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Higher physics 2014/15

Hey guys I noticed that I hadn't seen a forum for higher physics so I thought I would make one incase any of us need help this year.

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Laaaveed higher physics (Well.. not the electricity parts... :yawn: )

Quote me if you want help :cool:
Reply 3
Hi, can anyone help me with these questions, thanks :smile:
Original post by ch3mz
Hi, can anyone help me with these questions, thanks :smile:


75) The relationship it is asking for is the one using pressure and volume (P1V1=P2V2). Remember it is in cm and the standard unit for volume is in meters.

76) You know mg = 700N, and the relationship of the component weight on a slope is mg x sin(theta), where theta = 20 (the angle of the slope.) For this one Id encourage you to go back to your notes and learn how the triangles work with the slopes (how you get mgsin theta etc).

79) Firstly, get out your tuvas table.
You know the initial vertical velocity is 0. The vertical distance is 2, the horizontal distance is 5, the vertical acceleration is -9.8.
You can work out the time taken via t=v-u/a, using the Vertical distance, vertical acceleration and final vertical speed (found using v=sqrt(2gh) )
Then, using these values, you have enough to sub it into (starting horizontal velocity)=horizontal distance/time.
Theres alot to do in this question, which is suprising for a MC.
Reply 5
Thanks for the quick reply! Any idea how you would do this one??
Original post by ch3mz
Thanks for the quick reply! Any idea how you would do this one??


If air resistance is negligable then horizontal velocity will be constant (newtons first law).

Think about the vertical velocity logically. The acceleration due to gravity is 9.8 m/s^2 in a down ward direction. Another way to think about that is that it is losing 9.8 m/s of upward vertical velocity every second. So after 5 seonds it will be 90-5(9.8)

(although g=10 in this question)
Good luck to everyone doing the new higher in 2014-15! :smile: how is it? I sat the old higher this year, new one sounds a lot more theory-intense
Reply 8
Original post by Brandon_97
Good luck to everyone doing the new higher in 2014-15! :smile: how is it? I sat the old higher this year, new one sounds a lot more theory-intense


Basically? Awful.
Loving the theory and applications for the first two units but I'm absolutely dreading the third, electricity unit :afraid:
Its alright really if you've done your nat 5s as this lays the foundation: for nat 5 the course consisted of roughly 1 part standard grade course, 1 part higher, and 1 part AH- at least thats what we were told.
Unfortunately this means the CfE higher is also much harder, supposedly comprising of: higher, AH and University levels coursework!
(edited 9 years ago)
Hi guys, I'm really stuck on this question and I don't get why the mark scheme well... Says what it says! Really grateful for any help :smile:
Original post by Heisenberg___
Hi guys, I'm really stuck on this question and I don't get why the mark scheme well... Says what it says! Really grateful for any help :smile:


Mfw I've never done moments :colondollar:
A pile is driven into the ground by means of a large mass of 450kg being raised to a height of 1.2m and released.

a) calculate the velocity of the mass just as it hits the piles (air resistance is negligible). The impact of the mass with the pile transfers 3500J of energy to the 1200kg pile. The impact drives the pile 0.14m into the ground.

I have 4.8 m/s as the answer to part a, which according to the answer scheme is correct.

b) Give a value for the force of friction between the pile and the earth (ignore potential energy of the pile in this example).

Answer for this part is 25 000N. No idea how to get there. Help, please?
Original post by WithoutMuchHope
A pile is driven into the ground by means of a large mass of 450kg being raised to a height of 1.2m and released.

a) calculate the velocity of the mass just as it hits the piles (air resistance is negligible). The impact of the mass with the pile transfers 3500J of energy to the 1200kg pile. The impact drives the pile 0.14m into the ground.

I have 4.8 m/s as the answer to part a, which according to the answer scheme is correct.

b) Give a value for the force of friction between the pile and the earth (ignore potential energy of the pile in this example).

Answer for this part is 25 000N. No idea how to get there. Help, please?



W=Fd

The earth does 3400 J of work in slowing down the pile over a distance of 0.14 m.

F = W/d = 3400/0.14 = 25000 N
Original post by langlitz
W=Fd

The earth does 3400 J of work in slowing down the pile over a distance of 0.14 m.

F = W/d = 3400/0.14 = 25000 N


Ah, so simple. Thanks very much.
Has there been any questions about this?
Original post by Higherdude
Has there been any questions about this?


Do you mean whether or not it would come up in an exam? Or do you want examples questions to do?

It's quite a common question I think. Would normally be a small 2 mark part of a larger question or multi choice. You'll defo need to know it though. Are you doing new or old higher?

I do quite miss me some higher physics. ^^
Original post by GoldenAge
Do you mean whether or not it would come up in an exam? Or do you want examples questions to do?

It's quite a common question I think. Would normally be a small 2 mark part of a larger question or multi choice. You'll defo need to know it though. Are you doing new or old higher?

I do quite miss me some higher physics. ^^

Old higher :smile:
I am confused at how to get the answer in Q9, anyone know how to do it?
Original post by Higherdude
I am confused at how to get the answer in Q9, anyone know how to do it?




Posted from TSR Mobile

First find the resistors in series: Rt=R1+R2 for top and bottom line
The find the resistors in parallel using 1/Rt=1/R1+1/R2

Hope this helps
Thanks
I over thought it.

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