in your previous post, is there any way you found the marble mass = 4.5 g? calculations i meanhutchphd said:Your eyeball and ruler will draw the "best" line pretty well. Gives the best fit to all the data.
sure, but as Charles mentioned, y=total measured mass vs. x=number of marbles. To find the number of marbles I need the mass of one so I wanted to know how you can say its 4.5 ghutchphd said:Rough graph in my head. Do the graph. Excel?
The increments are about 4.5 other than the missing 16. This is a very simple graph to draw by hand=if you have some graph paper it helps. Otherwise EXCEL is often used, but that can take some work to learn.orangegalaxies said:in your previous post, is there any way you found the marble mass = 4.5 g? calculations i mean
Hello, after graphing, I got a slope of 4.49992 and a y-intercept of 3.00000061. Is this reasonable?hutchphd said:You only need the integer number of marbles to plot the data. I guessed the 4.5 because it seemed to give integers within reasonable error limits. The error in the mass obtained from the best fit line (slope) from the graph should be good to a few percent. This slope in the Millikan oil drop experiment gave Millikan the charge increment for little balls of oil, electron by electron as he irradiated them). A truly beautiful experiment.
Incidentally it is possible that all the changes in the graph involved always moving 2 marbles (or3 or4) but it is less and less likely More data would make it more certain.
Please draw the graph. I don't think you graphed it, because I just graphed it, and I got round numbers, but not real close to 4.5, and certainly not 3 for the other number. The accuracy is only one decimal place or thereabouts.orangegalaxies said:Hello, after graphing, I got a slope of 4.49992 and a y-intercept of 3.00000061. Is this reasonable?
The graph shows this answer very clearly, and I'm hoping the OP will draw the graph like I did to see this for himself.hutchphd said:And I think the first data cluster is for n=1 marble so the mass of the container is ~6g
To be honest, in the 63 posts so far, there have not been many contributions from you. You need to show a better grasp of the problem yourself.orangegalaxies said:thanks guys, i don't have any graph paper on me so i tried to graph it digitally which didn't work apparently. i will get some today to finish it off. would you be able to help me answer these questions?
- Would having more containers make this easier?: i think so since it would give more data right?
- Would having a large variety of quantity (eg: having max of 7-11 marbles instead of 5-6) make this easier?
- How does this relate to Millikan's experiment?
- What are some sources of error?: damage to either the container/marbles which would alter their mass?
okay thanks. i was told to solve it a different way by my teacher so i tried doing that but the other contributors said to do it another way which i obviously don't know how to do therefore i am unable to contribute. i can only grasp the problem if someone explains which is not the case since it seems more like everyone is asking me why I'm not doing this a certain way instead of telling me WHY AND HOW i am supposed to do it that way.PeroK said:To be honest, in the 63 posts so far, there have not been many contributions from you. You need to show a better grasp of the problem yourself.
You need to take responsibility for your own academic progress. I'm not convinced that it's your teacher's fault - or the people who've posted ideas or suggestions here.orangegalaxies said:okay thanks. i was told to solve it a different way by my teacher so i tried doing that but the other contributors said to do it another way which i obviously don't know how to do therefore i am unable to contribute
Please read my post 58. There is a shortcut presented there that you can do without graph paper that is reasonably accurate.orangegalaxies said:okay thanks. i was told to solve it a different way by my teacher so i tried doing that but the other contributors said to do it another way which i obviously don't know how to do therefore i am unable to contribute. i can only grasp the problem if someone explains which is not the case since it seems more like everyone is asking me why I'm not doing this a certain way instead of telling me WHY AND HOW i am supposed to do it that way.
i'm not saying it's their fault. my teacher doesn't offer any support to their students, i can't teach myself everything through youtube videos and tutorials. i need someone to explain things to me which is why i came here. i am new to these concepts, if someone tells me something once, i won't just take it to be correct, i need to understand why that method was used which takes time.PeroK said:You need to take responsibility for your own academic progress. I'm not convinced that it's your teacher's fault - or the people who've posted ideas or suggestions here.
Here's the first thing I said on this thread:orangegalaxies said:i'm not saying it's their fault. my teacher doesn't offer any support to their students, i can't teach myself everything through youtube videos and tutorials. i need someone to explain things to me which is why i came here. i am new to these concepts, if someone tells me something once, i won't just take it to be correct, i need to understand why that method was used which takes time.
PeroK said:This is perhaps a tough question, as you may actually have to think about what to do.
yes i don't need you to, thanks.PeroK said:Here's the first thing I said on this thread:
We can't think for you: you have to do that yourself.
i got the marble to be around 5.17 g and used this to find the empty container which was 5.83 gCharles Link said:What did you get for an average/estimated mass of the marble? The graph shows our "guess/first look" of 4.5 to be somewhat on the low side. Using the ## n=1 ## and ##n=5 ## points as mentioned in post 58 will give a much more accurate number. This method then computes the mass of 4 marbles using these two points.
You could also similarly compute the mass of two marbles or three marbles using the other points.
The graph is perhaps one of the best ways to compute this, but you need graph paper. I'd be glad to compare your answer to what I got from the graph.
Sounds good. Looks like you took ((31.6+31.9)/2-(11.0+11.1)/2)/4 =5.17, and 11.0-5.17=5.83, but in the future please try to show us a little work.orangegalaxies said:i got the marble to be around 5.17 g and used this to find the empty container which was 5.83 g
i actually found the average mass for the containers that might have had the same number of marbles (11,11.1 and 21.4, 21.5 and 25.7, 25.8, 25.9 and 31.6, 31.9), found their differences, divided the jump from 11 to 21 by 2 and then averaged the three differences. this probably makes no sense lol but yeah your method does work too.Charles Link said:Sounds good. Looks like you took ((31.6+31.9)/2-(11.0+11.1)/2)/4 =5.17, and 11.0-5.17=5.83, but in the future please try to show us a little work.
The graph gives very nearly 5.0 grams for the marble mass and 6.0 grams for the container, but your answer is perfectly valid.
You might want to try the other points and see what they give as well, e.g. n=1 vs. n=3 for the mass of two marbles.
Yes, but as others have also said, we really need more effort from you. e.g. You describe a method, but we need to see the calculations.orangegalaxies said:is it okay if i run what my guesses for the additional questions are by you for feedback?
i ran my method by a classmate and they had an answer a decimal point off by mine. i did my work by myself in a way that made sense to me, I'm sorry, i should have explained it for everyone too. as for the questions:Charles Link said:Yes, but as others have also said, we really need more effort from you. e.g. You describe a method, but we need to see the calculations.
If you averaged all of the other points as well, I would expect it to be closer to the graphical result.
This is where you need to make an effort, and not have me or the other Homework Helpers writing out the steps that you should be coming up with.
yes I've shown my work in the assignment, i just didn't type it all out here, sorry for that. i can't go to the shop now because it's night and my paper is due in a few minutes. i went this morning and there was no graph paper there. i made the best of the situation and used the average method. the only question I'm having trouble with now is "Would having a large variety of quantity (eg: having max of 7-11 marbles instead of 5-6) make this easier?" my guess is that it would be harder but i don't really know why. thanks anyway for your help. i will make something up and hope it is rightCharles Link said:We need you to show calculations. Your teacher should also make you show your work, and not simply give credit for supplying a number.
For sources of error, the biggest one is probably inaccuracy of the scale.
Here's where like the other Homework Helpers have stated, we need more effort from you.
For starters, I still think you would do well to do what you need to do to obtain a couple pieces of graph paper, and graph the data.
