VCE Physics Question Thread!

[odeaa]

What acts as the commutator in a brushless DC motor?

[odeaa]

What acts as the commutator in a brushless DC motor?

[paper-back]

Thanks lzxnl

When asked for the average EMF induced during a full rotation, am I just finding the EMF induced during a quarter of a revolution?

[odeaa]

Thanks lzxnl

When asked for the average EMF induced during a full rotation, am I just finding the EMF induced during a quarter of a revolution?

Wouldnt the average emf of a full turn be 0, because it goes from positive to negative? that being said i think i've done a similar question and gotten it wrong ahah

[lzxnl]

Wouldnt the average emf of a full turn be 0, because it goes from positive to negative? that being said i think i've done a similar question and gotten it wrong ahah

It is zero, because of the symmetry of the sine function. If it rotates at constant speed, the emf/time relation is sinusoidal.

Thanks lzxnl

When asked for the average EMF induced during a full rotation, am I just finding the EMF induced during a quarter of a revolution?

You're finding the average EMF, which is the average rate of change of the flux, which is the change in flux over the time interval.

[paper-back]

When sketching the EMF graph for a rotating coil do I use NABW to find the maximum and minimum voltage?

[Adequace]

I've started Unit 2 motion by myself and I'm not getting the answer for this question in the textbook. Part b. of the attached image,it looks rather straight forward but I'm not entirely sure what the first step is.


(Sorry about posting this here, but I think it's best to post it here since motion overlaps)

[Maths Forever]

I've started Unit 2 motion by myself and I'm not getting the answer for this question in the textbook. Part b. of the attached image,it looks rather straight forward but I'm not entirely sure what the first step is.


(Sorry about posting this here, but I think it's best to post it here since motion overlaps)

Here is what I think:

For part (a), a quick prediction would be 90 km/h, since it is half way between 80 km/h (from Melbourne to Wodonga) and 100 km/h (back from Wodonga to Melbourne).

For part (b), find the total distance and total time covered throughout the journey.

Total Distance = 300 km + 300 km = 600 km

Now speed = distance / time, so time = distance / speed

Time covered between Melbourne to Wodonga = (300 km) / (80 km/h) = 3.75 hours

Time covered between Wodonga to Melbourne = (300 km) / (100 km/h) = 3.00 hours

Therefore, total time = 3.75 hours + 3.00 hours = 6.75 hours

Hence, average speed = total distance / total time = 600 km / 6.75 h = 89 km/h (to 2 significant figures)

Reason for difference in predicted and calculated speeds:

90 km/h would be correct if we were considering a case where the initial speed was 80 km/h and the final speed was 100 km/h (i.e. using the formula v (av) = [u + v] / 2.

However, this is not the case. We are told the average speed for one interval of the journey, and the average speed for another interval. So the formula v (av) = Δ x / Δ t must be used, which yields 89 km/h.

I hope this helps! 

[Adequace]

Here is what I think:

For part (a), a quick prediction would be 90 km/h, since it is half way between 80 km/h (from Melbourne to Wodonga) and 100 km/h (back from Wodonga to Melbourne).

For part (b), find the total distance and total time covered throughout the journey.

Total Distance = 300 km + 300 km = 600 km

Now speed = distance / time, so time = distance / speed

Time covered between Melbourne to Wodonga = (300 km) / (80 km/h) = 3.75 hours

Time covered between Wodonga to Melbourne = (300 km) / (100 km/h) = 3.00 hours

Therefore, total time = 3.75 hours + 3.00 hours = 6.75 hours

Hence, average speed = total distance / total time = 600 km / 6.75 h = 89 km/h (to 2 significant figures)

Reason for difference in predicted and calculated speeds:

90 km/h would be correct if we were considering a case where the initial speed was 80 km/h and the final speed was 100 km/h (i.e. using the formula v (av) = [u + v] / 2.

However, this is not the case. We are told the average speed for one interval of the journey, and the average speed for another interval. So the formula v (av) = Δ x / Δ t must be used, which yields 89 km/h.

I hope this helps! 

Thanks for the reply, greatly appreciated!

Edit: your answer is correct but I don't quite understand why you need to use the total distance / the total time taken?

I interpreted the question as the return leg of the journey only, not both legs combined? Did I just interpret the question wrong?

[Floatzel98]

Thanks for the reply, greatly appreciated!

Edit: your answer is correct but I don't quite understand why you need to use the total distance / the total time taken?

I interpreted the question as the return leg of the journey only, not both legs combined? Did I just interpret the question wrong?

Yeah, the way they ask it does seem weird. I'm guessing they meant to write calculate the average speed for the whole return journey, like they worded it in part a. Otherwise you wouldn't need to do much calculating because they give you the average speed of the return trip haha.

You need to use  Δx /  Δt because you aren't dealing with constant acceleration. When you get up to that, hopefully a bit more will make sense. If you look at  this link, and read all the way down to the 3rd box, it might help answer your question.

[stockstamp]

Motion Question:

An F1 racing car is travelling around a banked corner. The angle of the banking is 15^o and the radius of the circular corner is 300m. The mass of the car is 850 kg.
If the friction between the road and car's tyres can apply a maximum sideways acceleration of 10 ms^-2 parallel to the slope of the road, how fast can the car travel around the corner now before sliding off the road?

Not sure I completely understand the wording of the question, and therefore don't know how to solve it.
Any help appreciated 

[silverpixeli]

Hey guys, I'm taking over the physics lecture next Wednesday for Alwin and I'd love to get an idea for where you guys are up to in class. So to anyone who is coming along (or even if you're not but you feel like answering - btw there are still tickets available) please let me know where your class is at right now.

Thanks!

[odeaa]

Hey guys, I'm taking over the physics lecture next Wednesday for Alwin and I'd love to get an idea for where you guys are up to in class. So to anyone who is coming along (or even if you're not but you feel like answering - btw there are still tickets available) please let me know where your class is at right now.

Thanks!

Electric power sac first week back next term

[Floatzel98]

Hey guys, I'm taking over the physics lecture next Wednesday for Alwin and I'd love to get an idea for where you guys are up to in class. So to anyone who is coming along (or even if you're not but you feel like answering - btw there are still tickets available) please let me know where your class is at right now.

Thanks!

We are also up to Electric power. We just went over magnetic flux before the holidays.

[Floatzel98]

Just had a question, when we are looking at pictures like these (attached), is the magnetic field one created from the current carrying wire or has a current carrying wire been placed into a magnetic field? I was just a bit confused as to how you can actually cause a magnetic field to not be perpendicular to the wire (As in the 3rd and 4th diagrams). Is it just maybe interacting with another magnetic field and that changes the direction of the original one?

And with motors and generators, the difference is one moves with a current through it and then one moves/gets moved to make a current? How do you increase current produced from a generator and how high can it get? .Why don't cars make use of those generator principles? I might be correct to say I'm probably not the first person to suggest that though....

Also, with magnetic flux, how exactly is a current and EMF induced induced? Does the magnetic field like use the force and bring electrons up from the ground and put them into the wire? Also, since the rate of change of flux is equal to the induced EMF, how do you get there? Can you derive or something? I know it is probably more complicated but I feel it involes differentiation somehow since it is the rate of change of the flux.

With V(RMS) or I(RMS), is it just supposed to be the positive side of the graph because it is DC or is is positive and negative values of it?

Thanks