Recent content by ravsterphysics

Vx = I/nqA and Vy = 2[I/nqA] = 2Vx so doesn't that mean that for every 1 part of Vx we have half a part of Vy? So that 2 parts of Vx gives 1 part of Vy so the ratio is still C, 2:1?? Argh!

I've done this question a few times now and ended up with the same answer, but the question is from an official exam paper in the UK so I don't believe there's been a mistake. FYI, here's what the mark scheme says: How can it still be B?

Homework Statement Homework EquationsThe Attempt at a Solution Current is I = nqvA so drift velocity V is: V = I/nqA Drift velocity for x is: Vx = I/nqA Drift velocity for y is: Vy = 2I/nqA So the ratio of Vx : Vy should be 2:1 since Vy is equal to 2 lots of Vx?? (but correct answer is B)

Yep, I totally disregarded the effects on the wire's resistance. It's clear now why current goes up by a factor of 4. thanks for the help.

Homework Statement Homework EquationsThe Attempt at a Solution I know I = nqvA When the pd is applied the surface that is 8cm long, the cross sectional area is 32cm (8x4) but when the pd is applied across the 4cm side, the cross sectional area is now 16cm (4x4) so I has decreased by a...

Homework Statement Homework EquationsThe Attempt at a Solution [/B] Slightly confused here. Since this is a displacement-time graph, is it correct to say that at X, since the displacement is zero that it can't be a point of compression because that would imply the air molecules are moving...

Homework Statement Homework EquationsThe Attempt at a Solution Since resistance of the NTC falls we can say that the total resistance of the circuit has decreased so there is now more current in the circuit, so it's definitely answer B or D. And since X's resistance has fallen, PD across X...

2. Relevant equation The Attempt at a Solution I can see straight away that the waves are 90 degrees out of phase so pie/2. But how is X ahead of Y? It looks like Y is ahead of X by pie/2.

Here's a diagram I've drawn based on what I've read so far. So the object is actually at 'object 1' but due to refraction the observer ( and because the brain judges the image location to be where where light rays appear to originate from) deems the object to be at 'object 2'. Is that correct?

Homework Statement https://www.highlightskids.com/media/kids/highlightskids/images/thumbs/sciQuestions/sq1012_put-a-straw-in-a-glass-of-water_main.jpg Homework EquationsThe Attempt at a Solution From my understanding, this has to do with refraction. Here's my answer: ----- Light usually...

Argh such a silly mistake! Thanks for your help : )

Ah, in that case if X has a stiffer spring, 2K, then its extension (x) will only be half, so we'd have (0.5)(2K)(0.5x) so the energy stored is the same for both??

Homework Statement Homework EquationsThe Attempt at a Solution EPE is (0.5)kx^2 Let Y have spring constant of K then X has spring constant of 2K So EPE of Y is (0.50kx^2 which is E So EPE of X must be (0.5)2kx^2 which is kx^2 which is 2E? But correct answer is E/2??

But max vertical displacement occurs when time spent in air is t/2 (compared to horizontal time) and when v=0, if you take a look at this video you can see the guy did the same thing; he's working out vertical displacement using v=0 which is the same as total air time divided by 2?

Homework Statement Homework EquationsThe Attempt at a Solution So (i) was easy enough and I got a time of 0.67 seconds. For (ii), since the horizontal time is 0.67 seconds this means the TOTAL time spent in air is also 0.67 seconds, so to calculate max height we split this time in half...