HSC Physics Question Thread

[blasonduo]

Cool - So we know that in an AC Motor, the changing direction of current is what maintains the constant direction of torque (the job that the split ring commutator does in a DC motor). What this means is that the AC frequency (how fast it switches) will determine how quickly the motor will spin. It will complete one half spin, then the current will switch direction, so it can complete another half spin, and so on. In a DC motor, it is purely the size of the current that determines how quickly it will spin - In an AC motor this is more or less irrelevant, because the frequency will apply the strictest conditions on its motion!

It's a little hard to explain admittedly, does that make sense though?

Cool, Thank you, but this does make me ask;

Let's just go to the extreme, and assume we supply a voltage of 1million hertz. The change of flow in the AC motor is WAYY too fast for it to spin correctly. (as once the forces due to the flow of current let's say make it spin clockwise, but before it can really move, the current swaps so it is now spinning anti-clockwise, so it can't spin unidirectional)

From this, how can I deduce that B is incorrect?

I appreciate your help!

[julies]

Hey! This is right on the borderline of what you are expected to know, but roughly:

- When you connect a voltage source such that the positive terminal is connected to the P-type, and the negative terminal is connected to the N type, this pushes the positive holes in the P-type towards the junction, and the negative electrons in the N-type towards the junction. This is a forward bias connection. This reduces the size of the depletion region. Another way to think of it is that the applied voltage sort of 'cancels' the potential difference caused by the depletion region in the first place. This allows current to flow!
- If we connect it the other way around, the opposite occurs! We don't take away from the voltage, we make it worse, so to speak! This increases the size of the region and thus makes it very difficult for current to flow. Practically, eventually, you would get current - It happens when the diode goes "nope" and electrons start being ripped away from their atoms. But this is reverse bias, essentially a no current scenario!

thanks jamon, this was super helpful

[jamonwindeyer]

Cool, Thank you, but this does make me ask;

Let's just go to the extreme, and assume we supply a voltage of 1million hertz. The change of flow in the AC motor is WAYY too fast for it to spin correctly. (as once the forces due to the flow of current let's say make it spin clockwise, but before it can really move, the current swaps so it is now spinning anti-clockwise, so it can't spin unidirectional)

From this, how can I deduce that B is incorrect?

I appreciate your help!

Yeah cool, so let's go with your scenario. The motor won't spin, it will basically vibrate at 1MHz (I think you'd basically turn your motor into a radio antenna at that frequency lol), but you are right, too quick. My response is that, even if you turned up the voltage, it still wouldn't spin. So what that tells us is that whether the motor spins or not is dependent on frequency in that scenario, not voltage! So, that leads you back to A. You'd conclude B is incorrect from this because voltage has no effect on whether the motor spins or not at that frequency

[blasonduo]

Yeah cool, so let's go with your scenario. The motor won't spin, it will basically vibrate at 1MHz (I think you'd basically turn your motor into a radio antenna at that frequency lol), but you are right, too quick. My response is that, even if you turned up the voltage, it still wouldn't spin. So what that tells us is that whether the motor spins or not is dependent on frequency in that scenario, not voltage! So, that leads you back to A. You'd conclude B is incorrect from this because voltage has no effect on whether the motor spins or not at that frequency

Hmm, There's still something I don't quite get, sorry for being a bother. What is wrong with my logic?

So, when I have a set AC frequency which causes the motor to rotate, If I change the voltage, it WILL spin slower/faster, but it will cause the motor to "be out of whack" and not spin, So a new frequency will need to be set to fix the motor, setting this new frequency will always make the motor spin again.

HOWEVER, If I have a differing frequency which causes the motor to not rotate, fixing the voltage WILL fix the motor spin, as Increasing the voltage does increase the speed of a motor, (i.e its RPM)

So, whats wrong?

Thanks

[jamonwindeyer]

Hmm, There's still something I don't quite get, sorry for being a bother. What is wrong with my logic?

No bother at all! This is a little tough - First, remember that voltage (more appropriately, current) isn't directly related to speed. It is related to torque, \(\tau=BAIn\). While torque can translate into greater speed, it does depend a little what we've got the thing attached to. You can actually analyse what's called the torque/speed characteristics of a motor, the torque it can provide at a given speed. This is beyond syllabus but the idea is that we can't assume a direct relationship between voltage and speed, especially for an AC motor.

The logic you've used is definitely correct in principle. I think the simplest way to think about how to apply it to this question is this. Consider the simple AC motor that we like to draw (which we never use in the real world, mind you, which is why this all seems a bit wishy washy):

- If the AC frequency is 50Hz, we literally can't make the motor spin at more than 50 revolutions per second. We can up the voltage all we like, but 50 revs per second is the limit.
- Now what if we dropped to 45Hz without changing the voltage. Would we still spin at all? Maybe, maybe not - But what's the new limit? 45 revolutions per second. If we do spin, that's what we spin at.

I suppose the point I'm driving is, you can't go faster than your AC frequency allows. It sets the speed limit. You need to be in tune with it to get your spins. Varying voltage means nothing if you aren't working with your AC frequency.

I liken it to running on an electronic treadmill (where the belt runs at a speed you set on a control panel). Once you set that speed, that determines how quickly you are running. Sure, you can run 'harder,' pump harder with your legs and do more work - That's kind of like increasing your voltage. But you aren't going to move any faster - If you try to use that energy to move faster, you crash into the front of the treadmill and you stop. So it doesn't work!

