Physics % uncertainty (updated) HELP HELP HELP!!!!!!!
I know I've made this thread before but this my updated solution:
Question is attached on the thumbnail below:
Answer:
For part a)
i) Mean diameter = (10.2 + 10.7 + 10.1 + 10.5 ) = 10.4mm
Uncertainty = (10.7 - 10.1) = 0.6/2 = 0.3. 10.4 +- 0.3
ii) Percentage uncertainty for diameter = (0.3)/ 10.4 x 100 = 2.88%
Percentage uncertainty for radius = 0.3/5.2 x 100 ?? ( not sure what the scale resolution is 0.03?)
iii) Volume = 4/3 pi Radius cubed = 4/3 x pi x (5.2mm)^3 = 588.98 mm cubed
0.58898 cm cubed
For part b)
% uncertainty in total mass = 0.01/11.16 x 100 = 0.0896...% = 0.09%
% uncertainty for one ball (mass) = 0.001/1.116 x 100 = 0.896...= 0.9%
For part c )
D= mass/ volume = 1.116g/ 0.58898cm^3 =1.89 g/cm^3 and 1890 kg/m^3
d) uncertainty of density would be... ???
Question is attached on the thumbnail below:
Answer:
For part a)
i) Mean diameter = (10.2 + 10.7 + 10.1 + 10.5 ) = 10.4mm
Uncertainty = (10.7 - 10.1) = 0.6/2 = 0.3. 10.4 +- 0.3
ii) Percentage uncertainty for diameter = (0.3)/ 10.4 x 100 = 2.88%
Percentage uncertainty for radius = 0.3/5.2 x 100 ?? ( not sure what the scale resolution is 0.03?)
iii) Volume = 4/3 pi Radius cubed = 4/3 x pi x (5.2mm)^3 = 588.98 mm cubed
0.58898 cm cubed
For part b)
% uncertainty in total mass = 0.01/11.16 x 100 = 0.0896...% = 0.09%
% uncertainty for one ball (mass) = 0.001/1.116 x 100 = 0.896...= 0.9%
For part c )
D= mass/ volume = 1.116g/ 0.58898cm^3 =1.89 g/cm^3 and 1890 kg/m^3
d) uncertainty of density would be... ???
Scroll to see replies
Bump!!!!!!!!!!!!!
ANYONE!!!!!!!!
Original post by Wasuppdude
Question is attached on the thumbnail below:
Answer:
For part a)
i) Mean diameter = (10.2 + 10.7 + 10.1 + 10.5 ) = 10.4mm
Uncertainty = (10.7 - 10.1) = 0.6/2 = 0.3. 10.4 +- 0.3mm
ii) Percentage uncertainty for diameter = (0.3)/ 10.4 x 100 = 2.88%
OK so far
Percentage uncertainty for radius = 0.3/5.2 x 100 ?? ( not sure what the scale resolution is 0.03?)
If you divide (or multiply) a value by a number, it doesn't change the % uncertainty.
The % uncertainty in the radius is the same as the diameter.
However it does change the actual (absolute) uncertainty.
If you half the value the uncertainty in the radius is half what it was for the diameter.
So the radius is probably best given as 5.20 ±0.15mm (that is still 2.88%)
iii) Volume = 4/3 pi Radius cubed = 4/3 x pi x (5.2mm)^3 = 588.98 mm cubed
0.58898 cm cubed
Yes 588.98mm3 is the volume of the sphere.
To get the % error in this value, you need to know the % error in the value of r that is used to calculate it.
This is explained on the sheet I posted in the other thread.
Have you read it?
If you cube a number (here radius cubed) the % error is 3 times the amount in the final result. So the volume is expressed to ± about 8.5% (3 x 2.88%)
(If you square a value with a % error, the % error in the final result is doubled etc.
If you multiply or divide two values each with a % error, the error in the final result is the sum of the % errors in those values.)
For the other parts the technique is the same.
If you haven't ever studied % errors it is difficult to know where to begin.
This is not the place to try to teach you. This should be done in school/college by your teacher.
The sheet I posted in the other thread is a good place to start.
My best advice is to read it and come back with specific questions about the parts you don't understand.
(edited 12 years ago)
Original post by Stonebridge
OK so far
If you divide (or multiply) a value by a number, it doesn't change the % uncertainty.
The % uncertainty in the radius is the same as the diameter.
However it does change the actual (absolute) uncertainty.
If you half the value the uncertainty in the radius is half what it was for the diameter.
So the radius is probably best given as 5.20 ±0.15mm (that is still 2.88%)
Yes 588.98mm3 is the volume of the sphere.
