|
Okay, I really don't like homework questions but these actually count a lot.
I'm using Mastering Physics (an online course.)
Here's the problem:
As you eat your way through a bag of chocolate chip cookies, you observe that each cookie is a circular disk with a diameter of (8.50 + or - 0.02)cm and a thickness of (8.5 x 10^-3) + or - 0.005 cm.
1. The question asks you to find the average volume of the cookie (in two sig figs)
How do you do it? I keep getting 2.8 but the answer says it is wrong.I have 5 attempts left...
2. Find the uncertainty in the volume of a cookie.
I got 0.3 and it is right. So no help needed here.
3. Find the ratio of the diameter to the thickness.
Express your answer using two significant figures.
I got 0.10 but it's wrong. IDK
4. Find the uncertainty in this ratio.
Express your answer using one significant figure
I got the uncertainty as 20. But it's wrong 
Please help!
|
webassign? haha
sorry cant help you though..forgot all my physics :[
|
On September 09 2008 11:07 BaDayOri wrote:
webassign? haha
webassign? O_o?
lol yeah i forgot ALL my physics and calc during the whole summer. go me!
and wow wtf we're doing this already after first lecture.
the things we learned in lecture and hw is COMPLETELY different and way harder.
|
8.5 : 0.0085
The ratio isn't 0.1, it is 1000:1.
I'm not too sure about what uncertainty is, but I'm going to assume:
(Max of diameter:max of thickness)-(Min of diameter:min of thickness)
8.52:0.00825 - 8.48:0.00815 = 1032.73:1 - 1040.49:1.
Difference in the ratios of ~7.7:1
That is just a guess.
|
On September 09 2008 11:08 DeathTray wrote:
8.5 : 0.0085
The ratio isn't 0.1, it is 1000:1.
I'm not too sure about what uncertainty is, but I'm going to assume:
(Max of diameter:max of thickness)-(Min of diameter:min of thickness)
8.52:0.00825 - 8.48:0.00815 = 1032.73:1 - 1040.49:1.
Difference in the ratios of ~7.7:1
That is just a guess.
Godamn I put 1000:1 and it's wrong
I hate this so much wtf
|
United States24741 Posts
You propagated the uncertainty in the volume correctly but need help with the sig figs?
|
LOL how do you correctly get the hardest problem but not know how to do the easy ones? Volume = base * height, right? base is a circle, pi R^2, r = D/2. You can just ignore the error here since it asks for the average.
And for the ratio, I think maybe you have to write 1000. with the decimal to show that the last 0 is significant. Or just write 10.0 *10^2 haha that's how i always handled sig fig problems.
|
let me show you the
BTW the 1000. doesn't fucking work WTFFFFFFFFFFFFFFFFFFFFFFfffffffffffffff
|
Isn't " 1000. " with the decimal 4 sig figs... -_-;;
1000 w/o decimal is 1 sig fig afaik.
Yah I'm pretty sure the decimal point makes all the 0's become significant.
1.0 * 10^3 is two sig figs I think.
|
CORRECT! Okay two more to go Part 1 and Part 4
i hate this.
|
|
|
On September 09 2008 11:32 viceroy wrote:
average volume = 0.48cm3
It's wrong
|
http://www.online-calculators.co.uk/volumetric/cylindervolume.php
you have a cylinder with a radius of 4.25 and a height of 0.0085 cm => volume = 0.48252678571428537 cm3
I don't see how I could be wrong, since it asks for the average value that means you are using the values without the uncertainty ,,,, so I don't know what's wrong...maybe it should be 0.4825 ? that would be wrong since in the diameter you only have 2 sig... figs
|
Rule: A zero standing alone to the left of a decimal point is not significant.
0.421 g [3] 0.5 m [ ]figs)
|
On September 09 2008 11:45 viceroy wrote:http://www.online-calculators.co.uk/volumetric/cylindervolume.phpyou have a cylinder with a radius of 4.25 and a height of 0.0085 cm => volume = 0.48252678571428537 cm3 I don't see how I could be wrong, since it asks for the average value that means you are using the values without the uncertainty ,,,, so I don't know what's wrong...maybe it should be 0.4825 ? that would be wrong since in the diameter you only have 2 sig... figs
It's 4.8
i got it thanks
|
Now how to calculate the uncertainty?
|
if I take the errors into considerations:
D = 8.52 and H = 0.009 => Volume = 0.51
|
ok that was because you gave me the wrong information height is 8.5x10^-2
|
I don't understand how you've got 0.3 uncertainty for part 2 without knowing the volume 
anyway the uncertainty int the ratio is 60 (not 60. and not 60.0, just 60 because is one sig fig)
|
On September 09 2008 12:17 viceroy wrote:I don't understand how you've got 0.3 uncertainty for part 2 without knowing the volume anyway the uncertainty int the ratio is 60 (not 60. and not 60.0, just 60 because is one sig fig)
That is my friend helped me!
|
|
|
|
|
|
EPT![[image loading]](http://i13.photobucket.com/albums/a255/obstrukt/myhw.jpg)
