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M1 Numbers, units and estimation

handle huge and tiny numbers, units and quick estimates with confidence.

Before you begin: school arithmetic. Start here after the preface.

99 steps about 124 min

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Play: M1 · Numbers, Units and Estimation
Chapters
  • 0:00 From a proton to the universe
  • 1:11 Powers of ten
  • 2:34 Scientific notation
  • 4:27 A ladder of prefixes
  • 6:22 A number needs a unit
  • 7:42 Where the metre came from
  • 9:09 The kilogram in a vault
  • 10:47 Converting units safely
  • 12:44 Dimensional analysis
  • 14:21 A formula from units alone
  • 15:57 Fermi's paper scraps
  • 17:05 How to estimate anything
  • 18:27 Estimates for everyday life
  • 19:47 Accuracy and precision
  • 21:14 Random and systematic errors
  • 22:34 Significant figures
  • 23:46 Common mistakes
  • 25:00 Recap and what's next

Notes

In this chapter

  • Write very large and very small numbers in scientific notation, and multiply and divide them by adding and subtracting powers.
  • Use the metric prefixes, from nano to giga.
  • Convert units safely with conversion factors, including squared and cubed units.
  • Use dimensional analysis to check an equation, and even to find the form of a formula.
  • Make Fermi estimates: break a "how many?" question into steps and land within a factor of ten.
  • Tell accuracy from precision and random from systematic error, and report results with error bars and sensible significant figures.

Powers of ten

FIRST, MAKE A PREDICTION

Which number does 10 to the power −3 represent?

Show the answer

0.001

A negative exponent takes the reciprocal: 10⁻³ = 1/1000. The value is positive.

TRY IT / FOLLOW THE REASONING

Move the power down one step at a time. Describe what happens to the value, then move through zero into positive powers.

Move one power at a time

Power of ten: 0

Value
1
Model assumptions

A reference value of 1. Negative exponents still give positive numbers.

MAKE THE CONNECTION

Key idea

A power of ten counts zeros. A negative power is a small number, not a negative one.

\(10^n\) means ten multiplied by itself \(n\) times. For whole positive powers, \(n\) is the number of zeros after the 1: \(10^6 = 1\,000\,000\). Each step down divides by ten, so \(10^0 = 1\), \(10^{-1} = 0.1\) and \(10^{-3} = 0.001\).

Powers of ten

The graph of ten to the power x for x from −3 to 3. It is always above zero: for negative x it is a tiny positive number, and it climbs to 1, 10, 100 at x = 0, 1, 2, then 1000 at x = 3.−3−2−1123250500750100010⁰10¹10²10ˣ
The curve of ten to the power x. It climbs steeply to the right and only creeps towards zero on the left: it never goes below zero.

Checkpoint: Powers of ten (1 of 2)

Question 1 of 2 Commit to an answer first, then look.

Write \(10^{-2}\) as an ordinary decimal number.

Show the answer

0.01

\(10^{0}=1\), \(10^{-1}=0.1\) and \(10^{-2}=0.01\). A negative power makes a small positive number, not a negative one.

Checkpoint: Powers of ten (2 of 2)

Question 2 of 2 Commit to an answer first, then look.

Which number is the largest?

Show the answer

\(10^{3}\)

Positive powers of ten grow with the power and negative ones shrink towards zero, so \(10^3 = 1000\) is the largest.

Scientific notation

FIRST, MAKE A PREDICTION

Which expression represents 300 000 000 in standard scientific notation?

Show the answer

3 × 10⁸

The leading number is between 1 and 10; moving the decimal eight places reconstructs the original number.

TRY IT / FOLLOW THE REASONING

Write 300 000 000 as 3, then record how many multiplications by ten restore the original. Try the same reasoning with 0.003 before opening the explanation.

MAKE THE CONNECTION

A number between 1 and 10, times a power of ten: \(300\,000\,000 = 3 \times 10^8\). To convert, move the decimal point until one digit is in front of it and count the places: left gives a positive power, right gives a negative one.

If the front number comes out at 10 or more, tidy it up: \(20\times10^5 = 2\times10^6\).

Checkpoint: Scientific notation (1 of 2)

Question 1 of 2 Commit to an answer first, then look.

Which of these is 0.00072 in scientific notation?

Show the answer

\(7.2\times10^{-4}\)

Moving the decimal point four places right gives 7.2, and a right move is a negative power, so \(0.00072 = 7.2\times10^{-4}\).

Checkpoint: Scientific notation (2 of 2)

Question 2 of 2 Commit to an answer first, then look.

A sheet of paper is \(1\times10^{-4}\) m thick. How many sheets make a stack 1 cm, or \(1\times10^{-2}\) m, high?

Show the answer

100

\(\dfrac{10^{-2}}{10^{-4}} = 10^{-2-(-4)} = 10^{2} = 100\) sheets.

Multiplying and dividing

FIRST, MAKE A PREDICTION

What is (2 × 10³) × (3 × 10²)?

Show the answer

6 × 10⁵

Multiply 2 by 3, then add the exponents because the powers of ten multiply.

