Physics I course sequence · Interactive lesson
SI Quantities, Units & Prefixes
Many first-year errors are not physics errors at all. They are a mass left in grams, a cm² converted as if it were a cm, or a number that lost its unit halfway through. Let SI do that bookkeeping, and the physics is the only thing left to get right.
UPI-01.01 · Lesson 1 of 9 in this unit. Position is not a completion score.Build the model
A physical quantity is a number multiplied by a unit.
Simple definition
A physical quantity is a numerical value multiplied by a unit.
Read the full explanation
Write a length as 1.85 km and you have written a product: the number 1.85 times the unit km. Only the product is physical. Change the unit and the number must change by exactly the inverse factor, which is why 1.85 km, 1850 m and 1.85 × 10⁵ cm are one length, not three. A dimensionless quantity may be a pure number, but 9.81 is not an acceleration until m s⁻² is attached, and a calculation that drops its units halfway has thrown away the only record of what its numbers were. Treat unit symbols as algebraic factors — multiply them, cancel them, raise them to powers — and they carry that record to the end, where the surviving unit is the first check on what you actually computed.
Example
A force of 4.7 kN has the same size as:
4.7 kN = 4.7 × 10³ N = 4700 NSame quantity, different units. A larger unit gives a smaller number.
Changing units changes the number inversely
1.85 km = 1850 m = 1.85 × 10⁵ cm
Larger units → smaller numbers
Smaller units → larger numbers
The physical quantity stays the same.
The seven SI base units
| Quantity | Unit | Symbol |
|---|---|---|
| Time | second | s |
| Length | metre | m |
| Mass | kilogram | kg |
| Electric current | ampere | A |
| Thermodynamic temperature | kelvin | K |
| Amount of substance | mole | mol |
| Luminous intensity | candela | cd |
Read the full explanation
SI chooses seven base quantities by convention — time, length, mass, electric current, thermodynamic temperature, amount of substance and luminous intensity — with base units s, m, kg, A, K, mol and cd. Every other quantity is derived from them. Since May 2019 no base unit rests on a physical artefact. Instead seven constants are given exact values and the units follow. The caesium-133 hyperfine frequency is exactly 9 192 631 770 Hz, which defines the second. The speed of light is exactly 299 792 458 m s⁻¹, so the metre is the distance light travels in 1/299 792 458 s. The Planck constant is exactly 6.626 070 15 × 10⁻³⁴ J s, and since J s = kg m² s⁻¹, fixing h once the metre and second are defined fixes the kilogram. Those values carry no uncertainty, because they are no longer measured; what carries uncertainty is every real clock and balance that realises them.
Modern SI is defined by fixed constants of nature
Caesium frequencyΔνCs = 9 192 631 770 HzDefines the second
Speed of lightc = 299 792 458 m s⁻¹Defines the metre with the second
Planck constanth = 6.626 070 15 × 10⁻³⁴ J sDefines the kilogram with m and s
The complete physical model
A physical quantity is a number multiplied by a unit, and only the product is physical: 72 km h⁻¹ and 20 m s⁻¹ are the same speed, so any change of unit must be paid for exactly by the number. SI makes that bookkeeping reliable in three ways. It fixes seven base units — in mechanics the second, the metre and the kilogram do all the work — by giving exact values to constants of nature, so a unit is a statement about the universe rather than an object kept in a vault. It gives every other quantity a coherent unit, a product of powers of the base units with a numerical factor of exactly one: put coherent inputs into a correct equation and the coherent unit comes out, with no conversion factor anywhere in the algebra. And it allows prefixes, exact powers of ten attached to a unit symbol, which are safest converted away before the algebra starts and which must be raised to the same power as the unit they sit on. The protection ends in two places. SI cannot stop a non-coherent unit such as g, km h⁻¹, L or min from entering an equation, where it silently plants a factor in the result. And a unit does not name the quantity: a torque and an energy reduce to the same kg m² s⁻², which is why SI writes torque in N m and keeps the joule for energy.
Coherent derived units
Derived units are formed from base units with no extra conversion factors.
Common derived units
| Force | N = kg m s⁻² |
|---|---|
| Energy | J = N m = kg m² s⁻² |
| Power | W = J s⁻¹ = kg m² s⁻³ |
| Pressure | Pa = N m⁻² = kg m⁻¹ s⁻² |
A coherent system
Use coherent SI inputs and a valid equation: the result comes out in coherent SI.
