HSC Physics Question Thread

[jamonwindeyer]

Thanks for the solid advice

A quite "cliche" question but why is GPE negative? I know it has to do with the reference point being infinity or something and that when an object is being brought closer to the earth it has "negative work done" but it still kind of confuses me how GPE is negative

Basically, it's a matter of definition! We choose GPE to be zero at an infinite distance from the centre of the field - That choice is convenient and makes the most sense for large scale applications.

We need GPE to decrease as we move closer to the centre of the field - As something loses 'altitude' it loses GPE. So, if we have chosen GPE to be zero at an infinite distance from the field, then every distance closer than that should be less. The only way to be less than zero is to be negative

That's the best explanation of WHY it happens. You can also talk about work. When we do work on an object, we give it energy. Bringing an object closer to the earth requires negative work (which makes sense, gravity pulls things closer), so the energy it acquires will be negative

Note that you'll never be required to give a detailed explanation of why GPE is negative - You just need to know it is the case

[kiwiberry]

Hey, I'm a bit confused about the power loss formula.
So I know to derive it, you sub V=IR into P=VI, but why does power turn into power loss when you do so?
Also, why can't you sub in I=V/R instead to get P_loss=V²/R? I know that this formula doesn't work because stepping up the voltages significantly reduces power loss, so it wouln't make sense to have P_loss proportional to V². I just don't understand why subbing in the same formula in a different way doesn't work. Could someone please clarify why this is the case?

[f_tan]

Having trouble with these questions:

1. The power of the sun is approximately 3.86 x 10^26 W. Calculate the mass conversion that must take place if all of this power is the result of the conversion of mass into energy.  -Is power just the same thing as energy?

2. A cyclotron can be used to accelerate protons to relativistic velocities as they travel repeatedly around a circular, evacuated tube. Explain why the ultimate speed of the protons is limited and account for the fact that the kinetic energy of such a proton has no theoretical limit. - I thought the first bit might be because of mass dilation, but i'm not sure about kinetic energy?

Thank you!

[strawberriesarekewl]

In HSC physics what are the hardest aspects in the following Modules, Space, motors and generators, ideas to implementation and quanta to quarks (option topic)

[jamonwindeyer]

Hey, I'm a bit confused about the power loss formula.
So I know to derive it, you sub V=IR into P=VI, but why does power turn into power loss when you do so?
Also, why can't you sub in I=V/R instead to get P_loss=V²/R? I know that this formula doesn't work because stepping up the voltages significantly reduces power loss, so it wouln't make sense to have P_loss proportional to V². I just don't understand why subbing in the same formula in a different way doesn't work. Could someone please clarify why this is the case?

Step into my office

Legit, the moment in my first year ELEC lecture when I finally figured out why this is, best moment ever. Not difficult! Just a little intricate.

Right, so \(P=VI\) gives the power dissipated/lost in a circuit element (EG - the wires of a transmission network), as a product of the current through the element and the voltage across the element. The formula is identical to \(P=I^2R\) and \(P=\frac{V^2}{R}\). So why does only one work?

Notice what I emphasised in the text above - Across the element. Meaning, the voltage at one end minus the voltage at the other, the potential difference. This goes right back to Electrical Energy in the Home, circuit analysis in Year 11 involved analysing voltage drops across resistors.

The problem is that we are almost always given the voltage going into the transmission wires, never the voltage ACROSS those wires. We might put 100,000V into the wires, but that's not the voltage across them. The voltage across them is the difference between 100,000V and the voltage at the other end, which is calculable as \(V=IR\), Ohm's Law!

If we use the voltage across the wires, those other formulas work - \(P=VI\) and \(P=\frac{V^2}{R}\). If we just use the input voltage, they break. THIS is why we usually use \(P=I^2R\), because we don't need to look at the other end of the wires. We know how much current goes in, we know how much resistance there is - And that is all we need

This is really hard until you have a click moment, then it is really easy. If it is still a little confusing let me know, in which case I'll do a numerical example using Year 11 techniques to show you the difference

[jamonwindeyer]

Having trouble with these questions:

1. The power of the sun is approximately 3.86 x 10^26 W. Calculate the mass conversion that must take place if all of this power is the result of the conversion of mass into energy.  -Is power just the same thing as energy?

2. A cyclotron can be used to accelerate protons to relativistic velocities as they travel repeatedly around a circular, evacuated tube. Explain why the ultimate speed of the protons is limited and account for the fact that the kinetic energy of such a proton has no theoretical limit. - I thought the first bit might be because of mass dilation, but i'm not sure about kinetic energy?

Thank you!

Hey!

1. Almost - Power is a rate of change of energy! 1 Watt = 1 Joule per second. So your answer will be a rate of change of mass, \(kg/s\)

2. Both ideas are linked to mass dilation. Say we keep pushing the protons around the ring, meaning we keep giving them extra energy. The speed will approach but never reach the speed of light - Mass dilation is the reason for that! So if speed isn't increasing, where does that energy go? It goes to mass, that's the idea of mass dilation in the first place. Unlike the speed, the mass can actually increase as much as it likes, approaching infinity! Theoretically, if we keep pushing the proton, it will keep getting heavier. Since kinetic energy is defined as \(\frac{1}{2}mv^2\), if mass keeps increasing without a limit, so does kinetic energy

Does that help?

