VCE Specialist 3/4 Question Thread!

[Shadowxo]

Awesome, thanks for the help! If anyone is willing to help with the other parts of the dynamics question I would be forever grateful...

What parts of it are you stuck on? If you post your working I can show you where you went wrong

[tinagranger]

What parts of it are you stuck on? If you post your working I can show you where you went wrong

I've figured out how to do a) but am getting the wrong answer for b i) and ii) 
For i) I solved F - 6888 = 0.1 x 60000
ii) I solved F - 360 - 12,000g x 1/200 = 0.1 x 60000

Thanks!

[Shadowxo]

I've figured out how to do a) but am getting the wrong answer for b i) and ii) 
For i) I solved F - 6888 = 0.1 x 60000
ii) I solved F - 360 - 12,000g x 1/200 = 0.1 x 60000

Thanks!

You've almost got it
i) So, sum of all forces = ma
F-6888=0.1*(60000+12000) as the force is the total force on the truck + engine, and you're looking at the whole system.
The total force acting on truck + engine is F-6888, not on the engine individually.
If you were just using 60000, you'd be looking only at the forces on the engine (ie tension weight resistance and tractive force)

Try ii) with that

[tinagranger]

Hi, urgent!! Can someone please tell me if friction is at a maximum (i.e. Fr = mew N) when the block is ALREADY MOVING?

[Sine]

Hi, urgent!! Can someone please tell me if friction is at a maximum (i.e. Fr = mew N) when the block is ALREADY MOVING?

Fr_MAX =  μN  is the technically the "correct" formula

Friction is always at it's maximum if the block is moving on a non-smooth surface. However you can have max friction when the block is stationary. This would mean the block is "on the verge of moving" i.e. An extra bit of force would be too much for the maximum friction hence net force and therefore an acceleration occurs.

[Rieko Ioane]

Hello, I need help on this MCQ. Why does testing each option in to the equation not work? Option E gives LHS = RHS but the answer is D http://imgur.com/a/AV0bN

Ty

[VanillaRice]

Hello, I need help on this MCQ. Why does testing each option in to the equation not work? Option E gives LHS = RHS but the answer is D http://imgur.com/a/AV0bN

Ty

I think you've gotten a few negatives mixed up. It might help to test each option somewhere with more space, or even on your calculator.
For E, LHS should be:
3(6 - 2i) - (6 + 2i)
= 18 - 12i -6 - 2i

Hope this helps

[Sine]

Hello, I need help on this MCQ. Why does testing each option in to the equation not work? Option E gives LHS = RHS but the answer is D http://imgur.com/a/AV0bN

Ty

find solution attatched

sorry if it isn't that neat (only used my laptop trackpad)

[Rieko Ioane]

Could someone check my method for this question? I think it's different to what the answer has. http://imgur.com/a/aJ2Z0

Pretty much point A = (500,450) and OP(plane's path) = 30(0.5t^2 + t)i + 20(0.5t^2 - t)j is the info required to do this q.

Thanks

[Syndicate]

Could someone check my method for this question? I think it's different to what the answer has. http://imgur.com/a/aJ2Z0

Pretty much point A = (500,450) and OP(plane's path) = 30(0.5t^2 + t)i + 20(0.5t^2 - t)j is the info required to do this q.

Thanks

Your method is sufficient.

(btw you have used 400 instead of 500, but you should still get t1 =/= t2)

[Rieko Ioane]

Your method is sufficient.

(btw you have used 400 instead of 500, but you should still get t1 =/= t2)

Ah dammit, oops. Thanks though!

[Rieko Ioane]

Hello, for part a https://ibb.co/fyrsta
The question found the average value of the confidence interval (ie the mean of the sample mean). But the mean of the sample =/= the sample mean? So why can we do this?

Thanks

[Sine]

Hello, for part a https://ibb.co/fyrsta
The question found the average value of the confidence interval (ie the mean of the sample mean). But the mean of the sample =/= the sample mean? So why can we do this?

Thanks

so remember when you create a C.I you use the sample mean that has been found which is x-bar.

The "end values " of the C.I is a result of plus/minus 1.96 * standard error to the sample mean.

That is why you can find sample mean from "end values" of the C.I.

[Rieko Ioane]

so remember when you create a C.I you use the sample mean that has been found which is x-bar.

The "end values " of the C.I is a result of plus/minus 1.96 * standard error to the sample mean.

That is why you can find sample mean from "end values" of the C.I.

Thanks I see.

For example, if this question did not give us a confidence interval, then we would say that the sample mean equals mew, as this is what we would expect it to be? Because we have no other info anymore?

[Sine]

Thanks I see.

For example, if this question did not give us a confidence interval, then we would say that the sample mean equals mew, as this is what we would expect it to be? Because we have no other info anymore?

you will either get
1) sample mean given explictly to you
2a) implicitly given to you via a confidence interval
2b) implicitly given to you through raw data