Area

Area is ameasure of the size of a surface. The S.I unit is a squaremetre (m2).
Other units of area include: π‘π‘š2,π‘šπ‘š2,π‘˜π‘š2 𝑒.𝑑.𝑐

Example 1:
Convert the following as instructed
(i) 15 mm2 to cm2
(ii) 20 m2 to mm2
(iii) 16.4 mm2 to m2

Solution

(i) 15 π‘šπ‘š2 π‘‘π‘œ π‘π‘š2

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(ii) 20 π‘š2 π‘‘π‘œ π‘šπ‘š2

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Solution
(iii) 16.4 π‘šπ‘š2 π‘‘π‘œ π‘š2

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Types of areas
(i) Cross-sectional area
(ii) Surface area

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VOLUME:
Volume is the space occupied by matter. The S.I unit of volume is cubic metre (m3).
Other units of volume include: π‘π‘š3,π‘šπ‘š3,π‘˜π‘š3,π‘šπ‘–π‘™π‘™π‘–π‘™π‘–π‘‘π‘Ÿπ‘’(π‘šπ‘™),π‘™π‘–π‘‘π‘Ÿπ‘’(𝑙).𝑒.𝑑.𝑐

Instruments for measuring Volume include:

  • Measuring cylinder
  • Volumetric flask
  • Burette
  • Pipette
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Experiment to determine the volume of an irregular object

The volume of an irregular object can be obtained by the Displacement method.
Pour water into a measuring cylinder up to a certain level. Record the volume of water (V1).

Tie a thread on the irregular object and gently lower it into the water in the measuring cylinder. Note the new volume of water in the cylinder (V2).

The Volume of the irregular object is then equal to the volume of displaced water; Thus V = (V2 – V1).
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OR
Pour water in an over flow can up to the level of the spout. Place a measuring cylinder just below the spout.

Tie a thread around the irregular object and gently lower it into the overflow can.

Note the volume of water, V that collects in the measuring cylinder. It is equal to the volume of the irregular object.
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Question:
A measuring cylinder containing some water stands on a scale pan. A solid ball is lowered into the water. The water level rises from the 30cm3 mark to 40cm3 mark. The scale reading increases from 100g to 180g.

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What is the density of the material of the ball?
A. 2.0 gcm-3. B. 4.5 gcm-3.
B. 8.0 gcm-3. D. 18 gcm-3.

Example 1:
Convert the following as instructed
(i) 250 ….3 …. ..3 (……)0.032….3 …. ..3
(ii) 500…. …. ..3 (….)10,000 ………… …. ..3

Solution
(i) 250 π‘π‘š3 π‘‘π‘œ π‘š3

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Example 2:
Use the match box below to answer questions that follow.
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Find the volume
(i) in cm3 [400cm3 ]
(ii) in m3 [0.0004m3]

Exercise:

  1. A cuboid has dimensions 2cm by 10cm. Find its width in metre if it occupies a volume of 80cm3. [ 0.04m]
  2. (a) Find the volume of water in a cylinder of water radius 7cm if its height is 10cm.[ 1540cm3]
  3. (b) The volume of the cylinder was 120m3.When a stone was lowered in the cylinder filled with water the volume increased to 15cm3. Find the height of the cylinder of radius 7cm. [0.078 cm]
  4. A Perspex box has 10cm square base and contains water to a height of 10cm. A piece of rock of mass 600g is lowered gently into the water and the level rises to 12cm. Find the;
  5. (i) Volume of water displaced by the rock.
  6. (ii) volume of the rock in π’„π’ŽπŸ‘ and π’ŽπŸ‘
  7. (iii) density of the rock in π’ˆπ’„π’Žβˆ’πŸ‘ and π’Œπ’ˆπ’Žβˆ’πŸ‘

SCIENTIFIC NOTATION AND SIGNIFICANT FIGURES.

Scientific notation or Standard form

A number is in scientific form, when it is written as a vnumber between 1 and 9 which is multiplied by a power of 10. i.e when it is written in the form π€Γ—πŸπŸŽπ§.
Where1≀A<10; i.e A lies between 1 and 10 with 1 inclusive but 10 exclusive. n is an integer (….-2,-1,0,1,2…).
Scientific notation is used for writing down very large and very small measurements.

Example:
(i) 598,000,000m = 5.98 x 108 m
(ii) 0.00000087m = 8.7 x 10-7 m
(iii) 60220m = 6.022 x 104 m

Questions:
Convert the following to scientific form.
(a) 0.048 = 4.8 x 10-2
(b) 34 = 0.75 = 7.5 x 10-1
(c) 1000 = 1.0 x 103
(d) 8.72 = 8.72 x 100
(e) 18 = 0.125 = 1.25 x 10-1

Significant figures

Decimal Places
The number of decimal places (dp) is the number of digits to the right end of a decimal point. E.g. the number 3.6420 is given to 4dp.Thus 3.6420 β‰ˆ 3.642(3dp), 3.6420 β‰ˆ 3.64(2dp), 3.6420 β‰ˆ 3.6(1dp), 3 .6420 β‰ˆ 4 (0dp).

Significant Figures
a) None zero digits (1, 2, 3, 4, 5, 6, 7, 8 and 9) are significant figures.
b) Zeros
Leading zeros (i.e. zeros at the left end of a number) i.e zeros before the first significant figure; are not significant figures e.g. 0.000456 (3s.f), 0.017 (2s.f).

Tapped zeros; zeros between significant figures i.e. zeros between non zero digits are significant figures e.g. 6.0037 (5s.f), 0.0100034 (6 s.f).

Trailing zeros (zeros at the right end of a number);
(i) Trailing after a decimal point: These are significant figures. E.g 2.00 (3s.f), 0.0020 (2s.f), 0.0120700 (6s.f)

Normally these values are obtained by using an instrument.
(ii) Trailing before a decimal point: These are NOT significant figures. E.g 20 (1s.f), 2400 (2s.f), 580100 (4s.f)

Normally these values are obtained as a result of rounding off certain numbers to the nearest tens, fifties, hundreds, thousands, ten thousands e.t.c.

For example, if a number 348 is rounded off to 1 s.f, we get 300 and if it`s rounded off to 2 s.f we get 350. The trailing zeros in these approximations (i.e. 300 and 350) are due to rounding off and therefore are not significant.

Questions

Write the following to the stated significant figures
a) 28.8 to 3 s.f b) 2/7 to 2 s.f c) 4.027 x10*2 to 3 s.f

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Example:
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