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YEAR 9 · INDICES AND SCIENTIFIC NOTATION · 15 MIN

Measurement and time-scale applications

Use scientific notation with compatible measurement units and sensible interpretations of time and scale.

In this lesson

Before you start
  • Scientific calculations
  • Metric units and speed
01 / LEARN

Understand the idea.

Numbers and units must travel together

A numerical value needs its unit to describe a measurement. For example, 1 mm=10−3 m and 1 micrometre=10−6 m. Changing units changes the numerical value but preserves the physical length.

For magnification, divide drawing length by actual length after converting both to the same unit. The units then cancel. For travel time, match the distance unit to the distance unit in the speed before dividing distance by speed.

Interpret the result, not just the exponent

Interpret the calculated scale in familiar terms: 2.5×103 s is 2500 s, or approximately 41.7 min. Retain enough digits during conversion and round the final quantity only to the requested precision.

A formula also carries assumptions. Travel time from distance divided by speed assumes that speed applies throughout the journey. Multiplying sheet thickness by a sheet count assumes identical sheets and no extra gaps; the model may differ from a physical stack.

02 / WORKED EXAMPLES

See the method in action.

EXAMPLE 01

Compare a drawing with a real object

A cell is 8×10−6 m long. Its drawing is 4 cm long. Find the drawing magnification.

Worked solution

  1. 1

    Convert the drawing length to metres.

    4 cm=4×10−2 m

    One centimetre is one hundredth of a metre.

  2. 2

    Form the ratio in matching units.

    M=4×10−2 m8×10−6 m

    Magnification is drawing length divided by actual length; metres cancel.

  3. 3

    Divide coefficients and subtract exponents.

    M=48×10−2−(−6)=0.5×104

    Division by a power subtracts its exponent, including the negative sign.

  4. 4

    Interpret the dimensionless result.

    M=5×103=5000

    The drawing is 5000 times as long as the cell.

Answer

5000 times

EXAMPLE 02

Find travel time

A signal travels 9×107 m at 3×108 m/s. How long does the journey take?

Worked solution

  1. 1

    Use the constant-speed relationship.

    t=dv

    Distance equals speed multiplied by travel time.

  2. 2

    Substitute compatible quantities.

    t=9×107 m3×108 m/s

    The metres match, leaving seconds after division.

  3. 3

    Simplify the coefficient and power.

    t=93×107−8 s=3×10−1 s

    The coefficients divide to 3 and the exponent difference is -1.

  4. 4

    Express the time as a decimal.

    t=0.3 s

    Multiplication by 10 to the power -1 divides 3 by 10.

Answer

0.3 s

03 / YOUR TURN

Try it yourself.

Check your understanding

A sheet is 9×10−5 m thick. Find the thickness of 1000 sheets, ignoring gaps.

Show answer & reasoning

0.09 m

The total is (9×10−5)⁢(103) m=9×10−2 m=0.09 m, assuming all 1000 sheets have the stated thickness and there are no gaps.

Keep practising

6 worksheet questions · 15 marks

Read the questions online
  1. Write 3.5 km in metres using scientific notation.
    2 marks
  2. Write 0.0042 m in millimetres.
    2 marks
  3. A particle is 2.5×10−6 m wide. Express its width in micrometres.
    2 marks
  4. A beam travels 1.8×106 m in 6×10−3 s. Find its constant speed.
    3 marks
  5. A model is 15 cm long and represents a real object 30 m long. Find the scale factor from real object to model.
    3 marks
  6. A sensor records 2×104 readings each second. How many readings does it record in 5 minutes?
    3 marks
Read the worked solutions

Equivalent correct methods earn the corresponding marks. Include units where asked.

  1. Write 3.5 km in metres using scientific notation.
    2 marks

    1 km =103 m, so multiply 3.5 by 103.

    3.5×103 m
  2. Write 0.0042 m in millimetres.
    2 marks

    1 m =1000 mm.

    0.0042×1000=4.2.

    4.2 mm
  3. A particle is 2.5×10−6 m wide. Express its width in micrometres.
    2 marks

    1 micrometre =10−6 m.

    Divide 2.5×10−6 m by 10−6 m per micrometre.

    2.5 micrometres
  4. A beam travels 1.8×106 m in 6×10−3 s. Find its constant speed.
    3 marks

    v=d÷t=1.86×106−(−3)=0.3×109.

    Normalise to 3×108 m/s.

    3×108 m/s
  5. A model is 15 cm long and represents a real object 30 m long. Find the scale factor from real object to model.
    3 marks

    Convert 30 m to 3000 cm.

    15÷3000=1200.

    1200
  6. A sensor records 2×104 readings each second. How many readings does it record in 5 minutes?
    3 marks

    5 minutes =300=3×102 seconds.

    (2×104)⁢(3×102)=6×106.

    6×106 readings
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Curriculum reference & lesson scope

Australian Curriculum v9 / SA Curriculum

Year 9 numerical integer exponent laws and their extension to variables; scientific notation and practical measurement applications. Excludes fractional-index and full surd algebra taught in Year 10.

Official curriculum source ↗

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