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# A2-1 physics Ccea

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2. how did you find that? data analysis was tricky
3. (Original post by harrytqo)
how did you find that? data analysis was tricky
Thought it was hard enough, what did you get for the percentage increase question?
4. Did E=3/2.kT for both then compared.

Think you had to find the root of the % difference since E=1/2. m . c^2 where 1/2.m is constant. So c^2 would've had the same increase as energy meaning c would've had root of the % difference.
5. (Original post by MrManGuy)
Did E=3/2.kT for both then compared.

Think you had to find the root of the % difference since E=1/2. m . c^2 where 1/2.m is constant. So c^2 would've had the same increase as energy meaning c would've had root of the % difference.
Similar principal to how you did but instead i equated pv= 1/3Nm<c> where (1/3Nm)/v is a constant so i ended up with the expression <c^2>/p1 = <c^2>\ p2 and the c squareds are differnt . I jsut hope i read the prior question correctly so that all the values are correct!!!

I just quickly did your method and the answers (using my values) are the same

Overall what did you think of it? Didnt think it was too bad myself
6. Don't think it was too bad, the nuclear parts where pretty straight forward. Graph was simple to do too.

Wasn't too sure on the circular motion gymnast question, whether there are more forces on the bar than just the force of the gymnast so F=mg therefore mg=mvw
7. 3.67% for percentage increase question?

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8. (Original post by TirnanF)
3.67% for percentage increase question?

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I think you had to sqaure root it because it was the percentage difference between root mean square speed.
EDIT i think i ****ed up part of that question do by using the pressures in kilo pascals unless the ten to the power of three cancel out i hope
9. What about the last question with percentage error of a?
10. And what was the minimum ke?
11. (Original post by Sunflower654321)
What about the last question with percentage error of a?
Everyones will be slightly different as it depends on the gradient you drew. All i know is you had to multiply the percentage uncertainty by 2 because it was a^2
12. Yeh you drew the min or max line of best fit and compared the two so if grad = 2x and grad(min) = x then grad = 2x +/- x where % uncertainty was (x/2x)*100 then since c = a^2, % uncertainty a was double that of percent uncertainty in c.
13. (Original post by Mr.bob)
I think you had to sqaure root it because it was the percentage difference between root mean square speed.
EDIT i think i ****ed up part of that question do by using the pressures in kilo pascals unless the ten to the power of three cancel out i hope

I'm pretty sure for that question in particular it is pretty standard to use E=3/2kT

Which is derived from pV = NkT and pV=1/3 Nmc^2

so NkT = 1/3 Nm c^2

therefore kt = 1/3 mc^2

which we get to be like kinetic energy by multiplying by 3/2 so 3/2kt = 1/2 mc^2 = kinetic energy.
14. (Original post by MrManGuy)
I'm pretty sure for that question in particular it is pretty standard to use E=3/2kT

Which is derived from pV = NkT and pV=1/3 Nmc^2

so NkT = 1/3 Nm c^2

therefore kt = 1/3 mc^2

which we get to be like kinetic energy by multiplying by 3/2 so 3/2kt = 1/2 mc^2 = kinetic energy.
Yes but there is an alternative method were you use the pressure rather than temperature(see above). Like the way you described is perfect as well.
15. I can't quite remember the specific numbers for that question, if you have them I can try to show what I think would be the answer. I think I didn't convert from celcius to kelvin so my number would be a bit off in my actual answer too ;/
16. since c=a^2 then would a not be the square root of C and hence % uncertainty should be halved
17. (Original post by MrManGuy)
I can't quite remember the specific numbers for that question, if you have them I can try to show what I think would be the answer. I think I didn't convert from celcius to kelvin so my number would be a bit off in my actual answer too ;/
I just made up values for testing but you end up with the same % increase for c squared for both ways(yours and mine). But you do have to square root the answer still
18. (Original post by niamhus123)
since c=a^2 then would a not be the square root of C and hence % uncertainty should be halved
Think so yeh, was just typing in a blur
19. (Original post by niamhus123)
since c=a^2 then would a not be the square root of C and hence % uncertainty should be halved
I thought this originally but then i visualised the equation as D^2 = a^2 x n and it seemed to me that you doubled it. Honestly i really dont know which it is i just went with my gut instinct
20. I divided my percentage uncertainty by 2 since %C = 2 x %A
so %A = %C x 0.5 (divided by 2)

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