This document discusses the system of units used in physics. It defines fundamental and derived quantities, and explains that derived quantities are combinations of fundamental quantities. The seven base SI units are listed as the meter, kilogram, second, ampere, kelvin, mole, and candela. Prefixes are used to modify unit names to indicate multiplicative factors of powers of ten. The document also covers converting between different units, and how to determine the number of significant figures in calculations. Three examples at the end demonstrate applying concepts like unit conversions and significant figures to physics problems.

1.  System of Units
 Fundamental Quantities
 Derived Quantities
 Units
 Prefixes
 Conversion of Units
 Significant Figure

2.  State the definition and differences of based and
derived quantities.
 able to list down the SI Prefixes.
 know how to apply the significant figures
 make a conversion of any units given by using simple
& common method of rational number method

3. SYSTEM OF UNITS
 Basic Quantities
 Derived Quantities
 Units
 Prefixes

4. Physical
Quantities
* Basis of physical
quantities * Combination of
one or more basic
quantity quantities
BASIC
QUANTITIES
DERIVED
QUANTITIES
•Basis of physical quantities
•Example :
Length (m)
Mass (kg)
Time (s)
Temperature (K)
Electric current (A)
•Combination of one or more
basic quantities.
•Example :
Area (m2)
Volume (m3)
Velocity (ms-1)
Acceleration (ms-2)

5. Also known as Base Quantities
5
5
Fundamental Quantities
Quantity Unit Abbreviation
Length (l) meter m
Time (t) second s
Mass (m) kilogram kg
Electric Current (I) ampere A
Temperature (T) kelvin K
Amount of Substance Mole mol
Luminous Intensity candela cd
Table 1: SI Base Quantity and Units

6.  Other quantities which defined in term of seven (7)
fundamental quantities.
 Example:
 Speed
 Work
 Force
 Electric Potential
 Power
 Frequency
 Angle
6
6
Derived Quantities

7. BASIC
QUANTITIES
COMBINATION OF
QUANTITIES
DERIVED
QUANTITIES
Length (Length)2 Area(m2)
Length (Length)3 Volume(m3)
Length, time Length/time Speed(ms-1)
Length, time Length/(time)2 Acceleration(ms-2)
Length, mass Mass/(length)3 Density(kgm-3)
Mass, time (Mass x
length)/(time)2
Force(kgms-2)

8.  Physical quantities measured by using unit.
 Example: Length is a physical quantity.
 1960 – General Conference on Weights and Measures decided on a
universal system of unit called the International System or SI based
on the metric system.
UNITS
Physical Quantity Unit of Measurements Symbol
Length Metre m
Mass Kilogram kg
Time Second s
Electric current Ampere A
Thermodynamic temperature Kelvin K
Amount of substance Mole mol
Luminous intensity Candela cd

9. A way of writing numbers that accommodates values too large or
small to be conveniently written in standard decimal notation.
In scientific notation, numbers are written in the form:
Example:
An electron's mass is about
0.000 000 000 000 000 000 000 000 000 000 910 938 22 kg.
In scientific notation, this is written 9.1093822 10−31 kg.

10.  Used to simplify big numbers.
 Replace powers of ten.
 To make the calculation easier.
 Y, Z, E, h, da, a, z, and y are
rarely used.
PREFIXES

11.  2000 m = 2 x 103 m = 2 km
 0.005 m = 5 x 10-3 m = 5 mm
 45 000 000 bytes = 45 x 106 bytes
= 45 Mbytes
 0.00000008 s = 80 x 10-9 s = 80 ns
 200 mA = 200 x 10-3 A
PREFIXES
Example :

12.  Any quantity can be measured in several different units.
 Hence it is important to know how to convert from one unit to
another.
 Multiplying or dividing an equation by a factor of 1 does not
alter an equation.
 Example: Length: foot / inch / metre
12
12
CONVERSION OF UNITS

13.  3 km = ? m
 1 km = 1000 m
 3 km = 3 x 1000 m
=3000 m
OR
3 km = 3 km x 1000 m
1 km
= 3000 m
Conversion of Units

14.  45 cm = ? km
km4.5x10cm45
km45x10cm45
m1000
1km
cm100
1m
xcm45cm45
4
5
CONVERSION OF UNITS

15.  35 km.hr-1 = ? m.s-1
11
ms9.72km.hr35
s
m
60x60
35x1000
1hr
km35
s60
1min
min60
1hr
km1
m1000
1hr
km35
hr1
km35
CONVERSION OF UNITS

16.  20 kg.m-3 = ? g.cm-3
323
33
3
3
33
3
33
cm.g10x2m.kg20
cm
g
100x100x100
1000x20
m1
kg20
cm100
m1
kg1
g1000
m1
kg20
m1
kg20
cm100
m1
kg1
g1000
m1
kg20
m1
kg20
CONVERSION OF UNITS

17.  The digits that carry meaning contributing to its precision.
 Retain all figures during calculation.
 The leftmost non-zero digit is sometimes called the most
significant digit or the most significant figure.
 The rightmost digit of a decimal number is the least
significant digit or least significant figure.
 Numbers having three significant figures:
587 0.777 0.000999 121000
 Numbers having two significant figures:
16 8.9 0.12 0.0082
17
17
Significant Figures

18. 1. Non zero integers always count as significant
figures.
2. Zeros: There are three classes of zeros.
• Leading zeros
• Captive zeros
• Trailing zeros
18
18
Rules for Significant Figures

19. a) Leading zeros
 Zeros that precede all the non zeros digit
 They do not count as significant figures
 Ex: 0.000562 [3 s.f]
b) Captive zeros
 Zeros between non zeros digits. They always count as
significant figures
 Ex: 13.009 [5 s.f.]
c) Trailing zeros
 Zeros at the end of numbers. They count as significant figures
only if the number contains a decimal point.
 Ex: 200 [1 s.f.]
2.00 [3 s.f]
19
19
Rules for Significant Figures

20. Multiplying or Dividing
• Ex: 16.3 x 4.5 = 73.35
(but the final answer must have 2 s.f.)
Therefore, 16.3 x 4.5 = 73 (2 s.f.)
20
20
Significant figure
for final answer
= the quantity which has the least
number of significant figures
Mathematical Operation For
Significant Figures

21. Adding or Subtracting
• Ex: 12.11 + 8.0 + 1.013 = 31.123
The final answer is 31.1 (1 decimal places)
21
21
Number of decimal
places for final
answer
= the smallest number of decimal
places of any quantity in the
sum
Mathematical Operation For
Significant Figures

22. 1. Ohms law states that V = IR. If V = 3.75 V and I = 0.45 A,
calculate R and express your answer to the correct number
of significant figures.
2. If the resultant force on an object of mass 260 kg is 5.20 x
102 N, use equation F = ma to find acceleration.
3. If a car is traveling at a constant speed 72 km/h for a time
35.5 s, how far has the car traveled? (use distance = speed x
time)
22
22
Exercise 1

23. 1. R = V/I = 3.75/0.45 = 8.3333333Ω
Due to the least s.f. (0.45 = 2 s.f.), thus the answer is 8.3 Ω
2.
Due to the least s.f. (260 = 2 s.f. ), thus the answer is 2.0ms-2
3. Change v=72km/h to m/s => 72km/3600s=20m/s
23
23
2
2
2
260
1020.5
/ ms
x
mFa
mssmtvl 7105.35/20
Due to the least s.f. (72x103m/h = 2 s.f.), thus the answer is
0.71 km or 7.1x102m.
Solutions

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