VCE Physics Question Thread!

[Bri MT]

I didn't really understand what you meant by this:
"the component of BC/AD which is perpendicular to F = the component of BC/AD which is horizontal"

By the right hand rule (slap version), we know that the force acting on the wire is acting in a vertical direction.
When calculating torque, we are interested in the component of  r perpendicular to F (or equivalently the component of F perpendicular to r)
Perpendicular refers to being 90 degrees away, and anything is the horizontal direction is perpendicular to anything in the vertical direction (& vice versa)

[Seno72]

Indeed. The only possible opposing force is air resistance, which IS ignored in VCE Physics.

I think you need to know the effects of air resistance on a projectile, but you are right in that we don't need to know about air resistance in calculations.

[Yertle the Turtle]

I think you need to know the effects of air resistance on a projectile, but you are right in that we don't need to know about air resistance in calculations.

Unless you are doing the Sport Module in U2 Physics, which covers drag coefficient and air resistance etc. Otherwise, no, it is just a variable that occurs in real life but is discounted in VCE.

[geek123456]

Hi,
I was wondering if anyone could help me solve the following two questions attached.

Cheers

[Bri MT]

Hi,
I was wondering if anyone could help me solve the following two questions attached.

Cheers

For the second one, split it into 3 segments:
1) acceleration from rest to 12 (find distance and time)
2) acceleration from 12 to rest (find distance and time)
3) left over distance and time associated with travelling at 12 m/s (use distance to find time)


Does that help?

[geek123456]

For the second one, split it into 3 segments:
1) acceleration from rest to 12 (find distance and time)
2) acceleration from 12 to rest (find distance and time)
3) left over distance and time associated with travelling at 12 m/s (use distance to find time)


Does that help?

thank you!

[banksduyvu]

I need help with special relativity - simultaneity
Suppose Anna is moving towards a space station (that is not moving relative to an observer on Earth). When she is half way between Earth and the space station, she sends an electromagnetic pulse out. I don't understand how the signal reaches the space station first, when at the instant the light signal is sent out, she is exactly half way between Earth and the spacelab.

Special relativity kills me ~ . ~ Thank yoou !

[Bri MT]

I need help with special relativity - simultaneity
Suppose Anna is moving towards a space station (that is not moving relative to an observer on Earth). When she is half way between Earth and the space station, she sends an electromagnetic pulse out. I don't understand how the signal reaches the space station first, when at the instant the light signal is sent out, she is exactly half way between Earth and the spacelab.

Special relativity kills me ~ . ~ Thank yoou !

The pulse moves at a speed relative to her.  Let's say it moves at a speed of P, and Anna is moving at a speed of A.
Relative to Earth, the pulse is moving forwards at P + A
Relative to Earth, the pulse is m moving backwards at P - A.


It's like if you were on a train and threw a ball upwards you wouldn't need to run backwards as fast as the train is going to catch it,  because the ball also has the train's velocity,  just like you.

[S200]

Hey,
Can anyone help with part B of this checkpoint question? (Question 12 of Chapter 6)
The answers say 0.31N, but I consistently get -0.09N.
I am using the formula \(N+mg=\frac{mv^2} {r}\), which is what the answer's say to use, but I still can't get the 0.31

Any help would be appreciated...
(and sorry for terrible pic quality)

[Seno72]

Hey,
Can anyone help with part B of this checkpoint question? (Question 12 of Chapter 6)
The answers say 0.31N, but I consistently get -0.09N.
I am using the formula \(N+mg=\frac{mv^2} {r}\), which is what the answer's say to use, but I still can't get the 0.31

Any help would be appreciated...
(and sorry for terrible pic quality)

Fnet= mv^2/r
Fg= mg
R = ? (reaction force)

We know that both the reaction and gravity force is facing downwards so for vector calculations, we will take downwards as positive, so:

Fnet= mg + R (take R to the subject side)
R = Fnet - mg
R = mv^2/r - mg
R = (0.05 x 2^2)/(0.25) - (9.8 x 0.05)
R = 0.31 N downwards.

This is what I got. You had the right equation, now you just need to transpose so to make R the subject. Then put in the correct values inside. (Make sure it's 0.25 metres not 0.50 metres as it's radius and make sure that 50/1000 equals 0.05) Maybe you made a mistake up there. Hope it helps:

[S200]

Thanks heaps! I used 0.5 as the radius, so that screwed it up!

Thanks heaps.

[Seno72]

Hey guys. Does the work function of a metal only refer to the surface electrons or does it extend to all electrons on the metal. Shouldn't it just be electrons on the surface of the metal (where the work function is constant) as deeper down the metal, the minimum energy required to free electrons is greater and so isn't constant?

[S200]

I'm pretty sure that it refers to just surface, yeah.

[skrt skrt]

Hey AN, really struggling to understand how electron's actually form standing waves, do they move around the nucleus like a standing wave or is it just one of those concepts where you can't imagine it abstractly.

Thanks

[Bri MT]

Hey AN, really struggling to understand how electron's actually form standing waves, do they move around the nucleus like a standing wave or is it just one of those concepts where you can't imagine it abstractly.

Thanks

I wouldn't try to visualise the electron actually moving around in a wave shape (an definitely not in a neat circular orbit) - instead I think about it as the wave telling you about the probability of the electron being there.