Hog Dog
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#1001
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#1001
(Original post by Adamcfc)
I meant that Ns-1 is a unit for impulse, and im sure answer A was Ns-1?
Ns and Ns^-1 are different. Edit: many thanks for the rep! The answers did include Ns^-1 however this is not the right answer.

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Qwertish
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#1002
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#1002
(Original post by benethcheese)
You would get the same dY/dX so its k cause you'd draw it from the zero point to the maximum value
No, you draw a tangent at any point where the flux linkage graph crosses the x-axis. You don't get the same value from dy/dx:



N\Phi = -0.55\cos\left(\dfrac{2\pi}{40 \times 10^{-3}}t\right)



\dfrac{d(N\Phi)}{dt} = 0.55\left(\dfrac{2\pi}{40 \times 10^{-3}}\right)\sin\left(\dfrac{2\pi}  {40 \times 10^{-3}}t\right)

\dfrac{d(N\Phi)}{dt} is max when \sin(\theta) = 1, so:



\dfrac{d(N\Phi)}{dt}_{max} = 0.55\left(\dfrac{2\pi}{40 \times 10^{-3}}\right)



\dfrac{d(N\Phi)}{dt}_{max} = 86.3938...
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MisterE1
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#1003
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#1003
M/C in no particular order:

1)t root2, <--- it was something to do with the time period of a mass on two springs (if I remember right..)
2)460 and current its greater that 0.26,
3)path of electron was A the arrow pointing upwards,
4)potential at mars was -13,
5)maximum value of tension was 24
6) K.E: P>Q, G.P.E: Q<P
7) capacitance graph question was D,
8) fe>fs>fm I just thought that would mean something would be more attracted to the sun over the moon so it would leave the moons orbit? many of us also got:fe>fm>fs
9) ratio of y/d=1/3I got 2/3 :/
10) R earth^2/Rmoon^2=14 ----- >As I remember the question:

Mass of the earth is 81 x mass of the moon
Gravitational field strength on the earth = 9.8
Gravitational field strength on the moon = 1.8

g =GM/r^2.

9.8 = 81GM/re^2

1.8 = GM/rm^2

Re-arrange and you get re^2/rm^2 = ~14.88
11)which one is incorrect: when spring is compressed &suspended,it has minimum PE
12) which one is incorrect: change of momentum=zero
13)moving horizontally,Blv …
14)rotating coil was A q14 from http://papers.xtremepapers.com/AQA/P...W-QP-JAN03.pdf MS - http://papers.xtremepapers.com/AQA/P...W-MS-JAN03.pdf
15) Capacitance value of 0.02F for one question forgot which was, answer choice was B
16) units of impulse = kgms-1
17) momentum question when the 2 trolleys collided: 12000Ns
18) negative ion moving through 2 plates, the lower earthed the upper at + 50V, the ion moved upwards towards the positive plate
19)Something about max angular speed when the friction on a turntable of radius r is = mg/2

6 MORE MULTIPLE CHOICE QUESTIONS THEN DONE!
SECTION B IS COMPLETE
1a)define shm:
acceleration is proportinal to displacement, and acceleration acts toward equilibrium (2)
acceleration was -0.4
frequency of pendulum = 0.503 3s.f (3)
time for oscilations to be in phase again = 38 (3) -- if you think about it,
after 1 oscilation - difference on 0.1s
after 2 oscillations - difference of 0.2s....
after 19 oscillations difference of 1.9 seconds
so on the 19th osciation of the 1.9s pendulum, the 2s pendulum will arrive 1.9s after it, so on the 20th oscillation of the 1.9s pendulum they are in phase. again, on the 19th oscillation of the 2s pendulum they will be in phase.
19 x 2 = 38s
20 x 1.9 = 38s
TOTAL = 12

2) voltage = 30,000 (1)
b) i think this was the capacity was increase from X to Y.time taken for the spark to be produced after the last discharge?
then Q=CV, then Q/Current = time was ~ 3.46s
c) time taken for discharge will increase because time constant increases (2)
d) flash is brighter because more energy stored (2)
TOTAL = 7

4) 6 marker on fields
1. gravity always attractive
2. electric attractive/repulsive
3. both follow inverse square for Force
4. both follow inverse for potential
5. gravity uses mass
6. electric uses charge
7. only electric fields can be shielded
8. only electric fields depend on medium between charges
9. gradient of potential vs distance graph gives field strength
10. potential definition for both revolves around bringing a mass/charge from infinity to a certain point in a field and is zero at infinity
TOTAL = 6

3) 2 conditions when no force is exerted on the particle (2)
field is parallel to velocity
particle is stationary

b) out of the plane of the paper
b) prove momentum is proportional to radius= mv^2/r = BQv , mv = BQr --> mv = kr
the speed of particle=8.7x107
c)time taken for partile to go through 1 dee = ~6.8x10^-8
d)show the time taken is independent of radius = 2pim/bq (3)
kinetic energy im MEV = 2.44 (3)
TOTAL = 15

5)
1500 revs per minute (2) [I'm sorry MSI_10 :'( ]

angle for max emf = 60 degrees -- this is at 90 degrees to the plane as emf = BANw sin(wt) therefore max emf is when sin(wt) =1, there for wt (which is the angle) = 90, 90-30 = 60 degrees
angle for max flux linkage = 5pi/6 = 2.62 radians -- this is at 0 degrees = 180 degrees, and N(PHI) = BAN cos(THETA), so max is when cos(THETA) = 1 so theta = 180, 180-39 = 150 degrees = 5pi/6
gradient of graph = emf (1)


drawing a graph (2) Positive Sine graph
peak voltage (3) 86.4V http://www.thestudentroom.co.uk/show...6#post43151006 THANK YOU SO MUCH QWERTISH!
calculate flux density = 0.26 (2)
TOTAL= 10
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cooldudeman
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#1004
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#1004
Why am I the only one who got 0.27 for the last one????