TRY IT / FOLLOW THE REASONING

Expand 10³ × 10² into individual factors of ten. Count the factors, then compare multiplication with cancelling factors in a division.

\[ \begin{gathered} (a \times 10^m)\times(b \times 10^n) \\ = (a\times b)\times 10^{m+n} \\[.6em] \frac{a\times10^m}{b\times10^n} = \frac{a}{b}\times10^{m-n} \end{gathered} \]

MAKE THE CONNECTION

  • \(a, b\): the front numbers, between 1 and 10
  • \(m, n\): the powers of ten

Checkpoint: Multiplying and dividing (1 of 3)

Question 1 of 3 Commit to an answer first, then look.

Light travels at \(3\times10^8\) m/s. How far does it go in \(2\times10^3\) s?

m

Show the answer

6×10^11 m

Distance is speed times time. Multiply the front numbers, 3 × 2 = 6, and add the powers of ten, 8 + 3 = 11, to get \(6\times10^{11}\) m.

Checkpoint: Multiplying and dividing (2 of 3)

Question 2 of 3 Commit to an answer first, then look.

Work out \((5\times10^{3})\times(4\times10^{2})\), and give the value as a plain number.

Show the answer

2×10^6

\(5\times4=20\) and \(3+2=5\), so the product is \(20\times10^{5}\). Tidying up the front number gives \(2\times10^{6}\).

Checkpoint: Multiplying and dividing (3 of 3)

Question 3 of 3 Commit to an answer first, then look.

Work out \(\dfrac{6\times10^{10}}{2\times10^{4}}\).

Show the answer

3×10^6

\(6\div2=3\) and \(10-4=6\), so the answer is \(3\times10^{6}\).

Metric prefixes

FIRST, MAKE A PREDICTION

Which is longer: 2 millimetres or 500 micrometres?

Show the answer

2 millimetres

2 mm is 2000 micrometres, so it is four times as long as 500 micrometres.

TRY IT / FOLLOW THE REASONING

Use the prefix table to write both lengths in metres. Circle the exponent before comparing the leading numbers. Find another pair where the bigger written number is the smaller length.

Prefix Symbol Power Example
giga G \(10^9\) a gigabyte
mega M \(10^6\) a megawatt
kilo k \(10^3\) a kilometre
milli m \(10^{-3}\) a millimetre
micro µ \(10^{-6}\) a micrometre (a hair is about 70 µm wide)
nano n \(10^{-9}\) a nanometre

MAKE THE CONNECTION

2 mm is 2000 micrometres, so it is four times as long as 500 micrometres.

Checkpoint: Metric prefixes (1 of 2)

Question 1 of 2 Commit to an answer first, then look.

Which of these is the longest length?

Show the answer

2 mm

Compare the powers of ten first. \(10^{-3}\) beats \(10^{-4}\), \(10^{-6}\) and \(10^{-8}\), so 2 mm is the longest.

Checkpoint: Metric prefixes (2 of 2)

Question 2 of 2 Commit to an answer first, then look.

How many nanometres are there in one micrometre?

Show the answer

1000

\(\dfrac{10^{-6}}{10^{-9}} = 10^{3} = 1000\), so one micrometre is a thousand nanometres.

Forty-two powers of ten

FIRST, MAKE A PREDICTION

From a length of 10⁻¹⁵ m to 10²⁷ m, how many factors of ten separate the two?

Show the answer

42

Divide the lengths: 10²⁷/10⁻¹⁵ = 10⁴², so the exponents differ by 42.

TRY IT / FOLLOW THE REASONING

Place the proton, a metre and the observable universe on an exponent line. Compare the ratio of the largest to the smallest without writing out either length.

MAKE THE CONNECTION

From a proton (about \(10^{-15}\) m) to the observable universe (about \(10^{27}\) m) is a factor of \(10^{42}\).

Forty-two powers of ten

A number line of sizes in powers of ten, from −15 to 27, so 42 powers of ten. Marked: proton at about 10⁻¹⁵ m, hair at about 10⁻⁴ m, person at about 10⁰ m, Earth at about 10⁷ m, universe at about 10²⁷ m.−15−9−339152127protonhairpersonEarthuniverse
Physics spans 42 powers of ten. Each tick is six powers of ten, so equal steps along the line are enormous jumps in size.

Checkpoint: Forty-two powers of ten

Question Commit to an answer first, then look.

Each tick on the line of sizes is six powers of ten apart. How many ticks does it take to go from a proton, \(10^{-15}\) m, to the observable universe, \(10^{27}\) m?

Show the answer

7

From \(-15\) to \(27\) is 42 powers of ten, and \(42\div6=7\) gaps between the ticks.

Units

FIRST, MAKE A PREDICTION

A measurement is reported as ‘12’. Can you tell whether it describes a length, a time or a mass?

Show the answer

No; the unit is needed

The number tells how many units were counted; the unit identifies the physical quantity and scale.

TRY IT / FOLLOW THE REASONING

Give three different physical meanings to the number 12 by changing its unit. Use the SI list to identify which measurements are lengths, times and masses.

MAKE THE CONNECTION

Key idea

A number without a unit is incomplete.

The international system (SI) has seven base units: the second, metre, kilogram, ampere, kelvin, mole and candela. Since 20 May 2019, all seven have been defined by fixing seven constants of nature, such as the speed of light and the Planck constant. Every other unit (the newton, the watt, the pascal…) is built from these seven.

Checkpoint: Units (1 of 2)

Question 1 of 2 Commit to an answer first, then look.

Which of these is not one of the seven SI base units?

Show the answer

newton

The seven base units are the second, metre, kilogram, ampere, kelvin, mole and candela. The newton is built from three of them, as kg m/s².

