question 11 graph vcaa 2011?

[dfgjgddjidfg]

i dont get how to draw the graph?

[panicatthelunchbar]

plot the points given, because maximum kinetic energy=voltage (when in eV)

since the line is linear, u simply have to extend the line given by the points all the way down till u get to the intercept which gives u the work function of the metal

not clear, but hope it helps

[datfatcat]

And when you extend the line, use dotted lines once u passed the horizontal axis

[dfgjgddjidfg]

sorry question 11 of electric power section

[TheRajinator]

yea i'd like to know was well. for this question is the inflection point on the x-axis neccessary to gain full marks??

[panicatthelunchbar]

yes, the current has to be zero at one point to be negative doesn't it?

[supermanflyaway]

Confused by this too... how do you get this graph? And what would the flux/time graph look like?

[BigAl]

it looks like a sin(x) graph..remember the needle of ampere metre? going right and left depending on the direction of the magnetic flux..

[rife168]

yea i'd like to know was well. for this question is the inflection point on the x-axis neccessary to gain full marks??

Yes, that's when the magnet's poles are either side of the coil, if you just had a line that went more-or-less straight through the t axis, then that implies that the flux is constant for only an instant, which would imply that the poles of the magnet were infinitely close together. The flat part of the stationary point of inflection in the V/t graph is simply due to the fact that the flux is constant for the time when the magnet 'threads' the coil.

Confused by this too... how do you get this graph? And what would the flux/time graph look like?

The flux/time graph would look a bit like an upside-down trapezium with rounded vertices

It starts at zero, then the flux slowly decreases from zero to the point of minimum flux ( max), then it is constant for a little while, then it goes back up to zero, in a way mirroring the first part of the graph.
Then for the EMF/t graph, you simply sketch the negative version of the gradient function.