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Absolute/Percentage Uncertainties

Equation: T2= kH

Where T2 (seconds2) and H (metres) are the variables, k is a constant.


I know the percentage uncertainty of T2, 0.01%, and i'm trying to calculate the value of H and its absolute uncertainty.

Does this mean it will also have the same percentage uncertainty of 0.01%, but i will have to calculate its absolute uncertainty by doing "Value of H divided multiplied by 0.01" ?

Sorry if badly worded
Original post by minnigayuen
Equation: T2= kH

Where T2 (seconds2) and H (metres) are the variables, k is a constant.


I know the percentage uncertainty of T2, 0.01%, and i'm trying to calculate the value of H and its absolute uncertainty.

Does this mean it will also have the same percentage uncertainty of 0.01%, but i will have to calculate its absolute uncertainty by doing "Value of H divided multiplied by 0.01" ?

Sorry if badly worded


The % uncertainty in H will be two times the % uncertainty in T (because it's T2)
So in your case it will be 0.02%
Convert this back to an absolute uncertainty.
The absolute uncertainty will be the value you have for H times the % uncertainty divided by 100

0.01% seems very small for an uncertainty. Is this correct?
Reply 2
Original post by Stonebridge
The % uncertainty in H will be two times the % uncertainty in T (because it's T2)
So in your case it will be 0.02%
Convert this back to an absolute uncertainty.
The absolute uncertainty will be the value you have for H times the % uncertainty divided by 100

0.01% seems very small for an uncertainty. Is this correct?


Oh sorry, i meant 0.1%!

Also, the percentage uncertainty is 0.1% for (T2), not T on its own, just to be clear. Should i still double the percentage uncertainty for H?
Original post by minnigayuen
Oh sorry, i meant 0.1%!

Also, the percentage uncertainty is 0.1% for (T2), not T on its own, just to be clear. Should i still double the percentage uncertainty for H?


The % uncertainty for T2 is double the % uncertainty for T
Yes, if you already have the % uncertainty for T2 then you don't need to do anything.
The % uncertainty for H will be whatever the % uncertainty in T2 is.
This is assuming there is no uncertainty in k.
(edited 10 years ago)
Reply 4
Original post by Stonebridge
The % uncertainty for T2 is double the % uncertainty for T
Yes, if you already have the % uncertainty for T2 then you don't need to do anything.
The % uncertainty for H will be whatever the % uncertainty in T2 is.
This is assuming there is no uncertainty in k.


Thank you very muchly!! :smile:
Original post by Stonebridge
The % uncertainty for T2 is double the % uncertainty for T
Yes, if you already have the % uncertainty for T2 then you don't need to do anything.
The % uncertainty for H will be whatever the % uncertainty in T2 is.
This is assuming there is no uncertainty in k.

Can you please tell me what happens if there will be an uncertainty in k ?
It's all in this thread at the top of the forum.
http://www.thestudentroom.co.uk/showthread.php?t=2661762
Original post by Stonebridge
It's all in this thread at the top of the forum.
http://www.thestudentroom.co.uk/showthread.php?t=2661762

Thanks a lot!

I need your help in this question: http://pastpapers.edexcel.com/content/dam/pdf/A%20Level/Physics/2013/Question%20papers%20and%20mark%20schemes/6PH04_01_que_20120611.pdf
Q17 b
I did get some of it, but still not quite confident at all.

Many thanks!
Original post by Daniel Atieh
Thanks a lot!

I need your help in this question: http://pastpapers.edexcel.com/content/dam/pdf/A%20Level/Physics/2013/Question%20papers%20and%20mark%20schemes/6PH04_01_que_20120611.pdf
Q17 b
I did get some of it, but still not quite confident at all.

Many thanks!


Well if you look at that equation in part b it has two parts

Ee/m and l/v

The time the electron is between the plates depends on its horizontal speed which is v. The plates are length l so the time it's in there is l/v (time = distance / velocity)

The electrons experience a force F due to the electric field E which equals Ee where e is the charge on the electron.
So their acceleration, from F=ma will be a = F/m which gives Ee/m

I leave it to you to put the two ideas together.
Original post by Stonebridge
Well if you look at that equation in part b it has two parts

Ee/m and l/v

The time the electron is between the plates depends on its horizontal speed which is v. The plates are length l so the time it's in there is l/v (time = distance / velocity)

The electrons experience a force F due to the electric field E which equals Ee where e is the charge on the electron.
So their acceleration, from F=ma will be a = F/m which gives Ee/m

I leave it to you to put the two ideas together.


Thanks a lot! you made it look so easy.

The thing that confused me is saying the vertical component. So the equation for getting the horizontal component as they leave is the same ?
Original post by Daniel Atieh
Thanks a lot! you made it look so easy.

The thing that confused me is saying the vertical component. So the equation for getting the horizontal component as they leave is the same ?



What's happening to the electron in the electric field is just the same as happens to a projectile with gravity. If you understand the gravitational case you already understand 90% of the electrical case.

If you throw a stone horizontally from the top of a cliff it
a) carries on horizontally with the same horizontal component it started with
b) falls vertically downwards with an acceleration a where a is F/m and F is the gravitational force pulling it down.

In the case of the electron in the tube, it
a) moves horizontally with the velocity component it attained from the anode earlier on.
b) "falls" in the direction of the force due to the electric field which is at right angles to the initial velocity in a)

The maths is identical, just the terms are gravitational in the one and electrical in the other. The path is the same shape, a parabola, in both cases.

A lot of things in physics are interconnected and interrelated. When you start to see this it all becomes much clearer and a lot easier.
(edited 9 years ago)
Original post by Stonebridge
What's happening to the electron in the electric field is just the same as happens to a projectile with gravity. If you understand the gravitational case you already understand 90% of the electrical case.

If you throw a stone horizontally from the top of a cliff it
a) carries on horizontally with the same horizontal component it started with
b) falls vertically downwards with an acceleration a where a is F/m and F is the gravitational force pulling it down.

In the case of the electron in the tube, it
a) moves horizontally with the velocity component it attained from the anode earlier on.
b) "falls" in the direction of the force due to the electric field which is at right angles to the initial velocity in a)

The maths is identical, just the terms are gravitational in the one and electrical in the other. The path is the same shape, a parabola, in both cases.

A lot of things in physics are interconnected and interrelated. When you start to see this it all becomes much clearer and a lot easier.

Well explained! Thank you so much really!


Last thing I ll ask you for now... Well, I know i can compare it straight away with y=mx + c, but what if i divided both sides by h, and said plot T^2 against h? Why it won't work?
qq.JPG
Original post by Daniel Atieh
Well explained! Thank you so much really!


Last thing I ll ask you for now... Well, I know i can compare it straight away with y=mx + c, but what if i divided both sides by h, and said plot T^2 against h? Why it won't work?
qq.JPG



I need the rest of this question. This is part b) and I suspect there is something important in part a)

It would be better to post this (the whole question) in a new thread.
Original post by Stonebridge
I need the rest of this question. This is part b) and I suspect there is something important in part a)

It would be better to post this (the whole question) in a new thread.

http://www.thestudentroom.co.uk/showthread.php?t=2672560

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