Checkpoint: Units (2 of 2)

Question 2 of 2 Commit to an answer first, then look.

A force is mass times acceleration. What force, in newtons, accelerates a 2.0 kg mass at 3.0 m/s²?

N

Show the answer

6 N

\(F = ma = 2.0\times3.0 = 6.0\) kg m/s², and one kg m/s² is one newton. A derived unit is a product of base units.

Converting units

FIRST, MAKE A PREDICTION

When 2 kilometres is written in metres, what happens to the number and the actual length?

Show the answer

The number becomes 2000; the length stays the same

Metres are smaller units, so more of them describe the same length.

TRY IT / FOLLOW THE REASONING

Write 1000 m / 1 km as a fraction equal to one. Choose its orientation so kilometres cancel in 2 km, then reverse the conversion as a check.

MAKE THE CONNECTION

Key idea

A conversion factor is a fraction equal to one, like \(\dfrac{1000\ \text{m}}{1\ \text{km}}\). Multiply by it, and cancel units like algebra.

Converting units

\[ \begin{gathered} 100\ \frac{\text{km}}{\text{h}} \times \frac{1000\ \text{m}}{1\ \text{km}} \times \frac{1\ \text{h}}{3600\ \text{s}} \\ \approx 28\ \text{m/s} \end{gathered} \]

Checkpoint: Converting units

Question Commit to an answer first, then look.

Convert each quantity to SI units (metres, or metres per second).

2.5 km in metres.

m

350 mm in metres.

m

90 km/h in metres per second.

m/s

Show the answer
  1. 2500 m
  2. 0.35 m
  3. 25 m/s

Kilo means \(\times1000\) and milli means \(\div1000\). For speed, \(90\ \text{km/h} = 90\times1000\ \text{m} / 3600\ \text{s} = 25\) m/s.

Squared and cubed units

FIRST, MAKE A PREDICTION

A square is 1 metre on each side. What is its area in square centimetres?

Show the answer

10 000 cm²

Each side is 100 cm, so the area is 100 × 100 square centimetres.

TRY IT / FOLLOW THE REASONING

Change the length multiplier and compare the area and volume multipliers. Then use the same reasoning for a 100-fold change of length unit.

Scale the same shape

Length multiplier: 1 ×

Area multiplier
1 ×
Volume multiplier
1 ×
Model assumptions

Similar shapes: every length changes by the same factor. Bars use a common scale.

MAKE THE CONNECTION

For squared or cubed units, square or cube the factor: \(1\ \text{m}^2 = 10\,000\ \text{cm}^2\), and \(1\ \text{m}^3 = 10^6\ \text{cm}^3 = 1000\) litres.

Checkpoint: Squared and cubed units (1 of 2)

Question 1 of 2 Commit to an answer first, then look.

How many square millimetres are there in one square centimetre?

Show the answer

100

\(1\ \text{cm} = 10\ \text{mm}\), so \(1\ \text{cm}^2 = 10^2\ \text{mm}^2 = 100\ \text{mm}^2\). Squaring the factor is the whole trick.

Checkpoint: Squared and cubed units (2 of 2)

Question 2 of 2 Commit to an answer first, then look.

A tank holds 3 m³ of water. How many litres is that?

L

Show the answer

3000 L

\(1\ \text{m}^3 = 1000\) litres, so \(3\ \text{m}^3 = 3000\) litres.

Density: a derived unit

FIRST, MAKE A PREDICTION

Two solid blocks are made from the same uniform material. One has twice the volume. What happens to its mass and density?

Show the answer

Twice the mass, the same density

Mass and volume grow together for the same uniform material, so their ratio stays constant.

TRY IT / FOLLOW THE REASONING

Imagine 1 litre of water, then join it to an identical litre. Calculate mass divided by volume before and after joining them, keeping the units visible.

MAKE THE CONNECTION

Many useful quantities are ratios of two others, and their units are built the same way. Density is mass per unit volume, \(\rho = m/V\), in kilograms per cubic metre (kg/m³).

  • Water is about 1000 kg/m³ (one litre has a mass of one kilogram).
  • Air is about 1.2 kg/m³.
  • Steel is about 7850 kg/m³.

Density: a derived unit

A density-units story: the Gimli Glider

The Gimli Glider (see History) is a density-units story: the fuel was worked out with 1.77 pounds per litre where 0.803 kilograms per litre was needed, so the plane took off with about half the fuel it needed.

Density: a derived unit

Density is taught properly, with practice, in P6 · Matter in bulk.

Checkpoint: Density: a derived unit (1 of 2)

Question 1 of 2 Commit to an answer first, then look.

A block of wood of volume 0.50 m³ has a mass of 350 kg. What is its density?

kg/m^3

Show the answer

700 kg/m^3

\(\rho = m/V = 350/0.50 = 700\) kg/m³. That is less than water's 1000 kg/m³, so the wood floats.

Checkpoint: Density: a derived unit (2 of 2)

Question 2 of 2 Commit to an answer first, then look.

Water has a density of 1000 kg/m³. What is the mass of 3.0 litres of water, in kilograms?

kg

Show the answer

3 kg

\(3.0\ \text{L} = 3.0\times10^{-3}\ \text{m}^3\), so \(m = \rho V = 1000\times3.0\times10^{-3} = 3.0\) kg. A litre of water is a kilogram.