Again, I know this is wish-washy. You are critiquing a model of an AC motor, and you are right to do it, because the model you learn isn't practical and doesn't actually make sense. Don't worry, the AC motors that actually move stuff around in our world are far more sophisticated than a coil in between a couple of bar magnets it might not rest quite right, and that's okay. Just try and remember that frequency is king/queen for an AC motor

[blasonduo]

No bother at all! This is a little tough - First, remember that voltage (more appropriately, current) isn't directly related to speed. It is related to torque, \(\tau=BAIn\). While torque can translate into greater speed, it does depend a little what we've got the thing attached to. You can actually analyse what's called the torque/speed characteristics of a motor, the torque it can provide at a given speed. This is beyond syllabus but the idea is that we can't assume a direct relationship between voltage and speed, especially for an AC motor.

The logic you've used is definitely correct in principle. I think the simplest way to think about how to apply it to this question is this. Consider the simple AC motor that we like to draw (which we never use in the real world, mind you, which is why this all seems a bit wishy washy):

- If the AC frequency is 50Hz, we literally can't make the motor spin at more than 50 revolutions per second. We can up the voltage all we like, but 50 revs per second is the limit.
- Now what if we dropped to 45Hz without changing the voltage. Would we still spin at all? Maybe, maybe not - But what's the new limit? 45 revolutions per second. If we do spin, that's what we spin at.

I suppose the point I'm driving is, you can't go faster than your AC frequency allows. It sets the speed limit. You need to be in tune with it to get your spins. Varying voltage means nothing if you aren't working with your AC frequency.

I liken it to running on an electronic treadmill (where the belt runs at a speed you set on a control panel). Once you set that speed, that determines how quickly you are running. Sure, you can run 'harder,' pump harder with your legs and do more work - That's kind of like increasing your voltage. But you aren't going to move any faster - If you try to use that energy to move faster, you crash into the front of the treadmill and you stop. So it doesn't work!

Again, I know this is wish-washy. You are critiquing a model of an AC motor, and you are right to do it, because the model you learn isn't practical and doesn't actually make sense. Don't worry, the AC motors that actually move stuff around in our world are far more sophisticated than a coil in between a couple of bar magnets it might not rest quite right, and that's okay. Just try and remember that frequency is king/queen for an AC motor

<3 It has finally clicked! Thank you so much!

[jamonwindeyer]

<3 It has finally clicked! Thank you so much!

Sahweeet - No worries at all!!

[Dragomistress]

May someone help me with this question?

A heater uses 2.2x10^6 J of energy in three hours and 20 minutes. How much energy, in kilowatt-hours, is used by this heater?

[Shadowxo]

May someone help me with this question?

A heater uses 2.2x10^6 J of energy in three hours and 20 minutes. How much energy, in kilowatt-hours, is used by this heater?

First you have to figure out how many joules a kilowatt hour is.
It's basically if you ran a 1000 Watt (1 kilowatt) device for an hour. Remember a Watt is 1 Joule per Second
So 1 kWh = 1000 * 60 * 60 (as an hour is 60*60 seconds) = 3.6*10⁶J

So, it uses

[austv99]

Hi,
Can I get help on answering part ii? (Quanta to Quark)

[Iminschool]

I said D, the answers say B. How'd they get B?

[jamonwindeyer]

I said D, the answers say B. How'd they get B?

Hey! So without the coil and magnet, we'd expect the speed to be:



Now obviously it ends up moving more slowly than that - Some of the kinetic energy has been taken and converted into electrical energy. How much? Well it is just the kinetic energy corresponding to the difference in speed:



Now since the resistance is small, the light will convert all of that to other forms of energy - Issue of course being this doesn't match the answers hmm, can't spot what I did wrong right this second (multitasking, lol) - Could you upload your working to get D? Could anyone else chime in?

[kiwiberry]

I said D, the answers say B. How'd they get B?

Hey! So without the coil and magnet, we'd expect the speed to be:



Now obviously it ends up moving more slowly than that - Some of the kinetic energy has been taken and converted into electrical energy. How much? Well it is just the kinetic energy corresponding to the difference in speed:



Now since the resistance is small, the light will convert all of that to other forms of energy - Issue of course being this doesn't match the answers hmm, can't spot what I did wrong right this second (multitasking, lol) - Could you upload your working to get D? Could anyone else chime in?

You forgot the square the difference in velocities in the KE formula! But even then, you get 0.0198 J I used GPE=mgh to find the difference in GPE after the 1m drop and then subtracted the final KE of the magnet from that, and I got D - I'm stumped too haha

[jamonwindeyer]

You forgot the square the difference in velocities in the KE formula! But even then, you get 0.0198 J I used GPE=mgh to find the difference in GPE after the 1m drop and then subtracted the final KE of the magnet from that, and I got D - I'm stumped too haha

Ahhhh thank you so much. So it should have been:



Yep, I'm feeling like the answer is D!

[kiwiberry]

Hi,
Can I get help on answering part ii? (Quanta to Quark)

Hi there, sorry for the late reply, I must have missed this!
So the binding energy is the amount of energy required to separate a nucleus into its individual nucleons, and is equal to the mass defect (mass of individual nucleons - mass of nucleus) using E=mc². Hence, atoms with a higher binding energy per nucleon will have a larger mass defect, and each nucleon in the nucleus will have less mass. When two elements below iron undergo nuclear fusion, they produce an element with a higher binding energy per nucleon as shown by the graph - each nucleon in this larger nucleus will have less mass than ones in the reactants, so the extra mass is converted into energy and released. Similarly, when an element above iron undergoes nuclear fission, it produces two elements with higher binding energies per nucleon, producing energy. Let me know if that doesn't make sense - hope this helps!