VCE Physics Question Thread!

[Einstein]

how and what is the answer to this question, from electricity sac.

Thanks

[b^3]

how and what is the answer to this question, from electricity sac.

Thanks

They need to be a multiple of the elementary charge, which is approximately (I can't remember how many sig figs you do it to in physics but it doesn't matter here).
So if you divide each of the charges by the elementarycharge, you should get an integer.
Doing this gives
A. 20
B. 5
C. 37.5
D. 90

So option C is the only one that isn't an integer multiple of the elementary charge, and so that charge could not exist. i.e. Option C.

[Einstein]

how is this derived?

[RKTR]

how is this derived?

already forgot about this. had to refer to my year 11 textbook. Sv is the unit for dose equivalent while Gy is the unit for absorbed dose.
the symbol before the Sv and Gy is micro which is 10^-6 . To determine the most damaging, you have to convert all them to dose equivalent. dose equivalent = absorbed dose x quality factor.
A 20 x 10^-6
B 30 x 10^-6 x (1) =30 x 10^-6
C 30 x 10^-6
D 20 x 10^-6  x( quality factor of 10-20) =(200  to 400) x 10^-6

dose equivalent of D is highest so it is the most damaging.

[Stew_822]

Hi guys,

I'm really stuck with a voltage divider question, I thought I had them down but this one is confusing me.

It's question 32 and 33 of A+ Notes for physics.

I realise the photo isn't clear - Change in Vout = 6V and change in Vin = 0.4V

So here's the answers from the book and my answers afterwards:
Answer to Q30 is 0.6V and 4V - no worries.
Answer to Q31 is 15V - wtf? Isn't it negative 15V because the amplifier is inverting?
Answer to Q32 is 2.5V - That's reading from the graph but when you do Vout = Vgain * Vin = -15 * 0.7 = 10.5V, it doesn't agree with the graph. Can anyone explain why this is the case?
Answer to Q33 is 5.5V - Same problem as Q32.

Thanks,
Stewart

[speedy]

Hi guys,

I'm really stuck with a voltage divider question, I thought I had them down but this one is confusing me.

It's question 32 and 33 of A+ Notes for physics.

I realise the photo isn't clear - Change in Vout = 6V and change in Vin = 0.4V

So here's the answers from the book and my answers afterwards:
Answer to Q30 is 0.6V and 4V - no worries.
Answer to Q31 is 15V - wtf? Isn't it negative 15V because the amplifier is inverting?
Answer to Q32 is 2.5V - That's reading from the graph but when you do Vout = Vgain * Vin = -15 * 0.7 = 10.5V, it doesn't agree with the graph. Can anyone explain why this is the case?
Answer to Q33 is 5.5V - Same problem as Q32.

Thanks,
Stewart

Gain is an absolute value and is usually given as positive regardless of it inverting or not. However the assessors accept either positive or negative values.

What you must realise is that this amplifier is biased - its transfer characteristics graph doesn't pass through the origin. When you simply multiply V_in by the gain, you are treating the amplifier as if it is unbiased. To take the biasing into account, the graph must be used.

[Stew_822]

Ah, I see - I didn't realise there was a difference (or more to the point didn't realise it made a difference). The Heinemann textbook only seems to contain non-biased amplifiers, lucky then that I got these questions now rather than in an exam.

Is that the only way you can find V_out, by reading it off the graph? Or is there a mathematical method also that takes into account the biased nature of the amplifier. Final question, if there is a mathematical method do I need to know it or will the graph be given in the exam?

Many thanks for your help,
Stewart

[Rishi97]

Power lines carry an electric current in the Earth's magnetic field. Which would experience the greater magnetic force: a north-south power line or an east-west power line? Explain
Many thanks

[lzxnl]

Earth's magnetic field runs south to north (at least along the Earth's surface), so any current parallel or antiparallel to this would experience no force. Hence your power line needs to be east-west for there to be a force.

[Orb]

When we're calculating the co-efficient of friction, formula Fr= m x N
where m is the co-efficient of friction,

What unit do we attribute to N, the normal reaction force? Do we use the mass provided or do we convert it into newtons?

[Jacyan]

When we're calculating the co-efficient of friction, formula Fr= m x N
where m is the co-efficient of friction,

What unit do we attribute to N, the normal reaction force? Do we use the mass provided or do we convert it into newtons?

Fr is in newtons, N is in newtons, and the co-efficient of friction has no units.

[Orb]

So how should I approach this question...

I get a result that's 10x more than the actual answer :/

[Ancora_Imparo]

So how should I approach this question...

I get a result that's 10x more than the actual answer :/

As we are told the system is moving left, let left be positive. Let the tension in the left rope be and the tension in the right rope be .

Draw the forces the acting on each block.
For M1: mg is down and T1 is up
For M2: mg is down, N is up, T1 is left, T2 is right, friction is right (as we are told block is moving to the left)
For M3: mg is down, T2 is up

Since acceleration of whole system is the same, we can apply Newton's second law to each block individually.
For M1:

--- Eq. 1

For M2:
You also know that
Thus:

--- Eq. 2

For M3:

--- Eq. 3

Adding Eq. 1 and Eq 3. gives:
--- Eq. 4

Adding Eq. 2 and Eq 4. gives:

[EDIT: Whoops, got the fraction the wrong way around the first time]
(negative sign shows that it's decelerating)

Hope that helps.

[Orb]

As we are told the system is moving left, let left be positive. Let the tension in the left rope be and the tension in the right rope be .

Draw the forces the acting on each block.
For M1: mg is down and T1 is up
For M2: mg is down, N is up, T1 is left, T2 is right, friction is right (as we are told block is moving to the left)
For M3: mg is down, T2 is up

Since acceleration of whole system is the same, we can apply Newton's second law to each block individually.
For M1:

--- Eq. 1

For M2:
You also know that
Thus:

--- Eq. 2

For M3:

--- Eq. 3

Adding Eq. 1 and Eq 3. gives:
--- Eq. 4

Adding Eq. 2 and Eq 4. gives:

(negative sign shows that it's decelerating)

Hope that helps.

I'm sorry, i'm not sure what you did wrong but apparently that wasn't what the answer is...

Solution for the question was 5.6 ms ^-2

[Zealous]

I'm sorry, i'm not sure what you did wrong but apparently that wasn't what the answer is...

Solution for the question was 5.6 ms ^-2

I'll give this a shot. Hopefully my explanation makes sense.

First let's look at the forces due to gravity. We have a force due to gravity of M3 which is and the force due to gravity on M1 which is .

Now let's consider the frictional force of M2:




Friction acts against a motion, so since M2 was initially moving left, the force of friction will be acting in the right/clockwise direction.

If we consider moving right/clockwise to be positive, the sum of the forces will be:

M3 and M2 are causing the system to move right so they are positive, and M1 is working against both of them so is negative.

Hence: