Introduction to Physics

Posted on July 9, 2019 by user

Introduction to Physics

1.1.1 Understanding Physics

Physics is a branch of science that studies the

natural phenomena
properties of matter
energy

Field of study in Physics

The field of studies in Physics including

Motion
Pressure
Heat
Light
Waves
Electricity
Magnetism and electromagnetism
Electronics
Nuclear Physics

8.5.1 Alloy (Structured Questions)

Posted on July 5, 2019 by user

Question 1:
Figure 1.1 shows the formation of bronze.

(a) On figure 1.1, name the atoms of metals, P and Q. [2 marks]

(b) Name process W. [1 mark]

(c)(i) State one property of bronze. [1 mark]

(ii) State the effect of the atoms of metal Q in bronze. [1 mark]

(d) The medal in Figure 1.2 is made of bronze.
Give one property of the medal if it is made of metal P only. [1 mark]

Answer:
(a)
P: Copper
Q: Tin

(b)
Alloying

(c)(i) Harder/ can withstand corrosion better than copper.

(c)(ii) The atoms of metal Q in bronze prevent the layers of atoms of metal W in the bronze from sliding easily over one another.

(d) Softer, more malleable, less resistant to corrosion.

Question 2:
Diagram 2.1 shows an experiment to study the resistance of steel alloy towards corrosion.

After 3 days, the result obtained is shown in Diagram 2.2.

(a) Based on Diagram 2.2, write down one observation for this experiment. [1 mark]

(b) Write down one hypothesis for this experiment. [1 mark]

(c) State the variables in this experiment.
(i) Constant variable [1 mark]
(ii) Manipulated variable [1 mark]

(d) Predict the condition of the steel nail on the 5^th day. [1 mark]

(e) Based on this experiment, state the operational definition for an alloy. [1 mark]

Answer:
(a) The iron nail has corroded while the steel nail has not.

(b) Steel is more resistant to corrosion than iron.

(c)(i) Distilled water, number of days the nails are immersed in water.
(any one)

(c)(ii) Type of metal (steel on iron)

(d) The steel nail will remain not corroded

(e) An alloy is a mixture of metal which is harder and more resistant to corrosion than a pure metal.

SPM Science Forecast Paper

Posted on January 2, 2019 by user

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8.4.1 Radio Communication (Structured Questions)

Posted on October 2, 2018 by user

Question 1:
Diagram 1.1 shows three types of electronic components used in a radio receiver.

Diagram 1.1

(a) Name electronic components W and Y. [2 marks]

(b) State the function of electronic component X. [1 mark]

(c) Diagram 1.2 shows a black diagram of a radio receiver.

Diagram 1.2

(i) Write any two electronic components, W, X or Y from Diagram 1.1 into the corresponding blocks provided in Diagram 1.2. [2 marks]

(ii) What type of wave is received by T?
Mark (\/) your answer in the box provided in Diagram 1.3. [1 mark]

Diagram 1.3

Answer:
(a)
W: Diode
Y: Transistor

(b)

Stores electric charges and discharges them at regular intervals when required
To channel the flow of radio frequency carrier waves into the Earth.
When connected in parallel with the inductor, it determines the frequency of the radio wave that will be detected.

(any one)

(c)(i)

(c)(ii)

Question 2:
Table 1 shows the symbols of electronic components.
(a) Complete Table 1 using the name of electronic components given. [3 marks]

Table 1

Diagram 2 shows a block diagram of a radio receiver system.

(b) What is the function of P? [1 mark]

(c) What is Q? [1 mark]

(d) State the energy change that occurs at Q? [1 mark]

Answer:
(a)

(b) Receives modulated radio waves from stations.

(c)(i) Loudspeaker

(c)(ii) Electrical energy → sound energy

7.4.3 Practising Responsible Attitudes in the Disposal of Synthetic Polymers (Structured Questions)

Posted on August 6, 2018 by user

Question 1:
Diagram 1 shows containers which are made of synthetic polymer.

Diagram 1

(a) Name the type of the synthetic polymer shown in Diagram 1. [1 mark]

(b) State two characteristics of the synthetic polymer in Diagram 1. [2 marks]

(c) What is the effect of improper disposal of the synthetic polymer in Diagram 1 on the environment? [1 mark]

(d) State two correct ways to dispose of the synthetic polymer in Diagram 1 to preserve the environment. [2 marks]

Answer:
(a)
Polythene or polyethene

(b)
- Soft when heated and hard when cooled
- Can be moulded more than once

(c)
Improper disposal of the synthetic polymers such as the burning of synthetic polymers releases toxic gases and acidic gases that cause air pollution and acid rain.