F = ma → kg × m s⁻² = N
Units also carry meaning: torque is written N m, while energy is written J.Read the full explanation
Every coherent derived SI unit is a product of powers of the base units with a numerical factor of exactly 1, and that property, coherence, is what lets a calculation run without conversion factors. Newton's second law gives 1 N = 1 kg m s⁻²; work gives 1 J = 1 N m = 1 kg m² s⁻²; power gives 1 W = 1 J s⁻¹ = 1 kg m² s⁻³; pressure gives 1 Pa = 1 N m⁻² = 1 kg m⁻¹ s⁻². So coherent inputs guarantee a coherent output. A 75 kg climber who gains 12 m of height in 20 s delivers mgh/t = (75 kg)(9.81 m s⁻²)(12 m)/(20 s) = 4.4 × 10² kg m² s⁻³, which is 4.4 × 10² W with nothing to convert. Enter the mass in grams or the time in minutes and coherence is gone: the same arithmetic returns a number in a unit you never intended, and the equation looks perfectly respectable while it does so.
Prefix explorer
A prefix multiplies a unit by an exact power of ten.
- Case matters: k ≠ K, m ≠ M.
- Use one prefix, never stacked prefixes.
- Mass prefixes attach to grams, not kilograms.
- 1 mg = 10⁻⁶ kg, not 10⁻³ kg.
Read the full explanation
A prefix multiplies a unit by an exact power of ten, and the working set steps by 10³: f (10⁻¹⁵), p (10⁻¹²), n (10⁻⁹), μ (10⁻⁶), m (10⁻³), k (10³), M (10⁶), G (10⁹) and T (10¹²), with c (10⁻²) kept mainly for centimetres. Choose the prefix that leaves a number between about 0.1 and 1000, so 0.000 047 m reads as 47 μm. Three rules stop prefixes misleading. Case matters: m is milli and M is mega, so 3 mW and 3 MW differ by 10⁹. Prefixes never stack: 10⁻⁹ m is 1 nm, not 1 mμm. Mass prefixes attach to the gram, because the kilogram already carries one, so 10⁻⁶ kg is 1 mg, never 1 μkg. The list grew in 2022 to ronna (10²⁷) and quetta (10³⁰) at the top and ronto (10⁻²⁷) and quecto (10⁻³⁰) at the bottom, which makes Earth's mass of 5.97 × 10²⁴ kg exactly as easy to write as 5.97 Rg.
Common working prefixes shown. View the complete official SI prefix list ↗
Interactive SI power lab
Explore prefixes, powers and compound units. Compare the correct result with a common slip.
Converting 1.85 km
Change one input and explain why the answer moves. All displayed results are rounded; calculations use unrounded values.
Explore the original powers-of-ten diagram
Start at 2.5 cm², with k = −2 and p = 2: the ring sits two decades right of the dot, the factor-of-100 slip. Step p to 3 and the gap doubles to four. Then set k = 3 for km² and km³: the boxes flip right, and the once-only answer falls short by 10³ or 10⁶.
SI POWER OF TEN, k × p-4
LOG₁₀ OF VALUE IN SI-3.60
PREFIX-ONCE ERROR2 decades
Live interpretationSI POWER OF TEN, k × p: −4. LOG₁₀ OF VALUE IN SI: −3.60. PREFIX-ONCE ERROR: 2 decades
Catch the common trap
The power belongs to the whole prefixed unit.
× Wrong
1 cm³ = 10⁻² m³
The prefix was used only once.✓ Correct
1 cm³ = (10⁻² m)³ = 10⁻⁶ m³
The factor is cubed too.Density example
2.70 g cm⁻³ = 2.70 × 10³ kg m⁻³
Divide 2.70 × 10⁻³ kg by 10⁻⁶ m³.Read the full explanation
The classic slip is converting a squared or cubed unit with the prefix factor used only once, writing 1 cm³ = 10⁻² m³, or turning aluminium's 2.70 g cm⁻³ into 0.270 kg m⁻³. It treats the prefix as a label that converts on its own, but cm³ means (cm)³ = (10⁻² m)³ = 10⁻⁶ m³: the exponent acts on the whole prefixed symbol, factor included. Both wrong answers fail at once against the world. If a cubic centimetre were 10⁻² m³, a 1 L bottle holding 1000 cm³ would contain 10 m³, the air of a small room, and aluminium at 0.270 kg m⁻³ would float on air at 1.2 kg m⁻³. Done properly, 2.70 g cm⁻³ = (2.70 × 10⁻³ kg)/(10⁻⁶ m³) = 2.70 × 10³ kg m⁻³.
Convert by multiplying by one
A conversion factor is a ratio equal to one. Put the unwanted unit on the opposite side so it cancels.
