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Advanced higher physics help please? thanks :)
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ah4p
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#1
could someone please explain question 4a)ii) and 4b) from here
http://mrmackenzie.wikispaces.com/fi...tpaper2001.pdf
the answers are here
http://mrmackenzie.wikispaces.com/fi...iPhysicsAH.pdf
thankyou v much in advance I really appreciate any help
http://mrmackenzie.wikispaces.com/fi...tpaper2001.pdf
the answers are here
http://mrmackenzie.wikispaces.com/fi...iPhysicsAH.pdf
thankyou v much in advance I really appreciate any help
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Joinedup
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#2
(Original post by ah4p)
could someone please explain question 4a)ii) and 4b) from here
http://mrmackenzie.wikispaces.com/fi...tpaper2001.pdf
the answers are here
http://mrmackenzie.wikispaces.com/fi...iPhysicsAH.pdf
thankyou v much in advance I really appreciate any help
could someone please explain question 4a)ii) and 4b) from here
http://mrmackenzie.wikispaces.com/fi...tpaper2001.pdf
the answers are here
http://mrmackenzie.wikispaces.com/fi...iPhysicsAH.pdf
thankyou v much in advance I really appreciate any help
f=-kx
if the force is zero when x is not zero then you can't have SHM
----------
do you get given formula sheets for scottish exams?
if not you'll just have to remember them...
for circular and SHM
ω=2πf
for SHM
speed=2πf sqrt(A^2 - x^2)
since max speed occurs at x=0
max_speed=2πf A = ωA
(can't attach a picture of the formula today
)
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ah4p
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#3
(Original post by Joinedup)
4b is just restating the requirement for SHM - a restoring force proportional to displacement from the equilibrium point.
f=-kx
if the force is zero when x is not zero then you can't have SHM
----------
do you get given formula sheets for scottish exams?
if not you'll just have to remember them...
for circular and SHM
ω=2πf
for SHM
speed=2πf sqrt(A^2 - x^2)
since max speed occurs at x=0
max_speed=2πf A = ωA
(can't attach a picture of the formula today
)
4b is just restating the requirement for SHM - a restoring force proportional to displacement from the equilibrium point.
f=-kx
if the force is zero when x is not zero then you can't have SHM
----------
do you get given formula sheets for scottish exams?
if not you'll just have to remember them...
for circular and SHM
ω=2πf
for SHM
speed=2πf sqrt(A^2 - x^2)
since max speed occurs at x=0
max_speed=2πf A = ωA
(can't attach a picture of the formula today
)
I have a prelim on Wednesday and I'm very worried :/
I thought it was just me who was having problems uploading pictures :P
I still don't really understand 4b because there's no froe mentioned in the question?
we do get a formula sheet
how would you sketch the graph of velocity?
Thanks again
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Joinedup
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#4
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#4
(Original post by ah4p)
thank you very much for replying
I have a prelim on Wednesday and I'm very worried :/
I thought it was just me who was having problems uploading pictures :P
I still don't really understand 4b because there's no froe mentioned in the question?
we do get a formula sheet
how would you sketch the graph of velocity?
Thanks again
thank you very much for replying
I have a prelim on Wednesday and I'm very worried :/
I thought it was just me who was having problems uploading pictures :P
I still don't really understand 4b because there's no froe mentioned in the question?
we do get a formula sheet
how would you sketch the graph of velocity?
