- #36
jbriggs444
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Note the caveats I've edited into the previous post.Ok . thanks again
Note the caveats I've edited into the previous post.Ok . thanks again
Yes, almost certainly. The units I would expect area to be reported in would be square meters.
However... The rules for significant figures suggest that less precision should be reported. And the calculated result you give is incorrect in the last two digits.
##\pi## is not equal to 3.142 -- at least not to the six significant figures that you quoted in the result.I did , 0.025m x 0.025m = 0.000625 m2
0.000625 x 3.142 = 0.00196375m2
##\pi## is not equal to 3.142 -- at least not to the six significant figures that you quoted in the result.
##\pi## is not equal to 3.142 -- at least not to the six significant figures that you quoted in the result.
You've not told us that someone told you that you were mistaken. We are not mind readers.so that could be were I m mistaken ?
You've not told us that someone told you that you were mistaken. We are not mind readers.
When in doubt, check it out.I mean I am asking if it is that what I am doing wrong. If it is because I am using only till 3.142 than 6 significant figures.
thanks for your time I will study more to what you wrote , it will help for sure.
So from you wrote , the best thing is that they tell you to how many numbers should be rounded , the pi and the answer.
No, it is not.When you make 3.14 x ( 0.025m x 0.025m) = 0.0019625m2 converted to mm is 1.9625mm2
No, it is not.
As I'd already pointed out in post #27, the conversion factor between square meters and square millimeters is 1,000,000. There are one million squares, one millimeter on a side in a square that is one meter on a side.
You are not converting from meters to millimeters. You are converting from square meters to square millimeters.
$$0.0019625 \text{ m}^2 \times \frac{1,000,000 \text{ mm}^2}{1 \text{ m}^2} = 1962.5 \text{ mm}^2$$
Given an input measurement of 25 millimeters radius and desiring a result for the area of a corresponding circle, there are two ways to proceed.oh I think I get it now so the best is convert in the question , because they told me that the answer have to be in Newtons and meters.
Who is "they"? Has this been for your homework all along?because they told me that the answer have to be in Newtons and meters.
Who is "they"? Has this been for your homework all along?
Given an input measurement of 25 millimeters radius and desiring a result for the area of a corresponding circle, there are two ways to proceed.
1. Convert 25 millimeters to meters and compute ##\pi r^2## with the result in square meters.
2. Compute ##\pi r^2## with the result in square millimeters and convert the result to square meters.
Let us proceed with the first approach.
$$r=25\text{ mm} \times \frac{1\text{ m}}{1000\text{ mm}}=0.025\text{m}$$ $$a=\pi r^2=\pi (0.025\text{ m})^2=0.0019635\text{ m}^2$$ Round to 0.0020 m2
Let us try the second approach instead.
$$r=25\text{ mm}$$ $$a=\pi r^2=\pi\ (25\text{ mm})^2=1963.5\text{ mm}^2=1963.5\text{ mm}^2 \times (\frac{1\text{ m}}{1000\text{ mm}})^2=\frac{1963.5}{1000 \times 1000}\text{ m}^2 = 0.0019635\text{ m}^2$$ Round to 0.0020 m2
Six of one. Half dozen of the other.
That's the same result, just written in a different format. The e-5 at the end of the first number is a standard computer and calculator way of writing ##×10^{-5}##. So that number is ##3.333×10^{-5}##, which is ##3.333×\frac 1{10^5}## or ##3.333×\frac 1{100,000}## or 0.00003333, which is what the other calculator says.
This stuff is like learning where to put your fingers on a guitar or something. It's slow going and you get a lot of nasty sounds, but you have to learn it if you want to play. It does eventually become second nature and it's really difficult to remember not knowing how to do it.Oh so much to learn I guess .
Thanks
This stuff is like learning where to put your fingers on a guitar or something. It's slow going and you get a lot of nasty sounds, but you have to learn it if you want to play. It does eventually become second nature and it's really difficult to remember not knowing how to do it.![]()
Use the one for which you have the required variables.E is equal to two different formulas
Online converters work. Just have to use the right units. I Googled "convert square millimeters to square meters". Without leaving the Google page, a converter was displayed. I keyed in 1963.5.Ok so you did convert from square mm to square m but in the proper way , I used an online converter.
Online converters work. Just have to use the right units. I Googled "convert square millimeters to square meters". Without leaving the Google page, a converter was displayed. I keyed in 1963.5.
View attachment 255555
Use the one for which you have the required variables.
Multiply what by what or work what division? Are you seriously asking how to evaluate ##\frac{a}{b} \times \frac{c}{d}##?Yeah and is it that you multiply first or work the division separated and the multiply the results ?
Multiply what by what or work what division? Are you seriously asking how to evaluate ##\frac{a}{b} \times \frac{c}{d}##?
Here in the U.S. we are taught to add and multiply fractions at about age eight.Yes is that so basic ?
Here in the U.S. we are taught to add and multiply fractions at about age eight.
For ##\frac{a}{b} \times \frac{c}{d}## there are many ways to evaluate the result and get the same right answer. The canonical "right" way is to:
1. Divide a by b giving the result ##\frac{a}{b}##.
2. Divide c by d giving the result ##\frac{c}{d}##.
3. Multiply those two results together.
Another way is to:
1. Multiply a by c giving the result ##ac##.
2. Multiply b by d giving the result ##bd##.
3. Divide the result from 1) by the result from 2).
It is a rule of arithmetic and of real algebra that$$\frac{a}{b} \times \frac{c}{d} = \frac{ac}{bd}$$
One rule of algebra that they didn't teach us until age 16 or so was that putting two variable names side by side ("juxtaposition") is a notation that conventionally means "multiply them together".
[As it turns out my first exposure to the juxtaposition notation was in a standardized test -- they tested on it before my school had ever presented the content. It is over 40 years later now and that incident still cheeses me off]
Maybe I missed it in the thread, but this chart of metric prefixes and corresponding powers of ten may come in handy for you. http://www.ibiblio.org/units/prefixes.html