VCE Physics Question Thread!

[JellyBeanz]

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

The total time distance is equal to 2t because (ignoring air resistance) and the only force acting on the liquid is the force of gravity, the motion of the liquid is identical whether it is coming up or it's coming down. since it takes 4 seconds for the liquid to reach it's maximum height, it must take twice as much time to reach the ground.


EDIT: the quickest way to find the time in this case, when giving the angle and initial velocity is through the formula t=2usin(45)/g

notice the 2, it's basically a derivation of newton's formula v=u+at but multiplied by two.

[lzxnl]

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

Assume all lengths given are proper lengths.

[Maz]

Assume all lengths given are proper lengths.

right thanks- so l₀ would be 600

[Gogo14]

For 1/2 physics, the q is
In a solution, a total positive charge of 15 C moved past a point to the right in 5s. At the same time, a total negative charge of 30 C moved to the left. What is the current through the solution, and which direction.

So I got that there is a current of 6A to the left and 3A to the right. So that should cancel out to be 3A to the left, right? The answers says it is 9A to the right. Help

[lzxnl]

For 1/2 physics, the q is
In a solution, a total positive charge of 15 C moved past a point to the right in 5s. At the same time, a total negative charge of 30 C moved to the left. What is the current through the solution, and which direction.

So I got that there is a current of 6A to the left and 3A to the right. So that should cancel out to be 3A to the left, right? The answers says it is 9A to the right. Help

Currents have directions given by the sign of the moving charge. An electron moving to the right at 10000 m/s is mathematically the same current as a proton moving to the left at 10000 m/s. Hence you add the currents there.

[Gogo14]

Currents have directions given by the sign of the moving charge. An electron moving to the right at 10000 m/s is mathematically the same current as a proton moving to the left at 10000 m/s. Hence you add the currents there.

But if they are travelling in opposite directions, why would you add them together?But even if you add them, how do you know which way the current will flow. In the q, it has more current flowing left than right, but the final total current is flowing right anyway according to the answers

[lzxnl]

But if they are travelling in opposite directions, why would you add them together?But even if you add them, how do you know which way the current will flow. In the q, it has more current flowing left than right, but the final total current is flowing right anyway according to the answers

They're opposite charges. That's why you'd add them even though they're in opposite directions

[Alwin]

For 1/2 physics, the q is
In a solution, a total positive charge of 15 C moved past a point to the right in 5s. At the same time, a total negative charge of 30 C moved to the left. What is the current through the solution, and which direction.

So I got that there is a current of 6A to the left and 3A to the right. So that should cancel out to be 3A to the left, right? The answers says it is 9A to the right. Help

Another way to think of it is: "a total negative charge of 30 C moved to the left" is equivalent to +30 C moving to the right as lzxnl stated in his post.
So then we would have +15 C moving to the right and another +30 C moving towards the right which will give a total of 45 C moving towards the right. Then we could find current by dividing by time and get 9A.

Also: "So I got that there is a current of 6A to the left and 3A to the right" ... This doesn't really sit well with me (sorry) because conventional current is defined as the movement of positive charge. That's why we had to consider -30 C moving left as equal to +30 C. Hope it makes sense

[lach3087]

Hi, i have a question about projectile motion that i was wondering if i could get some help with. I need to know why the angle at which a projectile is fired at affects its range and how to relate that answer to the formula R = (v^2sin(2theta))/g. I know the optimum angle is 45 degrees but apart from it meaning that sin(2theta) will be equal to the highest value (1), i am unsure why. 

[Syndicate]

Hi, i have a question about projectile motion that i was wondering if i could get some help with. I need to know why the angle at which a projectile is fired at affects its range and how to relate that answer to the formula R = (v^2sin(2theta))/g. I know the optimum angle is 45 degrees but apart from it meaning that sin(2theta) will be equal to the highest value (1), i am unsure why.

*haven't been here since a long time  *
well sin(2x45) = sin(90)

sin(90) = 1

[lzxnl]

Hi, i have a question about projectile motion that i was wondering if i could get some help with. I need to know why the angle at which a projectile is fired at affects its range and how to relate that answer to the formula R = (v^2sin(2theta))/g. I know the optimum angle is 45 degrees but apart from it meaning that sin(2theta) will be equal to the highest value (1), i am unsure why.

On the one hand, you need to throw your projectile high so that it has enough time to travel far.
On the other hand, you need to throw it flat enough so that its horizontal velocity is decent.

[lach3087]

On the one hand, you need to throw your projectile high so that it has enough time to travel far.
On the other hand, you need to throw it flat enough so that its horizontal velocity is decent.

Can any of that be explained using W=mg?

[Maz]

Hey
i have a relativity question i was wandering if you could please help me in?
two protons in an accelerator are moving towards each other at 0.75c. At what speed are the protons approaching each other at relative to a stationary observer in the laboratory and relative to each other?

one more thing please...how would you do it if instead of then particles moving towards each other, they were moving in the same direction?

thankyou soo much 

[lzxnl]

Can any of that be explained using W=mg?

Yes, you can.

Hey
i have a relativity question i was wandering if you could please help me in?
two protons in an accelerator are moving towards each other at 0.75c. At what speed are the protons approaching each other at relative to a stationary observer in the laboratory and relative to each other?

one more thing please...how would you do it if instead of then particles moving towards each other, they were moving in the same direction?

thankyou soo much 

Consider the S frame as the lab frame. Then the velocity of one of the particles is 0.75c.
Pretend the S' frame is the other particle, which is moving at -0.75c relative to the S frame. This -0.75c is your v.
Then the velocity u' = (u - v)/(1 - uv/c^2) = (0.75c + 0.75c)/(1.5625) = 1.5c/1.5625 = 0.9375c

I'm assuming, of course, that you mean both particles are moving at 0.75c towards each other. If that's not the question, please say so.

If they're moving in the same direction, then both particles are moving at the same velocity in the S frame and they must necessarily be perceived to be moving at the same velocity in any other reference frame.

[Maz]

Consider the S frame as the lab frame. Then the velocity of one of the particles is 0.75c.
Pretend the S' frame is the other particle, which is moving at -0.75c relative to the S frame. This -0.75c is your v.
Then the velocity u' = (u - v)/(1 - uv/c^2) = (0.75c + 0.75c)/(1.5625) = 1.5c/1.5625 = 0.9375c

I'm assuming, of course, that you mean both particles are moving at 0.75c towards each other. If that's not the question, please say so.

If they're moving in the same direction, then both particles are moving at the same velocity in the S frame and they must necessarily be perceived to be moving at the same velocity in any other reference frame.

thankyou!