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A2-1 physics Ccea

Any answers??

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Reply 1
Avatar for harrytqo
harrytqo
1
how did you find that? data analysis was tricky
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?
Reply 3
Avatar for MrManGuy
MrManGuy
1
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.
Reply 4
Avatar for Mr.bob
Mr.bob
16
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 :biggrin:

Overall what did you think of it? Didnt think it was too bad myself
(edited 7 years ago)
Reply 5
Avatar for MrManGuy
MrManGuy
1
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
Reply 6
Avatar for TirnanF
TirnanF
3
3.67% for percentage increase question?


Posted from TSR Mobile
Reply 7
Avatar for Mr.bob
Mr.bob
16
Original post by TirnanF
3.67% for percentage increase question?


Posted from TSR Mobile


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
(edited 7 years ago)
What about the last question with percentage error of a?
And what was the minimum ke?
Reply 10
Avatar for Mr.bob
Mr.bob
16
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
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.
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.
Reply 13
Avatar for Mr.bob
Mr.bob
16
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.
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 ;/
since c=a^2 then would a not be the square root of C and hence % uncertainty should be halved
Reply 16
Avatar for Mr.bob
Mr.bob
16
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
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
Reply 18
Avatar for Mr.bob
Mr.bob
16
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
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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