rintangan, Resistance, R / Ω Panjang, Length, ℓ / cm

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Eksperimen / Experiment Inferens Inference

Luas keratan rentas dawai mempengaruhi rintangannya.

Cross- sectional area of wire affect its resistance.

Jenis bahan dawai mempengaruhi rintangannya.

The type of material wire affect its resistance.

Hipotesis Hypothesis

Apabila luas keratan rentas bertambah, rintangannya berkurang.

When the cross-sectional area of wire increases, its resistance decreases.

Apabila jenis bahan dawai berubah, rintangannya turut berubah.

When the type of material of the wire changes, its resistance also changes.

Tujuan Aim

Untuk mengkaji hubungan antara luas keratan rentas dawai dan rintangannya.

To investigate the relationship between the cross-sectional area of wire and its resistance.

Untuk mengkaji hubungan antara jenis bahan dawai dan rintangannya.

To investigate the relationship between the type of material of the wire and its resistance.

Pemboleh ubah Variables

Luas keratan rentas dawai, A

Cross-sectional of wire, A

Rintangan, R

Resistance, R

Panjang, jenis dawai dan suhu dawai

Length, type and temperature of wire

Jenis bahan dawai

Type of material of the wire

Rintangan, R

Resistance, R

Ketebalan, panjang dawai dan suhu dawai

Thickness, length of the wire and temperature of the wire Senarai radas dan bahan List of apparatus and materials Dawai konstantan sepanjang 30 cm dengan luas keratan rentas 0.02 mm2, 0.04 mm2, 0.06 mm2, 0.08 mm2, 0.10 mm2, ammeter, voltmeter, dawai penyambung, sel kering, suis dan reostat.

Constantan wire of length 30 cm with cross-sectional area of 0.02 mm2, 0.04 mm2, 0.06 mm2, 0.08 mm2, 0.10 mm2, ammeter, voltmeter, connecting wires, dry cells, a switch and rheostat.

50 cm dawai nikrom, 50 cm dawai konstantan, 50 cm dawai kuprum, ammeter, voltmeter, dawai penyambung, sel kering, suis dan reostat.

50 cm nichrome wire, 50 cm constantan wire, 50 cm copper wire, ammeter, voltmeter, connecting wires, dry cells, a switch and rheostat.

Prosedur Procedure

1 Litar elektrik disusun seperti yang ditunjukkan.

The electric circuit is set up as shown.

2 Dawai konstantan dengan luas keratan rentas 0.02 mm2 disambung merentasi terminal P dan Q. Constantan wire with cross-sectional area of 0.02 mm2 is

connected across the terminal P and Q.

3 Suis dihidupkan dan reostat dilaraskan sehingga bacaan ammeter ialah 0.5 A. Rekod bacaan voltmeter.

The switch is turned on and the rheostat is adjusted until the reading of ammeter is 0.5 A. The voltmeter reading is recorded. 4 Langkah 2 dan 3 diulang dengan menggunakan dawai konstantan dengan luas keratan rentas 0.04 mm2, 0.06 mm2, 0.08 mm2, 0.10 mm2.

Step 2 and 3 are repeated by using constantan wire of cross-sectional area 0.04 mm2, 0.06 mm2, 0.08 mm2, 0.10 mm2. 5 Semua keputusan direkodkan.

All the results are recorded.

1 Litar elektrik disusun seperti yang ditunjukkan.

The electric circuit is set up as shown. 2 Dawai nikrom 50 cm disambung merentasi terminal P dan Q. 50 cm nichrome wire is connected across the terminal P and Q.

3 Suis dihidupkan dan reostat dilaraskan sehingga bacaan ammeter ialah 0.5 A. Bacaan voltmeter direkodkan. The switch is on and the rheostat is adjusted until the reading of ammeter is 0.5 A. The voltmeter reading is recorded. 4 Langkah 2 dan 3 diulang dengan menggunakan dawai konstantan 50 cm dan dawai kuprum 50 cm.

Step 2 and 3 are repeated by using 50 cm constantan wire and 50 cm copper wire. 5 Semua keputusan direkodkan.

