Riemann surface
matlab

American football is played on a 100 -yd-long field, excluding the end zones. How long is the field in meters? (Assume that 1 meter equals 3.281 feet.)

Solution Since 3 feet = 1 yard and 3.281 feet $=1$ meter, multiply 100 yards by these conversion factors to cancel the units of yards, leaving the units of meters:
$$100 \mathrm{yd}=100 \mathrm{yd} \times \frac{3 \mathrm{ft}}{1 \mathrm{yd}} \times \frac{1 \mathrm{~m}}{3.281 \mathrm{ft}}=91.4 \mathrm{~m}$$
A football field is $91.4 \mathrm{~m}$ long.

Solution (a) The average speed of the earth’s orbit around the sun is calculated by dividing the distance traveled by the time it takes to go one revolution:
\begin{aligned} \text { average speed } & =\frac{2 \pi(\text { average dist of Earth to sun })}{1 \text { year }} \ & =\frac{2 \pi\left(10^8 \mathrm{~km}\right)}{365.25 \mathrm{~d}} \times \frac{1 \mathrm{~d}}{24 \mathrm{~h}} \times \frac{1 \mathrm{~h}}{3600 \mathrm{~s}}=20 \mathrm{~km} / \mathrm{s} \end{aligned}
The earth travels at an average speed of $20 \mathrm{~km} / \mathrm{s}$ around the sun.
(b) To convert the average speed into units of $\mathrm{m} / \mathrm{s}$, use the conversion factor: $1000 \mathrm{~m}$

$=1 \mathrm{~km}$ :
average speed $=\frac{20 \mathrm{~km}}{\mathrm{~s}} \times \frac{1000 \mathrm{~m}}{1 \mathrm{~km}}=20 \times 10^3 \mathrm{~m} / \mathrm{s}$

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