(d)
- Recycle or process again
- Burning in special incinerators which does not release toxic gases into the environment

7.4.2 Plastics (Structured Questions)

Posted on August 4, 2018 by user

Question 1:
Table 1 shows the characteristics of plastic A and plastic B.

Table 1

(a) State the type of plastic A and B. [2 marks]

(b) State one example of plastic A and B. [2 marks]

(c) Which plastic can be recycled? [1 mark]

(d) Which plastic is suitable to make as the handle of an iron? [1 mark]

Answer:
(a)
A: Thermoplastic
B: Thermosetting plastic (Thermoset)

(b)
A: Perspex
B: Bakelite

(c)
Perspex

(d)
Bakelite

7.4.1 Synthetic Polymers (Structured Questions)

Posted on August 2, 2018 by user

Question 1:
Table 1.1 shows two types of product made of rubber.
(a) Complete Table 1.1. [4 marks]

Table 1.1

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(b) Table 1.2 shows the characteristics of plastic A and plastic B.

Table 1.2

In Table 1.2, state the type of plastic A and plastic B. [2 marks]

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Answer:
(a)

(b)

15.2.3 Trigonometry, PT3 Focus Practice

Posted on July 11, 2018 by user

Question 10:
A technician needs to climb a ladder to repair a street lamp as shown in Diagram below.

(a) What is the length, in m, of the ladder?
(b) Find the height, in m, of the lamp post.

Solution:

(a)

\begin{array}{l} cos 70^{o} = \frac{1}{h} \\ h = \frac{1}{cos 70^{o}} \\ h = \frac{1}{0.342} \\ h = 2.924 m \\ Hence, the length of the ladder \\ = 2.924 m \end{array}

(b)

\begin{array}{l} tan 70^{o} = \frac{T}{1} \\ T = tan 70^{o} \times 1 \\ = 2.747 \times 1 \\ = 2.747 m \\ The height of the lamp post \\ = 2.747 + 1.2 \\ = 3.947 m \end{array}

15.2.2 Trigonometry, PT3 Focus Practice

Posted on July 10, 2018 by user

Question 6:
In diagram below, AEC and BCD are straight lines. E is the midpoint of AC.

Given \cos x = \frac{5}{13} and \sin y = \frac{3}{5}

(a) find the value of tan x.
(b) Calculate the length, in cm, of BD.

Solution:

\begin{array}{l} (a) \\ Given cos x = \frac{5}{13}, therefore B C = 5, A B = 13 \\ A C = \sqrt{13^{2} - 5^{2}} \\ = \sqrt{169 - 25} \\ = \sqrt{144} \\ = 12 cm \\ tan x = \frac{A C}{B C} = \frac{12}{5} \end{array}

\begin{array}{l} (b) \\ For Δ D C E : \\ \sin y = \frac{3}{5} \\ \frac{E C}{D E} = \frac{3}{5} \\ \frac{E C}{10} = \frac{3}{5} \\ E C = \frac{3}{5} \times 10 = 6 cm \\ D C^{2} = 10^{2} - 6^{2} \\ = 64 \\ D C = 8 cm \\ For Δ A B C : \\ A C = 2 \times 6 = 12 cm \\ \tan x = \frac{12}{5} \\ \frac{12}{C B} = \frac{12}{5} \\ C B = 5 cm \\ B D = D C + C B \\ = 8 cm + 5 cm \\ = 13 cm \end{array}

Question 7:
In diagram below, T is the midpoint of the line PR.

(a) Find the value of tan x^o.
(b) Calculate the length, in cm, of PQ.

Solution:

\begin{array}{l} (a) \\ T R^{2} = 13^{2} - 12^{2} \\ = 169 - 144 \\ = 25 \\ T R = \sqrt{25} \\ = 5 cm \\ \tan x^{o} = \frac{12}{5} \end{array}

\begin{array}{l} (b) \\ P R = 2 \times 5 cm \\ = 10 cm \\ P Q^{2} = 10^{2} - 8^{2} \\ = 100 - 64 \\ = 36 \\ P Q = \sqrt{36} \\ = 6 cm \end{array}

Question 8:
In diagram below, ABE and DBC are two right-angled triangles ABC and DEB are straight lines.

It is given that \cos y^{o} = \frac{3}{5} .

(a) Find the value of tan x^o.
(b) Calculate the length, in cm, of DE.