All the results are recorded. Menganalisis data Analysis of the data A (mm2) 0 R (Ω) V (V) 0 I (A) Nikrom Nichrome Konstantan Constantan Kuprum Copper Latihan / Exercises 1 A = 0.01 cm2 = 0.01 × 10–6 m2 R = pℓ A = (1.724 × 10 –6 Ω m)(100 m) (0.01 × 10–6 m2) R = 1.724 × 104 Ω 2 A = 0.05 cm2 = 0.05 × 10–6 m2 R = pℓ A 0.5 Ω = 0.05 × 10ρ(50 m)–6 m2 ρ = 5 × 10–10 Ω

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3.3

Perbandingan antara daya gerak elektrik dan beza keupayaan Comparison between electromotive force and potential difference

• terbuka; daya gerak elektrik (d.g.e)

open; electromotive force (e.m.f.) • 1.5 V • kecil / smaller • haba / heat • kurang / less kecerunan / gradient V = –rI + Ԑ Eksperimen / Experiment Hipotesis Hypothesis

Apabila arus, I meningkat, pengurangan tenaga semakin bertambah dan menyebabkan beza keupayaan, V, menurun.

When the current, I, increases, the energy being dissipated increases and causes the potential difference, V, to decrease.

Pemboleh ubah Variables

Arus yang mengalir melalui sel, I

Current flowing through the cell, I

Beza keupayaan merentasi sel, V

Potential difference across the cell, V

Suhu wayar / Temperature of wire

Senarai radas dan bahan List of apparatus and materials

Sel kering, suis, voltmeter, ammeter, reostat dan wayar penyambung

Dry cell, switch, voltmeter, ammeter, rheostat and connecting wires

Susunan radas Arrangement of apparatus V A Reostat Rheostat R Sel kering Dry cell Suis /Switch Prosedur Procedure

1 Litar elektrik disediakan seperti dalam rajah.

The electric circuit is set up as shown in diagram.

2 Suis ditutup dan bacaan ammeter, I = 0.2 A dan voltmeter, V dicatatkan dengan melaraskan reostat.

The switch is closed and the reading of the ammeter, I = 0.2 A and voltmeter, V is recorded by adjusting the rheostat.

3 Eksperimen diulangi dengan nilai I yang berbeza iaitu I = 0.3 A, 0.4 A, 0.5 A, 0.6 A dengan melaraskan reostat.

The experiment is repeated with different values of I = 0.3 A, 0.4 A, 0.5 A, 0.6 A, by adjusting the rheostat.

Menjadualkan data

Tabulation of data

Arus melalui sel, I / A

Current flowing through cell, I / A 0.2 0.3 0.4 0.5 0.6

Beza keupayaan merentasi sel, V / V Potential difference across cell, V / V

Menganalisis data Analysis of the data 0 V / V I / A Pengiraan rintangan dalam Calculation of internal resistance Apabila / When I = 0 A,

V = Ԑ (dalam Volt / in Volt)

Kecerunan graf / Gradient of the graph = – r ∴ r = – kecerunan graf (dalam Ω) – gradient of the graph (in Ω) Latihan / Exercises 1 (a) Ԑ = 1.5 V (b) Ԑ = V + Ir 1.5 V = 1.35 V + (0.3 A)r (0.3 A)r = (1.5 – 1.35)V r = 0.15 V0.3 A = 0.5 Ω (c) V = IR 1.35 V = (0.3 A)R R = 1.35 V0.3 A = 4.5 Ω 2 Diberi / Given Ԑ = 1.5 V, V = 1.0 V, R = 5 Ω Arus / Current, I = VR = 1.0 V5 A = 0.2 A Ԑ = V + Ir 1.5 V = 1.0 V + (0.2 A)r r = (1.5 – 1.0) V0.2 A = 2.5 Ω 3 Apabila nilai I = 0, didapati V = Ԑ.

When the value of I = 0, V = Ԑ is obtained.

