Worked examples · Sheet 1

Unit conversions and scientific notation

Theory used: Multiples and submultiples Unit conversion Scientific notation Significant figures Significant figures in calculations

Go to: Exercise 1 Exercise 2 Exercise 3 Exercise 4

Exercise 1

A car travels on the highway at $108 km/h$ and a cyclist at $7.5 m/s$ . Express the speed of the car in $m/s$ and that of the cyclist in $km/h$ .

1 km = 1 0^{3} m, 1 h = 3600 s

v_{car} = 108 \frac{km}{h} = 108 \cdot \frac{1 0 ^{3} m}{3600 s} = \frac{108}{3.6} m/s = 30 m/s

v_{bike} = 7.5 \frac{m}{s} = 7.5 \cdot \frac{1 0 ^{- 3} km}{\frac{1}{3600} h} = 7.5 \cdot 3.6 km/h = 27 km/h

For a unit conversion each unit is replaced by its value in the new unit. A kilometer equals $1 0^{3} m$ and an hour equals $3600 s$ , so one kilometer per hour is $\frac{1000}{3600} m/s = \frac{1}{3.6} m/s$ , and to go from $km/h$ to $m/s$ one divides by $3.6$ . In the opposite direction a meter equals $1 0^{- 3} km$ and a second equals $\frac{1}{3600} h$ , therefore one multiplies by $3.6$ . Hence the car travels at $30 m/s$ and the cyclist at $27 km/h$ . As a check, a speed expressed in $m/s$ is always a smaller number than the same speed in $km/h$ , because the meter per second is the larger unit.

Exercise 2

The density of aluminum is $2.70 g/c m^{3}$ . Express it in $kg/ m^{3}$ and compute the mass of a solid aluminum cube with an edge of $5.0 cm$ .

1 g = 1 0^{- 3} kg, 1 c m^{3} = (1 0^{- 2} m)^{3} = 1 0^{- 6} m^{3}

ρ = 2.70 \frac{g}{c m ^{3}} = 2.70 \cdot \frac{1 0 ^{- 3} kg}{1 0 ^{- 6} m ^{3}} = 2.70 \cdot 1 0^{3} kg/ m^{3}

V = ℓ^{3} = (5.0 \cdot 1 0^{- 2} m)^{3} = 1.25 \cdot 1 0^{- 4} m^{3}

m = ρ V = 2.70 \cdot 1 0^{3} kg/ m^{3} \cdot 1.25 \cdot 1 0^{- 4} m^{3} = 0.3375 kg \approx 0.34 kg

The cubic centimeter is the cube of the centimeter, so the conversion factor $1 0^{- 2}$ must be cubed and gives $1 0^{- 6} m^{3}$ . Dividing $1 0^{- 3}$ by $1 0^{- 6}$ gives $1 0^{3}$ , therefore the density in SI units is $2.70 \cdot 1 0^{3} kg/ m^{3}$ . The volume of the cube is the cube of the edge expressed in meters, and the mass is the product of the density and the volume, from the definition $ρ = \frac{m}{V}$ . The edge $5.0 cm$ has two significant figures, and in a product the result has the figures of the least precise datum, so the mass of the cube is $0.34 kg$ .

Exercise 3

A faucet delivers $12$ liters of water per minute. Express the flow rate in $m^{3} /s$ and compute the time needed to fill a tub of $0.18 m^{3}$ .

1 L = 1 d m^{3} = 1 0^{- 3} m^{3}, 1 min = 60 s

q = 12 \frac{L}{min} = \frac{12 \cdot 1 0 ^{- 3} m ^{3}}{60 s} = 2.0 \cdot 1 0^{- 4} m^{3} /s

t = \frac{V}{q} = \frac{0.18 m ^{3}}{2.0 \cdot 1 0 ^{- 4} m ^{3} /s} = 9.0 \cdot 1 0^{2} s = 15 min

The liter is the cubic decimeter, and since $1 dm = 1 0^{- 1} m$ one has $1 L = 1 0^{- 3} m^{3}$ . The flow rate is the volume that flows out per unit time, so the volume and the time are converted separately and the quotient is computed. The filling time is the volume of the tub divided by the flow rate; in the quotient the cubic meters cancel and the second remains, as must happen for a time. The tub fills in $9.0 \cdot 1 0^{2} s$ , that is, in a quarter of an hour.

Exercise 4

The mean distance between the Earth and the Sun is $1.496 \cdot 1 0^{11} m$ and light propagates in vacuum at $3.00 \cdot 1 0^{8} m/s$ . Compute the time light takes to travel from the Sun to the Earth, in seconds and in minutes, with the correct number of significant figures.

t = \frac{d}{c}

t = \frac{1.496 \cdot 1 0 ^{11} m}{3.00 \cdot 1 0 ^{8} m/s} = \frac{1.496}{3.00} \cdot 1 0^{11 - 8} s = 0.4987 \cdot 1 0^{3} s \approx 4.99 \cdot 1 0^{2} s

t = \frac{498.7 s}{60 s/min} \approx 8.31 min

Light moves at constant speed, so the time is the distance divided by the speed. In scientific notation the coefficients are divided, $\frac{1.496}{3.00} = 0.4987$ , and the exponents of ten are subtracted, $11 - 8 = 3$ . The coefficient $0.4987$ is less than $1$ , therefore the result is rewritten as $4.987 \cdot 1 0^{2} s$ . The data have four and three significant figures, and in a quotient the least precise datum prevails, so the time is $4.99 \cdot 1 0^{2} s$ . For the conversion to minutes the unrounded value $498.7 s$ is used and rounding is done only at the end, giving about $8.31 min$ ; the factor $60 s/min$ is exact and does not limit the figures.