Dimensional analysis

FIRST, MAKE A PREDICTION

Which expression could describe a distance if v is speed and t is time?

Show the answer

v × t

Speed has dimension length/time; multiplying by time leaves length.

TRY IT / FOLLOW THE REASONING

Write the units beside each candidate expression. Cross out the impossible ones, then ask whether having the right units is enough to guarantee the right numerical factor.

MAKE THE CONNECTION

Key idea

Every quantity has a dimension: length \(L\), mass \(M\), time \(T\) and so on. You can only add or compare things with the same dimensions, so both sides of a correct equation have the same dimensions.

  • \(d = \tfrac12 g t^2\): \([g t^2] = \dfrac{L}{T^2}\times T^2 = L\). ✓
  • \(d = g t\): \([g t] = \dfrac{L}{T}\), a speed, not a length. ✗

Checkpoint: Dimensional analysis (1 of 2)

Question 1 of 2 Commit to an answer first, then look.

A ball is dropped from height \(h\) in gravity \(g\). Using only dimensions, how must the fall time \(t\) depend on \(h\) and \(g\)? Write your guess.

Show the solution

\(h/g\) has dimension \(L \div (L/T^2) = T^2\), so \(t \propto \sqrt{h/g}\). Dimensions cannot tell you the numerical factor; the full result is \(t=\sqrt{2h/g}\).

Checkpoint: Dimensional analysis (2 of 2)

Question 2 of 2 Commit to an answer first, then look.

A distance \(d\) is covered in a time \(t\). Which of these could be a formula for a speed?

Show the answer

\(d/t\)

Only \(d/t\) has dimension \(L/T\). Dimensions rule options out quickly, though they cannot fix a numerical factor.

A formula from units alone

FIRST, MAKE A PREDICTION

For a simple pendulum at small amplitude, which combination of length L and acceleration g has units of time?

Show the answer

Square root of L/g

L/g has units of time squared, so its square root has units of time.

TRY IT / FOLLOW THE REASONING

Replace L and g by their dimensions. Find the combination that leaves only time, then compare with the full pendulum formula to identify what units alone could not supply.

MAKE THE CONNECTION

Worked number: the pendulum

The period \(T\) can depend only on \(L\), \(m\) and \(g\). Nothing else contains mass, so \(m\) can't appear; and \(\sqrt{L/g}\) has dimension \(T\). So \(T \propto \sqrt{L/g}\).

The full result is \(T = 2\pi\sqrt{L/g}\), about 2 s for a 1 m pendulum. Units can't supply the \(2\pi\).

A formula from units alone

The period of a pendulum against its length. It rises as a square root: a 1 m pendulum takes 2 s, and four times the length takes twice as long, 4 s.123412341 m: 2 speriod T (s) against length L (m)
The period of a pendulum grows as the square root of its length: four times the length takes only twice as long.

A formula from units alone

Myth: a heavier pendulum swings more slowly

The mass drops out. For small swings, heavy and light pendulums of the same length keep time together, as long as air resistance is small.

Checkpoint: A formula from units alone (1 of 2)

Question 1 of 2 Commit to an answer first, then look.

Use \(T = 2\pi\sqrt{L/g}\) with \(g = 9.81\) m/s². What is the period of a pendulum 0.25 m long, in seconds?

s

Show the answer

1.00303334036 s

\(L/g = 0.25/9.81 \approx 0.0255\), its square root is about 0.160, and \(2\pi\times0.160\approx1.0\) s. A quarter of the length takes half the time, as the graph says.

Checkpoint: A formula from units alone (2 of 2)

Question 2 of 2 Commit to an answer first, then look.

The bob of a pendulum is swapped for one twice as heavy, on the same string. What happens to the period?

Show the answer

It stays the same.

The only combination of \(L\), \(m\) and \(g\) with dimension \(T\) is \(\sqrt{L/g}\), so the mass cannot matter.

Fermi estimates

FIRST, MAKE A PREDICTION

To estimate the number of piano tuners in a city, which is the more useful first move?

Show the answer

Split the question into population, piano ownership and tuning workload

A chain of roughly known factors turns an unfamiliar total into estimates you can inspect and improve.

TRY IT / FOLLOW THE REASONING

Build your own factor chain for the tuners question. Attach a unit to each factor, cancel down to people, and test which assumption changes your result most.

MAKE THE CONNECTION

Key idea

Break a question you can't answer into factors you can guess, round each, keep the units, multiply with powers of ten, and check that the answer is reasonable. The errors partly cancel, so the answer usually lands within a factor of ten.

For example:

  • Piano tuners in 1940s Chicago: about 100.
  • The air in a 4 × 5 × 2.5 m room: about 60 kg.
  • Breaths in a lifetime: about \(6\times10^8\).
  • Heartbeats in a lifetime: about \(3\times10^9\), roughly five per breath.

Fermi estimates

Two Fermi estimates on a line of powers of ten: breaths in a lifetime about 6×10⁸, heartbeats in a lifetime about 3×10⁹, five heartbeats per breath, so about 0.7 of a power of ten apart.88.599.510breathsheartbeats
Breaths and heartbeats in a lifetime on a line of powers of ten. Five times as many heartbeats is less than one tick further along.

Checkpoint: Fermi estimates (1 of 2)

Question 1 of 2 Commit to an answer first, then look.

A resting adult breathes about 15 times a minute. About how many breaths is that in one day?

