HSC Physics Question Thread

[Rathin]

I know I've asked this question before. I'm having trouble with this dot point: Describe ways in which applications of reflection of light, radio waves and microwaves have assisted information transfer. What do I write for Light and Microwaves.
Thanks

- Reflection of light is used in optic fibres which allows massive amounts of information transfer.
- Radio waves reflects off the ionosphere which allows the information to be transferred long distances.
- Microwaves are used in radars which reflect off objects to determine the distance of that object.

[kiwiberry]

I know I've asked this question before. I'm having trouble with this dot point: Describe ways in which applications of reflection of light, radio waves and microwaves have assisted information transfer. What do I write for Light and Microwaves.
Thanks

Some more for light:
- Forms virtual images of objects placed in front of plane mirror, both real and virtual from curved mirrors
- Telescopes use parabolic concave mirrors to reflect light from stars
- Torches and driving lights have parabolic reflectors – adjust light source for flood or spot beams

[bluecookie]

I don't really get the relative velocity formula...

V(a rel b) = V(a rel C) – V(b rel c)

[jamonwindeyer]

I don't really get the relative velocity formula...

V(a rel b) = V(a rel C) – V(b rel c)

Hey! What the formula actually does mechanically is a little tricky to explain. Essentially, it lets you measure the relative velocity between two frames of references. Best done with an example I think.

Say that you're sitting down, watching two people run past you in the same direction. You measure the first runner as 9 kilometres an hour, and the second at 7 kilometres an hour. For the formula, that's \(v_{ac}\) and \(v_{bc}\); the two velocities relative to a third reference frame - You! So in the formula, C represents the frame of reference of the measurer, and A and B represents two moving bodies.

The formula lets you calculate the velocity of A relative to B, meaning, the second runner is watching the first runner. How fast does he measure? The answer is, \(v_{ab}=v_{ac}-v_{bc}=9-7=2\) kilometres an hour. This makes sense, the first runner is moving 2 kilometres faster than the second runner, so from the frame of reference of the second runner, they'd say the first runner is moving away at 2 kilometres per hour.

The trick here is the frames of reference, and recognising that we don't always have to measure velocities with respect to the ground. We can measure them with respect to any inertial frame of reference. In this case, we choose the second runner; the first runner is moving 2 kilometres faster than that, so the relative velocity is 2 kilometres an hour.

The formula should only be used once you've got an intuitive understanding of how it works, otherwise it leads to trouble I hope this helps!!

[Mayalily]

Hey, I'm doing my HSC this year, and in relation to the option topic, my teacher wants to do medical physics, which I'm really uninterested in. He said it would be fine if I just studied for the option I want to do (Quanta to Quarks) but I'm worried that this may make it significantly harder to get the marks that I want in it.

[jamonwindeyer]

Hey, I'm doing my HSC this year, and in relation to the option topic, my teacher wants to do medical physics, which I'm really uninterested in. He said it would be fine if I just studied for the option I want to do (Quanta to Quarks) but I'm worried that this may make it significantly harder to get the marks that I want in it.

Hey Mayalily! I think its fair enough that you want to study another option (although I did Medical Physics and it is awesome )

There are no options that are significantly harder than others directly; the difficulty will come from needing to teach everything to yourself. If you are self motivated and driven, why not! Quanta is a popular option so there are a HEAP of resources for it; you won't have trouble in that regard. You just need to ask yourself; can you self teach an entire Option? Will you do the work to find the resources? Not doing the work with your class is definitely a disadvantage but if it means you can study something you are more interested in, I'm all for it!

Of course we will help you every step of the way too if you were super keen you could even get a tutor to help you with that Option maybe?

[bsdfjnlkasn]

Hey is gravitational field strength the same as the gravitational acceleration?

[jamonwindeyer]

Hey is gravitational field strength the same as the gravitational acceleration?

Hey! It definitely is, we define gravitational field strength as 'force experienced per unit mass,' so \(\text{Strength}=\frac{F}{m}\), which is also gravitational acceleration!

[bsdfjnlkasn]

Hey! It definitely is, we define gravitational field strength as 'force experienced per unit mass,' so \(\text{Strength}=\frac{F}{m}\), which is also gravitational acceleration!

Thanks Jamon, one more: whigh formula would you use to find the weight of an object on Earth's surface if you're given a value for the Earth's radius?

[jamonwindeyer]

Thanks Jamon, one more: whigh formula would you use to find the weight of an object on Earth's surface if you're given a value for the Earth's radius?

No worries! If you are given the radius of the earth, chances are it will be the full derivation using \(F=\frac{GMm}{d^2}\), rather than \(W=mg\), because you will probably be dealing with an object significantly higher than the surface of the earth. Remember, the value of \(g\) is only valid at or near the earth's surface, too far above and you need the full version

[bsdfjnlkasn]

No worries! If you are given the radius of the earth, chances are it will be the full derivation using \(F=\frac{GMm}{d^2}\), rather than \(W=mg\), because you will probably be dealing with an object significantly higher than the surface of the earth. Remember, the value of \(g\) is only valid at or near the earth's surface, too far above and you need the full version

Soo that F expression is actually equivalent to weight on Earth?

[jamonwindeyer]

Soo that F expression is actually equivalent to weight on Earth?

If you substitute the earths radius and mass \(M\) into that formula, with the weight of the object as \(m\), then yep! You'll get the weight force on the object

[bsdfjnlkasn]

If you substitute the earths radius and mass \(M\) into that formula, with the weight of the object as \(m\), then yep! You'll get the weight force on the object

That's sick! Thank you for all your help 

[jamonwindeyer]

That's sick! Thank you for all your help 

Gotta love it when stuff fits together no worries at all!

[bsdfjnlkasn]

Hey, could I get some help with the following:

If a 10kg mass were to be released from 1000km above the surface of Earth, initially being stationary, determine its velocity just before it hits the surface of the Earth.