In the American engineering system of units, the viscosity can gave the units of (lbf)(hr)/(ft^2), while in the handbook the units are (g)/(cm)(s). Convert a viscosity of 20.0 (g)/(m)(s) to the given American Engineering units?
1 gram = 0.0022046 lbs 1m = 39.37 ins
1 m = 3.28083 ft
You can figure out any conversion factor this way, for instance: \[1 \text{ hour} = 60 \text{ min}\] So if you want to convert hours to minutes, like \[120 \text{ minutes}\] you want to divide out the minutes, so we take our first equation and divide both sides by minutes to get: \[\frac{1 \text{ hour}}{60 \text{ minutes} }=1\] See, we can multiply anything by 1 and it doesn't change it, and this thing on the left is equal to 1! \[120 \text{ minutes}*1 =120 \text{ minutes}*\frac{1 \text{ hour}}{60 \text{ minutes} } = 2 \text{ hours} \] The minutes divide out just like anything divided by itself is also 1. \[\frac{7}{7}=1\] \[\frac{minutes}{minutes}=1\] \[\frac{elephants}{elephants}=1\]
\[\frac{ 20g }{ (m)(s) } \times \frac{ 0.0022046lbs}{ 1g } \times \frac{ 1m }{ 3.28083 ft. } \times \frac{ 60s }{ 1 \min.} \times \frac{ 60 mins.}{ 1 hr }\]
is it correct? :)
how can convert it into \[ft^2\] for ft? if the given length is m not\[m^2\]
@Empty
and how can i put hr into the numerator?
can you help me @ganeshie8 and @iGreen ? :)
have another look at your question - you have ft^2 in american system but m in the metric - is this correct?
looks like thers a typo somewhere
yeah :(, i'll just ask my professor about this one. Thank you so much :)
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