To get the % error in this value, you need to know the % error in the value of r that is used to calculate it.
This is explained on the sheet I posted in the other thread.
Have you read it?
If you cube a number (here radius cubed) the % error is 3 times the amount in the final result. So the volume is expressed to ± about 8.5% (3 x 2.88%)
(If you square a value with a % error, the % error in the final result is doubled etc.
If you multiply or divide two values each with a % error, the error in the final result is the sum of the % errors in those values.)
For the other parts the technique is the same.
If you haven't ever studied % errors it is difficult to know where to begin.
This is not the place to try to teach you. This should be done in school/college by your teacher.
The sheet I posted in the other thread is a good place to start.
My best advice is to read it and come back with specific questions about the parts you don't understand.
If you divide (or multiply) a value by a number, it doesn't change the % uncertainty.
The % uncertainty in the radius is the same as the diameter.
However it does change the actual (absolute) uncertainty.
If you half the value the uncertainty in the radius is half what it was for the diameter.
So the radius is probably best given as 5.20 ±0.15mm (that is still 2.88%)
Yes 588.98mm3 is the volume of the sphere.
To get the % error in this value, you need to know the % error in the value of r that is used to calculate it.
This is explained on the sheet I posted in the other thread.
Have you read it?
If you cube a number (here radius cubed) the % error is 3 times the amount in the final result. So the volume is expressed to ± about 8.5% (3 x 2.88%)
(If you square a value with a % error, the % error in the final result is doubled etc.
If you multiply or divide two values each with a % error, the error in the final result is the sum of the % errors in those values.)
For the other parts the technique is the same.
If you haven't ever studied % errors it is difficult to know where to begin.
This is not the place to try to teach you. This should be done in school/college by your teacher.
The sheet I posted in the other thread is a good place to start.
My best advice is to read it and come back with specific questions about the parts you don't understand.
But I don't need to calculate percentage error in these questions, maybe thats why we haven't covered it yet.
Original post by Stonebridge
OK so far
If you divide (or multiply) a value by a number, it doesn't change the % uncertainty.
The % uncertainty in the radius is the same as the diameter.
However it does change the actual (absolute) uncertainty.
If you half the value the uncertainty in the radius is half what it was for the diameter.
So the radius is probably best given as 5.20 ±0.15mm (that is still 2.88%)
Yes 588.98mm3 is the volume of the sphere.
To get the % error in this value, you need to know the % error in the value of r that is used to calculate it.
This is explained on the sheet I posted in the other thread.
Have you read it?
If you cube a number (here radius cubed) the % error is 3 times the amount in the final result. So the volume is expressed to ± about 8.5% (3 x 2.88%)
(If you square a value with a % error, the % error in the final result is doubled etc.
If you multiply or divide two values each with a % error, the error in the final result is the sum of the % errors in those values.)
For the other parts the technique is the same.
If you haven't ever studied % errors it is difficult to know where to begin.
This is not the place to try to teach you. This should be done in school/college by your teacher.
The sheet I posted in the other thread is a good place to start.
My best advice is to read it and come back with specific questions about the parts you don't understand.
If you divide (or multiply) a value by a number, it doesn't change the % uncertainty.
The % uncertainty in the radius is the same as the diameter.
However it does change the actual (absolute) uncertainty.
If you half the value the uncertainty in the radius is half what it was for the diameter.
So the radius is probably best given as 5.20 ±0.15mm (that is still 2.88%)
Yes 588.98mm3 is the volume of the sphere.
To get the % error in this value, you need to know the % error in the value of r that is used to calculate it.
This is explained on the sheet I posted in the other thread.
Have you read it?
If you cube a number (here radius cubed) the % error is 3 times the amount in the final result. So the volume is expressed to ± about 8.5% (3 x 2.88%)
(If you square a value with a % error, the % error in the final result is doubled etc.
If you multiply or divide two values each with a % error, the error in the final result is the sum of the % errors in those values.)
For the other parts the technique is the same.
If you haven't ever studied % errors it is difficult to know where to begin.
This is not the place to try to teach you. This should be done in school/college by your teacher.
The sheet I posted in the other thread is a good place to start.
My best advice is to read it and come back with specific questions about the parts you don't understand.
Also what will be the scale resolution for density??
Original post by Wasuppdude
But I don't need to calculate percentage error in these questions, maybe thats why we haven't covered it yet.
Qu a) 2 and Qu b) both ask for % uncertainty.
In order to calculate the actual uncertainty as asked in the other part, you first need to find the % uncertainty.
I suggest you wait until you have covered this in class before attempting these questions.
