Rolls of aluminum foil are 305 mm wide and 0.015 mm thick. What maximum length of aluminum foil can be made from 0.83 kg of aluminum?( d(aluminum)=2.70 g/cm3)
Turn the mass into a volume using the density and then use the volume equation to find the length.
Though you will need to convert everything so that the units match.
V=M/D D=2.70g/cm^3 M=0.83kg \[\left(\begin{matrix}0.83Kg \\ 1\end{matrix}\right)*\left(\begin{matrix}1000g \\ 1kg\end{matrix}\right)=830g\] \[\left(\begin{matrix}830g \\ 270g/cm^3\end{matrix}\right)=307.41cm^3\]
Looks good so far.
im stuck here...how do i account for width and thickness?
V=LWH
Where H would be the thickness
to confirm \[307.41=L*305mm*0.015mm\]
Not quite once again you will need to do conversions
your volume is cm^3 while those measurements are in mm
\[305mm \rightarrow 305000mm^3 \rightarrow 305cm^3\] \[0.015mm \rightarrow 15mm^3 \rightarrow 0.015cm^3\] \[307.41cm^3=L*305cm^3*0.015cm^3\] L=67.2cm^3
67.2cm^3 = 4.07 cm = 0.0407meters
Conversions appear to be off, we can put it all in cm and cm^3 if you like
to go from mm to cm you would divide by 10.
and then you can leave the volume alone
so 305mm /10 =30.5cm and 0.015mm /10 = 0.0015cm
Yes
307.41cm^3 = 6.75cm 6.75=L*30.5*0.0015 L=147.54cm or 1.48 Meters
you can leave the volume as it is, it is already in cm^3.
307.41=L*30.5*.0015 where L will be in cm
answer needs to be in meters so wouldn't i need to convert the cm^3 to cm?
well when you are finished I would convert the length to meters by dividing by 100.
You could turn everything to meters but much easier to just solve for L and then convert that answer to meters.
Oh I see because L W and H are all cm when they multiply the become cm^3 so If I do 307.41=L*30.5*.0015 L=6719.34
Yes
67.2 meters
Looks good
Whew... thank you very much for sticking with me on that one. I really appreciate it
No problem, all the conversions do make it a bit tricky
Join our real-time social learning platform and learn together with your friends!