Daripada graf, Ԑ = 3.0 V

From the graph, Ԑ = 3.0 V

Maka, rintangan dalam, r (kecerunan graf) Hence, the internal resistance, r (gradient of graph)

r = Ԑ – VI = (3.0 – 1.7) V0.74 A = 1.76 Ω 4 Apabila akumulator disambungkan kepada perintang 2 Ω, When the accumulator is connected to the 2 Ω resistor, Ԑ = V + Ir = IR + Ir = (4 A)(2 Ω) + (4 A)r Ԑ = 8 V + (4 A)r ———(i) Apabila akumulator disambungkan kepada perintang 3 Ω, When the accumulator is connected to the 3 Ω resistor, Ԑ = V + Ir = IR + Ir = (3 A)(3 Ω) + (3 A)r Ԑ = 9 V + (3 A)r ———(ii)

Persamaan (i) = Persamaan (ii), / Equation (i) = Equation (ii), 8 V + (4 A)r = 9 V + (3 A)r

(1 A)r = 1 V

r = 1 V1 A

∴ r = 1 Ω

Dari (i), / From (i), Ԑ = 8 V + (4 A)(1 Ω) = 12 V ∴ e.m.f. = 12 V

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5 (a) (i) Apabila I meningkat, V berkurang.

When I increases, V decreases.

0.5 1.0 1.5 0.4 0.2 0 V / V I / A 1.2 1.0 0.8 0.6 (c)(i)

(ii) Berdasarkan ekstrapolasi pada graf, apabila

Based on extrapolation of the graph, when

I = 0.0 A, V = 1.5 V

(iii) Daya gerak elektrik, Ԑ

Electromotive force, Ԑ

(b) r = –m = (1.5 – 1.0) V(0 – 1.0) A = 0.5 Ω

(c) (i) Daripada graf, apabila / From the graph, when I = 0.8 A, V = 1.1 V

(ii) r = 1.1 V0.8 A = 1.38 Ω

(d) Betulkan ralat sifar bagi voltmeter dan ammeter

Correct zero errors in the voltmeter and ammeter

atau / or

Elakkan ralat paralaks semasa mengambil bacaan voltmeter dan ammeter

Avoid parallax errors when taking the voltmeter and ammeter readings

3.4

Formula

E = Pt P = VI

Tenaga elektrik / Electrical energy

• tenaga bunyi / sound energy

Latihan / Exercises 1 Diberi / Given V = 240 V, I = 5 A, t = 10 × 60 s E = Pt = (VI)t = 240 V × 5 A × (10 × 60 s) = 720 000 J

2 Pertama, kira rintangan berkesan RXY

Firstly, calculate the effective resistance, RXY

R1 XY = 12 Ω + 1 6 Ω R1 XY = 46 Ω ∴ RXY = 6 Ω4 = 1.5 Ω ∴ RWZ = 8 Ω + 1.5 Ω + 2.5 Ω = 12 Ω Dari / From V = IR I = VR WZ = 24 V12 Ω = 2 A

Kemudian, gunakan rumus tenaga elektrik

Then, using the formula of electrical energy

E = VIt = 24 V × 2 A × (5 × 60 s) = 14 400 J 3 (a) P = VI 3 000 W = (240 V)I I = 3 000 W240 V = 12.5 A (b) P = VI = (IR)I P = I2 R 3 000 W = (12.5 A)2R R = 3 000 W156.25 A2 = 19.2 Ω

Mengira kos tenaga elektrik Calculating the cost of electrial energy

Pemanas air / Water heater 2 × 1 kW × 60 j = 120 kWj Lampu / Lamps 6 × 0.04 kW × 200 j = 48 kWj Pengisar makanan / Food blender 1 × 0.06 kW × 30 j = 1.8 kWj Kipas angin / Fan 5 × 0.06 kW × 100 j = 30 kWj Jumlah penggunaan elektrik

Total electricity consumption

= 739.8 kWj / kWh

Untuk 200 kWj yang pertama:

For the first 200 kWh:

200 × 0.2180 = RM43.60 Untuk 100 kWj yang berikutnya:

For the next 100 kWh:

100 × 0.3340 = RM33.40 Untuk 300 kWj yang berikutnya:

For the next 300 kWh:

300 × 0.5160 = RM154.80

Bayaran bil untuk bulan Ogos 2020:

