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 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)
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...
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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...
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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...
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 [Broken]
Homework Equations
The Attempt at a Solution
From my understanding, this has to do with refraction. Here's my answer:
-----
Light...
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??
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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??
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?
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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...
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...
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??
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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...
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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...
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...
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??
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??
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??
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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...
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...