VCAA 2007 physics exam help

[rk]

VCAA 2007 physics exam:

- AREA OF STUDY 1 -

Why is question 3's magnitude 1.6N?

why is question 4 anticlockwise [pretty much the answer to 4b]

For question 6 I wrote: 1/200 instead of 5*10^-3 as in the solutions...my teacher [this was the exam we did as our school trial] gave me only 2 marks not 3. Would this be the case in the actual exam? What about 0.005?

For question 7 I had the right working out except I multiplied it by (1/4) in light of the 'quarter turn' bit. Don't we have to do that?

I don't get question 8 either? The solution says the period of rotation is 80ms. What is the period of rotattion and whete is this number from based on the gaph? I can see that ththe frequency is 80 but what aout 12.5Hz as well how do you get that?

Question 11.i, ii and iii as well as 12i,ii and iii is something I can't do and the solutions give no indication for me. As you guys may or may no I have little to know understanding of physics so could you explain this to me in a basic way?

[Whatlol]

For q 3 . F = IBL = 2000x20 x 4.x10^-5 = 1.6

q4. The current flows in the direction JKLM. The field runs from the north pole to the south. Using the right hand slap rule you will see that on side JK the force is down, therefore the coil will rotate anticlockwise.

q.6 i wouldnt write anything in fractions, best to just write it in either scientific notation, 5x10^-3 will be fine, also 0.005 is fine too im fairly sure.

q7. they have already told you that the time taken for a quater revolution is 0.01 s. In the equation for calculating EMF , this will be the change in time. It is only if they tell you a coil has rotated a quater turn that has a period of x seconds, where you need you quater the period in order to find the average EMF induced.

q8. The graph provides you with the period, this is the time taken FOR ONE ROTATION. the frequency is rotations per second i.e 1/T . Therefore you simply do 1/ 80ms = 12.5 Hz

q11. The value for the currents can be found using the turns ratio of the transformer.
Since they are ideal P₁ = P₂
                                V₁I₁ = V₂I₂
                                 therefore V₁/ V₂ = I₂ / I₁ = N₁ / N₂

Since we know that N1 / N2 = 5

the current at A2 will be given by 5 x 0.5 = 2.5A


To figure out the voltage at V1 you must first find the voltage drop across the transmission lines.
This can be found using the relationship V =IR
V = 0.5 x 4 = 2V
therefore the voltage left will be 12 -2 = 10 V

Now we also know that the ratio of voltages equals the ratio of the windings on the primary and secondary coils. so the voltage at V2 will be given by :


V₁ V₂  = N₁ / N₂

V₂ =  N₂ x V₁ / N₁
                    = 1/5 x 10 =2 V

Question 12, hopefully you can understand how to do question 11 now and will figure this out. If not ill show you:

To calculate the power output, all you need is a direct aplication of P = VI
therefore P = 12 x 0.5 = 6W

At the input of the primary coil, we know the voltage which was 10 V
so P = 10 x 0.5 = 5W
Since this transformer is ideal, we assume the power on both sides of the transformer are equal so iii. is 5 W too

Question 14. First thing you need to identify is that we are dealing with a DC OUTPUT, since it is a battery. the consequence of this is that once the current flows for a small amount of time i.e the current flow is established in the coil there wont be a changing magnetic field and hence no emf induced in the other coil . So this means there will only be a change in flux when the battery is turned off or on. This leaves you with only options A and B. Now when the current flows, a magnetic field is created such that there is a south pole towards the right end of the first coil. To oppose this change in flux, a south pole will also be created in the secondary coil to the left end of it, for this to happen the current must flow from X to Y.
          
For light you just use c = lambda x frequency therefore 3x10^8 / 400x10^-9 =7.5x10^14Hz.
anyways i gotta go, ill finish the rest off tomorrow, or someone else might. hope you can understand this and its all correct.

[Whatlol]

Look since the exam is tommorow, i would just advice you remember the y intercept provides you with the work function, and the x intercept with the threshold frequency.

just simply convert 400nm to frequency by using c = lambda x frequency.

so 3x10^8 / 400x10^-9 = 7.5 x10^14 Hz.

[Blakhitman]

Which question?

[Whatlol]

Which question?

*more like... what question isnt here...