OpenStudy (yacoub1993):

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

4 years ago
OpenStudy (yacoub1993):

@hartnn

4 years ago
hartnn (hartnn):

ok

4 years ago
hartnn (hartnn):

thanks for the info :P

4 years ago
hartnn (hartnn):

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

4 years ago
OpenStudy (yacoub1993):

do you want me to capture it

4 years ago
OpenStudy (yacoub1993):

and attach it here

4 years ago
hartnn (hartnn):

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

4 years ago
hartnn (hartnn):

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

4 years ago
OpenStudy (yacoub1993):

4 years ago
OpenStudy (yacoub1993):

can you see the question more clear now @hartnn

4 years ago
hartnn (hartnn):

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' ?

4 years ago
OpenStudy (yacoub1993):

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

4 years ago
hartnn (hartnn):

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

4 years ago
OpenStudy (yacoub1993):

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.)

4 years ago
OpenStudy (yacoub1993):

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

4 years ago
hartnn (hartnn):

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

4 years ago
hartnn (hartnn):

v =12 lies in which range of v ?

4 years ago
hartnn (hartnn):

1st, 2nd or 3rd ?

4 years ago
OpenStudy (yacoub1993):

2nd

4 years ago
hartnn (hartnn):

correct!

4 years ago
hartnn (hartnn):

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

4 years ago
OpenStudy (yacoub1993):

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

4 years ago
OpenStudy (yacoub1993):

@phi Can you continue Solving Please

4 years ago
OpenStudy (phi):

There was a typo in the second equation \[W(t)= t \text{ if}\ 0\le v <1.79\\ W(t)= 33-\frac{ (10.45+10\sqrt{v}-v)(33-t) }{ 2204 }\ if\ 1.79\le v <20 \\ W(t)=33-1.5958(33-t)\ if\ v \ge20\]

4 years ago
OpenStudy (phi):

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.

4 years ago
OpenStudy (yacoub1993):

what do you mean by "order of operation"

4 years ago
OpenStudy (phi):

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

4 years ago
OpenStudy (yacoub1993):

@phi 32.91 is the correct answer

4 years ago
OpenStudy (phi):

how did you get that number?

4 years ago
OpenStudy (yacoub1993):

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

4 years ago
OpenStudy (yacoub1993):

the way i did is correct or not

4 years ago
OpenStudy (phi):

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

4 years ago
OpenStudy (yacoub1993):

sqrt(12-12) is sqrt0

4 years ago
OpenStudy (phi):

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)

4 years ago
OpenStudy (phi):

for example sqr(12)= 3.464 10* sqr(12) = 10*3.464= 34.64 now you have \[ W(t) =33-\frac{ (10.45+34.64 - 12)(18) }{ 2204 } \]

4 years ago
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