Solution:

(a) \tan x^{o} = \frac{7}{24}

\begin{array}{l} (b) \\ \cos y^{o} = \frac{B C}{20} \\ \frac{3}{5} = \frac{B C}{20} \\ B C = \frac{3}{5} \times 20 \\ = 12 cm \\ ∴ B D^{2} = 20^{2} - 12^{2} \\ = 400 - 144 \\ = 256 \\ B D = \sqrt{256} \\ = 16 cm \\ D E = 16 - 7 \\ = 9 cm \end{array}

Question 9:
Diagram below shows a vertical pole, PQ. At 2.30 p.m. and 5.00 p.m., the shadow of the pole falls on QR and QS respectively.

Calculate
(a) the height, in m, of the pole.
(b) the value of w.

Solution:
(a)

\begin{array}{l} tan 55^{o} = \frac{Height of the pole}{3.2} \\ Height of the pole = tan 55^{o} \times 3.2 \\ = 1.428 \times 3.2 \\ = 4.57 m \end{array}

(b)

\begin{array}{l} tan w = \frac{4.57}{3.20 + 2} \\ = \frac{4.57}{5.20} \\ = 0.879 \\ w = 41^{o} 18' \end{array}

15.2.1 Trigonometry, PT3 Focus Practice

Posted on July 5, 2018 by user

15.2.1 Trigonometry, PT3 Focus Practice

Question 1:

Diagram below shows a right-angled triangle ABC.

It is given that

\cos x^{o} = \frac{5}{13}

, calculate the length, in cm, of AB.

Solution:

\begin{array}{l} \cos x^{o} = \frac{A B}{A C} \\ \cos x^{o} = \frac{5}{13} \\ \frac{A B}{39} = \frac{5}{13} \\ A B = \frac{5}{13} \times 39 \\ = 15 cm \end{array}

Question 2:

In the diagram, PQR and QTS are straight lines.

It is given that

\tan y = \frac{3}{4}

, calculate the length, in cm, of RS.

Solution:

\begin{array}{l} In △ P Q T, \\ \tan y = \frac{P Q}{Q T} \\ \frac{3}{4} = \frac{6}{Q T} \\ Q T = 6 \times \frac{4}{3} \\ = 8 cm \\ In △ Q R S, Q S = 8 + 8 = 16 cm \\ ∴ R S^{2} = 12^{2} + 16^{2} \leftarrow pythagoras' Theorem \\ = 144 + 256 \\ = 400 \\ R S = \sqrt{400} \\ = 20 cm \end{array}

Question 3:

In the diagram, PQR is a straight line.

It is given that

\cos x^{o} = \frac{3}{5}

, hence sin y^o =

Solution:

\begin{array}{l} \cos x^{o} = \frac{P Q}{P S} \\ \frac{P Q}{10} = \frac{3}{5} \\ P Q = \frac{3}{5} \times 10 \\ = 6 cm \\ Q R = P R - P Q \\ = 21 - 6 \\ = 15 cm \end{array}

\begin{array}{l} Q S^{2} = 10^{2} - 6^{2} \leftarrow pythagoras' Theorem \\ = 100 - 36 \\ = 64 \\ Q S = \sqrt{64} \\ = 8 cm \\ R S^{2} = 15^{2} + 8^{2} \\ = 225 + 64 \\ = 289 \\ R S = \sqrt{289} \\ = 17 cm \\ \sin y^{o} = \frac{15}{17} \end{array}

Question 4:

Diagram below consists of two right-angled triangles.

Determine the value of cos x^o.

Solution:

\begin{array}{l} A C = \sqrt{13^{2} - 12^{2}} \\ = \sqrt{25} \\ = 5 cm \\ C D = \sqrt{5^{2} - 3^{2}} \\ = \sqrt{16} \\ = 4 cm \\ \cos x^{o} = \frac{C D}{A C} \\ = \frac{4}{5} \end{array}

Question 5:

Diagram below consists of two right-angled triangles ABE and DBC.

ABC and EBD are straight lines.

It is given that

s i n x^{o} = \frac{5}{13} and \cos y^{o} = \frac{3}{5} .

(a) Find the value of tan x^o.

(b) Calculate the length, in cm, of ABC.

Solution:

(a)

\begin{array}{l} s i n x^{o} = \frac{5}{13}, D C = 13 cm \\ B C = \sqrt{13^{2} - 5^{2}} \\ = \sqrt{144} \\ = 12 cm \\ Thus, \tan x^{o} = \frac{5}{12} \end{array}

(b)

\begin{array}{l} \cos y^{o} = \frac{A B}{15} \\ \frac{3}{5} = \frac{A B}{15} \\ A B = 9 cm \\ Thus, A B C = 9 + 12 \\ = 21 cm \end{array}