Math Snap

STEP 1

Assumptions1. The wind chill factor formula is given as$$W=\left\{\begin{array}{ll} 33-\frac{(10.45+10 \sqrt{v}-v)(33-t)}{22.03} &0 \leq v<1.77 \\ 33-1.5957(33-t) & v>20\end{array}\right.
$$. The air temperature $t$ is $10^{\circ} \mathrm{C}$.
3. The wind speed $v$ is $1 \mathrm{~m} / \mathrm{sec}$.

STEP 2

Since the wind speed $v$ is $1 \mathrm{~m} / \mathrm{sec}$, which is in the range $0 \leq v<1.77$, we use the first part of the formula to calculate the wind chill factor.
$$W =33-\frac{(10.45+10 \sqrt{v}-v)(33-t)}{22.03}
$$

STEP 3

Now, plug in the given values for $t$ and $v$ into the formula.
$$W =33-\frac{(10.45+10 \sqrt{1}-1)(33-10)}{22.03}
$$

STEP 4

implify the expression inside the square root and the parentheses.
$$W =33-\frac{(10.45+10-1)(23)}{22.03}
$$

STEP 5

Perform the operations inside the parentheses.
$$W =33-\frac{19.45 \times23}{22.03}
$$

STEP 6

Perform the multiplication operation.
$$W =33-\frac{447.35}{22.03}
$$

STEP 7

Perform the division operation.
$$W =33-20.32$$

STEP 8

Subtract to find the value of $W$.
$$W =12.68$$

Since we need to round to the nearest degree, we round $12.68$ to $13$.
$$W \approx13^{\circ} \mathrm{C}
$$The wind chill for an air temperature of $^{\circ} \mathrm{C}$ and a wind speed of $ \mathrm{~m} / \mathrm{sec}$ is approximately $13^{\circ} \mathrm{C}$.