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MisterE1
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#1005
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#1005
(Original post by cooldudeman)
Why am I the only one who got 0.27 for the last one????

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Most people got 0.26, a few people got 0.25, I've not heard anyone say 0.27 sorry!

Oh I thought you said 'Am I the only one...' now why :P
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FO12DY
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#1006
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#1006
(Original post by MisterE1)
18) negative ion moving through 2 plates, the lower earthed the upper at + 50V, the ion moved downwards towards the earthed plate (answer B)
No.

\vec{F} = q\cdot\vec{E}

The E-field is directed downward, q is negative, thus the force acting on it is directed upward.
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cooldudeman
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#1007
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#1007
(Original post by FO12DY)
No.

\vec{F} = q\cdot\vec{E}

The E-field is directed downward, q is negative, thus the force acting on it is directed upward.
I know it went upwards to the positive but I dont get how its 25V in the middle or something

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jonnyb123
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#1008
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#1008
(Original post by FO12DY)
No.

\vec{F} = q\cdot\vec{E}

The E-field is directed downward, q is negative, thus the force acting on it is directed upward.
(Original post by cooldudeman)
I know it went upwards to the positive but I dont get how its 25V in the middle or something

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It's question 13 on here. The answer is B, downwards

http://papers.xtremepapers.com/AQA/P...W-QP-JAN03.pdf

http://papers.xtremepapers.com/AQA/P...W-MS-JAN03.pdf
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cooldudeman
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#1009
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#1009
(Original post by Qwertish)
No, you draw a tangent at any point where the flux linkage graph crosses the x-axis. You don't get the same value from dy/dx:



N\Phi = -0.55\cos\left(\dfrac{2\pi}{40 \times 10^{-3}}t\right)



\dfrac{d(N\Phi)}{dt} = 0.55\left(\dfrac{2\pi}{40 \times 10^{-3}}\right)\sin\left(\dfrac{2\pi}  {40 \times 10^{-3}}t\right)

\dfrac{d(N\Phi)}{dt} is max when \sin(\theta) = 1, so:



\dfrac{d(N\Phi)}{dt}_{max} = 0.55\left(\dfrac{2\pi}{40 \times 10^{-3}}\right)



\dfrac{d(N\Phi)}{dt}_{max} = 86.3938...
So are you saying 86 is the correct ans or not? ?

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FO12DY
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#1010
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#1010
Yup. That's right.

But I was talking about the negative ion question. It goes to the +plate, not down to earth as MisterE1 said.
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Qwertish
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#1011
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#1011
(Original post by cooldudeman)
So are you saying 86 is the correct ans or not? ?

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Correct
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MisterE1
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#1012
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#1012
(Original post by FO12DY)
No.

\vec{F} = q\cdot\vec{E}

The E-field is directed downward, q is negative, thus the force acting on it is directed upward.
well spotted! thanks
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FO12DY
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#1013
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#1013
(Original post by cooldudeman)
I know it went upwards to the positive but I dont get how its 25V in the middle or something

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Well, that's just a fundamental property of uniform fields - it's essentially acting as a resistance.

The air creates a potential divider, where the distances between the point under test and each plate d1, d2 are effectively resistors, R1, R2.

So the pd (50-0=50) is divided by the resistors 50*R2/(R1+R2).

As it's in the middle for this question, d1 = d2, R1 = R2 => V = 50/2 = 25.
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jonnyb123
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#1014
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#1014
(Original post by FO12DY)
Yup. That's right.

But I was talking about the negative ion question. It goes to the +plate, not down to earth as MisterE1 said.
Oh right yeah I got that one too, sorry thought you were talking about this one!
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FO12DY
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#1015
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(Original post by MisterE1)
well spotted! thanks
No worries. Thanks for compiling the list.
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cooldudeman
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#1016
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#1016
(Original post by Qwertish)
Correct
So you get 86 if you do it the max gradient way too?

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Qwertish
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#1017
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#1017
(Original post by cooldudeman)
So you get 86 if you do it the max gradient way too?

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Yea, you get 86 if you use calculus and if you do the emf=BANw.
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Jack93o
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#1018
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#1018
guys, I think the emf graph is supposed to be minus sine

(assuming that the flux graph was minus cosine)
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cooldudeman
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#1019
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#1019
(Original post by Gernick)
For the peak emf I calculated the gradient at 0 flux linkage and got 100V.. Is that right?


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Thats what I got (102) but I think we are wrong cuz everyone us saying is 86. Wtf

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kingm
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#1020
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#1020
(Original post by Jack93o)
guys, I think the emf graph is supposed to be minus sine

(assuming that the flux graph was minus cosine)
It wants to be!
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