anonymous
  • anonymous
convert 0.004 g/(min.)(m^3)cm to lbm/(hr)(ft^3)
Mathematics
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schrodinger
  • schrodinger
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anonymous
  • anonymous
can you help me @ganeshie8 and @Michele_Laino :) i can do it if the given is g/(min.)(m^3) ,but there's a cm in the denominator :/
Michele_Laino
  • Michele_Laino
is your measure like this? \[\Large 0.04\frac{g}{{\min \times {m^3} \times cm}}\]
anonymous
  • anonymous
yes :)

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Michele_Laino
  • Michele_Laino
"lbm" stands for?
anonymous
  • anonymous
pound (mass)
Michele_Laino
  • Michele_Laino
g stands for grams? right?
anonymous
  • anonymous
yes :)
Michele_Laino
  • Michele_Laino
please wait, I'm working on your question
anonymous
  • anonymous
it's okay and take your time, thank you for helping me :)
Michele_Laino
  • Michele_Laino
please note that, in the first unit measure, we divide by a length^4 whereas in the second unit of measure we divide by a length^3
Michele_Laino
  • Michele_Laino
since the second unit of measure is: \[\Large \frac{{{\text{lbm}}}}{{{\text{hour}} \cdot {\text{ fee}}{{\text{t}}^{\text{3}}}}}\]
anonymous
  • anonymous
uhmm i'm sorry I didn't quite get it about dividiing by a length^4 part, can you show it to me. i'm really sorry :(
Michele_Laino
  • Michele_Laino
the first unit of measure is: \[\Large \frac{{{\text{grams}}}}{{{\text{min }} \times {{\text{m}}^{\text{3}}}{\text{ }} \times {\text{cm}}}}{\text{ = }}\frac{{{\text{mass}}}}{{{\text{time }} \times {\text{lengt}}{{\text{h}}^{\text{4}}}}}\]
Michele_Laino
  • Michele_Laino
m stands for meter, right?
anonymous
  • anonymous
yes :)
Michele_Laino
  • Michele_Laino
so: m^3*cm = length^4
anonymous
  • anonymous
\[4x10^-5 kg/(\min)(m^4) \times 60 \min./1hr \times 1lbm/0.4536 kg \times (0.3048m)^3/1ft.^3\]
Michele_Laino
  • Michele_Laino
so, there is not cm at the denominator?
anonymous
  • anonymous
oops sorry i'll do it again i forgot to convert cm to m. am I doing it right?
anonymous
  • anonymous
nope, it says on our worksheet that it's placed in the denominator :(
Michele_Laino
  • Michele_Laino
if the dimensions are not equal, we can not change from one unit of measure to another
Michele_Laino
  • Michele_Laino
maybe there is m^2 instead of m^3?
anonymous
  • anonymous
i'll just ask my professor about this quetion, maybe it's a typographical error, it would be easier if cm is not present in the question :)))
Michele_Laino
  • Michele_Laino
ok!
anonymous
  • anonymous
thank you so much for helping me :D
Michele_Laino
  • Michele_Laino
:)

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