9.0 L min⁻¹ × (10⁻³ m³ / 1 L) × (1 min / 60 s) = 1.5 × 10⁻⁴ m³ s⁻¹
Read the full explanation
A conversion factor is a fraction equal to one, such as (10⁻³ m³)/(1 L) or (1 min)/(60 s). Multiplying by it changes how a quantity is written and nothing else, and orienting it so the unwanted unit cancels makes the direction automatic. A tap delivering 9.0 L min⁻¹ becomes 9.0 L min⁻¹ × (10⁻³ m³/1 L) × (1 min/60 s) = 1.5 × 10⁻⁴ m³ s⁻¹, with L and min cancelling in plain sight. Because a factor equal to one is still one when raised to any power, a unit with an exponent takes its factor to that same power: 1 km² = (10³ m)² = 10⁶ m², and 1 L = 1 dm³ = (10⁻¹ m)³ = 10⁻³ m³. Finish with a direction check. Moving to a smaller unit must produce a larger number, so if converting metres to millimetres made your number smaller, the fraction went in upside down.
Practice & worked examples
Convert first. Calculate second. Check the units and scale.
Easy · Laser pulse power
A laboratory laser emits pulses that each carry 2.8 mJ of energy and last 35 fs. Find the average power during one pulse, P = E/t, in watts, and restate it with a sensible prefix.
View worked solution →
- Replace each prefix with its power of ten: E = 2.8 mJ = 2.8 × 10⁻³ J and t = 35 fs = 35 × 10⁻¹⁵ s = 3.5 × 10⁻¹⁴ s.
- Both inputs are now coherent SI, so the quotient comes out in J s⁻¹ = W with no extra factor: P = (2.8 × 10⁻³ J)/(3.5 × 10⁻¹⁴ s).
- Divide the numbers and combine the powers: (2.8/3.5) × 10⁻³ × 10¹⁴ = 0.80 × 10¹¹ W = 8.0 × 10¹⁰ W.
- Choose the prefix that leaves a number between about 0.1 and 1000: 8.0 × 10¹⁰ W = 80 × 10⁹ W = 80 GW.
P = 8.0 × 10¹⁰ W = 80 GW. Each pulse carries only 2.8 mJ; the power is enormous because the time is tiny.
Medium · Car braking
A car of mass 1.35 × 10³ kg brakes steadily from 108 km h⁻¹ to 72.0 km h⁻¹ over 125 m. Using a = (v² − u²)/(2s) and F = ma, find the size of the braking force in kilonewtons, then show what happens if the speeds are left in km h⁻¹.
View worked solution →
- Convert the speeds before they meet the formula: 1 km h⁻¹ = (10³ m)/(3600 s) = (1/3.6) m s⁻¹, so u = 108/3.6 = 30.0 m s⁻¹ and v = 72.0/3.6 = 20.0 m s⁻¹.
- Everything is now coherent SI: a = (20.0² − 30.0²) m² s⁻² ÷ (2 × 125 m) = (−500 m² s⁻²)/(250 m) = −2.00 m s⁻². The negative sign says the car is slowing.
- The force has size F = (1.35 × 10³ kg)(2.00 m s⁻²) = 2.70 × 10³ kg m s⁻² = 2.70 × 10³ N, and writing 10³ as its prefix gives 2.70 kN.
- With unconverted speeds, a = (72.0² − 108²)/250 = −25.9 and the force comes out as 3.50 × 10⁴, which a hurried solution labels N, or 35.0 kN. That is too large by exactly 3.6² = 12.96, because each squared speed carried the factor 3.6 twice.
- Sense check: the car's weight is (1.35 × 10³ kg)(9.81 m s⁻²) = 13.2 kN, so 2.70 kN is about a fifth of it, firm but ordinary braking.
F = 2.70 kN, with a = −2.00 m s⁻². Leaving the speeds in km h⁻¹ inflates the result by 3.6² ≈ 13, to a meaningless 35.0.
Hard · Rainfall and reservoir
A storm leaves 12 mm of rain over a 2.5 km² hillside, and all of it drains into a reservoir 45 m lower whose surface area is 0.60 km². Taking the density of water as 1.00 g cm⁻³ and the drainage time as 3.0 h, find the mass of water, the rise in reservoir level, and the average rate at which gravitational energy mgh is released.
