[cooldudeman]

the way that ums works is by percentages,
they want about 10% of the people who took the exam to get full ums so they set the boundary for the lowest raw mark to equal 120ums
they want about 10% to get 96ums and so on...

[blahblahh]

(Original post by AhiTribianni)
for the phase relationships both out of phase because X was half way up for the wave, Y was somewhere down below and Z was at maximum displacement so X couldnt have been in phase with Y or Z?

On a stationary wave if there is an even number of nodes between them they are in phase, so one was in phase one was out of phase

[Kirby711]

(Original post by blahblahh)
On a stationary wave if there is an even number of nodes between them they are in phase, so one was in phase one was out of phase

I put that they were in phase and out of phase too, but I definitely didn't get that actual difference right... Oh well.

As for 4c) if anyone is curious a thinner core prevents multipath dispersion by preventing waves propagating at various angles. This means that data isn't adjusted (the wavelength).

EDIT - Did the phase question ask for the relationship? Because if it did technically they should ignore the other answer I put and mark me right?

[milliezhao]

Where is the paper?


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[Bruno]

(Original post by Amirrryy)
lol i'm not good with raw marks and ums etc i dont understand how they do it very well...so you think its 2 UMS for 1 exam mark?

This is the AQA UMS mark converter:

http://web.aqa.org.uk/UMS/index.php

It's really useful when worrying about grade boundaries, despite everyone telling you not to bother.

[milliezhao]

And I think for the velocity time graph it should be a straight line up as it is free fall from the hill, it has no air resistance or friction, there is no force to decelerate it, how steep the slop doesnt affect the acceleration of 9.81


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[Amirrryy]

(Original post by Bruno)
This is the AQA UMS mark converter:

http://web.aqa.org.uk/UMS/index.php

It's really useful when worrying about grade boundaries, despite everyone telling you not to bother.

yea used it thanks

[Amirrryy]

(Original post by Kirby711)
I put that they were in phase and out of phase too, but I definitely didn't get that actual difference right... Oh well.

As for 4c) if anyone is curious a thinner core prevents multipath dispersion by preventing waves propagating at various angles. This means that data isn't adjusted (the wavelength).

EDIT - Did the phase question ask for the relationship? Because if it did technically they should ignore the other answer I put and mark me right?

no no it only asked for relationship, so you're fine
uum for 4c) i said prevents signal loss due to light leakage from the core which causes the signal of data being transmitted to be weaker....will i get any marks? :/

[Benniboi1]

(Original post by milliezhao)
And I think for the velocity time graph it should be a straight line up as it is free fall from the hill, it has no air resistance or friction, there is no force to decelerate it, how steep the slop doesnt affect the acceleration of 9.81


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What's important is the acceleration parallel to the slope. Acceleration due to gravity yes is a constant but as the gradient of the slope is constantly decreasing the component of acceleration due to gravity parallel to the slope is also decreasing. Hence the graph is a curve with decreasing gradient.

[Amirrryy]

(Original post by Benniboi1)
What's important is the acceleration parallel to the slope. Acceleration due to gravity yes is a constant but as the gradient of the slope is constantly decreasing the component of acceleration due to gravity parallel to the slope is also decreasing. Hence the graph is a curve with decreasing gradient.

get in

[milliezhao]

(Original post by Benniboi1)
What's important is the acceleration parallel to the slope. Acceleration due to gravity yes is a constant but as the gradient of the slope is constantly decreasing the component of acceleration due to gravity parallel to the slope is also decreasing. Hence the graph is a curve with decreasing gradient.

There is no such thing as component of gravity because gravity is not acting from a single point, it is pulling all object downward no matter where they are and the force is constant. So if the slop has a decreasing gradient the acceleration will still be the same as the gravity is pulling it vertically downward at all times.


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[Little Wing]

(Original post by milliezhao)
There is no such thing as component of gravity because gravity is not acting from a single point, it is pulling all object downward no matter where they are and the force is constant. So if the slop has a decreasing gradient the acceleration will still be the same as the gravity is pulling it vertically downward at all times.


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Drop a rock. Get the same rock and let it roll down a hill from the same heigh. I guarantee you that the dropped rock will reach the ground first. Why do you think we use mgsin(theta) for slopes?

Edit: I meant drop the rock from a hilltop of same distance to the ground, not height

[FK_]

(Original post by ragre)
Judging by this, I'd say for 120 UMS you'd have to get around the 60 mark for this paper, which, hopefully, I've barely scraped. Frankly though, it's not about how hard a paper is per se, but about how badly people do. So if the rest of the country do badly, which I think they will, the boundaries may be that smallest bit lower.

