What Physics Formulas are not on - Data sheet (2 Viewers)

PandA_ · Mar 28, 2009

What physics formulas do we need to memorise core 1 + 2 (space motors and generators)

a list would be nice
thnx

study-freak · Mar 28, 2009

Nothing.

The board must give out any formula that you'll have to use in the HSC.

(I guess, if there are to be expections, they would be just escape velocity and orbital velocity formulae.)

undalay · Mar 28, 2009

study-freak said:
Nothing.

The board must give out any formula that you'll have to use in the HSC.

(I guess, if there are to be expections, they would be just escape velocity and orbital velocity formulae.)

Yep escape velocity and orbital velocity are not on the data sheet.
These however must be derived, and not quoted, as they are not provided.

k02033 · Mar 28, 2009

g force isnt given

helper · Mar 31, 2009

study-freak said:
Nothing.

The board must give out any formula that you'll have to use in the HSC.

(I guess, if there are to be expections, they would be just escape velocity and orbital velocity formulae.)

They are not obliged to provide any formula that is not published in the syllabus. The formula sheet is provided on the board of studies website.

Any thing that is defined in the syllabus but isn't provided with a formula won't be givem.

Eg. Magnetic Flux, Induced EMF

study-freak · Mar 31, 2009

helper said:
They are not obliged to provide any formula that is not published in the syllabus. The formula sheet is provided on the board of studies website.

Any thing that is defined in the syllabus but isn't provided with a formula won't be givem.

Eg. Magnetic Flux, Induced EMF

But I thought they cannot ask you to calculate something using those meant-to-be-known formulae? E.g. Induced emf is not asked to be calculated but just to be described in words.

helper · Mar 31, 2009

study-freak said:
But I thought they cannot ask you to calculate something using those meant-to-be-known formulae? E.g. Induced emf is not asked to be calculated but just to be described in words.

The syllabus point is:describe generated potential difference as the rate of change of magnetic flux through a circuit

So they can ask you to apply that defintion. It is a formula provided in word form and as such can still be asked to be used.

youngminii · Mar 31, 2009

Most institutions and teachers will just teach you the formula for emf induced anyway. You can use it in the HSC (-change in BA/change in time) and say 'oh since B, magnetic flux, is changing, there is an emf induced'.

clonestar · Apr 1, 2009

Don't forget EMF = -nBlv for Faraday's law where v = velocity of conductor experiencing change of flux.

Not quoted in the syllabus but certainly one to remember.

I would also remember

-Sy= Uyt - 1/2at^2
For total time of a projectile off a cliff. Then remember to use quadratic to derive the time and always take + value.

Miss Neutrino · Apr 6, 2009

clonestar said:
Don't forget EMF = -nBlv for Faraday's law where v = velocity of conductor experiencing change of flux.

Not quoted in the syllabus but certainly one to remember.

I would also remember

-Sy= Uyt - 1/2at^2
For total time of a projectile off a cliff. Then remember to use quadratic to derive the time and always take + value.

Just a question... from that equation and the quadratic equation you can derive that
$t = \frac{u_y + \sqrt{u_y^2 + 2 * a * \bigtriangleup y}}{a}$
... right?

I've tried it on a couple of questions and it seems to work, but I could be wrong.
However if it does work, would you be allowed to use that in an exam since it's not really an "official" equation?

clintmyster · Apr 6, 2009

Miss Neutrino said:
Just a question... from that equation and the quadratic equation you can derive that
$t = \frac{u_y + \sqrt{u_y^2 + 2 * a * \bigtriangleup y}}{a}$
... right?

I've tried it on a couple of questions and it seems to work, but I could be wrong.
However if it does work, would you be allowed to use that in an exam since it's not really an "official" equation?

Why would you even bother doing that though?

Miss Neutrino · Apr 6, 2009

Also, as for the formulas that aren't on the data sheet, I've made a list but it's only part for part of the Space topic so I'm not sure if there's any more from the rest of the topic that aren't on here...

Most of these are fairly easy to derive so it's probably only worth remembering the more difficult ones...

$g = \frac{Gm}{r^2}$

$u_x = u \cos \theta$

$u_y = u \cos \theta$

$\left | v \right | = \sqrt{v_y^2 + v_x^2}$

$\theta = \tan^{-1} \left ( \frac{v_y}{v_x} \right )$

$v_e = \sqrt{\frac{2Gm}{r}}$ (escape velocity)

$g force = \frac{g + a}{g}$

$a_c = \frac{v^2}{r}$

Orbital velocity:
$v = \frac{2 \pi r}{T}$ (given the period)

$v = \sqrt{\frac{GM}{R}}$ (given mass of object orbited and radius of orbit)

Miss Neutrino · Apr 6, 2009

clintmyster said:
Why would you even bother doing that though?