Bill payment for month of August 2020:

RM43.60 + RM33.40 + RM134.06 + RM76.33 = RM287.39 Untuk 300 kWj yang berikutnya:

For the next 300 kWh:

139.8 × 0.5460 = RM76.33

Kecekapan peralatan elektrik / Efficiency of electrical appliances

1 (a) Einput = VIt

= 12 V × 5.0 A × 2.5 s = 150 J

(b) Eoutput = Tenaga keupayaan graviti

Potential gravitational energy

= mgh

= 2 kg × 10 m s–2 × 3 m

= 60 J

(c) ∴ Pinput = 150 J2.5 s = 60 W

Poutput = 60 J2.5 s = 24 W

∴ Kecekapan motor / Efficiency of the motor

= POutput PInput × 100% = 24 W60 W × 100% = 40% Latihan / Exercises

1 Kuasainput / Powerinput = Pinput = 3 000 W Eoutput = mcθ

= 0.5 kg × 4 200 J kg–1ºC–1 × (100 – 20)ºC = 168 000 J

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∴ Kuasaoutput / Poweroutput = Poutput

= 168 000 J90 s

= 1866.67 W

Kecekapan cerek elektrik / The efficiency of the kettle = POutput

PInput × 100%

= 1866.67 W3 000 W × 100% = 62.22%

2 (a) Tenaga elektrik ditukarkan ke tenaga keupayaan graviti

Electrical energy is changed to gravitational potential energy

(b) (i) Eoutput = mgh

= 0.8 kg × 10 m s–2 × 1.5 m

= 12 J

(ii) Einput = VIt

= 5.0 V × 1.2 A × 4.0 s

= 24 J

(iii) Kecekapan / Efficiency = 12 J24 J × 100% = 50% (c) (i) Bertambah / Increases

(ii) Bertambah / Increases 3 Diberi / Given, F = mg = 60 N

Jika tenaga elektrik digunakan = tenaga keupayaan graviti yang diperoleh

Since electrical energy used = gravitational potential energy gained

VIt = mgh I = mghVt = 60 N × 2 m12 V × 4 s = 2.5 A

4 [1 kWj = 1 unit tenaga elektrik]

[1 kWh = 1 unit of electrical energy]

Jumlah tenaga yang digunakan = 0.8 kW × 8 j × 30 = 192 kWj Oleh itu, kos elektrik yang digunakan = 192 unit × RM0.22unit = RM42.24

Total energy used = 0.8 kW × 8 j × 30 = 192 kWh Hence, the cost of using electricity

= 192 units × RM0.22unit = RM42.24

5 60% daripada tenaga elektrik = tenaga cahaya

60% of the electrical energy = light energy

60 % × E = tenaga cahaya / light energy 100 × Pt = tenaga cahaya / light energy60 610 × 40 W × (7 × 60 s) = tenaga cahaya / light energy Oleh itu, tenaga cahaya / Hence, light energy = 10 080 J 6 Ei = Pt = 3 000 W × 90 s = 270 000 J Eo = mcθ = 0.5 kg × 4 200 J kg–1 °C–1 × (100 – 20)°C = 168 000 J Efficiency = Eo Ei × 100 = 168 000 J270 000 J × 100 = 62.22%

LATIHAN PENGUKUHAN / ENRICHMENT EXERCISE

Soalan Objektif / Objective Questions

1 D 2 C 3 A 4 A 5 C 6 D

7 B 8 B 9 D 10 B

Soalan Struktur / Structure Question

1 (a) Apabila beza keupayaan yang dibekalkan ialah 6 V, kuasa yang dihasilkan ialah 12 W.

When the voltage supplied is 6 V, the power produced is

12 W.

(b) Bersiri / Series Selari / Parallel

(c) Voltan untuk setiap mentol dalam Rajah (b) lebih daripada Rajah (a). Jumlah rintangan dalam Rajah (b) kurang daripada Rajah (a). Arus mengalir dalam setiap mentol dalam Rajah (b) lebih daripada Rajah (a).