Original post by Stonebridge
Qu a) 2 and Qu b) both ask for % uncertainty.
In order to calculate the actual uncertainty as asked in the other part, you first need to find the % uncertainty.
I suggest you wait until you have covered this in class before attempting these questions.
Qu a) 2 and Qu b) both ask for % uncertainty.
In order to calculate the actual uncertainty as asked in the other part, you first need to find the % uncertainty.
I suggest you wait until you have covered this in class before attempting these questions.
Yeah but it says percentage uncertainty of the mass? not the volume.
wouldn't that just be 0.01/11.16 x 100 (for total mass)
0.001/1.116 x 100 ( for mass of one ball)
Original post by Wasuppdude
Yeah but it says percentage uncertainty of the mass? not the volume.
wouldn't that just be 0.01/11.16 x 100 (for total mass)
0.001/1.116 x 100 ( for mass of one ball)
wouldn't that just be 0.01/11.16 x 100 (for total mass)
0.001/1.116 x 100 ( for mass of one ball)
The 10 balls mass was 11.16±0.01g
So yes, the % uncertainty is 100 x 0.01/11.16 (that's about 0.1% and is very small)
For one ball the mass is 1.116±0.001g (same % error)
The % uncertainty remains the same but the actual uncertainty is one tenth.
That's why it's better to take the mean of 10 readings that to take just 1.
If you then go on the calculate, say, density using the formula
density = mass/volume
the % error in the density is equal to the sum of the % errors in the mass and the volume
You can add % errors this way. It's explained on the sheet I posted.
You can't add the actual errors unless the formula is adding two quantities with the same units.
Once you have the final % error you need to convert it back to an actual error.
For example, if you had a value of 10.0 with a % error of ±5%
you find 5% of 10.0, which is 0.5, and give your answer as
10.0±0.5
Do the same for the density. You have the % errors for the mass and the volume.
In this case the error for the mass is very small compared with the error for the volume and can actually be ignored.
(edited 12 years ago)
Original post by Stonebridge
The 10 balls mass was 11.16±0.01g
So yes, the % uncertainty is 100 x 0.01/11.16 (that's about 0.1% and is very small)
For one ball the mass is 1.116±0.001g (same % error)
The % uncertainty remains the same but the actual uncertainty is one tenth.
That's why it's better to take the mean of 10 readings that to take just 1.
If you then go on the calculate, say, density using the formula
density = mass/volume
the % error in the density is equal to the sum of the % errors in the mass and the volume
You can add % errors this way. It's explained on the sheet I posted.
You can't add the actual errors unless the formula is adding two quantities with the same units.
Once you have the final % error you need to convert it back to an actual error.
For example, if you had a value of 10.0 with a % error of ±5%
you find 5% of 10.0, which is 0.5, and give your answer as
10.0±0.5
Do the same for the density. You have the % errors for the mass and the volume.
In this case the error for the mass is very small compared with the error for the volume and can actually be ignored.
So yes, the % uncertainty is 100 x 0.01/11.16 (that's about 0.1% and is very small)
For one ball the mass is 1.116±0.001g (same % error)
The % uncertainty remains the same but the actual uncertainty is one tenth.
That's why it's better to take the mean of 10 readings that to take just 1.
If you then go on the calculate, say, density using the formula
density = mass/volume
the % error in the density is equal to the sum of the % errors in the mass and the volume
You can add % errors this way. It's explained on the sheet I posted.
You can't add the actual errors unless the formula is adding two quantities with the same units.
Once you have the final % error you need to convert it back to an actual error.
For example, if you had a value of 10.0 with a % error of ±5%
you find 5% of 10.0, which is 0.5, and give your answer as
10.0±0.5
Do the same for the density. You have the % errors for the mass and the volume.
In this case the error for the mass is very small compared with the error for the volume and can actually be ignored.
Oh I get it, so will the % uncertainty for density be : %m + %r + %r = 0.09% + 2.88% + 2.88% = 5.85%
1.89 g/cm^3 +-(5.85%)
1.89 g/cm^3 +- 0.1g/cm^3??? (absolute value)
Original post by Wasuppdude
Oh I get it, so will the % uncertainty for density be : %m + %r + %r = 0.09% + 2.88% + 2.88% = 5.85%
1.89 g/cm^3 +-(5.85%)
1.89 g/cm^3 +- 0.1g/cm^3??? (absolute value)
1.89 g/cm^3 +-(5.85%)
1.89 g/cm^3 +- 0.1g/cm^3??? (absolute value)
If you check back to my earlier posts, you will see that I said that the % error in the volume will be 3 times the % error in the radius. This is because the radius is cubed to get the volume. In that post we agreed that the % error in the radius was 2.8% and the volume is 588.98mm cubed
So the % error in the volume is about 8.5%
The % error in the density will be this plus the % error in the mass.
