QuestionFind the wind chill $W$ for $t=5^{\circ} C$ and $v=19 \mathrm{~m/s}$ . Use the formula provided. $W \approx \square^{\circ} C$

Studdy Solution

STEP 1

Assumptions1. The wind chill factor, W, is given by the piecewise function $W=\left\{\begin{array}{ll} t &0 \leq v<1.78 \\ 33-\frac{(10.43+10 \sqrt{v}-v)(33-t)}{22.04} &1.78 \leq v \leq20 \\ 33-1.5958(33-t) & v>20\end{array}\right.$ . The air temperature, t, is $5^{\circ}C$
3. The wind speed, v, is $19m/s$

STEP 2

We need to determine which part of the piecewise function to use based on the given wind speed. Since the wind speed is $19m/s$ , we use the second part of the piecewise function, which applies when $1.78 \leq v \leq20$ .

STEP 3

Substitute the given values for v and t into the equation for the wind chill factor.
$W =33-\frac{(10.43+10 \sqrt{v}-v)(33-t)}{22.04}$ $W =33-\frac{(10.43+10 \sqrt{19}-19)(33-5)}{22.04}$

STEP 4

implify the equation by performing the operations in the numerator of the fraction.
$W =33-\frac{(10.43+10 \sqrt{19}-19)(28)}{22.04}$

STEP 5

Perform the operations in the denominator of the fraction.
$W =33-\frac{(10.43+43.58-19)(28)}{22.04}$ $W =33-\frac{(35.01)(28)}{22.04}$

STEP 6

Perform the multiplication in the numerator of the fraction.
$W =33-\frac{980.28}{22.04}$

STEP 7

Perform the division in the fraction.
$W =33-44.47$

STEP 8

Subtract the fraction from33 to find the wind chill factor.
$W = -11.47^{\circ}C$

STEP 9

Round the wind chill factor to the nearest degree.
$W \approx -11^{\circ}C$ The wind chill for an air temperature of $5^{\circ}C$ and a wind speed of $19m/s$ is approximately $-11^{\circ}C$ .

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