HSC Physics Question Thread

[beau77bro]

ok so question, i have a couple of sources saying x-rays have wavelengths the size of the atoms, then others saying they are the size of the distance between atoms in the lattice? which is it? could it be both since x-rays are portion of the spectrum?

[jakesilove]

ok so question, i have a couple of sources saying x-rays have wavelengths the size of the atoms, then others saying they are the size of the distance between atoms in the lattice? which is it? could it be both since x-rays are portion of the spectrum?

Both, because the distance between atoms in a lattice is often pretty close to the size of the atom itself (in atomic units, so like even if the distance between atoms is a couple of times bigger than the atom itself, it's still goddamned small).

[bsdfjnlkasn]

For b) we use



The final velocity is zero, as the rock is stationary for a moment. Thus,




Now, it takes 9.00-1.30=7.70s for the rock to go from the maximum height to the ground. Thus




However, this includes the height ABOVE the building. Subtracting this, 290.5-17.69=272.8m is our answer

Hey Jake,

SO sorry if i'm wrong here, but shouldn't the time for part c) be 9 - 1.9 = 7.1 seconds? Maybe i'm missing some info somewhere so just ignore me if i'm wrong haha. Thank you!!

[jakesilove]

Hey Jake,

SO sorry if i'm wrong here, but shouldn't the time for part c) be 9 - 1.9 = 7.1 seconds? Maybe i'm missing some info somewhere so just ignore me if i'm wrong haha. Thank you!!

Yep you're probably right, it's possible that I misread the question (whatever the real time is was stated in the question).

[itssona]

Heey, could someone help me with this:

so I have this velcoty time graph and my teacher gave notes for it but I am skeptical about it (maybe its just me not being good lol)

but so in the segment C, he said : deaccelerating, stopping, and then accelerating to the left
isnt that wrong?

also, for segment D, he said uniform decceleration?

and for G he said its accelerating while for H he said its deccelerating, but isnt it the OTHER way round?

[itssona]

sorry, heres the image

[ellipse]

Heey, could someone help me with this:

so I have this velcoty time graph and my teacher gave notes for it but I am skeptical about it (maybe its just me not being good lol)

but so in the segment C, he said : deaccelerating, stopping, and then accelerating to the left
isnt that wrong?

also, for segment D, he said uniform decceleration?

and for G he said its accelerating while for H he said its deccelerating, but isnt it the OTHER way round?

He is correct.
In segment C, the object initially has a high velocity, but it slows down (as seen by the magnitude of velocity deceasing). When the velocity is 0, it is not moving, thus it is stopping. Soon after that, although the velocity is getting negative, the magntitude still increases. This means that it is accelearting to the left (if the right is the positive direction)

In segment D, the velicity changes from -3 to 0. Again, if you look at the magntidue, it decreases, thus it is decelarating.

Similarly if you look at the magntiude of the velocity, it increases at G (goes from 0 to 3 in the left) thus it is accelerating (non-uniformly) and at H the magnitude decreases, thus it decelerates.

Hope that helps

[itssona]

He is correct.
In segment C, the object initially has a high velocity, but it slows down (as seen by the magnitude of velocity deceasing). When the velocity is 0, it is not moving, thus it is stopping. Soon after that, although the velocity is getting negative, the magntitude still increases. This means that it is accelearting to the left (if the right is the positive direction)

In segment D, the velicity changes from -3 to 0. Again, if you look at the magntidue, it decreases, thus it is decelarating.

Similarly if you look at the magntiude of the velocity, it increases at G (goes from 0 to 3 in the left) thus it is accelerating (non-uniformly) and at H the magnitude decreases, thus it decelerates.

Hope that helps

OMG thank you!!

I guess I was confused but now that you explained it in terms of magnitude, it makes sense! Thankss

[ellipse]

OMG thank you!!

I guess I was confused but now that you explained it in terms of magnitude, it makes sense! Thankss

No worries! Happy to help

[itssona]

hey also could someone please help with this:

all I can see is that its not in the centre :/ but im missing something since its 2 marks

[bsdfjnlkasn]

hey also could someone please help with this:

all I can see is that its not in the centre :/ but im missing something since its 2 marks

Hey!

