Need help we homeschool please - I need to know how to explain to my son

[geissingert]

I had a problem I'm trying to figure out so that I can show my son, we homeschool.

a gold coin has a mass of 9.25 grams. What is its volume if the density of gold is 19300 kg/ m3?

Can you do the work out so that I can actually see the answer so that I can explain it to him?

Here is what I have

m= 9.25 g or 9.25 x 10-^3kg

D= 19,300 kg/m3 or 1.93 x 10^4kg/m^3

V= 9.25 x 10^-3kg / 1.93 x 10^4 kg/m^3

I don't know what to do next, please help.

 

[HallsofIvy]

I had a problem I'm trying to figure out so that I can show my son, we homeschool.

a gold coin has a mass of 9.25 grams. What is its volume if the density of gold is 19300 kg/ m3?

Can you do the work out so that I can actually see the answer so that I can explain it to him?

Here is what I have

m= 9.25 g or 9.25 x 10-^3kg

D= 19,300 kg/m3 or 1.93 x 10^4kg/m^3

V= 9.25 x 10^-3kg / 1.93 x 10^4 kg/m^3

I don't know what to do next, please help.

You've set it up very nicely, density is defined as "mass over volume" or d= m/v. Multiply on both sides by v to get dv= m and then divide both sides by d to solve for v: v= m/d, just as you say. You can also do this by "dimensional analysis" (a fancy name for a very simple idea). Since density has units of "kg/m^3", mass has units of "kg", and you want a result in "m^3" (units for a volume) you want "(kg)(m^3/kg)= m^3" and that "inverting" "kg/m^3" to get "m^3/kg" means you are dividing.

Now, do the division: \(\displaystyle \left(\frac{9.25}{1.93}\right)\left(10^{-3+4}\right)\left(kg \frac{m^3}{kg}\right)\).

So divide 9.25 by 1.93, times \(\displaystyle 10^{-3+4}= 10\) with units of "m cubed= cubic meters" since the "kg" cancel.

 

[HallsofIvy]

I had a problem I'm trying to figure out so that I can show my son, we homeschool.

a gold coin has a mass of 9.25 grams. What is its volume if the density of gold is 19300 kg/ m3?

Can you do the work out so that I can actually see the answer so that I can explain it to him?

Here is what I have

m= 9.25 g or 9.25 x 10-^3kg

D= 19,300 kg/m3 or 1.93 x 10^4kg/m^3

V= 9.25 x 10^-3kg / 1.93 x 10^4 kg/m^3

I don't know what to do next, please help.

You've set it up very nicely, now do the division: \(\displaystyle \frac{9.25}{1.93} 10^{-3+4} kg \frac{m^3}{kg}\).

So divide 9.25 by 1.93, times \(\displaystyle 10^{-3+4}= 10\) with units of "m cubed= cubic meters" since the "kg" cancel.

 

[DrPhil]

You've set it up very nicely, now do the division: \(\displaystyle \left(\frac{9.25}{1.93}\right)\left(10^{-3+4}\right)\left(kg \frac{m^3}{kg}\right)\).

So divide 9.25 by 1.93, times \(\displaystyle 10^{-3+4}= 10\) with units of "m cubed= cubic meters" since the "kg" cancel.

Unfortunately that gives a result that is MUCH to large to be reasonable!

I would revise the units of Density two steps further:
D = 19,300 kg/m^3 = 1.93 x 10^4kg/m^3 = 1.93×10^7 g/(100 cm)^3 = 19.3 g/cm^3
That takes care of all the powers of 10.

Then mass (g) divided by density (g/cm^3) will give volume in (cm^3), which is a much more practical unit for the volume of a coin.

 

[JeffM]

I wish this kind of problem was not assigned before dimensional analysis is taught. The problem really has two pieces, understanding and manipulating a simple formula and making units consistent and appropriate. But if dimensional analysis has not been taught, the second part of the problem may seem difficult and is certainly prone to errors, which means that even the very simple manipulation of the formula required may be made confusing to the student.

I usually love the khan academy videos, but the one for dimensional analysis is not as good as most. Moreover, it is addressed to chemistry students, which was when I was taught it, but it really can and should be taught long before then.

http://www.bing.com/videos/search?q...E147B57123DA5490F9FCE1&view=detail&FORM=VIRE2

 

[HallsofIvy]

You've set it up very nicely, now do the division: \(\displaystyle \frac{9.25}{1.93} 10^{-3+4} kg \frac{m^3}{kg}\).

So divide 9.25 by 1.93, times \(\displaystyle 10^{-3+4}= 10\) with units of "m cubed= cubic meters" since the "kg" cancel.

As per Dr. Phil's correction, this should be
"divide 9.25 by 1.93, times \(\displaystyle 10^{-3-4}= 10^{-7}\) with units of "m cubed= cubic meters"..