Show the answer

21600

\(15\times60\times24 = 21\,600\), about \(2\times10^4\) breaths a day. Over 80 years that is roughly \(6\times10^8\), the figure in the Notes.

Checkpoint: Fermi estimates (2 of 2)

Question 2 of 2 Commit to an answer first, then look.

How many grains of rice would fill a bathtub? What is the best first step?

Show the answer

Estimate the volume of the bath and the volume of one grain, then divide.

Bath volume is about 0.2 m³ and a grain is about 0.02 mL, so the answer is of order \(10^7\). The steps are what make it believable.

Accuracy and precision

FIRST, MAKE A PREDICTION

A scale repeatedly reads 1.20 kg for a known 1.00 kg mass. How would you describe the readings?

Show the answer

Precise, but inaccurate

The readings agree closely with one another, but they are displaced from the known value.

TRY IT / FOLLOW THE REASONING

Sketch repeated measurements as dots around a target. Make one cluster tight but off-centre, and another spread around the centre. Match the sketches to accuracy and precision.

MAKE THE CONNECTION

Key idea

Accuracy: how close a result is to the true value. Precision: how close repeated results are to each other.

Accuracy and precision

accurateand precise precise,not accurate accurate,not precise neither a systematic errorrandom errors
Precise means tightly grouped; accurate means centred on the truth. A systematic error shifts the whole group; random errors spread it out.

Checkpoint: Accuracy and precision

Question Commit to an answer first, then look.

Five readings of a true length of 10.00 cm come out as 10.51, 10.50, 10.52, 10.51 and 10.50 cm. What do they show?

Show the answer

Precise but not accurate, which points to a systematic error.

Tight grouping means precise. Every reading is high by about 0.5 cm, so something shifts them all the same way, such as a zero error on the ruler. Averaging will not fix it, but calibration will.

Random and systematic errors

FIRST, MAKE A PREDICTION

A scale has a fixed +0.20 kg offset. Will averaging many readings remove that offset?

Show the answer

No; the offset remains

A systematic shift moves every reading the same way, so averaging preserves the bias.

TRY IT / FOLLOW THE REASONING

Write three readings with positive and negative random deviations, then three with the same offset. Compare their means and decide which problem calls for calibration.

MAKE THE CONNECTION

  • Random errors scatter results either way; averaging repeated measurements reduces them.
  • Systematic errors push every result the same way; averaging doesn't help, but calibration does.

Checkpoint: Random and systematic errors (1 of 2)

Question 1 of 2 Commit to an answer first, then look.

A kitchen scale reads 5 g too high every time, whatever you put on it. What kind of error is that?

Show the answer

A systematic error, which calibration will remove.

The same shift on every reading is the signature of a systematic error. Weighing more times will not help, but checking the scale against a known mass will.

Checkpoint: Random and systematic errors (2 of 2)

Question 2 of 2 Commit to an answer first, then look.

Four timings of a swing are 9.6 s, 9.8 s, 9.7 s and 9.9 s. What is their average, in seconds?

s

Show the answer

9.75 s

\((9.6+9.8+9.7+9.9)/4 = 39.0/4 = 9.75\) s. Averaging smooths the random scatter, though it can never fix a systematic shift.

Error bars

FIRST, MAKE A PREDICTION

A result is 9.7 ± 0.2 m/s². Is 9.81 m/s² inside the quoted interval?

Show the answer

Yes; the interval runs from 9.5 to 9.9

The accepted value lies inside the stated range, so the result is consistent with it at that quoted uncertainty.

TRY IT / FOLLOW THE REASONING

Draw the interval on a number line and place the accepted value on it. Compare this result with a tighter interval centred at the same value; state what the quoted uncertainty means before treating it as a probability.

MAKE THE CONNECTION

Error bars: \(g = 9.7 \pm 0.2\ \text{m/s}^2\) means the true value probably lies between 9.5 and 9.9. The accepted 9.81 is inside, so the measurement agrees.

Error bars

Two results for g in m/s². Ana: 9.6 ± 0.3, from 9.3 to 9.9, which contains the accepted 9.81. Ben: 10.4 ± 0.1, from 10.3 to 10.5, a narrower bar that misses it.accepted 9.8199.51010.511AnaBen
Two students measure g. Ana's bar reaches the accepted 9.81, so her result agrees. Ben's bar is narrower, so he is more precise, but it stops short: a sign of a systematic error.

Checkpoint: Error bars (1 of 2)

Question 1 of 2 Commit to an answer first, then look.

A student reports \(g = 9.7 \pm 0.2\ \text{m/s}^2\). What is the largest value the result allows, in m/s²?

m/s^2

Show the answer

9.9 m/s^2

The range is 9.5 to 9.9 m/s². The accepted 9.81 sits inside it, so the measurement agrees.

Checkpoint: Error bars (2 of 2)

Question 2 of 2 Commit to an answer first, then look.

Ben reports \(g = 10.4 \pm 0.1\ \text{m/s}^2\). Does his result agree with the accepted 9.81?

Show the answer

No, because his range of 10.3 to 10.5 does not contain 9.81.

His range, 10.3 to 10.5, stops short of 9.81, so his result does not agree. He is precise but not accurate, a hint of a systematic error.

Significant figures

FIRST, MAKE A PREDICTION

How many significant figures are there in 0.00420?

