Physics question thread....

fatal

Im going through the mocks and some questions I cant seem to find the answers to,so i am making this thread because i am sure than in the next day or two i will come across **** that i dunno and i cant find in the book....


2005 mock
how is the threshold of hearing defined?(i know what it is but dont know the definition)

what is meant by self induction?

in the cathode ray tube,what effect would increaing the distance between the cathode and anode have on the speed of the electron?(im guessing that it would be faster because it gets more time to accelerate)

O and why EXACTLY does the resistance of a filament bulb increase....apart from saying that more electrons are released i dunno why exactly it increases?

niamhy101

the resistance of a filament bulb increases because the increased temperature when the current increases (P=I^2R) causes the atoms to vibrate faster so the electrons which are flowing hit more atoms, i.e they meet more resistance so teh current doesn't have the same increase with voltage as it did when the temperature was lower

Edit: with non semiconductors(i.e filament bulb) increased temp doesn't release more conduction electrons, so this has no effect on the resistance, just the extra vibration of the atoms

Threshold of hearing is the smallest sound intensity detectable by a normal human ear at a frequency of 1kHz (1x10^-12 Wm^-2)

Self induction is the same as back emf. When a d.c supply is connected to a coil of wire around a soft iron core, and is switched on , the magnetic field in the coil changes, inducing an emf, and according to Lenz's law the direction of the induced current is such as to oppose the change causing it so it flows against the d.c current causing the current to increase over the first few seconds ( if you have a bulb in series with it, it will light up gradually)

I have no idea what the anode cathode thing will do, cos the acceleration of the cathode rays only depends on the voltage they are accelerated through, not the distance

fatal

niamhy101 wrote:

I have no idea what the anode cathode thing will do, cos the acceleration of the cathode rays only depends on the voltage they are accelerated through, not the distance

yea but if you accelerate something and give it a bigger distance its velocity will be more than if the distance is reduced

Kovik

Well, the impetus for it to accelerate would be less since the anode and cathode are further apart. The electric field would be weaker, or something. Like when the X/Y plates are further apart, less bending happens.

(I suck at physics, so there's a very real possibility that I've no idea what I'm talking about.)

nedward

eV=1/2mv^2

It would have no difference to the speed (kinetic energy), as p.d. between the two remains the same. I think.

Something like that shouldn't come up, though? It's the kind of ****y questions in which mock papers specialise.

Raphael

Well, by definition potential difference is the work done in moving from point A to point B, righty? Distance doesn't come into it. So if your anode is on mercury and your cathode is on jupiter, as long as the p.d is constant, the work done is constant.

But Work = Force by displacement. So if work is constant, and distance increases, then the force must decrease. That force = ma. SO as distance goes up, acceleration goes down. (as m is constant - mass of the electron)

Nyow, let's deal with v^2 = 2as. Let's assume that the speed at the end is constant. We want to prove 2AS=2as. That the product of acceleration and displacement is constant.

W=Fs=Mas. Mass is constant, weight is constant, therefore a.s is constant. Therefore the speed at the end is always the same

And there as GOT to be a better way to do that.

fatal

Raphael wrote:

Well, by definition potential difference is the work done in moving from point A to point B, righty? Distance doesn't come into it. So if your anode is on mercury and your cathode is on jupiter, as long as the p.d is constant, the work done is constant.

But Work = Force by displacement. So if work is constant, and distance increases, then the force must decrease. That force = ma. SO as distance goes up, acceleration goes down. (as m is constant - mass of the electron)

Nyow, let's deal with v^2 = 2as. Let's assume that the speed at the end is constant. We want to prove 2AS=2as. That the product of acceleration and displacement is constant.

W=Fs=Mas. Mass is constant, weight is constant, therefore a.s is constant. Therefore the speed at the end is always the same

And there as GOT to be a better way to do that.

that makes some sense...thanks for that

Raphael

No worries. Just glad you can follow my train of thought.

fatal

when finding the max number of dots on the screen when doing the diffraction grating experiment....you let the angle be 90 degree yea?im guessing because 90 degrees is the largest angle that you can make and any higher angle wouldnt show anything on the screen

Raphael

Max number of dots? You mean what the nth bright line is? Tbh I usually find that by looking for it, and finding the furthest one I can see.

fatal

Raphael wrote:

