Search results

Asin emf from flux? You could consider emf = -n*change in flux / change in time Usually in multiple choice, you're given a flux curve and asked to determine which one of the options is the emf? So if you have a sine curve, think about differentiating to a cosine curve and flipping it.

+1

Reduce the size of eddy currents and so reduces energy converted into heat.

It's a side on view

It doesn't if it's parallel, but there are a few field lines higher up which are perpendicular to the current.

The key to Mathematics is getting a lot of exposure and practice. If you're one of those people that isn't very motivated, then a tutor will definitely help. Tutoring gives you an extra push that school might not provide, and it also gives you another batch of techniques on how to solve...

Just skipping the boring algebra, you agree that the equation of the normal at P is: x + py = -4p(p²+1) P(-8p,-4p²) M (0 , -4(p²+1)) mid(PM) = (-4p , -4(2p² + 1)) Let L(x,y) be the general point on the required locus as P varies: x = -4p y = -4(2p² + 1) p = -x/4 sub into u y = -4(2(x²/16) +...

Just curious, if it's easy then what's the problem?

You can put $ around the text as if they were brackets eg. $thanks a lot mate that helped! as you can see =)$ (Inside the brackets of course [tex]$thanks a lot mate that helped! as you can see =)$

I think ohexploitable means 0< k <1. For the 2nd image, part ii) For part i), you get a max and min turning point. To get 3 real roots, the y value of the max has to be above or on the x axis, and the y value of the min has to be below or on the x-axis. Then you look at the other keyword...

Wait hang on. y = - a is parallel to the x-axis...

Both seem fine.

It's an open book exam with about 8 questions or so. Very challenging and moneys for a distinction

Look at it this way: If you do 4 unit next year, >3 unit will count for 2 units and >4 unit will count for 2 units If you don't do 4 unit next year, but do 3 unit >3 unit will count for 1 unit >2 unit will count for 2 units So if you do 4 unit: >2 unit won't count >3 unit will...

We're doing Quanta to Quarks which is like a history lesson.

Thanks for the rep =]

I think this is right If you think about it, light travels faster in air than in glass (you can probably research this somewhere), and in the question you have light bending closer to the normal as it slows down.

\frac{sin\theta1}{sin\theta2}=\frac{v1}{v2}(Snell's Law)\\$rearranging this we get$ \frac{v2}{sin\theta2}=\frac{v1}{sin\theta1}\\$From the question, for both green and yellow light $\theta1=60$ and assuming v1 is the same (speed of light)\\So $ \frac{v1}{sin\theta1}$ is the same for both green...

It seems HSC mark is different from examination mark and assessment mark, but isn't the average as you said. So what exactly is HSC mark if not the "examination mark"? I'm confused.