Show the answer

Three

The digits 4, 2 and the final zero carry information; the leading zeros only locate the decimal point.

TRY IT / FOLLOW THE REASONING

Compare 0.0042 with 0.00420 in scientific notation. Decide which digits locate the scale and which tell a reader how precisely the value was recorded.

MAKE THE CONNECTION

Significant figures are the digits that carry real information. Leading zeros don't count. Give answers to about as many significant figures as your least precise measurement.

Significant figures

Myth: more digits means a more accurate answer

Digits beyond what you measured are false precision: they claim knowledge you don't have.

Checkpoint: Significant figures

Question Commit to an answer first, then look.

How many significant figures does 0.00560 have?

Show the answer

3

Leading zeros only place the decimal point. The 5, 6 and the final 0 are real digits, so 0.00560 has three significant figures.

Key terms

Key terms

Exponent (power)
The small raised number: how many times ten is multiplied (or, if negative, divided).
Scientific notation
A number from 1 to 10 times a power of ten.
Conversion factor
A fraction equal to one, used to change units without changing the quantity.
Dimension
The kind of quantity (length, mass, time…), whatever the units.

Key terms

Fermi problem
An estimate built from simple, rough steps.
Accuracy / precision
Closeness to the true value / closeness of repeated results to each other.
Random / systematic error
Scatter either way / a shift in the same direction every time.
Significant figures
The digits that carry real information.

Key terms

False precision
More digits than a measurement supports.

History and sources

Charles and Ray Eames (Chicago, 1977)

Charles and Ray Eames (Chicago, 1977). The film Powers of Ten zooms out from a picnic by a factor of ten every ten seconds, then back in to an atom. It was adapted from Kees Boeke's book Cosmic View (1957). Source: Eames Foundation.

The Mars Climate Orbiter (1999)

The Mars Climate Orbiter (1999). Ground software reported thruster impulse in pound-force seconds; the navigation software expected newton-seconds (1 lbf·s ≈ 4.45 N·s). The probe approached Mars far too low, and was lost. Source: Mars Climate Orbiter Mishap Investigation Board, Phase I Report, 10 November 1999.

The metre: Jean-Baptiste Delambre and Pierre Méchain (France, 1792–1798)

The metre: Jean-Baptiste Delambre and Pierre Méchain (France, 1792–1798). The metre was defined as one ten-millionth of the distance from the North Pole to the equator. The two astronomers measured the meridian arc from Dunkirk to Barcelona by triangulation. The metre was redefined in 1960 (krypton light) and again in 1983 (the distance light travels in 1/299 792 458 s). Source: BIPM, 17th CGPM Resolution 1 (1983).

The kilogram: Le Grand K and the 2019 redefinition

The kilogram: Le Grand K and the 2019 redefinition. A platinum–iridium cylinder defined the kilogram from 1889. On 16 November 2018 the 26th General Conference on Weights and Measures voted to define it by fixing the Planck constant; the change took effect on 20 May 2019. The kilogram is realised with the Kibble balance, invented by Bryan Kibble at the UK's National Physical Laboratory in 1975. Sources: BIPM, 26th CGPM Resolution 1 (2018) · NIST: SI redefinition · BIPM: the SI.

The Gimli Glider (Air Canada Flight 143, 1983)

The Gimli Glider (Air Canada Flight 143, 1983). The fuel for an all-metric Boeing 767 was worked out with a density in pounds per litre (1.77) instead of kilograms per litre (about 0.8). The plane had less than half the fuel it needed, ran out at 41,000 feet, and was glided to a safe landing at Gimli, Manitoba. Source: Canadian Aviation Safety Board report, 1985.

Enrico Fermi at Trinity (New Mexico, 1945)

Enrico Fermi at Trinity (New Mexico, 1945). Fermi dropped scraps of paper as the blast wave passed; they moved about 2½ metres, from which he estimated about 10,000 tons of TNT. Later measurements put the yield at roughly twice that. Sources: Fermi, "My Observations During the Explosion at Trinity on July 16, 1945" · J. Katz, "Fermi at Trinity" (2021).

The Hubble Space Telescope (1990)

The Hubble Space Telescope (1990). The main mirror was polished very precisely to the wrong shape, about 2.2 micrometres too flat at the edge, because a test instrument had a lens out of place. Corrective optics were fitted in 1993. Sources: NASA: Hubble's mirror flaw · the Allen report, Hubble Space Telescope Optical Systems Failure Report, 1990.

Worked example: weighing the atmosphere

Watch the worked example

Play: M1 Worked Example · Weighing the Atmosphere
Chapters
  • 0:00 The question
  • 0:39 What air pressure means
  • 2:05 The size of the Earth
  • 3:03 Putting it together
  • 4:32 What we used

The problem

Problem. Air pressure at sea level is about 100,000 Pa and the Earth's radius is about 6400 km. Estimate the mass of the Earth's atmosphere.

Solution

  1. Pressure is force per area. 100,000 Pa means 100,000 N pressing on each square metre: the weight of the air above it.
  2. Mass from weight: \(m = \dfrac{W}{g} = \dfrac{10^5\ \text{N}}{10\ \text{N/kg}} = 10^4\ \text{kg}\) above each square metre. Units: \(\dfrac{\text{N/m}^2}{\text{N/kg}} = \text{kg/m}^2\). ✓
  3. Area of the Earth: \(R = 6400\ \text{km} = 6.4\times10^6\ \text{m}\), so \(4\pi R^2 \approx 12.6 \times 4.1\times10^{13} \approx 5\times10^{14}\ \text{m}^2\).