You are right in that it does matter how bad/good the rest of the country does, but isn't that the same for every year of exams? In that sense, whether the 120 ums mark was as high as 70, or even when it was as low as 59, that factor would have been accounted for nonetheless. Therefore, taking into account what people have said about how they found the exam, I would optimistically guess that the 120 UMS boundary is around the 65-68 mark. However, I doubt/hope that it is any higher than that.

[Benniboi1]

(Original post by milliezhao)
There is no such thing as component of gravity because gravity is not acting from a single point, it is pulling all object downward no matter where they are and the force is constant. So if the slop has a decreasing gradient the acceleration will still be the same as the gravity is pulling it vertically downward at all times.


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You're right, there isn't, I meant the component of the weight.



(Apologies for the bad drawing, I cannot draw on a laptop)

If you look at 1), the ball is on the first part of the slope in which the gradient is quite steep. This means that the component of the weight parallel to the slope (which is sin(thetha1) is fairly large. Using F=ma, rearrange to find F/m=a and you have your acceleration at that point. Now look at 2), this is further on down the slope where the gradient isn't as great as in 1) which means theta 2 is smaller. Again using sin(theta2) you can work out the component of the weight parallel to the slope but as theta2 is less than theta1 then then sin(thetha2) is less than sin(theta1) meaning that there is a small force acting parallel to the slope. Use F/m=a again and a small force means a smaller acceleration. As theta2 tends towards 0 then the acceleration tends towards zero as sin0 is indeed 0.

As the gradient of a velocity/speed time graph gives acceleration then as the acceleration over time tends towards zero then the gradient will tend towards zero also, on the graph this is the point which should be labelled B.

Just think, if the gradient on the ground that the ball was on was 0, would the ball be accelerating?

[Tanmayee]

Sooo could someone very kindly help me to calculate what I'd need to have achieved in this exam (marks wise) to get above 90% overall? I'm not sure what 90% is in terms of UMS.
Considering: I got 9/9 practical marks, 35/41 in the ISA, I think I dropped 2-3 marks *at the most* in unit 1.
Thanks so much in advance


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[Kirby711]

(Original post by Amirrryy)
no no it only asked for relationship, so you're fine
uum for 4c) i said prevents signal loss due to light leakage from the core which causes the signal of data being transmitted to be weaker....will i get any marks? :/

Um... I honestly don't know sorry...

[milliezhao]

(Original post by Benniboi1)
You're right, there isn't, I meant the component of the weight.



(Apologies for the bad drawing, I cannot draw on a laptop)

If you look at 1), the ball is on the first part of the slope in which the gradient is quite steep. This means that the component of the weight parallel to the slope (which is sin(thetha1) is fairly large. Using F=ma, rearrange to find F/m=a and you have your acceleration at that point. Now look at 2), this is further on down the slope where the gradient isn't as great as in 1) which means theta 2 is smaller. Again using sin(theta2) you can work out the component of the weight parallel to the slope but as theta2 is less than theta1 then then sin(thetha2) is less than sin(theta1) meaning that there is a small force acting parallel to the slope. Use F/m=a again and a small force means a smaller acceleration. As theta2 tends towards 0 then the acceleration tends towards zero as sin0 is indeed 0.

As the gradient of a velocity/speed time graph gives acceleration then as the acceleration over time tends towards zero then the gradient will tend towards zero also, on the graph this is the point which should be labelled B.

Just think, if the gradient on the ground that the ball was on was 0, would the ball be accelerating?

Yes that make sense thank you... Like want I did in mechanic. Do u think 51 is enough for 89 ums? As I need 89 for an A...


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[CR95]

(Original post by Tanmayee)
Sooo could someone very kindly help me to calculate what I'd need to have achieved in this exam (marks wise) to get above 90% overall? I'm not sure what 90% is in terms of UMS.
Considering: I got 9/9 practical marks, 35/41 in the ISA, I think I dropped 2-3 marks *at the most* in unit 1.
Thanks so much in advance


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your mark on isa will likely be 48-50 ums
your mark on unit 1 will likely be 120ums
to achieve 90% overall you need 270 ums
therefore unit 2 mark must be atleast 100 ums
100 ums is likely to be about 55/70 marks

[internet tough guy]

(Original post by milliezhao)
And I think for the velocity time graph it should be a straight line up as it is free fall from the hill, it has no air resistance or friction, there is no force to decelerate it, how steep the slop doesnt affect the acceleration of 9.81


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but logically if you think about it, put a ball on a very small slope verus just on a much steeper slope, the ball will speed up much faster on the steeper slope.

[Tanmayee]

(Original post by CR95)
your mark on isa will likely be 48-50 ums
your mark on unit 1 will likely be 120ums
to achieve 90% overall you need 270 ums
therefore unit 2 mark must be atleast 100 ums
100 ums is likely to be about 55/70 marks

Thank you so much!


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