You mean what the nth bright line is? Tbh I usually find that by looking for it, and finding the furthest one I can see.

yea thats what i meant...and how do you do that?

niamhy101

to find the maximum value of n=DSinA/Lambda you can only max the angle and the max of SinA = 1 so n(max) = d/lambda

for the anode cathode thing, the kinetic energy of the electron is dependent on the p.d because p.d. = work done moving a charge from one place to another, so that work = the kinetic energy and kinetic energy = 1/2mv^2 = eV (V = p.d) i know its not the clearest but ya get me?

but i still cant see how moving the anode would change anything

stoopid ppl who set the mocks, half the time they aren't on our course anyway!!!

Kovik

This thread makes me sad. I'm not looking forward to Monday.

My very basic two cents: When we're discussing the speed of the electron, my instinct would be to say that the speed becomes greater because the attractive force between the the anode and cathode (along with the electrons) would be intensified by bringing them closer together. Hence, the electron would move quicker. The opposite circumstances, I assume, would have the opposite effect.

I have no formulae to back this up, but that's what I'd write on the exam.

Faerie

I have a question; Does anyone know if the actual syllabus changed in 2002? I know the paper changed but did the course itself change? I just printed off the 2001 paper, so I'm wondering if it would be helpful to do it.

X-SL

possible to learn all this in 6th year? or should i have worked in 5th?:eek:

:|

X-SL wrote:

possible to learn all this in 6th year? or should i have worked in 5th?:eek:

No, who works in 5th? not me anyway...

Nichololas

Well we spent the second half of 5th year trying to cover the stuff that the sub teacher had failed to teach us in the first half, so yeah it's possible.

fatal

stupid question but in circuit diagrams how do you know what way the circuit flows?and does it ALWAYS flow in this direction?

:|

it actually flows from the negative to the positive(the small side of the battery to the big) but they like to pretend it goes the other way

Or the other way around.....im not sure....i better go study....

Nichololas

From the 2005 LC:

What is the experiment that shows the ionising effect of radiation?

Faerie

Faerie wrote:

I have a question; Does anyone know if the actual syllabus changed in 2002? I know the paper changed but did the course itself change? I just printed off the 2001 paper, so I'm wondering if it would be helpful to do it.

yes it did change, we lost a lot of Maths and a huge portion of circular motion amongst other things.

fatal

wrote:

it actually flows from the negative to the positive(the small side of the battery to the big) but they like to pretend it goes the other way

what way does it actually flow then?????

Enlil_Nick wrote:

From the 2005 LC:

What is the experiment that shows the ionising effect of radiation?

you use the gold leaf electroscope....negatively charge it and bring the the radioactive source close to the cap using tongs and if its alpha the electroscope will lose its charge quickly because it is the most ionising,beta will lose it less quickly and gamma will have little or not effect on it because it has no charge

Conventional current flows from positive to negative

Electrons flow from negative to positive...

fatal

Deleted User wrote:

Conventional current flows from positive to negative

Electrons flow from negative to positive...

alright so in a circuit it flows from the small line to the big one yea?
O and arent the symbols for a variable resistor and variable rheostat the same?

niamhy101

fatal wrote:

alright so in a circuit it flows from the small line to the big one yea?
O and arent the symbols for a variable resistor and variable rheostat the same?

Conventional current flows from big line to small line

and a variable resistor is a rheostat, rheostat is just the name for the apparatus, a variable resistor is a rheostat when its in a circuit - - i think.....

caesar

If your asked to define p.d, what definition do you give? or does it matter because there is different ways of looking at it.

Btw if the definition of the ampere comes up make sure you give the long definition, p.g 308 in real world physics.

Tar.Aldarion

Potential difference is the amount of work needed to move a unit electric charge from the second point to the first or the amount of work that a unit charge flowing from the first point to the second can perform.

fatal

has anyone got an easy way to remember the long ampere definition?

whassupp2

think of space and 2 parallel conductors of infinate length a metre apart. thats all i know about it. i'll get a mark or two for that.

caesar

fatal wrote:

has anyone got an easy way to remember the long ampere definition?

Put it into a song, sorry thats the best thing i could think of.

Raphael

fatal wrote:

has anyone got an easy way to remember the long ampere definition?

Right it out 10 times before you go into the exam. RIght it out again as soon as you open your answer book