The wind chill factor on a certain planet is given by the following formula, where $v$ is the wind speed (in meters per second) and $t$ is the air temperature $\left({ }^{\circ} \mathrm{C}\right.$ ). Complete parts (a) through (c).
\[
W=\left\{\begin{array}{ll}
t & 0 \leq v< 1.79 \\
33-\frac{(10.43+10 \sqrt{v}-v)(33-t)}{22.03} & 1.79 \leq v \leq 25 \\
33-1.5957(33-t) & v> 25
\end{array}\right.
\]
(a) Find the wind chill for an air temperature of $5^{\circ} \mathrm{C}$ and a wind speed of $0.75 \mathrm{~m} / \mathrm{sec}$.
\[
\mathrm{w} \approx \square^{\circ} \mathrm{C}
\]
\[
\mathrm{C}
\]
(Round to the nearest degree as needed.)
Answer
\(\boxed{W = 5^\circ \mathrm{C}}\)
Step 1 ：Given that the air temperature $t$ is $5^\circ \mathrm{C}$ and the wind speed $v$ is $0.75 \mathrm{~m} / \mathrm{sec}$, we can see that $0 \leq v<1.79$. Therefore, we should use the first part of the formula, which is $W=t$.
Step 2 ：Substituting the given values into the formula, we get: $W = t$
Step 3 ：So, $W = 5^\circ \mathrm{C}$
Step 4 ：Therefore, the wind chill for an air temperature of $5^\circ \mathrm{C}$ and a wind speed of $0.75 \mathrm{~m} / \mathrm{sec}$ is approximately $5^\circ \mathrm{C}$.
Step 5 ：\(\boxed{W = 5^\circ \mathrm{C}}\)