Solution

  1. Multiply: \(10^4\ \text{kg/m}^2 \times 5\times10^{14}\ \text{m}^2 = 5\times10^{18}\ \text{kg}\).
  2. Check: careful studies give about \(5.15\times10^{18}\) kg (Trenberth & Smith, Journal of Climate, 2005). Using \(g = 10\) instead of 9.8 makes our answer slightly small, and ignoring high ground (with less air above it) makes it slightly big; the two roughly cancel. The air is less than a millionth of the Earth's mass (about \(6\times10^{24}\) kg).

Practice

★ Warm-up

  1. Write in scientific notation: (a) 93,000,000 (b) 0.000 052 (c) 602,000,000,000,000,000,000,000.

    Solution

    (a) \(9.3\times10^7\) (point moves 7 places left) · (b) \(5.2\times10^{-5}\) (5 places right) · (c) \(6.02\times10^{23}\).

Problem 2 (★ Warm-up)

  1. Work out (a) \((3\times10^4)\times(2\times10^{-6})\) and (b) \((8\times10^9)\div(4\times10^3)\).

    Solution

    (a) \(3\times2 = 6\); \(4 + (-6) = -2\): \(6\times10^{-2} = 0.06\). (b) \(8\div4 = 2\); \(9-3 = 6\): \(2\times10^6\).

Problem 3 (★ Warm-up)

  1. Convert 72 km/h to metres per second.

    Solution

    \(72\ \dfrac{\text{km}}{\text{h}}\times\dfrac{1000\ \text{m}}{1\ \text{km}}\times\dfrac{1\ \text{h}}{3600\ \text{s}} = \dfrac{72\,000}{3600}\ \text{m/s} = 20\ \text{m/s}\). Handy shortcut: divide km/h by 3.6.

Problem 4 (★ Warm-up)

  1. How many cubic metres are 2.5 litres?

    Solution

    \(1\ \text{m}^3 = 1000\ \text{L}\), so \(1\ \text{L} = 10^{-3}\ \text{m}^3\) and \(2.5\ \text{L} = 2.5\times10^{-3}\ \text{m}^3\).

Problem 5 (★ Warm-up)

  1. How many significant figures: (a) 0.00420 (b) \(3.00\times10^8\) (c) 50.06?

    Solution

    (a) 3: leading zeros don't count, but the trailing zero after the decimal point does. (b) 3. (c) 4: zeros between non-zero digits count.

Problem 6 (★ Warm-up)

  1. A room is 4 m × 5 m × 2.5 m, and air has a density of about 1.2 kg/m³. What is the mass of the air in the room?

    kg

    Solution

    Volume \(V = 4\times5\times2.5 = 50\ \text{m}^3\). Mass \(m = \rho V = 1.2\times50 = 60\ \text{kg}\): about the mass of a person.

Problem 7 (★ Warm-up)

  1. A 2.0 litre bottle is full of olive oil, of mass 1.84 kg. What is the oil's density in kg/m³? Will it float on water?

    kg/m^3

    Solution

    \(V = 2.0\ \text{L} = 2.0\times10^{-3}\ \text{m}^3\), so \(\rho = m/V = 1.84/0.0020 = 920\ \text{kg/m}^3\). That is less than water's 1000 kg/m³, so it floats on water ✓.

★★ Standard

  1. Check the dimensions of (a) \(v^2 = 2ad\) and (b) \(d = v t^2\), where \(v\) is a speed, \(a\) an acceleration, \(d\) a distance and \(t\) a time.

    Hint

    Write each side in L, M and T. Replace every symbol by its dimension, cancel, and compare what is left.

    Solution

    (a) \([v^2] = L^2/T^2\); \([a d] = (L/T^2)\times L = L^2/T^2\). They match: plausible. ✓ (b) \([v t^2] = (L/T)\times T^2 = L\,T\), not \(L\). Wrong. ✗

Problem 9 (★★ Standard)

  1. The speed \(v\) of a wave on a string depends on the tension \(F\) (a force, dimensions \(M L T^{-2}\)) and the mass per unit length \(\mu\) (dimensions \(M L^{-1}\)). Use dimensional analysis to find how \(v\) depends on them.

    Hint

    Mass appears in both \(F\) and \(\mu\), and \(v\) has none. Combine \(F\) and \(\mu\) so M cancels, then see what dimensions are left.

    Solution

    Mass must cancel, so divide: \([F/\mu] = \dfrac{M L T^{-2}}{M L^{-1}} = L^2 T^{-2}\). Its square root is \(L/T\), a speed. So \(v \propto \sqrt{F/\mu}\). (The full result is exactly \(v = \sqrt{F/\mu}\).)

Problem 10 (★★ Standard)

  1. A floor is 3 m by 4 m. What is its area in square centimetres?

    Solution

    \(12\ \text{m}^2\times\left(\dfrac{100\ \text{cm}}{1\ \text{m}}\right)^2 = 12\times10^4\ \text{cm}^2 = 1.2\times10^5\ \text{cm}^2\). (Square the factor!)

Problem 11 (★★ Standard)

  1. Light travels at \(3\times10^8\) m/s, and a year is about \(3.15\times10^7\) s. (a) How far is a light-year in metres? (b) How long does sunlight take to reach us from \(1.5\times10^{11}\) m away?