Voltage for each bulb in Diagram (b) is more than

Diagram (a). The total resistance in Diagram (b) is less than Diagram (a). Current flow in each bulb in Diagram (b) is more than that in Diagram (a).

(d) (i) R1 = 4 Ω + 4 Ω + 4 Ω = 12 Ω I = VR = 6.0 V12 Ω = 0.5 A (ii) 1R = 4 Ω + 1 4 Ω + 1 4 Ω = 1 4 Ω ,3 RT = 4 Ω3 = 1.33 Ω IT = VR T = 6.0 V 1.33 Ω = 4.5 A Arus mengalir dalam setiap mentol

Current flow in each bulb

= 4.53 = 1.5 A (e) (i) Rajah (b) / Diagram (b)

(ii) • Jika satu daripada mentol terbakar, mentol yang lain masih berfungsi.

If one bulb blow, another bulb can still function.

• Rintangan berkesan kurang // lebih banyak arus mengalir.

Less effective resistance // more current flow.

Teknik Menjawab [Format Kertas 2 : Bahagian B] Answering Technique [Paper 2 Format : Part B]

Kekonduksian yang lebih baik

Good conductivity Boleh mengalirkan arusCan conduct current

Nikrom

Nichrome Rintangan lebih tinggiHigher resistance

Lebih kecil

Smaller Rintangan lebih tinggiHigher resistance

Lebih panjang

Longer Rintangan lebih tinggiHigher resistance

Lebih banyak

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KEELEKTROMAGNETAN ELECTROMAGNETISM Unit

4

4.1 Latihan / Exercises U/N S/S

Wayar lurus / Straight wire

(c) arus di tengah / the current in the middle

Elektromagnet / Electromagnet

(a) elektromagnet / electromagnet (c) (ii) Bilangan lilitan / Number of turns (iii) Jenis teras besi / Type of iron core

(iv) Bentuk teras besi / The shape of the iron core

Kekuatan medan magnet bertambah dengan: The strength of the magnetic field is increased by: (ii) menambahkan / increasing

Latihan / Exercises

(a) Sama / Same Sama / Same

Sama / Same Sama / Same

Banyak / More Kurang / Less

Lebih rapat / Closer Kurang rapat / Less closer Lebih kuat / Stronger Kurang kuat / Less stronger (b) Bilangan lilitan gegelung dawai semakin bertambah, kekuatan

medan magnet di sekeliling dawai semakin bertambah.

As the number of turns of wire coil increases, the strength of magnetic field around the wire increases.

(c) Elektromagnet / Electromagnet (d) (i) Bilangan lilitan gegelung dawai

Number of turns of wire coil

(ii) Kekuatan medan elektromagnet / Strength of electromagnet (iii) Arus elektrik / Current

Latihan / Exercises 1 (a) U/N S/S F (b) U/N S/S F

2 Dengan menggunakan peraturan tangan kiri Fleming: D

By using Fleming’s left-hand rule: D

3 Dengan menggunakan peraturan tangan kiri Fleming: C

By using Fleming’s left-hand rule: C

4 Dengan menggunakan peraturan tangan kiri Fleming: A

By using Fleming’s left-hand rule: A

5 Dengan menggunakan peraturan tangan kiri Fleming: A

By using Fleming’s left-hand rule: A

Eksperimen / Experiment

Inferens Inference

Magnitud daya pada konduktor pembawa arus dalam medan magnet bergantung kepada magnitud arus yang mengalir.

The magnitude of the force on a current-carrying conductor in a magnetic field depends on the magnitude of the current.

Magnitud daya pada konduktor pembawa arus dalam medan magnet bergantung kepada kekuatan medan magnet yang kekal. The magnitude of a force on a current-carrying conductor in a magnetic field depends on the strength of the permanent magnetic field. Hipotesis Hypothesis

Magnitud daya pada konduktor yang membawa arus dalam medan magnet bertambah (ditentukan oleh jarak gerakan wayar kuprum pendek) apabila magnitud arus yang mengalir bertambah (ditentukan oleh magnitud beza keupayaan, V).