[jamonwindeyer]

In HSC physics what are the hardest aspects in the following Modules, Space, motors and generators, ideas to implementation and quanta to quarks (option topic)

Hmm, if I had to pick one thing based on personal opinion (didn't do Quanta though):

Space: Special Relativity explanations - The calculations not so much

Motors and Generators: Three phase induction motors

Ideas to Implementation: Deflection plates in CRT's

[RuiAce]

Step into my office

Legit, the moment in my first year ELEC lecture when I finally figured out why this is, best moment ever. Not difficult! Just a little intricate.

Right, so \(P=VI\) gives the power dissipated in a circuit element (EG - the wires of a transmission network), as a product of the current through the element and the voltage across the element. The formula is identical to \(P=I^2R\) and \(P=\frac{V^2}{R}\). So why does only one work?

Notice what I emphasised in the text above - Across the element. Meaning, the voltage at one end minus the voltage at the other, the potential difference. This goes right back to Electrical Energy in the Home, circuit analysis in Year 11 involved analysing voltage drops across resistors.

The problem is that we are almost always given the voltage going into the transmission wires, never the voltage ACROSS those wires. We might put 100,000V into the wires, but that's not the voltage across them. The voltage across them is the difference between 100,000V and the voltage at the other end, which is calculable as \(V=IR\), Ohm's Law!

If we use the voltage across the wires, those other formulas work - \(P=VI\) and \(P=\frac{V^2}{R}\). If we just use the input voltage, they break. THIS is why we usually use \(P=I^2R\), because we don't need to look at the other end of the wires. We know how much current goes in, we know how much resistance there is - And that is all we need

This is really hard until you have a click moment, then it is really easy. If it is still a little confusing let me know, in which case I'll do a numerical example using Year 11 techniques to show you the difference

Flabbergasted. What the hell.

Life actually makes sense!!!

[f_tan]

Hey!

1. Almost - Power is a rate of change of energy! 1 Watt = 1 Joule per second. So your answer will be a rate of change of mass, \(kg/s\)

2. Both ideas are linked to mass dilation. Say we keep pushing the protons around the ring, meaning we keep giving them extra energy. The speed will approach but never reach the speed of light - Mass dilation is the reason for that! So if speed isn't increasing, where does that energy go? It goes to mass, that's the idea of mass dilation in the first place. Unlike the speed, the mass can actually increase as much as it likes, approaching infinity! Theoretically, if we keep pushing the proton, it will keep getting heavier. Since kinetic energy is defined as \(\frac{1}{2}mv^2\), if mass keeps increasing without a limit, so does kinetic energy

Does that help?

Yes! Thanks so much!

[kiwiberry]

Step into my office

Legit, the moment in my first year ELEC lecture when I finally figured out why this is, best moment ever. Not difficult! Just a little intricate.

Right, so \(P=VI\) gives the power dissipated/lost in a circuit element (EG - the wires of a transmission network), as a product of the current through the element and the voltage across the element. The formula is identical to \(P=I^2R\) and \(P=\frac{V^2}{R}\). So why does only one work?

Notice what I emphasised in the text above - Across the element. Meaning, the voltage at one end minus the voltage at the other, the potential difference. This goes right back to Electrical Energy in the Home, circuit analysis in Year 11 involved analysing voltage drops across resistors.

The problem is that we are almost always given the voltage going into the transmission wires, never the voltage ACROSS those wires. We might put 100,000V into the wires, but that's not the voltage across them. The voltage across them is the difference between 100,000V and the voltage at the other end, which is calculable as \(V=IR\), Ohm's Law!

If we use the voltage across the wires, those other formulas work - \(P=VI\) and \(P=\frac{V^2}{R}\). If we just use the input voltage, they break. THIS is why we usually use \(P=I^2R\), because we don't need to look at the other end of the wires. We know how much current goes in, we know how much resistance there is - And that is all we need

This is really hard until you have a click moment, then it is really easy. If it is still a little confusing let me know, in which case I'll do a numerical example using Year 11 techniques to show you the difference

Oh my god that makes so much sense now!!! Thanks Jamon, that was a really good explanation

[jamonwindeyer]

Flabbergasted. What the hell.
Life actually makes sense!!!

Yes! Thanks so much!

Oh my god that makes so much sense now!!! Thanks Jamon, that was a really good explanation

WOO! Knowledge is power most welcome guys so glad I could be helpful!

[strawberriesarekewl]

What other stuff would you of had considered difficult other than those aspects jamon?
(Also anyone else willing to put some input so then i can get a somewhat better idea)

[jamonwindeyer]

What other stuff would you of had considered difficult other than those aspects jamon?
(Also anyone else willing to put some input so then i can get a somewhat better idea)

I'd say the whole course is fairly difficult - Those are standouts for me personally, but like, difficult questions can be asked on practically any dot point

[jakesilove]

What other stuff would you of had considered difficult other than those aspects jamon?
(Also anyone else willing to put some input so then i can get a somewhat better idea)

I second everything Jamon has said.

[strawberriesarekewl]

Jake

what is the hardest thing in quanta to quarks in your opinion