View worked solution →
- Convert every input to coherent SI first, raising each prefix factor to the unit's power: 12 mm = 1.2 × 10⁻² m; 2.5 km² = 2.5 × (10³ m)² = 2.5 × 10⁶ m²; 0.60 km² = 6.0 × 10⁵ m²; 1.00 g cm⁻³ = (10⁻³ kg)/(10⁻² m)³ = 1.00 × 10³ kg m⁻³; 3.0 h = 3.0 × 3600 s = 1.08 × 10⁴ s.
- Volume is depth times area: V = (1.2 × 10⁻² m)(2.5 × 10⁶ m²) = 3.0 × 10⁴ m³. Then m = ρV = (1.00 × 10³ kg m⁻³)(3.0 × 10⁴ m³) = 3.0 × 10⁷ kg, which is 3.0 × 10⁴ t.
- Spread over the reservoir, that volume raises the level by V/A = (3.0 × 10⁴ m³)/(6.0 × 10⁵ m²) = 5.0 × 10⁻² m, best reported as 5.0 cm.
- Energy released on the way down: mgh = (3.0 × 10⁷ kg)(9.81 m s⁻²)(45 m) = 1.32 × 10¹⁰ kg m² s⁻² = 1.32 × 10¹⁰ J, carrying a guard digit into the next step.
- Average power: P = (1.32 × 10¹⁰ J)/(1.08 × 10⁴ s) = 1.2 × 10⁶ W = 1.2 MW. The units confirm it: kg m² s⁻² per s is kg m² s⁻³, which is the watt.
- Had 2.5 km² been converted as 2.5 × 10³ m², with the prefix used once, every later result would be 10³ too small, a 5.0 × 10⁻⁵ m rise and 1.2 kW, and nothing inside the working would look wrong.
m = 3.0 × 10⁷ kg (3.0 × 10⁴ t); the level rises 5.0 cm; energy is released at an average 1.2 MW. Two significant figures, set by 12 mm, 2.5 km², 45 m and 3.0 h.
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Lesson summary
Write each physical quantity as a number × unit. Convert prefixes before substituting, raise the prefix to the unit’s power, and check that the final dimensions and magnitude make sense.
Full formula reference
A quantity is a number times a unit
Q = n × U = n′ × U′ → n′ = n × (U/U′)
Switch to a unit 10³ times larger and the number shrinks by 10³: 1850 m = 1.85 km.
n, n′ are pure numbers; U, U′ are units of one kind, such as m and km; Q itself never changesPrefixes are exact powers of ten
T 10¹² · G 10⁹ · M 10⁶ · k 10³ · c 10⁻² · m 10⁻³; μ 10⁻⁶ · n 10⁻⁹ · p 10⁻¹² · f 10⁻¹⁵
Replace each prefix by its power of ten before substituting, and the calculation is back in coherent SI.
case-sensitive: M is mega and m is milli, so 1 MW = 10⁹ mW; mass prefixes attach to g, never to kgAn exponent acts on the prefix too
(10ᵏ U)ᵖ = 10ᵏᵖ Uᵖ
The prefix belongs to the symbol, so cm² means (cm)²: the factor 10⁻² is squared along with the metre.
1 cm² = 10⁻⁴ m² · 1 mm³ = 10⁻⁹ m³ · 1 km² = 10⁶ m² · 1 g cm⁻³ = 10³ kg m⁻³Coherent derived units
N = kg m s⁻² · J = N m = kg m² s⁻² · W = J s⁻¹ · Pa = N m⁻²
Coherent units in, coherent unit out: F = ma, K = ½mv² and P = Fv return N, J and W with no conversion factor.
each named unit is a product of base-unit powers with factor exactly 1; Hz = s⁻¹ and rad = m m⁻¹ = 1Non-coherent units, converted at the door
1 km h⁻¹ = (1/3.6) m s⁻¹ · 1 L = 10⁻³ m³ · 1 t = 10³ kg · 1 min = 60 s
These conversion factors bring common units into coherent SI. A conversion factor does not by itself mean that a non-SI unit is officially accepted for use with SI.
54 km h⁻¹ = 15 m s⁻¹ · 1 bar = 10⁵ Pa · 1 eV = 1.602 176 634 × 10⁻¹⁹ J exactlyBase units fixed by exact constants (2019)
Δν(¹³³Cs) = 9 192 631 770 Hz · c = 299 792 458 m s⁻¹ · h = 6.626 070 15 × 10⁻³⁴ J s
A unit is now a statement about nature: the metre is the distance light travels in 1/299 792 458 of a second.
exact by definition; in turn they fix the second, the metre and the kilogram, and four more constants fix A, K, mol and cdOptional supporting lessons
These chapter resources can help with earlier ideas; they are not additional required Physics I completion tasks.