The magnitude of the force on a current-carrying conductor in a magnetic field (indicated by the distance of movement of short copper wire) increases as the magnitude of the current increases (indicated by magnitude of potential difference, V).

Magnitud daya pada konduktor yang membawa arus dalam medan magnet bertambah (ditentukan oleh jarak gerakan wayar kuprum pendek) apabila kekuatan medan magnet bertambah (ditentukan oleh bilangan magnet).

The magnitude of the force on a current-carrying conductor in a magnetic field (indicated by the distance of movement of short copper wire) increases as the strength of the magnetic field increases (indicated by the number of magnets).

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Pemboleh ubah Variables Magnitud beza keupayaan, V. Magnitude of the potential difference, V.

Jarak gerakan wayar kuprum pendek, L.

Distance of movement of short copper wire, L.

Kekuatan medan magnet kekal

The strength of the permanent magnetic field Bilangan magnet magnadur Number of magnadur magnets

Jarak gerakan wayar kuprum pendek, L

Distance of movement of short copper wire, L

Magnitud beza keupayaan, V Magnitude of potential difference, V Prosedur Procedure 1 Voltan bekalan kuasa a.t. yang digunakan dicatatkan;

V = 1.5 V.

Bekalan kuasa a.t. dihidupkan.

The voltage of the d.c. power supply used is recorded; V = 1.5 V. The d.c.

power supply is switched on.

2 Jarak gerakan wayar kuprum pendek di atas wayar kuprum tebal diukur dengan pembaris = L.

The distance of movement of short copper wire on the thick copper wire is measured by a ruler = L.

3 Eksperimen diulangi dengan bekalan beza keupayaan, V = 2.0 V, 2.5 V, 3.0 V dan 3.5 V. The experiment is repeated with different voltages of d.c. power supply, V = 2.0 V, 2.5 V, 3.0 V and 3.5 V. 1 Radas disediakan seperti yang ditunjukan pada rajah.

The apparatus is set up as shown in the diagram. 2 Dua magnet magnadur diletakkan pada dening besi berbentuk U dengan kutub bertentangan menghadap satu sama lain. Two magnadur magnets are placed on the U-shaped iron yoke with opposite poles facing each other.

3 Bekalan kuasa a.t dihidupkan. Jarak gerakan wayar kuprum diukur dengan pembaris = L. The d.c power supply is switched on. The distance of movement of copper wire is measured by a ruler = L. 4 Eksperimen diulang dengan menambah 1, 2, 3 dan 4 pasang magnet magnadur pada dening besi berbentuk U. Experiment is repeated with 1, 2, 3 and 4 pairs of magnadur magnets on the U-shaped iron yoke. Penjadualan data Tabulation of data Beza keupayaan Potential difference, V / V Jarak gerakan wayar Distance of movement, L / cm 1.5 2.0 2.5 3.0 3.5 Bilangan magnet Number of magnets Jarak gerakan wayar, L (cm) Distance of movement, L (cm) 1 pasang magnet 1 pair of magnet 2 pasang magnet 2 pairs of magnet 3 pasang magnet 3 pairs of magnet 4 pasang magnet 4 pairs of magnet Menganalisis data Analysis of the data 0 Jarak gerakan, Distance of movement, L / cm Beza keupayaan, Potential difference, V / V 0 Jarak gerakan, Distance of movement, L / cm Bilangan pasang magnet Number of pair of magnets

Motor arus terus / Direct current motor Penerangan

Explanation (e) inersia / inertia(f) • menambahkan / increasing • menambahkan / increasing

Latihan / Exercises

1 D 2 D 3 A 4 A 5 C

Latihan / Exercises

(a) Kurang / Less Lebih / More

Kurang / Less Lebih / More

Kurang / Less Lebih / More

(b) (i) Apabila arus elektrik yang mengalir melalui kabel semakin bertambah, jisim besi buruk yang dinaikkan semakin bertambah atau sebaliknya.

As the current flowing through the cable increases, the mass

of scrap metal lifted increases or vice versa.

(ii) Apabila arus elektrik yang mengalir melalui kabel semakin bertambah, kekuatan medan magnet yang terhasil semakin bertambah atau sebaliknya.