Thanks again
constant velocity means acceleration = 0
therefore there is no force in those parts of the motion where velocity is constant
---
The instantaneous velocity is the gradient of the displacement time graph... so if the displacement graph looks like a cosine, the velocity graph looks like a negative sine graph
(if doing any calculus you might notice something about the above but if not, don't worry about it)
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ah4p
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#5
(Original post by Joinedup)
force is mass x acceleration
constant velocity means acceleration = 0
therefore there is no force in those parts of the motion where velocity is constant
---
The instantaneous velocity is the gradient of the displacement time graph... so if the displacement graph looks like a cosine, the velocity graph looks like a negative sine graph
(if doing any calculus you might notice something about the above but if not, don't worry about it)
force is mass x acceleration
constant velocity means acceleration = 0
therefore there is no force in those parts of the motion where velocity is constant
---
The instantaneous velocity is the gradient of the displacement time graph... so if the displacement graph looks like a cosine, the velocity graph looks like a negative sine graph
(if doing any calculus you might notice something about the above but if not, don't worry about it)
Haha yes I do
I'm doing advanced higher maths too
But can you assume that even if you're not specifically told its a cosine graph?
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ah4p
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#6
(Original post by Joinedup)
force is mass x acceleration
constant velocity means acceleration = 0
therefore there is no force in those parts of the motion where velocity is constant
---
The instantaneous velocity is the gradient of the displacement time graph... so if the displacement graph looks like a cosine, the velocity graph looks like a negative sine graph
(if doing any calculus you might notice something about the above but if not, don't worry about it)
force is mass x acceleration
constant velocity means acceleration = 0
therefore there is no force in those parts of the motion where velocity is constant
---
The instantaneous velocity is the gradient of the displacement time graph... so if the displacement graph looks like a cosine, the velocity graph looks like a negative sine graph
(if doing any calculus you might notice something about the above but if not, don't worry about it)
Is that right?
Thank you
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Joinedup
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#7
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#7
(Original post by ah4p)
Ah ok so does that mean objects performing simple harmonic motion are never at a constant speed they're always accelerating? (Even neglecting air resistance)
Haha yes I do
I'm doing advanced higher maths too
But can you assume that even if you're not specifically told its a cosine graph?
Posted from TSR Mobile
Ah ok so does that mean objects performing simple harmonic motion are never at a constant speed they're always accelerating? (Even neglecting air resistance)
Haha yes I do
I'm doing advanced higher maths too
But can you assume that even if you're not specifically told its a cosine graph?
Posted from TSR Mobile
your dv graph needs to start from 0 and turn negative
but in general for SHM you could get the zeroes and max/min points and join them up with an approximately sinusoidal shaped line.
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ah4p
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#8
(Original post by Joinedup)
it's pretty clear the displacement graph starts at +1A with a gradient of 0 so to my way of thinking it's cosine... the first thing the gradient does is turn negative
your dv graph needs to start from 0 and turn negative
but in general for SHM you could get the zeroes and max/min points and join them up with an approximately sinusoidal shaped line.
it's pretty clear the displacement graph starts at +1A with a gradient of 0 so to my way of thinking it's cosine... the first thing the gradient does is turn negative
your dv graph needs to start from 0 and turn negative
but in general for SHM you could get the zeroes and max/min points and join them up with an approximately sinusoidal shaped line.
yeah that's a good method to try
A d.c circuit consists of a switch, lamp and inductor.
The inductor is replaced with an inductor with twice the number of turns in the coil of wire, nothing else about it is changed.
State the effect on
A) maximum current B) the time to reach max current
I thought increasing the turns would increase the current
and also increase the time taken since the back emf would also be greater
but the answer is
max current decreases
and time to reach it increases
can anyone explain why?
for question 8b)
http://mrmackenzie.wikispaces.com/fi...tpaper2006.pdf
answers are here
http://www.sqa.org.uk/files_ccc/06miPhysicsAH.pdf
I don’t understand what they’ve done in the answers because I thought
back emf = supply voltage – voltage across inductor
so from the graph I used when 6V is across inductor the back emf is 3V and dI/dt is 8 (from graph)
then I used back emf = -L dI/dt
subbing in the values above I got L = 0.375H
but the answers have just used values straight from the graph – when 9V is across the inductor then dI/dt = 12 and subbed these in
I don’t understand this
any help is appreciated
thanks v much in advance
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