# Search results

1. ### Charge carrier drift velocity of wire

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!
2. ### Charge carrier drift velocity of wire

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?
3. ### Charge carrier drift velocity of wire

Homework Statement Homework Equations The 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)
4. ### Current in a wire

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.
5. ### Current in a wire

Homework Statement Homework Equations The 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...
6. ### Compressions and rarefactions of a wave

Homework Statement Homework Equations The 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...
7. ### Circuit with NTC/Voltmeter/Ammeter

Homework Statement Homework Equations The 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...
8. ### Out of phase waves

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.
9. ### Why does a straw look bent in water?

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?
10. ### Why does a straw look bent in water?

Homework Statement https://www.highlightskids.com/media/kids/highlightskids/images/thumbs/sciQuestions/sq1012_put-a-straw-in-a-glass-of-water_main.jpg [Broken] Homework Equations The Attempt at a Solution From my understanding, this has to do with refraction. Here's my answer: ----- Light...
11. ### Spring constant and energy

Argh such a silly mistake! Thanks for your help : )
12. ### Spring constant and energy

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??
13. ### Spring constant and energy

Homework Statement Homework Equations The 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??
14. ### Kinematic equations for max height

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?
15. ### Kinematic equations for max height

Homework Statement Homework Equations The 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...
16. ### Perpendicular forces

okay now i'm lost, i thought the mg/cos20 IS the vertical component? if not, can you write it, it'll probably click for me that way.
17. ### Perpendicular forces

The vertical component is weight (mg) right? So to get lift, we use trig to end up with Lift = mg/cos20? Is that what you mean by vertical component?
18. ### Perpendicular forces

So only the 'top' part of the weight is what we're after? And the angle is also 20 degrees? So it looks like this? in which case lift would then be Lift = mg/cos20 I see! In that case, how does this differ to my notes that say the perpendicular force to the slope would be mgcos20? Is it...
19. ### Perpendicular forces

in the vertical direction we have weight acting downwards and lift acting upwards? so does it look like this?
20. ### Perpendicular forces

Homework Equations The Attempt at a Solution [/B] I was of the understanding that the slope can be calculated as mgsin20 whereas the force acting straight down through the lift is mgcos20 but the answer is mg divided by cos 20??
21. ### Acceleration and speed

Yep, you're right, I confused speed and velocity. So now I understand why speed V is the same. But for gravity, why has that not changed?
22. ### Stress/Strain and Youngs Modulus

Homework Statement Homework Equations The Attempt at a Solution The shaded area is a triangle so the area of a triangle for this particular graph is this: 0.5 (Stress)(Strain) which gives: 0.5(F/A)(e/L) so the top would give work done and the bottom would give volume but we're still left...
23. ### Acceleration and speed

Homework Statement Homework Equations The Attempt at a Solution Since at N the gravity is now heading down towards the person in a positive direction it's acceleration is G, where as velocity goes up then comes down towards the person so it's velocity is now negative, thus -V, but the...
24. ### Polarisation of light

Homework Statement Okay so this is a concept that I've been having a lot of problems with recently. I watched a few youtube vids to clear things up so now I'd like to make sure my notes are right. 1) For normal polarised light, the oscillations of a wave are in one plane only which include the...
25. ### Change in Velocity Question

so change in velocity is final velocity minus initial velocity = 8 - 2 = 6m/s ? which means in my example, final velocity - initial velocity = -4 - 4 = -8m/s but I don't get what the -8m/s represents? that is moved in the negative direction at 8m/s??
26. ### Change in Velocity Question

okay where am I going wrong here? the ball starts at point 1 with velocity of 4......it hits the wall and rebounds, also with velocity of 4 but since it's in the opposite direction, it's minus 4......so change in velocity is 4 + (-4) = zero??
27. ### Change in Velocity Question

Okay, with that line of thinking, the initial velocity is 4 and if the ball rebounds at the same speed (but in the opposite direction) isn't the final velocity also (-)4? so change in velocity would be 0??
28. ### Change in Velocity Question

Homework Statement Homework Equations The Attempt at a Solution [/B] Assuming the right direction is positive, it hits the wall and reaches point 2, so to get back to point 1, it's in a negative direction, so it should be MINUS 4m so option A but the correct answer is B?? Or is it to do...
29. ### Acceleration on the Moon

Yep, I completely glossed over that, thanks. Intuitively it makes sense now, but is there a formula I can use to get a better understanding?
30. ### Acceleration on the Moon

Homework Statement Homework Equations F=ma The Attempt at a Solution I know it's either B or D Weight = mg so let's say mass of ball is 11kg and gravity on moon is 6, so weight would be 66N since acceleration a = F/m and we can say that F=mg therefore a = mg / m which leaves: a=g and...