    Hint

    Distance is speed times time. For (b), time is distance divided by speed. Multiply or divide the front numbers, and add the powers of ten when multiplying but subtract them when dividing.

    Solution

    (a) \(3\times3.15 \approx 9.5\); \(8+7 = 15\): about \(9.5\times10^{15}\) m. (b) \(t = d/v = 1.5\times10^{11}/3\times10^8 = 0.5\times10^3 = 500\) s, about 8.3 minutes.

Problem 12 (★★ Standard)

  1. Two students measure \(g\): Ana gets \(9.6\pm0.3\ \text{m/s}^2\); Ben gets \(10.4\pm0.1\ \text{m/s}^2\). The accepted value is 9.81. Which result agrees with it? Which is more precise? What might have gone wrong for the other?

    Hint

    Write each result as a range, value minus error to value plus error. Does 9.81 fall inside it? A smaller error bar means more precise.

    Solution

    Ana's range, 9.3 to 9.9, contains 9.81: it agrees. Ben's range, 10.3 to 10.5, doesn't. Ben's is more precise (smaller error bar) but less accurate, a sign of a systematic error, such as a mis-measured length or a stopwatch habit that shifts every reading.

★★★ Stretch

  1. The "try it" from the video. How many litres of water do you drink in your lifetime? Give a range, not just one number.

    Hint

    Litres per day × days per year × years. What's a sensible low and high value for each?

    Solution

    About 2 L a day × 365 × 80 ≈ 58,000 L, about \(6\times10^4\) L (60 cubic metres, a small swimming pool). With 1 to 3 L a day, the range is about \(3\times10^4\) to \(9\times10^4\) L: comfortably within a factor of two either way.

Problem 14 (★★★ Stretch)

  1. Estimate the number of hairs on a human head.

    Hint

    Estimate the area of scalp with hair, and how far apart the hairs are (or how many grow per square centimetre).

    Solution

    Scalp area: roughly a hemisphere-ish patch of about \(600\ \text{cm}^2\) (0.06 m²). Hairs about 0.7 mm apart gives about 200 per cm². \(600\times200 \approx 1.2\times10^5\). Commonly quoted figures are around \(10^5\): our estimate is well within a factor of two.

Problem 15 (★★★ Stretch)

  1. A mass \(m\) on a spring of stiffness \(k\) (unit N/m, dimensions \(M T^{-2}\)) bounces up and down. Use dimensional analysis to find how its period depends on \(m\) and \(k\). Does a heavier mass bounce faster or slower?

    Hint

    Find the combination of \(m\) and \(k\) with dimension T squared. Then think about what happens to it when \(m\) grows.

    Solution

    \([m/k] = \dfrac{M}{M T^{-2}} = T^2\), so \(T \propto \sqrt{m/k}\). (The full result is \(T = 2\pi\sqrt{m/k}\).) A heavier mass bounces more slowly. Unlike the pendulum, here the mass does matter: gravity's pull grows with the mass, which cancels its extra inertia, but a spring's pull doesn't.

Problem 16 (★★★ Stretch)

  1. If you counted one number every second without stopping, how long would it take to reach a billion? Is your answer an overestimate or an underestimate?

    Hint

    Divide a billion seconds by the number of seconds in a year. Then ask whether anything in real counting makes it slower than one number per second.

    yr

    Solution

    \(10^9\ \text{s}\div3.15\times10^7\ \text{s/year} \approx 32\) years. It's an underestimate: big numbers like "seven hundred and forty-two million…" take several seconds to say, and you'd need to sleep.

Check yourself

Answer these out loud, from memory. Then try again in a few days.

Check yourself

  • How can order-of-magnitude estimates reveal whether an answer is physically reasonable?

Check yourself

  • What is a physical quantity, and how do a numerical value and a unit describe it?

Check yourself

  • What are the SI base units, how are they defined through fixed constants of nature, and how are derived units constructed?

Check yourself

  • How do scientific notation and metric prefixes describe very large or small quantities?

Check yourself

  • How can units be converted without changing the physical quantity being measured?

Check yourself

  • What is dimensional analysis, and how can it detect mistakes or suggest relationships?

Check yourself

  • How do accuracy and precision differ?

Check yourself

  • What are random and systematic measurement errors?

Check yourself

  • How should uncertainty, significant figures, and rounding be reported?

Check yourself

  • How do calibration, repeated measurements, and error bars help assess an experimental result?

The roadmap's other §01–§02 questions (models and theories, controlled experiments, conservation laws, resolution and sensitivity, combining uncertainties) are picked up later: P0, P3, P40 and M14.

Check yourself

Summary

Summary

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Further learning

Khan Academy: videos and refreshers on notation, exponents, units and significant figures.

Free textbook: OpenStax, University Physics Volume 1, chapter 1: 1.2 Units and Standards · 1.3 Unit Conversion · 1.4 Dimensional Analysis · 1.5 Estimates and Fermi Calculations · 1.6 Significant Figures

Simulation: PhET: Estimation: practise estimating lengths, areas and volumes.

Books: Sanjoy Mahajan, Street-Fighting Mathematics (estimation and dimensional analysis); John R. Taylor, An Introduction to Error Analysis (uncertainty).

Next: M2 · Algebra and graphs for physics · Progress