Solution
Part d
T= -2ºC, V=15 km/hr
We can convert to F and mi/h like this:
T= 1.8*(-2)+ 32= 28.4ºF
Ms= As/0.62= 15 km/hr /0.62= 24.19 mi/hr
Replacing we have:
[tex]WC=35.74+0.6215\cdot(28.4)-35.75(24.19^{0.16})+0.4275(28.4)(24.19^{0.16})[/tex]Solving we got:
[tex]WC=35.74+17.651-59.519+20.213=14.08[/tex]Rounded to one decimal we got:
14.1
Part e
T= 10ºC = 1.8(10)+ 32=50ºF
Ms = 34 km/hr
Ms= As/0.62= 34 km/hr/ 0.62= 54.84 mi/hr
Replacing we got:
[tex]WC=35.74+0.6215\cdot(50)-35.75\cdot(54.84^{0.16})+0.4275(50)(54.84^{0.16})[/tex][tex]WC=35.74+31.075-67.847+40.566=39.53[/tex]Rounded to the nearest tenth we got:
39,5
Althought the reported temperature is 50 ºFahrenheit, because of wind it feels like 39.5 Fahrenheit