HSC Physics Question Thread

[beau77bro]

can someone please explain what is meant by this graph of black body radiation, i thought id just started to understand blackbody radiation, but i dont get what the axises mean in the graph. very much appreciated if someone could explain the features of it thanks.

[jamonwindeyer]

can someone please explain what is meant by this graph of black body radiation, i thought id just started to understand blackbody radiation, but i dont get what the axises mean in the graph. very much appreciated if someone could explain the features of it thanks.

Hey! Sure thing, I'll try and make it super simple

So what we are looking at here is radiation curves for black bodies (which is, for the most part, the same thing as the thermal radiation given off by everyday objects, even you and I). Black bodies emit radiation at all frequencies - Meaning you and I (assuming we are black bodies, which we aren't, but you know, dramatic effect) are emitting X-rays right now. The issue is how much X-ray radiation is being emitted.

That's what these curves show. The horizontal axis is wavelength, the vertical is intensity, which is just (speaking basically) how much of each wavelength we get. So if we look at the red line, there is a HEAP of energy at the 0.5 micrometer wavelength, but not much at all at the 2 micrometer wavelength. This means that it emits a lot of 0.5 micrometer radiation, not a lot of 2 micrometer radiation.

Every black body has what is called a characteristic frequency/wavelength. This is just the wavelength/frequency where a black body emits the MOST radiation.

The different lines represent bodies of different temperatures - Indeed, for ideal black bodies, the only thing affecting the characteristic wavelength is the temperature. Objects with higher temperatures have lower characteristic wavelengths (and higher characteristic frequencies, by the \(c=f\lambda\) formula)

Let me know if I can clarify any of this for you! I tried to keep it as simple as I could but definitely happy to help pick any bits of it apart that you'd like

[winstondarmawan]

Hello! Another question I would like help with.
Thanks in advance.

[jamonwindeyer]

Hello! Another question I would like help with.
Thanks in advance.

Hey! Great question this one. So the first step is ALWAYS to resolve the velocity into its vertical and horizontal components. The launch velocity is known, \(V\), so:



You could also get decimal approximations to these values, or leave them in trig form - The exact values I've done you might only know if you do 2U So now let's examine what we know. For the horizontal axis, we know that:



So that's given us a formula linking velocity to time! Now in the vertical direction:



This is going to get messy - Replace \(V_y\) and \(t\) with the expressions involving \(V\) we've found, and we want a vertical displacement of -34 (34 metres BELOW our launch point):



Now if you solve this for \(V\), I think you get about \(21.42\) metres per second! does this working out make sense? It's a tough one!

[itssona]

hey this is a silly question but whats the difference between pos and negative acceleration??

Like negative is decceleration? or is it accelerartion in a negative DIRECTION

[RuiAce]

hey this is a silly question but whats the difference between pos and negative acceleration??

Like negative is decceleration? or is it accelerartion in a negative DIRECTION

'Deceleration' isn't a thing in physics. You're either accelerating in the positive or the negative direction. Deceleration is just a colloquial word for the act of slowing down, but we don't consider this in physics. We always consider it in that we are accelerating, just the other way around.

For an example, take the positive direction to be right, and negative to be left. You are accelerating in the positive direction if you are accelerating towards your right. You are accelerating in the negative direction if you are accelerating towards your left.

For another example, take the positive direction to be up and negative to be down. If you drop a ball, it will accelerate downwards due to gravity. So gravity is an example of negative acceleration in this scenario.

Note that the concept of acceleration only makes sense provided you can define a positive direction to begin with.

[beau77bro]

Hey! Sure thing, I'll try and make it super simple

So what we are looking at here is radiation curves for black bodies (which is, for the most part, the same thing as the thermal radiation given off by everyday objects, even you and I). Black bodies emit radiation at all frequencies - Meaning you and I (assuming we are black bodies, which we aren't, but you know, dramatic effect) are emitting X-rays right now. The issue is how much X-ray radiation is being emitted.

That's what these curves show. The horizontal axis is wavelength, the vertical is intensity, which is just (speaking basically) how much of each wavelength we get. So if we look at the red line, there is a HEAP of energy at the 0.5 micrometer wavelength, but not much at all at the 2 micrometer wavelength. This means that it emits a lot of 0.5 micrometer radiation, not a lot of 2 micrometer radiation.

Every black body has what is called a characteristic frequency/wavelength. This is just the wavelength/frequency where a black body emits the MOST radiation.

