VCE Physics Question Thread!

[Maz]

hey
could someone please help me with this question?
why does a photon have energy but not obey Einstein's equation E=mc^2

[lzxnl]

That is the rest mass energy of a particle. Photons have no rest mass, so E = mc^2 doesn't work for them. Rest mass energy means the energy the particle has in a stationary reference frame.
Rather, you have to use the total energy, which is given by E^2 = (pc)^2 + (mc^2)^2

[Maz]

That is the rest mass energy of a particle. Photons have no rest mass, so E = mc^2 doesn't work for them. Rest mass energy means the energy the particle has in a stationary reference frame.
Rather, you have to use the total energy, which is given by E^2 = (pc)^2 + (mc^2)^2

Thankyou

[Maz]

hello humans...again 
i was wandering if someone could please help me with this relativity question
while you are on earth you notice a space ship travel past you at 0.99c in 30ns. this spacecraft is normally stored in a hanger. what minimum length would this hanger need to be?
i tried doing it but my answer came out as 0.9785m and well that doesn't seem very realistic for a space ship 
id very much appreciate any help 

[lach3087]

My class is about to do a sac on propsion (as described in the picture below) and the variables we are testing are changing the angle at which the projectile is fired and changing its mass. We have to use the formula at the bottom of the photo and I was wondering if anyone could explain how you would incorporate mass into this formula. Thanks

[Gogo14]

I was doing a prac, but there is a variable resistor ( like the dial one) with 3 wires. 1 red, black and white. What does the white wire do and where do you connect it.

[lzxnl]

hello humans...again 
i was wandering if someone could please help me with this relativity question
while you are on earth you notice a space ship travel past you at 0.99c in 30ns. this spacecraft is normally stored in a hanger. what minimum length would this hanger need to be?
i tried doing it but my answer came out as 0.9785m and well that doesn't seem very realistic for a space ship 
id very much appreciate any help 

Well, you measure a distorted length; the length you measure is contracted by a factor of 1/(sqrt(1-0.99^2)) which turns out to be 7.1
The length you measure is 0.99c * 30 ns = 8.91 m
The actual length therefore is around 63.3 m

I was doing a prac, but there is a variable resistor ( like the dial one) with 3 wires. 1 red, black and white. What does the white wire do and where do you connect it.

How many places to plug wires are there in the resistor?

[Gogo14]

There are 2 places, but if the red wire is connected to the positive terminal and the black wire is connected to the negative terminal, where does the white wire go? Would that mean that the resistor would have to be put in a parallel circuit.

[Maz]

Hey
Can someone please help me out with this?
Two trains are moving at different but constant velocities. Are the any conditions under which they are
in the same inertial reference frame?
Do 2 non-inertial reference frames imply that one frame is accelerating?
i'm not really understanding relativity...
Thankyou so much in advance

[lzxnl]

Hey
Can someone please help me out with this?
Two trains are moving at different but constant velocities. Are the any conditions under which they are
in the same inertial reference frame?
Do 2 non-inertial reference frames imply that one frame is accelerating?
i'm not really understanding relativity...
Thankyou so much in advance

First question: no. If two trains aren't moving at the same velocity in one frame, you can't possibly find a reference frame in which they're moving at the same velocity. This is partly common sense (if one person thinks two things are moving at different speeds/directions, surely everyone else would think so too), but it can be proven quite easily because Lorentz transformations (i.e. changing from one reference frame to another) are reversible. If the transformation is reversible, then if you have two different velocities transforming into the same velocity, the reverse transformation isn't uniquely defined, which physically makes no sense.

Second question: yes. Although if they're both non-inertial, that means they're both accelerating. Inertial reference frame means obeys Newton's first law (i.e. if you let go of something in that frame of reference, does it appear to not accelerate). Curiously, this means that if you have someone in free-fall, the above test of whether a frame is inertial will tell you that this free-fall person is in an inertial frame because anything they drop will 'accelerate at the same rate' (because gravity is independent of mass) and thus they won't perceive any acceleration. This is the basis for Einstein's theory of general relativity which I'm hoping to look at next year

[Maz]

First question: no. If two trains aren't moving at the same velocity in one frame, you can't possibly find a reference frame in which they're moving at the same velocity. This is partly common sense (if one person thinks two things are moving at different speeds/directions, surely everyone else would think so too), but it can be proven quite easily because Lorentz transformations (i.e. changing from one reference frame to another) are reversible. If the transformation is reversible, then if you have two different velocities transforming into the same velocity, the reverse transformation isn't uniquely defined, which physically makes no sense.