So we must remember that validity is concerned with how the controlled variables/setup in an experiment is kept the same. The question only asks for a reason why the experiment is invalid so we'll need to explain for the second mark. Because it's asking why it is invalid, we must look at what isn't actually being sufficiently controlled

You could probably consider how the eventual direction of the bottom spring will be influenced by additional forces in the springs themselves therefore reducing the force's magnitude and (at least initially) changing the direction of the springs as they contract and stretch (this is all assuming that the springs are subject to gravity i.e. the set up is perpendicular to the ground). I'm not sure what impact the whole thing not being in the centre will be as we can see that the strings are of different lengths - this was most probably intentional as it will show how different forces (in this case tension in the string) acting in different directions will affect the bottom spring which will show the net force. OR we could just consider how the forces are not acting perpendicular to one another meaning a vector sum can't even be performed in the first place haha that would probably be the simplest answer.

Hopefully that helped a little bit, if someone could confirm/give a better answer that would be much appreciated 

[itssona]

Hey!

So we must remember that validity is concerned with how the controlled variables/setup in an experiment is kept the same. The question only asks for a reason why the experiment is invalid so we'll need to explain for the second mark. Because it's asking why it is invalid, we must look at what isn't actually being sufficiently controlled

You could probably consider how the eventual direction of the bottom spring will be influenced by additional forces in the springs themselves therefore reducing the force's magnitude and (at least initially) changing the direction of the springs as they contract and stretch (this is all assuming that the springs are subject to gravity i.e. the set up is perpendicular to the ground). I'm not sure what impact the whole thing not being in the centre will be as we can see that the strings are of different lengths - this was most probably intentional as it will show how different forces (in this case tension in the string) acting in different directions will affect the bottom spring which will show the net force. OR we could just consider how the forces are not acting perpendicular to one another meaning a vector sum can't even be performed in the first place haha that would probably be the simplest answer.

Hopefully that helped a little bit, if someone could confirm/give a better answer that would be much appreciated 

ohh spring force! I didnt consider that at all, thank you so much! !!!!

[itssona]

can someone please help with this:

a.Calculate how long it would take a boat that can travel at 6m/s in still water to cover a distance of 180m down river and back 180m. The water current is 4m/s.

b.The boat is then driven slightly angled against the current so that it travels exactly across the river, whose width is 120m.
Calculate how long it would take the boat to cover the river and back.

I know relative velocity but i never learnt stream questions :/

[jamonwindeyer]

can someone please help with this:

a.Calculate how long it would take a boat that can travel at 6m/s in still water to cover a distance of 180m down river and back 180m. The water current is 4m/s.

b.The boat is then driven slightly angled against the current so that it travels exactly across the river, whose width is 120m.
Calculate how long it would take the boat to cover the river and back.

I know relative velocity but i never learnt stream questions :/

Hey! So the first one, you just think of it as a constant \(4\) metres per second of velocity, downstream, being added to however fast the boat can travel. So, if it is travelling downstream, it can travel at \(6+4=10\text{ms}^{-1}\). If upstream it is going against the current, so \(6-4=2\text{ms}^{-1}\). So to travel 180m downstream then 180m upstream (in whatever order), we just use the idea that time is distance divided by speed:



The next bit is trickier - The boat driver angles themselves so that they travel directly across the river. What this means is, they make their velocity such a direction that the component of velocity parallel to the stream cancels with the flow of the river, leaving only the horizontal. You'll draw a right angled velocity triangle, with \(4\text{ms}^{-1}\) as the longer arm and \(6\text{ms}^{-1}\) as the hypotenuse (how fast the boat actually travels). Using pythag:



So to cover the 120 metre river:



This might be a little hard to picture without the diagram in front of you - Try to tackle it again having read this and see if it can help you guide your way through

[itssona]

Hey! So the first one, you just think of it as a constant \(4\) metres per second of velocity, downstream, being added to however fast the boat can travel. So, if it is travelling downstream, it can travel at \(6+4=10\text{ms}^{-1}\). If upstream it is going against the current, so \(6-4=2\text{ms}^{-1}\). So to travel 180m downstream then 180m upstream (in whatever order), we just use the idea that time is distance divided by speed:



The next bit is trickier - The boat driver angles themselves so that they travel directly across the river. What this means is, they make their velocity such a direction that the component of velocity parallel to the stream cancels with the flow of the river, leaving only the horizontal. You'll draw a right angled velocity triangle, with \(4\text{ms}^{-1}\) as the longer arm and \(6\text{ms}^{-1}\) as the hypotenuse (how fast the boat actually travels). Using pythag:



So to cover the 120 metre river:



This might be a little hard to picture without the diagram in front of you - Try to tackle it again having read this and see if it can help you guide your way through

the second part took me a while but I read what you said and re-attempted it, and I got it now! Thank you so much Jamon !!