G10-PSC-P1-JUn-2019-and-Memo - Physical Sciences Break 1.0 · 2019. 6. 1. · Physical Sciences P1...

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Transcript of G10-PSC-P1-JUn-2019-and-Memo - Physical Sciences Break 1.0 · 2019. 6. 1. · Physical Sciences P1...

Page 1: G10-PSC-P1-JUn-2019-and-Memo - Physical Sciences Break 1.0 · 2019. 6. 1. · Physical Sciences P1 3 June 2019 Common Test 2.2.1 Sound with a frequency range of between 20 kHz to
Page 2: G10-PSC-P1-JUn-2019-and-Memo - Physical Sciences Break 1.0 · 2019. 6. 1. · Physical Sciences P1 3 June 2019 Common Test 2.2.1 Sound with a frequency range of between 20 kHz to
Page 3: G10-PSC-P1-JUn-2019-and-Memo - Physical Sciences Break 1.0 · 2019. 6. 1. · Physical Sciences P1 3 June 2019 Common Test 2.2.1 Sound with a frequency range of between 20 kHz to
Page 4: G10-PSC-P1-JUn-2019-and-Memo - Physical Sciences Break 1.0 · 2019. 6. 1. · Physical Sciences P1 3 June 2019 Common Test 2.2.1 Sound with a frequency range of between 20 kHz to
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PHYSICAL SCIENCES P1 (PHYSICS)

COMMON TEST

MARKING GUIDELINE

JUNE 2019

NATIONAL SENIOR CERTIFICATE

GRADE 10

MARKS: 100

This marking guideline consists of 7 pages.

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Physical Sciences P1 2 June 2019 Common Test Grade 10 – Marking Guideline

QUESTION 1

1.1 D üü (2)

1.2 C üü (2)

1.3 C üü (2)

1.4 Aüü (2)

1.5 Düü (2)

1.6 Büü (2)

1.7 Cüü (2) [14]

QUESTION 2

2.1.1 No of complete waves passing a fixed point in one sec.üü (2)

2.1.2 f = 1

P T

= 1 0,06 P

= 16,67Hz OR

P

3 waves in 0,18 s P

x waves in 1 s P f = 16,67 s-1 P

(3)

2.1.3 Positive marking from 2.1.2

V = f x l P

= 16,67 x 0,21 P = 3,50 m.s-1 P

(3)

2.1.4 Halved P (1)

2.1.5 Negative Marking from 2.1.4

• Using v = f x l P • If frequency is doubled then wavelength must be halved • In order to keep speed constant P

P

(3)

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Grade 10 – Marking Guideline Physical Sciences P1 3 June 2019 Common Test

2.2.1 Sound with a frequency range of between 20 kHz to 100 kHz PP (2)

2.2.2 Part of the sound energy is absorbed by the sea bed. PP (2)

2.2.3

v

= Δx

P

Δt

1510 = Δx 1,5

Δx = 2265m = 2,27 km P

OR

(3)

v

1510

= Δx P

Δt

= Δx P

3 Dx = 4530

Depth = 4530 2

= 2265m = 2,27km P

2.2.4 Bat P OR Dolphin P (1) [20]

QUESTION 3

3.1 (ANY ONE) • Originate from accelerating electric charges P • Propagate as electric and magnetic fields that are perpendicular to each

other P • Can travel through a vacuum P • Have a speed of 3 x 108 m.s-1- P

(1)

3.2 • some of its behaviour is explained using a wave modelü

• while other aspects of its behaviour are explained using the particle model.P (2)

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P

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Physical Sciences P1 4 June 2019 Common Test Grade 10 – Marking Guideline

3.3.1 E = hf P = (6,63 x 10-34)x(97,60x106) P = 6,47x10-26J P (3)

3.3.2 OPTION 1 c = f x l P

P 8 3 3x 10 = f x (1,50 x 10 ) P f = 200 000Hz

f = 0,20 MHz P ≠ 2 MHz (4)

OPTION 2

Speed = f x l P = 2 x 106 P x 1,5 x 103 P = 3 x 109m.s-1

≠ 3 x 108 m.s-1 P

3.3.3 Increases P (1)

3.3.4 Negative marking from 3.3.3

• E a 1 since (c) remains constant P l

• When l decreases then (E) must increase.P (2) [13]

QUESTION 4

4.1 A region in space where a magnet/ferromagnetic material will experience a force PP (2)

4.2.1 It is ferromagnetic P (1)

4.2.2 P (1)

4.2.3 • Repulsion P • The north pole of A faces the north pole of B P (2)

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N S

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Grade 10 – Marking Guideline

Physical Sciences P1 5 June 2019 Common Test

4.2.4 Positive marking from 4.2.1

• Shape P • Direction P (2)

4.3 Region surrounding the earth in which charged particles are trapped.

PP

(2)

4.4 Streams of charged particles are given off by solar flares. PP (2)

[12]

QUESTION 5

5.1.1 Certain materials become electrically charged after they come into contact with different materials. PP (2)

5.1.2 ● electrons move P

● from the glass ball onto the cloth P (2)

5.1.3 Positively charged P (1)

5.1.4 To the left P (1)

5.2.1 n = Q

P Qe

- 3 x 10-6 -1,6 x 10-19

P

= 1,88 x 10 13 P

(3)

5.2.2 The net charge on an isolated system remains constant PP (2)

5.2.3 OR

Q = Q1 + Q2 P Total charge = -1 µC P P

2 -3 µC P+Qy = -1 µC P

- 0,5 = - 3 + Qy

P 2

Qy = 2 µC P

Qy = + 2µC P (4)

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=

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Physical Sciences P1 6 June 2019 Common Test Grade 10 – Marking Guideline

5.2.4 Gain P (1) [16]

QUESTION 6

6.1 As the number of resistors in series increases so will the current strength decrease PP (2)

6.2 Current strength P (1)

6.3 Electrical energy converted to light/heat energy PP (2)

[5]

QUESTION 7

7.1 emf P Since no current is flowing through the batteryP

(2)

7.2 8 P (1)

7.3.1 4V P (1)

7.3.2 4V P (1)

7.4 0,67 A P (1)

7.5 2C P (1)

7.6 1 = 1 + 1

Rp R1 R2 P

1 = 1 + 1 Rp 3 6 P

Rp = 2Ω P RT = 12Ω P

(4)

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Grade 10 – Marking Guideline Physical Sciences P1 7 June 2019 Common Test

7.7

V = W P

Q

20 = 4800 C P

Q = 240 C Q = IDt P 240 = 2 x Δt P Δt = 120 s

= 2 minutes P (5)

7.8 Apply Negative Marking

• Decreases P • total resistance will increase P • causing total current to decrease P • which causes V3 to decrease since V3 a IP (4)

[20]

TOTAL MARKS: 100

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