[DOWNLOAD] My Half-Yearly Examination (2 Viewers)

[bleakarcher]

1. B (obvious)
2. C (you switch engines off between the stages)
3. C (high altitude=lower g)
4. A (calculated)
5. C (Fc=Fg=Gm1m2/d^2)
6. C (Net EMF=0 for both, back EMF is double for the 20V, net force=0 for both)
7. D (if you consider friction and gravity)
8. B (Dodgy question - not clear enough, I'm assuming area means one loop)
9. A (net force=0)
10. B (Use RH rule, makes flux in the same direction - flux density increases)
11. D (Dodgy question - it's not clear with how everything in connected/measured - I feel that change to split-ring commutator makes it positive, rotating in the opposite direction makes the flux be measured as negative.)

With q11), since it says the EMF induced in the coils, wouldn't the correct answer be B? I'm thinking it would be D if you're talking about the EMF in the external circuit.

 

[Hypem]

Thanks

I just realised my calculator was in radians for quesiton 8... wow

 

[Hypem]

1. B (obvious)
2. C (you switch engines off between the stages)
3. C (high altitude=lower g)
4. A (calculated)
5. C (Fc=Fg=Gm1m2/d^2)
6. C (Net EMF=0 for both, back EMF is double for the 20V, net force=0 for both)
7. D (if you consider friction and gravity)
8. B (Dodgy question - not clear enough, I'm assuming area means one loop)
9. A (net force=0)
10. B (Use RH rule, makes flux in the same direction - flux density increases)
11. D (Dodgy question - it's not clear with how everything in connected/measured - I feel that change to split-ring commutator makes it positive, rotating in the opposite direction makes the flux be measured as negative.)

Damn... I did horribly.

4/11.

1) yeah
2) yeah
3) didn't even think of that
4) you should post the solution
5) yeah
6) eh
7) didn't even think of that
8) yeah
9) what?
10) I see now
11) eh

 

[someth1ng]

With q11), since it says the EMF induced in the coils, wouldn't the correct answer be B? I'm thinking it would be D if you're talking about the EMF in the external circuit.

Why would it need to be positive? It's a bit of a strange question to ask.

 

[someth1ng]

For question 4:

As you know, if they are to hit (and start at the same time), they must have the same V(x), right?
For the top object, we'll call that object 1 and the bottom object, we'll call that object 2, we calculate V(x)
V1(x)=20ms-1 (given)
V2(x)=40cos60=20ms-1

Therefore, you know they WILL collide at one point (this makes D wrong).

Now, calculate at what position y (measured from the ground) they will meet (x will be the same at all times).

For object 1:
S1(y)=100+(0)t+[1/2](-9.8)t^2
S1(y)=100-4.9t^2

S2(y)=40sin(60)t+[1/2](-9.8)t^2
S2(y)=40sin(60)t-4.9t^2

Since we're looking at equal values of y and these are both equations to show the location of either object at any instant...

S1(y)=S2(y)
100-4.9t^2=40sin(60)t-4.9t^2
100=40sin(60)t
t=100/40sin(60)
t=2.886...s
t=2.9s --> (A)

 

[Hypem]

For question 4:

As you know, if they are to hit (and start at the same time), they must have the same V(x), right?
For the top object, we'll call that object 1 and the bottom object, we'll call that object 2, we calculate V(x)
V1(x)=20ms-1 (given)
V2(x)=40cos60=20ms-1

Therefore, you know they WILL collide at one point (this makes D wrong).

Now, calculate at what position y (measured from the ground) they will meet (x will be the same at all times).

For object 1:
S1(y)=100+(0)t+[1/2](-9.8)t^2
S1(y)=100-4.9t^2

S2(y)=40sin(60)t+[1/2](-9.8)t^2
S2(y)=40sin(60)t-4.9t^2

Since we're looking at equal values of y and these are both equations to show the location of either object at any instant...

S1(y)=S2(y)
100-4.9t^2=40sin(60)t-4.9t^2
100=40sin(60)t
t=100/40sin(60)
t=2.886...s
t=2.9s --> (A)

Nice explanation and solution, thanks

That probably shouldn't have been a 1 mark multiple choice question though

 

[someth1ng]

No worries.

As Fizzy_Cyst said, it is hard for multiple choice and so, it would be natural for it to be a lot of work for a single mark but after all, these are the sort of discriminator questions that you will get in HSC and definitely worth knowing how to do.

 

[adii]

1.b
2.b
3.c
4.b
5.c
6.d
7.c
8.b
9.d
10.b
11.d

 

[iBibah]

1.b
2.b X
3.c
4.b X
5.c
6.d X
7.c X
8.b
9.d X
10.b
11.d X

.

 

[iBibah]

1.b
2.b X
3.c
4.b X
5.c
6.d X
7.c X
8.b
9.d X
10.b
11.d X

.

 

[adii]

but with Q 7, in hsc isnt friction ignored, so shudnt answer be C?
i know in reality friction will affect

 

[bleakarcher]

Why would it need to be positive? It's a bit of a strange question to ask.

Thinking about it again actually, I think it's A (call me crazy). I don't think either changes would cause a change in the V-t graph.

 

[adii]

yeah, but direction shouldnt affect if it is ac or dc yh? direction of rotation should only affect polarity
therefore, split ring commutator is the factor that makes it dc

 

[bleakarcher]

yeah, but direction shouldnt affect if it is ac or dc yh? direction of rotation should only affect polarity
therefore, split ring commutator is the factor that makes it dc

Yes, I agree but DC in the external circuit. The question says 'Which graph below best represents the EMF induced in the coils after the changes were made?', not the external circuit. The current induced in the rotating coils should remain AC.

 

[iBibah]

but with Q 7, in hsc isnt friction ignored, so shudnt answer be C?
i know in reality friction will affect

No because it states it's in low earth orbit and it also (in bold) asks you to consider all forces.

 

[iBibah]

Yes, I agree but DC in the external circuit. The question says 'Which graph below best represents the EMF induced in the coils after the changes were made?', not the external circuit. The current induced in the rotating coils should remain AC.

You were correct with B. The first change only effects the external circuit, however no effect on the coils. However the second change does effect the induced emf in the coils, making it negative (flipping it about the x axis).

 

[bleakarcher]

You were correct with B. The first change only effects the external circuit, however no effect on the coils. However the second change does effect the induced emf in the coils, making it negative (flipping it about the y axis).

So that means current flows in the opposite direction if turned anti-clockwise, right?

 

[Hypem]

One thing I've learnt from this: I need to read and analyse questions a lot more carefully.

 

[iBibah]

So that means current flows in the opposite direction if turned anti-clockwise, right?

According to Faradays law, direction of induced current depends on the direction of movement within the magnetic field. So yes, induced emf does flow in the opposite direction when rotated anti-clockwise.

 

[bleakarcher]

According to Faradays law, direction of induced current depends on the direction of movement within the magnetic field. So yes, induced emf does flow in the opposite direction when rotated anti-clockwise.

Here's what I was thinking. Since when t=0, V=0, the plane would initially be perpendicular to the direction of the magnetic field. If you turned it clockwise or anti-clockwise, the induced current would be induced in the same direction using right hand rule, and so as you said by Faraday's Law, the induced voltage should not change direction. Therefore, our answer would be A, if that makes any sense.