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gwiyomi17

no idea about this one

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Question :

If you are given the total mass of clips in each four containers and without knowing how many numbers of clips are there, how will you find out the mass of one clip?

First container: 7.5 g
Second container: 20 g
Third container: 50 g
Fourth container: 125 g

 

I have no idea how this relates to our topic in class. :confused:

Our topic in class is about the Millikan Oil drop and J.J thompson's charge-to-mass ratio.

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Question :

If you are given the total mass of clips in each four containers and without knowing how many numbers of clips are there, how will you find out the mass of one clip?

First container: 7.5 g

Second container: 20 g

Third container: 50 g

Fourth container: 125 g

 

I have no idea how this relates to our topic in class. :confused:

Our topic in class is about the Millikan Oil drop and J.J thompson's charge-to-mass ratio.

I have no idea about Millikan or Thompson, but I'll take a math stab.

 

Assuming there are no partial clips and all clips weigh the same have the same mass.

If the first container contains 1 clip then the masses in the others must be multiples. Try 20/7.5=2.666... Nope

Try 2 clips in #1. 7.5/2=3.75 then 20/3.75=5.333... Nope.

Try 3 clips in #1. 7.5/3=2.5 then 20/2.5=8. Yes. Then 50/2.5=20. Yes. Then 125/2.5=50. Yes

 

Assuming further that 'clips' mean paper clips, what's a reasonable estimate for what a paper clip weighs the mass of a paper clip? A large one might have mass 2.5 gms so we have our answer. Smaller ones would have to have masses that are even divisions of our answer or 2.5/2=1.25 gms or 2.5/4=.625 gms.

 

That's my best guess. :unsure:

Edited by Acme

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I have no idea about Millikan or Thompson, but I'll take a math stab.

 

Assuming there are no partial clips and all clips weigh the same have the same mass.

If the first container contains 1 clip then the masses in the others must be multiples. Try 20/7.5=2.666... Nope

Try 2 clips in #1. 7.5/2=3.75 then 20/3.75=5.333... Nope.

Try 3 clips in #1. 7.5/3=2.5 then 20/2.5=8. Yes. Then 50/2.5=20. Yes. Then 125/2.5=50. Yes

 

Assuming further that 'clips' mean paper clips, what's a reasonable estimate for what a paper clip weighs the mass of a paper clip? A large one might have mass 2.5 gms so we have our answer. Smaller ones would have to have masses that are even divisions of our answer or 2.5/2=1.25 gms or 2.5/4=.625 gms.

 

That's my best guess. :unsure:

But I think we have to assume that the mass of the paper clips does not have to fall within standard paper clip mass, so guessing in that sense wouldn't work though it would be a good logic basis.

 

Earlier I was trying to take a stab at the problem and I find it difficult to find the mass of a paper clip without more information available. If this is a homework problem I would assume it to be solvable though.

 

Also, I found some relationship between the oil drop experiment and this problem. It might be useful for solving it:

 

 

 

The experiment entailed balancing the downward gravitational force with the upward drag and electric forces on tiny charged droplets of oil suspended between two metal electrodes. Since the density of the oil was known, the droplets' masses, and therefore their gravitational and buoyant forces, could be determined from their observed radii. Using a known electric field, Millikan and Fletcher could determine the charge on oil droplets in mechanical equilibrium. By repeating the experiment for many droplets, they confirmed that the charges were all multiples of some fundamental value, and calculated it to be−1.5924(17)×10−19C, within 1% of the currently accepted value of −1.602176487(40)×10−19 C. They proposed that this was the charge of a single electron.

http://en.wikipedia.org/wiki/Oil_drop_experiment

Edited by Unity+

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Each container contains integer multiply of clips of the same mass.

 

m - mass of clip (might be fractional)

x,y,z,zz - quantity of clips (must be integer)

 

m*x=7.5 g

m*y=20 g

m*z=50 g

m*zz=125 g

 

so

m=7.5/x=20/y=50/z=125/zz

 

The first m fulfilling criteria is m=2.5 g, then 1.25 g, 0.625 g etc

Edited by Sensei

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But I think we have to assume that the mass of the paper clips does not have to fall within standard paper clip mass, so guessing in that sense wouldn't work though it would be a good logic basis.

No, you are mistaken.

 

Earlier I was trying to take a stab at the problem and I find it difficult to find the mass of a paper clip without more information available. If this is a homework problem I would assume it to be solvable though.

No; you are again mistaken.

Each container contains integer multiply of clips of the same mass.

 

m - mass of clip (might be fractional)

x,y,z,zz - quantity of clips (must be integer)

 

m*x=7.5 g

m*y=20 g

m*z=50 g

m*zz=125 g

 

so

m=7.5/x=20/y=50/z=125/zz

 

The first m fulfilling criteria is m=2.5 g, then 1.25 g, 0.625 g etc

Thank you for repeating my answer...I think. :unsure:

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I have no idea how this relates to our topic in class. :confused:

Our topic in class is about the Millikan Oil drop and J.J thompson's charge-to-mass ratio.

 

Let us assume that we do not know that the containers contain clips. Then it turns out when we do measurements the mass content of the containers come in multiples of 2.5 g. What would we conclude about the container contents?

Compare to the Millikan experiment. We measure the charge of the oil drops and find that they come in multiples of 1.6e-19 C. What do we conclude about the oil drop charges?

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No, you are mistaken. No; you are again mistaken.

Thank you for repeating my answer...I think. :unsure:

Enlighten me oh wise one.

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Enlighten me oh wise one.

I already have. Moreover, Sensei and Orodruin have succinctly -if not simply- echoed my explication.

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I already have. Moreover, Sensei and Orodruin have succinctly -if not simply- echoed my explication.

I was on tablet and didn't see those response. My apologies. :)

 

Also, I want to make sure you knew that I didn't thumb you down.

Edited by Unity+

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I was on tablet and didn't see those response. My apologies. :)

No worries. After the fact I rather expected some chastisement on my initial post for doing someone's homework, even though I left it to the original poster to figure out how to apply my method to the specifics on Millikan and Thompson. I did not anticipate folks repeating my post.

 

Also, I want to make sure you knew that I didn't thumb you down.

Thanks. I have no doubt now that it was someone else that took umbrage to my wisdom. Que sera sera. :)

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Somebody neg voted in this thread? :o

 

Why?

 

I see no single reason to do so...

 


Thank you for repeating my answer...I think. :unsure:

 

Sorry. I didn't read posts, just initial.

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Somebody neg voted in this thread? :o

Why?

I see no single reason to do so...

Who knows. No biggy. :cool:

 

Sorry. I didn't read posts, just initial.

So too with Unity. Having made the mistake myself in the past, I now make it a habit to read all posts before joining in.

I do appreciate your giving your take inasmuch as it agreed with my suggestion and I was a bit unsure if it was the right approach for the original poster. Perhaps that member will post again to comment on our responses. :)

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