masryboy94
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#121
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#121
(Original post by jonnyb123)
Well work done is energy, so m*deltav, grav potential is work done per unit mass (in bringing a test mass from infinity to that point) so deltav=deltaW/m, and gravitational potential difference is just the difference in grav potential between two points, so if one point from earth is -36MJkg^-1 (A) and another is -12Jkg^-1 (B), the grav potential difference in moving from A to B is -12--36 = +24 Jkg^-1, which makes sense because the further you get from the planet, the less negative the potential gets and it's an attractive force so you are doing work against it to move further away from it.
so difference from work done and gravitational potential is that work done is for the whole mass whereas gravitation potential is based on 1kg of the mass?
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jonnyb123
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#122
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(Original post by masryboy94)
so difference from work done and gravitational potential is that work done is for the whole mass whereas gravitation potential is based on 1kg of the mass?
yeah you could look at it like that I suppose, one is GPE per unit mass and the other is just the gravitational potential energy of the object.
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bobthebuilder
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#123
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(Original post by masryboy94)
ok so basically we know that  v = \frac{d}{t} but we also know the distance (circumference) around a circle is  2\pi r so sub that in to make  v = \frac{2\pi r}{T} ( it is T and not t because after one whole cycle around the circle in this case the planet, that is one time period) .... we also know that the gravitational force and centripetal force would equal each other because it says it only just remains on the surface , therefore  \frac{mv^2}{r} = \frac{GMm}{r^2} make v the subect to get  v = \sqrt \frac{GM}{r} make the 2 v's equal each other   \sqrt \frac{GM}{r} = \frac{2\pi r}{T} , then make T the subject to get  T = \sqrt \frac{4\pi ^2 r^3}{GM} which gives you  2\pi \sqrt \frac{r^3}{GM}
hey, thanks for answering. ok, i get it upto the point where you make the two v's equal. but when i rearrange to make T the subject i keep getting T = 4π^2r / GM
all under square root, just that i cant type it...
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masryboy94
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#124
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(Original post by bobthebuilder)
hey, thanks for answering. ok, i get it upto the point where you make the two v's equal. but when i rearrange to make T the subject i keep getting T = 4π^2r / GM
all under square root, just that i cant type it...
ok so  \sqrt \frac{GM}{r} = \frac{2\pi r}{T}

so then  T = \frac{2\pi r}{\sqrt \frac{GM}{r}}

square both sides to get  T^2 = \frac{4\pi^2 r^2}{\frac{GM}{r}}

this is the same as saying  T^2 = 4\pi^2 r^2\times \frac{r}{GM}

 T^2 = \frac{4\pi^2 r^3}{GM}

square root both sides to give you

so you get  T = \frac{2\pi \sqrt r^3}{\sqrt GM}
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StudentAnon
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#125
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Does anyone have good notes/videos on the theory of the simple pendulum and mass-spring system?
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The H
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#126
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Hope I don't f up the multiple choice section this time, I was consistently getting 20+ then I ended up with something like 14
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SDavis123
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#127
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#127
Just did two multiple choices one after the other and got 21/25 each time, getting my confidence


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The H
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#128
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#128
(Original post by SDavis123)
Just did two multiple choices one after the other and got 21/25 each time, getting my confidence


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Yeah I'm getting similar, loving them now
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masryboy94
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#129
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#129
whats the answer to this and why? help appreciated
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bobthebuilder
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#130
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(Original post by masryboy94)
ok so  \sqrt \frac{GM}{r} = \frac{2\pi r}{T}

so then  T = \frac{2\pi r}{\sqrt \frac{GM}{r}}

square both sides to get  T^2 = \frac{4\pi^2 r^2}{\frac{GM}{r}}

this is the same as saying  T^2 = 4\pi^2 r^2\times \frac{r}{GM}

 T^2 = \frac{4\pi^2 r^3}{GM}

square root both sides to give you

so you get  T = \frac{2\pi \sqrt r^3}{\sqrt GM}
Thankyou!
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The H
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#131
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#131
(Original post by masryboy94)
whats the answer to this and why? help appreciated
It's C
Because electric potential is scalar and field strength is a vector.
So if you had 2 positive charges with equal charge, the midpoint would have 0 field strength because you take one from the other, but you always add potential, so the electric potential would actually be doubled at the midpoint.
Hope that helps
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masryboy94
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#132
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(Original post by The H)
It's C
Because electric potential is scalar and field strength is a vector.
So if you had 2 positive charges with equal charge, the midpoint would have 0 field strength because you take one from the other, but you always add potential, so the electric potential would actually be doubled at the midpoint.
Hope that helps
how comes you add potentials?
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Olive123
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#133
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#133
Hey guys can someone help me on this question please.

So I answered this question as D, I was wrong the answer is C

So first off how would you obtain the acceleration of the ball at any given time from this graph ?

and why cant you obtain the speed of the ball as it leaves the racket ?

Thanks
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jonnyb123
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#134
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#134
(Original post by Olive123)
Hey guys can someone help me on this question please.

So I answered this question as D, I was wrong the answer is C

So first off how would you obtain the acceleration of the ball at any given time from this graph ?

and why cant you obtain the speed of the ball as it leaves the racket ?

Thanks
To get the speed of it you need to know its mass, as v=F*deltaT/m...[edit]wait actually it says of known mass....hmm....

:dontknow:

I'll think about it, doing chemistry right now, anyone else know?
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Olive123
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#135
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#135
(Original post by jonnyb123)
To get the speed of it you need to know its mass, as v=F*deltaT/m
So how would you get the acceleration of it then ? dont you need the mass too ?
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Olive123
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#136
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#136
(Original post by jonnyb123)
To get the speed of it you need to know its mass, as v=F*deltaT/m...[edit]wait actually it says of known mass....hmm....
Ohh so thats how you would get acceleration

I guess you would need final vel then thanks for pointing the known mass bit out haha
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UnknownOrigin
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#137
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#137
For anyone wondering, the paper in january 2013 topics examined were:

Q1 - 8 marks total
One Similarity & Difference of inelastic/elastic collisions
Momentum calc and kinetic energy "show that ..." question

Q2 - 12 marks total

Satellite in orbit - calculate the height above the earth of a satellite in geosynchronous orbit
Centripetal force calc.
Compare principle features of geosynchronous and polar orbits, explain consequences for possible uses of satellites in these orbits (6 marker)

Q3 - 5 marks total

Capacitors

Q4 - 12 marks total

Electric fields calculations and define Coulomb's law and drawing electric field lines surrounding charges

Q5 - 13 marks total

Define Lenz's law, electromagnetic induction q's - all written answers


So topics not on this paper were:

SHM
Magnetic fields
Transformers

So they might possibly come up next paper, transformers and SHM most likely, as they haven't been on for two papers now.
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Olive123
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#138
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#138
Ok so I have a problem with this question

I managed to guess the right answer which is D

What I did was I compared the two time periods by using T=2pir/v

Then I put that Rr/Vr < Rs/Vs and tried to compare the two

I only managed to guess D because I know that the further away a planet is from the sun the longer it takes to make one full orbit. So a greater radius = greater PE

However how would you explain this like with equations ?
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masryboy94
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#139
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#139
anyone have the phya4 january 2013 paper?? pleaseeee
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Jack93o
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#140
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ditto this ^
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