The mass error is so small it can safely be ignored.
So yes, you have the method correct, you just needed to take 3 times the % error in the radius, not 2.
So the system is, when you have a formula with values that contain uncertainties,
if the formula has multiply and/or division - add the % errors to get the total % error
if a quantity is squared it's 2 times the % error, if cubed it's 3 times etc
Covert the % errors back to actual errors in the final result.
This is all explained on the sheet I posted.
Original post by Stonebridge
If you check back to my earlier posts, you will see that I said that the % error in the volume will be 3 times the % error in the radius. This is because the radius is cubed to get the volume. In that post we agreed that the % error in the radius was 2.8% and the volume is 588.98mm cubed
So the % error in the volume is about 8.5%
The % error in the density will be this plus the % error in the mass.
The mass error is so small it can safely be ignored.
So yes, you have the method correct, you just needed to take 3 times the % error in the radius, not 2.
So the system is, when you have a formula with values that contain uncertainties,
if the formula has multiply and/or division - add the % errors to get the total % error
if a quantity is squared it's 2 times the % error, if cubed it's 3 times etc
Covert the % errors back to actual errors in the final result.
This is all explained on the sheet I posted.
So the % error in the volume is about 8.5%
The % error in the density will be this plus the % error in the mass.
The mass error is so small it can safely be ignored.
So yes, you have the method correct, you just needed to take 3 times the % error in the radius, not 2.
So the system is, when you have a formula with values that contain uncertainties,
if the formula has multiply and/or division - add the % errors to get the total % error
if a quantity is squared it's 2 times the % error, if cubed it's 3 times etc
Covert the % errors back to actual errors in the final result.
This is all explained on the sheet I posted.
Oh yeah sorry I forgot I was meant to cube it.
so % error for density would be 2.88 x 2.88 x 2.88 = 23.9%
1.89g/cm^3 +- ( 23.9%)
1.89g/cm^3 +- 0.45 g/cm^3
I got this from the sheet you posted. Now the last bit says how many significan figures you should give the density which I already have given to 3, so I leave it as 3 ??
Original post by Wasuppdude
so % error for density would be 2.88 x 2.88 x 2.88 = 23.9%
We agreed the % error in the volume was 3 times 2.88% = 8.5%
and the % error in the density will be the same because the error in the mass is so small. This is stated in my post you just quoted.
Where does 23.9% come from?
Regarding significant figures, you use as many as is sensible with due consideration to the error in the final result.
For example, if you have calculated a value to be 12.345s and you know it has an uncertainty of ±0.2s for example, there would be no pint in giving those last 2 significant figures (the 4 and 5) as the uncertainty lies in the third figure, the "3".
It would be 12.3±0.2 to 3 sig figs in this case.
The same rule applies to the question you are doing.
Original post by Stonebridge
We agreed the % error in the volume was 3 times 2.88% = 8.5%
and the % error in the density will be the same because the error in the mass is so small. This is stated in my post you just quoted.
Where does 23.9% come from?
Regarding significant figures, you use as many as is sensible with due consideration to the error in the final result.
For example, if you have calculated a value to be 12.345s and you know it has an uncertainty of ±0.2s for example, there would be no pint in giving those last 2 significant figures (the 4 and 5) as the uncertainty lies in the third figure, the "3".
It would be 12.3±0.2 to 3 sig figs in this case.
The same rule applies to the question you are doing.
and the % error in the density will be the same because the error in the mass is so small. This is stated in my post you just quoted.
Where does 23.9% come from?
Regarding significant figures, you use as many as is sensible with due consideration to the error in the final result.
For example, if you have calculated a value to be 12.345s and you know it has an uncertainty of ±0.2s for example, there would be no pint in giving those last 2 significant figures (the 4 and 5) as the uncertainty lies in the third figure, the "3".
It would be 12.3±0.2 to 3 sig figs in this case.
The same rule applies to the question you are doing.
Ahh damn you're right I multiplyed.
So The % error in density is 8.64%
1.89g/cm^3 +- (8.64%)
1.89g/cm^3 +- 0.2g/cm^3 (absolute)
But I already rounded the mean diamter to 3 sig fig therefore I would just leave it like that.
Original post by Wasuppdude
Ahh damn you're right I multiplyed.
So The % error in density is 8.64%
1.89g/cm^3 +- (8.64%)
1.89g/cm^3 +- 0.2g/cm^3 (absolute)
But I already rounded the mean diamter to 3 sig fig therefore I would just leave it like that.