Second question: yes. Although if they're both non-inertial, that means they're both accelerating. Inertial reference frame means obeys Newton's first law (i.e. if you let go of something in that frame of reference, does it appear to not accelerate). Curiously, this means that if you have someone in free-fall, the above test of whether a frame is inertial will tell you that this free-fall person is in an inertial frame because anything they drop will 'accelerate at the same rate' (because gravity is independent of mass) and thus they won't perceive any acceleration. This is the basis for Einstein's theory of general relativity which I'm hoping to look at next year

Thankyou so much

[Maz]

Hey humans...again
can i please get some help on another question? I've done part a...and the rest...i just don't know if it's right cos i
don't have any answers...
it's attached...my teachers are having a debate on what lo and l are- i think l₀ is 600...

any help would be appreciated

[soNasty]

Idk if this is the right place to post this but im in need of help with a gamsat physics question

'a terrorist is cornered down the end of a dead end street with a nuclear power plant behind him. a tank is advancing on him but he knows it will not fire and endanger the power plant. he tries to defend himself with a flame thrower. given that he can get maximum range from the flame thrower when the angle of projection is (theta=45 deg) and that, when fresh, the flame thrower can project liquid at a speed of 40root2 m/s, which of the following is closest to the maximum distance the tank can be from him for the weapon to be at all effective?

A. 50m
B. 100m
C. 200m
D. 300m

Answer is D.

Help pls

[JellyBeanz]

Idk if this is the right place to post this but im in need of help with a gamsat physics question

'a terrorist is cornered down the end of a dead end street with a nuclear power plant behind him. a tank is advancing on him but he knows it will not fire and endanger the power plant. he tries to defend himself with a flame thrower. given that he can get maximum range from the flame thrower when the angle of projection is (theta=45 deg) and that, when fresh, the flame thrower can project liquid at a speed of 40root2 m/s, which of the following is closest to the maximum distance the tank can be from him for the weapon to be at all effective?

A. 50m
B. 100m
C. 200m
D. 300m

Answer is D.

Help pls

So this is a projectile question.

We are given two things, the angle at which the liquid was projected (45) degrees and the initial velocity of the liquid. 40root2 m/s

1st thing you should do is find the vertical and horizontal components of the velocity.

Initial vertical velocity = initial velocity * sin (45)

= 40root2 * 1/root2
= 40 m/s

Initial horizontal velocity = initial velocity * cos (45)
= 40root2 * 1/root2, *Note you already know this because it is a right angled isosceles triangle

Since you have both components of velocity first thing you have to find is the total time it takes to reach x (distance).

using the initial vertical velocity (40m/s) use newton's Constant acceleration formula, v = u+at, substituting -10 m/s^2 for acceleration, v = 0 at top of the flight and u = 40 m/s, and solving for time, you get t = 4 seconds for half of the flight.

Therefore, total time it takes for the flight = 2* t
= 8 seconds for the total time.

Now what we do is use the horizontal component to find the distance the liquid will reach. we know that for the horizontal component x (distance) = (horizontal velocity) * (time)

using this distance = 40 m/s * 8
= 320 meters

hence the maximum distance the tank can be from him for the weapon to be effective is 320m, answer is D as 320m is closest to 300m


I hope it can get through to you somehow, i'm sure there may be easier ways to do this, and my explanation is not the best, a better explanation can be done with the aid of a diagram, i'm not great at diagrams XD

[soNasty]

So this is a projectile question.

We are given two things, the angle at which the liquid was projected (45) degrees and the initial velocity of the liquid. 40root2 m/s

1st thing you should do is find the vertical and horizontal components of the velocity.

Initial vertical velocity = initial velocity * sin (45)

= 40root2 * 1/root2
= 40 m/s

Initial horizontal velocity = initial velocity * cos (45)
= 40root2 * 1/root2, *Note you already know this because it is a right angled isosceles triangle

Since you have both components of velocity first thing you have to find is the total time it takes to reach x (distance).

using the initial vertical velocity (40m/s) use newton's Constant acceleration formula, v = u+at, substituting -10 m/s^2 for acceleration, v = 0 at top of the flight and u = 40 m/s, and solving for time, you get t = 4 seconds for half of the flight.

Therefore, total time it takes for the flight = 2* t
= 8 seconds for the total time.

Now what we do is use the horizontal component to find the distance the liquid will reach. we know that for the horizontal component x (distance) = (horizontal velocity) * (time)

using this distance = 40 m/s * 8
= 320 meters

hence the maximum distance the tank can be from him for the weapon to be effective is 320m, answer is D as 320m is closest to 300m


I hope it can get through to you somehow, i'm sure there may be easier ways to do this, and my explanation is not the best, a better explanation can be done with the aid of a diagram, i'm not great at diagrams XD

thanks! i get it all except how is the total time distance equal to 2t?