Question

The mass of a solid cube is 905 g, and each edge has a length of 5.05 cm. Find the density of the cube. Answer in units of kg/m 3

Answer 1

\(\huge d=\frac{m}{v}\)
m=?
v=?

Answer 2

i have tried using that but the only thing i need to know is how do i convert 5.05cm into m^3?

Answer 3

you can't. you can convert cm^3 to m^3 though

Answer 4

so am i just supposed to cube cm and then convert it to meters^3?

Answer 5

well first, you have to get the volume of a cube with side length 5.05cm, what would that be?

Answer 6

127.79?

Answer 7

I got 128.787625

Answer 8

hmm i must have messed up then. so then what would i do with that number?

Answer 9

we use dimensional analysis to get it into meters cubed.

\(\large \frac{905g}{128.787625cm^3}=\frac{905g}{128.787625\cancel{cm}*\cancel{cm}*\cancel{cm}} *\frac{100\cancel{cm}*100\cancel{cm}*100\cancel{cm}}{1m*1m*1m}\)

Answer 10

and what would happen next?

Answer 11

well what we have no is in terms of g/m^3 but we want kg/m^3

Answer 12

so convert g into kg?

Answer 13

yeah.

Answer 14

ok give me a sec.

Answer 15

ok so i ended up with .905 kg/ 128.787625 m^3 and i just do the division?

Answer 16

not exactly. Recall that we have to multiply by 100 3 times to get it into m^3

Answer 17

i mean yeah i remember

Answer 18

\(\large \frac{905g}{128.787625cm^3}=\frac{905g}{128.787625\cancel{cm}*\cancel{cm}*\cancel{cm}} \)
\(\large \frac{905\cancel{g}}{128.787625\cancel{cm}*\cancel{cm}*\cancel{cm}} *\frac{100\cancel{cm}*100\cancel{cm}*100\cancel{cm}}{1m*1m*1m}*\frac{1kg}{1000\cancel{g}}\) is the full work

Answer 19

so then i would do 905000000 kg/128787.625m^3 ?

Answer 20

well let's simplify it using algebra. we have 1000 in the denominator so we can cancel one 100 and a 10.
\[\large \frac{905\cancel{g}}{128.787625\cancel{cm}*\cancel{cm}*\cancel{cm}} *\frac{100\cancel{cm}*10\cancel{100}\cancel{cm}*\cancel{100}\cancel{cm}}{1m*1m*1m}*\frac{1kg}{\cancel{1000}\cancel{g}}\]

Answer 21

poopq

Answer 22

\[\frac{905\cancel{g}}{128.787625\cancel{cm}*\cancel{cm}*\cancel{cm}} *\frac{100\cancel{cm}*10\cancel{100}\cancel{cm}*\cancel{100}\cancel{cm}}{1m*1m*1m}*\frac{1kg}{\cancel{1000}\cancel{g}}\]

Answer 23

so 905000 kg/ 128.787625 m^3?

Answer 24

which is less ugly if we write
\(\huge \frac{905,000kg}{128.787625m^3}\) yeah

Answer 25

ohhhhh i got 7027.072671. would any more math have to be done or is that it?

Answer 26

http://www.wolframalpha.com/input/?i=7.0270726g%2Fcm%5E3+to+kg%2Fm%5E3
no. wolfram agrees with us

Answer 27

yessss thank you so much i have been stuck on this problem all day for ap physics thanks:)!

Answer 28

wat
this feels chem-y

oh well, np np