As the current flowing through the cable increases, the

strength of magnetic field produced increases or vice versa.

(c) Elektromagnet / Electromagnet (d) (i) Arus elektrik / Current (ii) Kekuatan medan elektromagnet

Strength of electromagnet field

(iii) Bilangan lilitan gegelung dawai pada teras besi lembut

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4.2

Gerakan relatif untuk menghasilkan arus teraruh Relative motion to produce induce currents

(a) pegun / stationary (b) wayar / solenoid

wire / solenoid

(d) berbeza / different (b) reostat R / rheostat R Mengayunkan / Oscillating

Penjana Arus Terus

D.C. Generator Penjana Arus Ulang-alik A.C. Generator (a) arus aruhan

induced current (b) maksimum / maximum (f) terus / direct

(b) garis medan magnet magnetic field lines (c) menegak / vertical (f) ulang-alik / alternating

Teknik Menjawab [Format Kertas 2 : Perbandingan] Answering Technique [Paper 2 Format : Comparison] (a)

Rajah (a) / Diagram (a) Rajah (b) / Diagram (b)

Lebih laju / Higher speed Kurang laju / Lower speed Lebih besar / Larger Lebih kecil / Smaller Lebih cerah / Brighter Kurang cerah / Dimmer (b) (i) Semakin bertambah laju kayuhan basikal, semakin cerah

nyalaan mentol.

As the cycling speed of bike increases, the bulb lights up

brighter.

(ii) Apabila kadar pemotongan fluks magnet oleh gegelung dawai dinamo semakin bertambah, arus elektrik yang terhasil semakin bertambah.

As the cutting rate of magnetic flux by the coil wire of

dynamo increases, the current produced increases.

(c) (i) Aruhan elektromagnet / Electromagnetic induction (ii) Hukum Lenz / Lenz’s law

Teknik Menjawab [Format Kertas 2 : Kefahaman / Esei Pendek] Answering Technique [Paper 2 Format : Comprehension / Short Essay]

1 • Magnet kekal diputarkan.

The permanent magnet is rotated.

• Fluks magnet dipotong oleh gegelung dawai secara gerakan relatif.

The magnetic flux is cut by the coil wire through the relative motion.

• Arus aruhan terhasil melalui aruhan elektromagnet.

The induced current produced through the electromagnetic induction.

• Tenaga kinetik → Tenaga elektrik → Tenaga cahaya

Kinetic energy → Electric energy → Light energy

2 Lebih kuat

Stronger Menghasilkan arus yang lebih besarProduce larger induced current

Besi lembut

Soft iron

Mudah dimagnet dan mudah dinyah-magnetkan

Easier to be magnetised and easier to be demagnetised Lebih banyak More Menambahkan kadar pemotongan fluks magnet To increase the cutting rate of magnetic flux Lebih besar

Larger Rintangan lebih kecilSmaller resistance

Kuprum

Copper Rintangan lebih kecilSmaller resistance

Lebih kecil

Smaller

Menambahkan kadar pemotongan fluks magnet

To increase the cutting rate of magnetic

Perbandingan antara Penjana Arus Terus dan Penjana Arus Ulang Alik

Differences between Direct Current Generator and Alternating Current Generator

satu arah yang tetap / one fixed direction

4.3

Struktur

Structure • gegelung primer / primary coil• gegelung sekunder / secondary coil

Prinsip kerja Working principle

• arus ulang-alik / alternating current • arus aruhan / induced current

Ciri-ciri

Characteristics Beza keupayaan sekunder Secondary potential difference

JENIS TRANSFORMER (Klasifikasi) TYPES OF TRANSFORMERS (Classifying)

VS > VP NS > NP NS < NP Gegelung primer Primary coil Gegelung sekunder Secondary coil VP VS Gegelung primer Primary coil Gegelung sekunder Secondary coil VP VS

Faktor yang mempengaruhi kecekapan transformer dan cara untuk meningkatkan kecekapannya

Factors that affect the efficiency of a transformer and ways to improve the efficiency

• tebal / thick • haba / heat Teras berlamina

Laminated core haba heat

Figur

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Referensi

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