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# OCR Advancing Physics G492 ~5 June 2013~ watch

1. (Original post by ArchieL)
Haha, you know it! Geoff's retiring this year

Also the section C notes were Tim's work this year, he rustled them out as soon as the prerelease came out! The physics dept. at csfc are so hyped on caffeine it's unreal.
I know, it sucks!
Ok awesome! Are there any answers for the section C notes around?
2. (Original post by ArchieL)
There are some great YT vids on the subject, but..
Resolution = Smallest observable change: In the case of a digital display, the amount of d.p, in a graph it's how many units 1 box represents.
Thankyou very much
3. liomeeringca,

Read the MS again, it says 169m^2s^-2 and the same for 900.

I just used v^2 = u^2 +2as, then rearrange for s.
I apologise, the reason they have those units is because they've been squared:
(m/s)^2 = (m^2/s^2)
4. (Original post by AndItsOllie)
I know, it sucks!
Ok awesome! Are there any answers for the section C notes around?
I'm afraid not mate, I've got a few if you need them! Just ask
5. (Original post by ArchieL)
I apologise, the reason they have those units is because they've been squared:
(m/s)^2 = (m^2/s^2)
I understand now, thank you for your help
6. (Original post by marseille_h)
This might seem straightforward but I'm a bit confused:

A car has an acceleration of 0.86g. Calculate the time it takes to reach a velocity of 27m.s-1.
g=9.81

v=u+at
so t= (v-u)/a

is g=acceleration in this case and the acceleration of the car they give we take it as u and v=27 ?

Acceleration would be 0.86 x g, so 0.86 x 9.81. I assume you would take u as 0 and v as 27 and then use v=u+at to find t.
7. (Original post by jacobfun)
Acceleration would be 0.86 x g, so 0.86 x 9.81. I assume you would take u as 0 and v as 27 and then use v=u+at to find t.
which formula do we use to get acceleration and do 0.86 x g ?
8. (Original post by marseille_h)
which formula do we use to get acceleration and do 0.86 x g ?
No formula. 'g' is shorthand for the acceleration due to gravity (on earth) ~ 9.81ms-2
=> an acceleration of 0.86g = (0.86)(9.81) = 8.4366 ~ 8.44
9. (Original post by ArchieL)
No formula. 'g' is shorthand for the acceleration due to gravity (on earth) ~ 9.81ms-2
=> an acceleration of 0.86g = (0.86)(9.81) = 8.4366 ~ 8.44
Oh i understand i thought it was a wierd unit. THANKS
10. How does everyone feel about potential dividers, thermisters and general electricity questions then? I'm bricking myself
11. (Original post by ArchieL)
How does everyone feel about potential dividers, thermisters and general electricity questions then? I'm bricking myself
None of this in unit 2 i think..
12. (Original post by liomeeringca)
How would one calculate gravity from a set of results like that?
V^2=U^2+2as. F=ma, and the component for force in the direction of the ramp is gsin(theta). Therefore you can rearrange that to a=gsin(theta)
V^2=U^2 + 2gsin(theta)s. U=0, so you can rearrange it to g=V^2 divided by 2sin(theta)s
13. I saw someone had already answered this but I'm still unsure how to determine the value for g from v2= u2+2as

Can someone explain step by step please?

I hate all these maths based questions
14. (Original post by ArchieL)
How does everyone feel about potential dividers, thermisters and general electricity questions then? I'm bricking myself
Wasn't this G491?
15. (Original post by marseille_h)
None of this in unit 2 i think..
As far as I was aware, section C is a synopsis unit (of sorts?)
Also I've got a question on it on some prelrelease questions set by the department at college, so I'd guess there might be a little on it? Don't get me wrong, I'd be over the moon if there wasn't! Haha
16. (Original post by King Hotpie)
V^2=U^2+2as. F=ma, and the component for force in the direction of the ramp is gsin(theta). Therefore you can rearrange that to a=gsin(theta)
V^2=U^2 + 2gsin(theta)s. U=0, so you can rearrange it to g=V^2 divided by 2sin(theta)s
Thank you so so much,
17. (Original post by ArchieL)
As far as I was aware, section C is a synopsis unit (of sorts?)
Also I've got a question on it on some prelrelease questions set by the department at college, so I'd guess there might be a little on it? Don't get me wrong, I'd be over the moon if there wasn't! Haha
I haven't looked a potential dividers in aaaaaaages! I really hope there's nothing on it, like you say D:
18. For the pre-release material on measuring the speed of light... why can it be assumed that the 'hotspot' where the chocolate melted is an antinode?
19. (Original post by seharnaznaz)
I saw someone had already answered this but I'm still unsure how to determine the value for g from v2= u2+2as

Can someone explain step by step please?

I hate all these maths based questions
With the trolley on the ramp, it has an acceleration, g, acting directly downwards. Construct a triangle with this force, another force c acting ϴ degrees to it originating from the floor (so they meet). Then add another from the top of grunning parallel to the slope. This cuts the line with c and forms a triangle. From the cosine rule, cosϴ=opp/hyp, so =a/g. Rearrange to get a=gsinϴ. (It's best to draw a diagram of all component forces, stripped down).

If we then substitute this into the equation v2=u2+2as with u=0, we get that result (v2 =2.g.l.sinϴ). Also it's probably worth noting that plotting a graph with v^2 against sinϴ gets us with a gradient of 2.g.l
20. (Original post by seharnaznaz)
For the pre-release material on measuring the speed of light... why can it be assumed that the 'hotspot' where the chocolate melted is an antinode?
Antinodes is where the intestity is the highest compared to nodes where it is =0

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