Question

The wind chill factor represents the equivalent air temperature at a standard wind speed that would produce the same heat loss as the given temperature and wind speed. One formula for computing the equivalent temperature is

\[W(t)= 33-\frac{ (10.45+10\sqrt{v-v})(33-t }{ 2204 }\]
\[W(t)=33-1.5958(33-t)\]
\[if 0\le v <1.79\]
\[if 1.79\le v <20\]
\[if v \ge20\]

Answer 1

@hartnn

Answer 2

ok

Answer 3

thanks for the info :P

Answer 4

i'll need a question to help you answer it :)

Answer 5

do you want me to capture it

Answer 6

and attach it here

Answer 7

you'll attach the same thing you posted?
i am asking, what do we need to find? or do ? or solve?

Answer 8

i just see an incorrect expression and definition of the terms in it

Answer 9

can you see the question more clear now @hartnn

Answer 10

i am able to see some information, but not the question.

what we need to do ? do we have to find W for a specific 'v' ?

Answer 11

idont know what they need
that is the way they gave it and said solve

Answer 12

anything after/before "solve" ?? because i cannot see 'solve' too
i can see W(t) expression though

Answer 13

sory my mistake here is the other part of the question

where v represents the wind speed (in meters per second) and t represents the air temperature . Compute the wind chill for an air temperature of 15°C and a wind speed of 12 meters per second. (Round the answer to one decimal place.)

Answer 14

@hartnn i made a mistake that is the rest part of the question

Answer 15

no problem :)
so we have t =15
with v =12, we need to select one of the 3 expression for W

Answer 16

v =12 lies in which range of v ?

Answer 17

1st, 2nd or 3rd ?

Answer 18

2nd

Answer 19

correct!

Answer 20

so just plug in v=12 and t=15
in the middle expression

Answer 21

@hartnn sory i am having some connection problems but can you continue solving Please

Answer 22

@phi Can you continue Solving Please

Answer 23

There was a typo in the second equation

\[W(t)= t \text{ if}\ 0\le v <1.79\\

W(t)= 33-\frac{ (10.45+10\sqrt{v}-v)(33-t) }{ 2204 }\ if\ 1.79\le v <20 \\
W(t)=33-1.5958(33-t)\ if\ v \ge20\]

Answer 24

Compute the wind chill for an air temperature of 15°C and a wind speed of 12 meters per second. (Round the answer to one decimal place.)

you have to find W(t) for v= 12 m/sec

as you know, you must use the 2nd equation.

replace v with 12 and t with 15
\[ W(t)= 33-\frac{ (10.45+10\sqrt{12}-12)(33-15) }{ 2204 } \]

now it is an "order of operations" problem.

Answer 25

what do you mean by "order of operation"

Answer 26

do the operations inside parens first. then exponents (or square roots)
then multiply/divide, and last add/subtract.

Answer 27

@phi 32.91 is the correct answer

Answer 28

how did you get that number?

Answer 29

\[W(t) =33-\frac{ (10.45)(18) }{ 2204 }\]
\[W(t)= 33-\frac{ 188.1 }{ 2204 }\]
W(t)=32.91

Answer 30

the way i did is correct or not

Answer 31

the first line should read
\[ W(t) =33-\frac{ (10.45+10\sqrt{12} - 12)(18) }{ 2204 }\]

I would first figure out sqrt(12)
then times 10

Answer 32

sqrt(12-12) is sqrt0

Answer 33

yes, but that is a typo. Look at the original equation
it says
10sqr(v) - v

or for v=12
\[ 10 \sqrt{12} - 12\]
that is different from sqr(12-12)

Answer 34

for example
sqr(12)= 3.464
10* sqr(12) = 10*3.464= 34.64

now you have
\[ W(t) =33-\frac{ (10.45+34.64 - 12)(18) }{ 2204 } \]