General Thoughts: Physics (3 Viewers)

[HecticLad]

does anybody know if my reasoning that the electric circuit wasn't closed (that 2 cm gap) means electrons cant be liberated from cathode and accel towards anode? i wrote that + electrons are - charge so its deflecting wrong way

what lol

 

[Sarah Jane 1]

What about the conductor that was circular and became rectangular (sorta) in the Multiple choice. Was the current anticlockwiseand the change in flux density? What about the forces on the parallel current carrying conductors? I think I got 3F/2 Newtons to the right

 

[jackleung34]

holy shit that last space question RIP. Was the question looking for uses of "G" as in 6.67x10^-11 because i was talking about Forces of Universal gravitational attraction

What about the conductor that was circular and became rectangular (sorta) in the Multiple choice. Was the current anticlockwiseand the change in flux density? What about the forces on the parallel current carrying conductors? I think I got 3F/2 Newtons to the right

I'm fairly certain that q the answer was F/2 to the right

 

[HecticLad]

there was too much of the other topics.

there was like projectile motion, relativity, that 6 mark one at the end, wouldve been at least 15-20 marks

 

[p0llex]

What about the conductor that was circular and became rectangular (sorta) in the Multiple choice. Was the current anticlockwiseand the change in flux density? What about the forces on the parallel current carrying conductors? I think I got 3F/2 Newtons to the right

Both wrong:

The flux density doesn't change when you change the shape, and Lenz's law says you create a magnetic field going in the same direction as the external one so it was clockwise.

P was attracting wire R while Q was pushing it away, but P was twice the distance, so it ended up as F/2 to the right.

 

[HecticLad]

What about the conductor that was circular and became rectangular (sorta) in the Multiple choice. Was the current anticlockwiseand the change in flux density? What about the forces on the parallel current carrying conductors? I think I got 3F/2 Newtons to the right

nope the conductor one was clockwise with change in flux because it will induce a current to oppose the motion of the wire, and the flux density wasnt actually change because it was the same field

the parallel conductor one was F/2 to the right, because the middle conductor repelled it F to the right while the far conductor attracted it F/2 to the left, giving a vector sum of F/2

 

[jackleung34]

Both wrong:

The flux density doesn't change when you change the shape, and Lenz's law says you create a magnetic field going in the same direction as the external one so it was clockwise.

P was attracting wire R while Q was pushing it away, but P was twice the distance, so it ended up as F/2 to the right.

aYYYyyyyYYYYYYYY lma0000000

 

[HecticLad]

holy shit that last space question RIP. Was the question looking for uses of "G" as in 6.67x10^-11 because i was talking about Forces of Universal gravitational attraction



I'm fairly certain that q the answer was F/2 to the right

question was talking about F=Gm1m2/r, which give equations such as escape velocity & orbital velocity

 

[malcolm21]

there was too much of the other topics.

there was dat 6 marker, 3-4 marks projectile, 5ish marks gravitational potential energy thing, seemed like a lot of space for me.. too much

 

[Crisium]

question was talking about F=Gm1m2/r, which give equations such as escape velocity & orbital velocity

Yeah I talked about how Newton's Law of Universal Gravitation is used in the derivation of orbital velocity and the significance of this in regards to ensuring that the probe can maintain the orbit around Saturn

 

[rhys-e]

It still applies for things which aren't projectiles, but in this case the electrons have some initial velocity which we don't know.

Best way of doing it I think was using qV=1/2mv^2, answer turned out to be somewhere between 7000 and 8000

What's qV? I equated work and kinetic energy (as work is equal to the change in kinetic energy). So Fs = 1/2mv^2 or qEs = 1/2mv^2. Answer came out to 4.1x10^7

 

[malcolm21]

question was talking about F=Gm1m2/r, which give equations such as escape velocity & orbital velocity

for this question did you have to equate these two? it wsa 3 marks and i did it in 1 line with v=root Gm/r

 

[Yeungster]

I dont remember integration in physics.

ur screwed

 

[malcolm21]

What's qV? I equated work and kinetic energy (as work is equal to the change in kinetic energy). So Fs = 1/2mv^2 or qEs = 1/2mv^2. Answer came out to 4.1x10^7

sick, i got this and i did the other formula, v^2 = u^2 + 2 (a) (s), im guessing we're right cause we used different formulas and still got the exact answer. Edit: I got 4.19 so ~ 4.2 actually

 

[atargainz]

for this question did you have to equate these two? it wsa 3 marks and i did it in 1 line with v=root Gm/r

you would have to derive it, I did it by subbing T=2pir/v into keplers, gives v=root Gm/r

 

[HecticLad]

for this question did you have to equate these two? it wsa 3 marks and i did it in 1 line with v=root Gm/r

that was a different question wasnt it, you wouldnt have to derive it
but for the 6 mark space one, it would have been good to explain that orbital & escape velocity are derived from the universal gravitation law

 

[jackleung34]

i think the problem with this test was how vague some questions were in terms of what they were looking for. But I'd assume that's to seperate those who rote learned and those who do actually conceptually understand content

for this question did you have to equate these two? it wsa 3 marks and i did it in 1 line with v=root Gm/r

Referring to the last 6 marker of the test

What's qV? I equated work and kinetic energy (as work is equal to the change in kinetic energy). So Fs = 1/2mv^2 or qEs = 1/2mv^2. Answer came out to 4.1x10^7

Couldn't you have just found acceleration first through a=f/m, then used it to find final velocity upon hitting anode through v^2=u^2+2a(delta y)

 

[p0llex]

What's qV? I equated work and kinetic energy (as work is equal to the change in kinetic energy). So Fs = 1/2mv^2 or qEs = 1/2mv^2. Answer came out to 4.1x10^7

Yep that's exactly what I did (qV is the same as qEs), but I got a different answer? Or maybe I'm thinking of the answer to a different question?

I'll have another go at it now

 

[leehuan]

Just.. no...

 

[HecticLad]

i think the problem with this test was how vague some questions were in terms of what they were looking for. But I'd assume that's to seperate those who rote learned and those who do actually conceptually understand content

RIP those who didnt learn hertz's experiment