How do I calculate the amount from the volume?

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Discussion Overview

The discussion revolves around calculating the area that gold leaf from a gold bar can cover, given its thickness and the volume of the bar. Participants explore the conversion of units and the application of formulas related to volume and area, with a focus on mathematical reasoning and unit conversions.

Discussion Character

  • Mathematical reasoning
  • Technical explanation
  • Homework-related

Main Points Raised

  • One participant states that the gold leaf is a rectangular solid and suggests using the formula V = Ah, where V is volume, A is area, and h is thickness.
  • Several participants emphasize the importance of converting all measurements to the same units before performing calculations, specifically converting dm to nm and calculating nm³ in a dm³.
  • A participant calculates the volume of a gold bar in nm³ and applies the formula to find the area covered by the gold leaf, leading to a discussion about unit conversions to m².
  • Another participant expresses doubt about an initial answer of 0.01 m², suggesting it seems too small for the area that gold leaf would cover.
  • There is a correction regarding the conversion of nm² to m², with participants discussing the necessary calculations and expressing uncertainty about the final result.
  • One participant arrives at a calculation of 6.5 e 21 nm² and discusses the conversion to m², leading to a final figure of 6500 m² after corrections.

Areas of Agreement / Disagreement

Participants generally agree on the need for unit conversions and the application of the volume-area relationship, but there is some uncertainty in the calculations and final results, particularly regarding the conversion process and the interpretation of the final area covered.

Contextual Notes

Some participants express difficulty with the mathematical aspects, and there are unresolved steps in the calculations, particularly in ensuring accuracy during conversions and the handling of large numbers.

Math_312 46
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The gold pavilion in the city of Kyoto is covered with gold leaf. Gold leaf has a thickness of 100 nm (nanometres) and is made from gold bars. A gold bar has a volume of 0.65 dm^3. How many square metres can the gold leaf from a gold bar cover?
 
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The gold leaf is (presumably) a rectangular solid, ie, flat. So V = Ah.

-Dan
 
Math_312 46 said:
The gold pavilion in the city of Kyoto is covered with gold leaf. Gold leaf has a thickness of 100 nm (nanometres) and is made from gold bars. A gold bar has a volume of 0.65 dm^3. How many square metres can the gold leaf from a gold bar cover?
Your first task should be to express everything in the same units. So how many nm are there in a dm?

Next step: how many nm^3 in a dm^3? (You'll be getting some large powers of 10 at this stage.)

If the volume of a gold bar is X nm^3 then the formula V=Ah (with H = 100nm) tells you that the area covered by the gold leaf will be given by 100A = X.

Finally, that area A is in units of nm^2. You need to convert that to m^2 to get the result in square metres.
 
Opalg said:
Your first task should be to express everything in the same units. So how many nm are there in a dm?

Next step: how many nm^3 in a dm^3? (You'll be getting some large powers of 10 at this stage.)

If the volume of a gold bar is X nm^3 then the formula V=Ah (with H = 100nm) tells you that the area covered by the gold leaf will be given by 100A = X.

Finally, that area A is in units of nm^2. You need to convert that to m^2 to get the result in square metres.
Is the answer 0,01m^2?
If this is wrong can you help me?
 
Math_312 46 said:
Is the answer 0,01m^2?
That really doesn't seem very likely, does it? In fact, 0.01m^2 is one hundredth of a square metre. That is less than the size of a sheet of paper. If you take a gold bar and hammer it out until it is microscopically thin (which is what gold leaf is), you would expect it to cover quite a large area, certainly many hundreds of square metres.
Math_312 46 said:
If this is wrong can you help me?
A decimetre (dm) is one tenth of a metre. A nanometre (nm) is one billionth of a metre. So there are $10^9$ nm in a metre, and therefore $10^8$ nm in a dm. When you cube that in order to get a measure of volume, you find that there are $10^{24}$ nm^3 in a dm^3. So a volume of $0.65$ dm^3 is equal to $0.65\times10^{24}$ (or equivalently $6.5\times10^{23}$) nm^3.

Can you continue from there?
 
Opalg said:
That really doesn't seem very likely, does it? In fact, 0.01m^2 is one hundredth of a square metre. That is less than the size of a sheet of paper. If you take a gold bar and hammer it out until it is microscopically thin (which is what gold leaf is), you would expect it to cover quite a large area, certainly many hundreds of square metres.

A decimetre (dm) is one tenth of a metre. A nanometre (nm) is one billionth of a metre. So there are $10^9$ nm in a metre, and therefore $10^8$ nm in a dm. When you cube that in order to get a measure of volume, you find that there are $10^{24}$ nm^3 in a dm^3. So a volume of $0.65$ dm^3 is equal to $0.65\times10^{24}$ (or equivalently $6.5\times10^{23}$) nm^3.

Can you continue from there?

Sorry in advance but I find maths very difficult. But according to my calculations based on your explanation, it came out like this:

6.5 x 10^23 / 100 = 65 e 21 (65 and 21 zeros)

And thus I get the answer that 65 e 21nm² can the gold leaf of a gold bar cover?
 
Math_312 46 said:
Sorry in advance but I find maths very difficult. But according to my calculations based on your explanation, it came out like this:

6.5 x 10^23 / 100 = 65 e 21 (65 and 21 zeros)

And thus I get the answer that 65 e 21nm² can the gold leaf of a gold bar cover?
You left out the decimal point, so it should be 6.5 e 21 nm^2. But the question asks for the answer in square metres, so you still have to divide by $\bigl(10^9\bigr)^2 = 10^{18}$ to get the answer in those units.
 
Opalg said:
You left out the decimal point, so it should be 6.5 e 21 nm^2. But the question asks for the answer in square metres, so you still have to divide by $\bigl(10^9\bigr)^2 = 10^{18}$ to get the answer in those units.

Ah, I see. so if I convert 6.5 e 21 nm² to m², which then becomes 65000 m². Is this the answer or is something wrong?
 
Math_312 46 said:
Ah, I see. so if I convert 6.5 e 21 nm² to m², which then becomes 65000 m². Is this the answer or is something wrong?
Nearly there! You have one too many zeros. It should be 6.5 e 3 = 6500 m^2.
 
  • #10
Opalg said:
Nearly there! You have one too many zeroes. It should be 6.5 e 3 = 6500 m^2.

So to be sure the answer is 6500 m^2 of the whole question?
 
  • #11
Math_312 46 said:
So to be sure the answer is 6500 m^2 of the whole question?
Yes. :)
 
  • #12
Opalg said:
Yes. :)

Thank you so much, you are a lifesaver:love:

I don't want to sound annoying, but when I was trying to solve another difficult task, I was stuck on how to figure out the lineup itself.

10.2 n + 10.8 2n / n + 2n

Question: In a survey, several measurements were taken and the average of the measurements was 10.2. In a new survey, twice as many measurements were taken and the mean of these measurements was 10.8. What is the mean of all the measurements?

It's all right if you don't want to help me with another question.
 

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