Well unless I've been doing something wrong, for the calculations of projectiles projected off cliffs at an angle you have to calculate it in 2 steps, one for the time it takes to be level with the cliff, and one for how long it takes to hit the ground from that level?

That way I figured could just get all the info and sub it in so there's less chance of making some stupid mistake along the way...

clonestar · Apr 8, 2009

Miss Neutrino said:
Well unless I've been doing something wrong, for the calculations of projectiles projected off cliffs at an angle you have to calculate it in 2 steps, one for the time it takes to be level with the cliff, and one for how long it takes to hit the ground from that level?

That way I figured could just get all the info and sub it in so there's less chance of making some stupid mistake along the way...

There is a quicker way... see my above post..

Use that equation above and go with

-Sy=Uyt -1/2at^2

Where the height of the cliff is placed in for Sy and the right side of equation becomes -4.9t^2

Then solve t using quadratic formula and take + value and viola you have total time of flight for a projectile thrown off the top of a cliff in one quick and easy step.

This way you avoid the two way process you stated in the quote which is time consuming and labor intensive but also works.

helper · Apr 8, 2009

I just can't work out why there is the debate over an equation that is on the formula sheet. Yes it is best to know them all but I can't work out why that one specifically has been mentioned.

TannouB · Aug 26, 2009

Lol just use what your given, should be enough, no need to cut corners, coz at least the calculations are straight forward. What i wanna know is what IS on the formula sheet LOL!

Uncle · Aug 26, 2009

I'll derive some.

Miss Neutrino said:
$g = \frac{Gm}{r^2}$

That can be derived from equating the weight of an object and the gravitational force.

F_weight = F_gravity

$mg = \frac{GMm}{r^2}$

$g = \frac{GM}{r^2}$

(The big M is for the mass of the planet while the small m is for an object on a planet)

QED - (quod erat demonstrandum) - means it is proven.

Miss Neutrino said:
$u_x = u \cos \theta$

$u_y = u \sin \theta$

$\theta = \tan^{-1} \left ( \frac{v_y}{v_x} \right )$

$\cos \theta = \frac{adjacent}{hypotenuse}$
$\cos \theta = \frac{u_x}{u}$

QED

$\sin \theta = \frac{opposite}{hypotenuse}$
$\sin \theta = \frac{u_y}{u}$

QED

$\tan \theta = \frac{opposite}{adjacent}$
$\tan \theta = \frac{v_y}{v_x}$
$\theta = \tan^{-1} \left ( \frac{v_y}{v_x} \right )$

QED

Comes from trigonometry.
Assumed knowledge from Year 10.

Miss Neutrino said:
$\left | v \right | = \sqrt{v_y^2 + v_x^2}$

No proof needed.
Pythagoras' Theorem.
Assumed knowledge from Year 8.

Miss Neutrino said:
$v_e = \sqrt{\frac{2Gm}{r}}$ (escape velocity)

From the principle of conservation of energy:

E_K + E_P = 0
(Kinetic Energy + Potential Energy = 0)

$\frac{1}{2}mv^{2}-\frac{GMm}{r} = 0$

$\frac{1}{2}mv^{2}= \frac{GMm}{r}$

$mv^{2}= \frac{2GMm}{r}$

$v^{2}= \frac{2GM}{r}$

$v = \sqrt{\frac{2GM}{r}}$

QED

In fact, the reason why those formulas are not included is because they assume you can derive them easily.

Miss Neutrino said:
Just a question... from that equation and the quadratic equation you can derive that
$t = \frac{u_y + \sqrt{u_y^2 + 2 * a * \bigtriangleup y}}{a}$
... right?

I've tried it on a couple of questions and it seems to work, but I could be wrong.
However if it does work, would you be allowed to use that in an exam since it's not really an "official" equation?

How else would you find the time of flight?

clonestar said:
Don't forget EMF = -nBlv for Faraday's law where v = velocity of conductor experiencing change of flux.

Not quoted in the syllabus but certainly one to remember.

I would also remember

-Sy= Uyt - 1/2at^2
For total time of a projectile off a cliff. Then remember to use quadratic to derive the time and always take + value.

Already on the formula sheet in a different form.

TannouB said:
Lol just use what your given, should be enough, no need to cut corners, coz at least the calculations are straight forward. What i wanna know is what IS on the formula sheet LOL!

.

What Physics Formulas are not on - Data sheet (2 Viewers)

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