So The % error in density is 8.64%
1.89g/cm^3 +- (8.64%)
1.89g/cm^3 +- 0.2g/cm^3 (absolute)
But I already rounded the mean diamter to 3 sig fig therefore I would just leave it like that.
I make the error ±0.16 [1.89 x 8.64/100]
The value would be better expressed as 1.89±0.16 in that case.
Original post by Stonebridge
I make the error ±0.16 [1.89 x 8.64/100]
The value would be better expressed as 1.89±0.16 in that case.
The value would be better expressed as 1.89±0.16 in that case.
Yes that makes sense, also do I need to put the units at the end of whats highlighted bold?
Thanks so so so much
Original post by Wasuppdude
Yes that makes sense, also do I need to put the units at the end of whats highlighted bold?
Thanks so so so much
Thanks so so so much
Yes of course - always include the units. I was just trying to save a bit of time when answering.
Original post by Stonebridge
Yes of course - always include the units. I was just trying to save a bit of time when answering.
Thankyou so much
Original post by Stonebridge
Yes of course - always include the units. I was just trying to save a bit of time when answering.
Can you just check the overall answers please:
a i) Mean diameter = 10.2 + 10.3 + 10.7 + 10.1 +10.5/5 = 10.4mm
Uncertainty =(10.7 - 10.1) = 0.6/2 = 0.3
10.4 mm +- 0.3mm
ii) % uncertaintys of diameter = (0.3/10.4) x 100 = 2.88%
% uncertainty of radius = 0.15/5.2 x 100 = 2.88%
iii) Volume = 4/3pir^3 = 4/3pi5.2^3= 588.98mm^3
= 0.58898cm^3
b) % uncertainty of total mass = 0.01/11.16 x 100 = 0.09%
% unceratinty mass of 1ball = 0.001/1.116 x 100 = 0.09%
c) density = mass/ vol = 1.116/0.58898 = 1.89g/cm^3 or 1890kg/m^3
d) i ) %uncertainty of density = %mass + %radius + %radius + %radius
= 0.09% + 2.88% + 2.88% + 2.88% = 8.73%
ii) You shoould give it to 3sig fig.
iii) 1.89 g/cm^3 +- 0.16g/cm^3
Original post by Wasuppdude
Can you just check the overall answers please:
a i) Mean diameter = 10.2 + 10.3 + 10.7 + 10.1 +10.5/5 = 10.4mm
Uncertainty =(10.7 - 10.1) = 0.6/2 = 0.3
10.4 mm +- 0.3mm
ii) % uncertaintys of diameter = (0.3/10.4) x 100 = 2.88%
% uncertainty of radius = 0.15/5.2 x 100 = 2.88%
iii) Volume = 4/3pir^3 = 4/3pi5.2^3= 588.98mm^3
= 0.58898cm^3
b) % uncertainty of total mass = 0.01/11.16 x 100 = 0.09%
% unceratinty mass of 1ball = 0.001/1.116 x 100 = 0.09%
c) density = mass/ vol = 1.116/0.58898 = 1.89g/cm^3 or 1890kg/m^3
d) i ) %uncertainty of density = %mass + %radius + %radius + %radius
= 0.09% + 2.88% + 2.88% + 2.88% = 8.73%
ii) You shoould give it to 3sig fig.
iii) 1.89 g/cm^3 +- 0.16g/cm^3
a i) Mean diameter = 10.2 + 10.3 + 10.7 + 10.1 +10.5/5 = 10.4mm
Uncertainty =(10.7 - 10.1) = 0.6/2 = 0.3
10.4 mm +- 0.3mm
ii) % uncertaintys of diameter = (0.3/10.4) x 100 = 2.88%
% uncertainty of radius = 0.15/5.2 x 100 = 2.88%
iii) Volume = 4/3pir^3 = 4/3pi5.2^3= 588.98mm^3
= 0.58898cm^3
b) % uncertainty of total mass = 0.01/11.16 x 100 = 0.09%
% unceratinty mass of 1ball = 0.001/1.116 x 100 = 0.09%
c) density = mass/ vol = 1.116/0.58898 = 1.89g/cm^3 or 1890kg/m^3
d) i ) %uncertainty of density = %mass + %radius + %radius + %radius
= 0.09% + 2.88% + 2.88% + 2.88% = 8.73%
ii) You shoould give it to 3sig fig.
iii) 1.89 g/cm^3 +- 0.16g/cm^3
That all looks fine to me.
You could even give that final density as
1890 ± 160 kgm-3 if you want it in the other units.
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