The different lines represent bodies of different temperatures - Indeed, for ideal black bodies, the only thing affecting the characteristic wavelength is the temperature. Objects with higher temperatures have lower characteristic wavelengths (and higher characteristic frequencies, by the \(c=f\lambda\) formula)

Let me know if I can clarify any of this for you! I tried to keep it as simple as I could but definitely happy to help pick any bits of it apart that you'd like

thanks so much jamon, cleared it up alot for me. the detail and extra info really helped as well. i think i get it alot more thankyou

[itssona]

'Deceleration' isn't a thing in physics. You're either accelerating in the positive or the negative direction. Deceleration is just a colloquial word for the act of slowing down, but we don't consider this in physics. We always consider it in that we are accelerating, just the other way around.

For an example, take the positive direction to be right, and negative to be left. You are accelerating in the positive direction if you are accelerating towards your right. You are accelerating in the negative direction if you are accelerating towards your left.

For another example, take the positive direction to be up and negative to be down. If you drop a ball, it will accelerate downwards due to gravity. So gravity is an example of negative acceleration in this scenario.

Note that the concept of acceleration only makes sense provided you can define a positive direction to begin with.

Ohh! Okay that makes sense :O

so is decceleration more like "decreasing rate of acceleration"

http://webs.mn.catholic.edu.au/physics/emery/assets/prelim60.gif
in a veloctity time graph, positive gradient is positive accleration but how would we know by just knowing velocity? how do they just conclude that negative gradient is negative acceleration when negative acceleration is based on direction

Anyways thank you sooo much for clearing that bit up!!

[jamonwindeyer]

Ohh! Okay that makes sense :O

so is decceleration more like "decreasing rate of acceleration"

http://webs.mn.catholic.edu.au/physics/emery/assets/prelim60.gif
in a veloctity time graph, positive gradient is positive accleration but how would we know by just knowing velocity? how do they just conclude that negative gradient is negative acceleration when negative acceleration is based on direction

Anyways thank you sooo much for clearing that bit up!!

I'd call de-acceleration, decreasing magnitude of acceleration. Meaning, whatever direction you are accelerating in, deceleration would mean reducing how much you are accelerating in that direction. That's how I'd interpret it at least

We define velocity and acceleration directions to be the same. Positive velocity and positive acceleration occur in the same direction. It doesn't make sense to define it any other way so, since we have a positive gradient, it just makes sense to define acceleration as positive. I suppose the answer is, why would you choose to define acceleration as negative when your rate of change of velocity (gradient) is positive? That would be seriously counterintuitive, which is why we never do it

[itssona]

I'd call de-acceleration, decreasing magnitude of acceleration. Meaning, whatever direction you are accelerating in, deceleration would mean reducing how much you are accelerating in that direction. That's how I'd interpret it at least

We define velocity and acceleration directions to be the same. Positive velocity and positive acceleration occur in the same direction. It doesn't make sense to define it any other way so, since we have a positive gradient, it just makes sense to define acceleration as positive. I suppose the answer is, why would you choose to define acceleration as negative when your rate of change of velocity (gradient) is positive? That would be seriously counterintuitive, which is why we never do it

Ohhhhh I finally get it :O Now I see how silly my thinking was LOL

Thank you a tonne Jamon!

[smile123]

HELP

[beau77bro]

not entirely sure what this means, mostly the 'overlapping in a continuous fashion' portion any help would be appreciated.

[jakesilove]

HELP

Hey!

We need to drive a total of 270km. The first 100km is fine, and as you've travelled at 100km/h, this distance will take 1 hour to travel. So, at this point it is 9:00am. Then, you slow down to 42km/h for 43km. This will take



So, this takes us through to 10:01am. We need to get to the interview by 11:15am, and have travelled 143 km. So, we have 270-143=127km to travel in 1 hour and fourteen minutes! We can convert this to 1.23333 hours. To find the minimum speed required to make this distance, we simply divide the distance by the time,

[jakesilove]

not entirely sure what this means, mostly the 'overlapping in a continuous fashion' portion any help would be appreciated.

Definitely a clumsy sentence, but to be honest, this entire area of the Physics curriculum is clumsy. All you need to know is that we define two bands when it comes to conduction; the valence band, and the conduction band. Electrons need to be able to jump from the valence band to the conduction band in order for conduction to occur. Thus, the bands are touches in Metals, closeish to each other in semi-conductors, and far away in insulators. In reality, there is crazy quantum mechanical stuff going on here, and so the description you've posted is just an attempt to avoid all of that (which is totally fair enough).

The below is all that is worth having in your head when you think about something like this



(Ignore the Fermi level)

[beau77bro]

hahah oh ok, sweet thankyou. so we dont need to know it? and i would honestly love to hear the crazy quantum stuff if u can be bothered to type it. quantum mechanics seems super interesting, i was the guy that kept asking u questions at the lectures btw. thanks for all the help and advice. you guys have an awesome site going