Post by: Saladin on May 08, 2010, 06:49:57 am
I do Edexcel AS Physics!
Post by: 7ooD on May 08, 2010, 03:02:34 pm
Post by: break freak on May 08, 2010, 03:18:00 pm
Range/2=(15.1-14.2)/2 Precision=0.1
=0.45
Therefore, %uncertainty= 0.5 (to 1 decimal place) ;D
Post by: break freak on May 08, 2010, 03:21:47 pm
Post by: 7ooD on May 08, 2010, 03:45:23 pm
Post by: 7ooD on May 08, 2010, 03:47:40 pm
Post by: Saladin on May 08, 2010, 03:56:42 pm
how to get the Precision though??
This is what you have to do.
1. Find the mean
2. Then find the difference between the mean and the highest value
3.
Post by: Saladin on May 08, 2010, 03:57:32 pm
look for the above question it should be D as they want to know the potential difference at which the diode will conduct and the arrangement is set so that the diode wont conduct electricity but when we apply a large p.d it will conduct
I still dont know why this answer is right.
Post by: Darth Austrakast on May 08, 2010, 04:05:09 pm
A student observes the diffraction of white light by holding a diffraction grating close to her eyes
a)draw a simple diagram to illustrate what the student observes
b) a tube of KMn04 solution is placed in front of white light source.Then the student views the filament through the solution with the grating close to her eyes. she notices that the blue, yellow and orange colors have vanished. Give an explanation.
Can you please help me asap? Thanks ! :)
Post by: Saladin on May 08, 2010, 04:09:46 pm
here's a question i need help solving:
A student observes the diffraction of white light by holding a diffraction grating close to her eyes
a)draw a simple diagram to illustrate what the student observes
b) a tube of KMn04 solution is placed in front of white light source.Then the student views the filament through the solution with the grating close to her eyes. she notices that the blue, yellow and orange colors have vanished. Give an explanation.
Can you please help me asap? Thanks ! :)
What paper is this?
Post by: thisishard on May 08, 2010, 04:22:26 pm
Post by: Saladin on May 08, 2010, 04:23:54 pm
where can i train for unit 3 physics other than the last 3 papers?
Try doing past IGCSE Physics paper 6.
Post by: Darth Austrakast on May 08, 2010, 04:28:30 pm
What paper is this?
it was given in our school mock
Post by: Ascer Zayan on May 10, 2010, 05:14:56 pm
Question 6 Parts C and D
Dont Give me the Marking Scheme Answer Cuz i cnt understand them
And How can u find the uncertainty of a Length Which is only taken Once. ???
Post by: Ascer Zayan on May 10, 2010, 05:19:11 pm
This is what you have to do.i tried doing wat Mysteriouse dude said
1. Find the mean
2. Then find the difference between the mean and the highest value
3.
But i got 2.95??
and wat is this 2.95%?
Post by: Saladin on May 10, 2010, 06:38:35 pm
Ok,
The mean:
Now find the maximum difference:
So, the maximum uncertainty is
Post by: Ascer Zayan on May 10, 2010, 07:23:07 pm
You are talking abt mj 2009 3b rite?Wat About part C in this exam i dont understand at allllllllllllll
Ok,
The mean:
Now find the maximum difference:
So, the maximum uncertainty is
Post by: T.Q on May 10, 2010, 08:16:08 pm
Wat About part C in this exam i dont understand at allllllllllllll
Value is in GPa, G means 10^9, so yes
10^(10) can mean anything from 5.0 x 10^9 to 4.9 x10^(10)so my answer is within range
Post by: Ascer Zayan on May 11, 2010, 06:19:47 am
Value is in GPa, G means 10^9, so yesWelll Do u Understand?
10^(10) can mean anything from 5.0 x 10^9 to 4.9 x10^(10)so my answer is within range
10^(10) can mean anything from 5.0 x 10^9 to 4.9 x10^(10
Post by: Saladin on May 11, 2010, 10:23:09 am
the precision is the least value that can be measured. Example: for these values 12.0,12.5,13.0 the precision is 0.5
but in our case 14.2 14.7 15.1 the precision is 0.1 8)
I am not too sure about that. The limits of accuracy for this is said in the mark-scheme to be 0.5
Post by: break freak on May 11, 2010, 10:25:34 am
how to get the Precision though??the precision is the least value that can be measured. Example: for these values 12.0,12.5,13.0 the precision is 0.5
but in our case 14.2 14.7 15.1 the precision is 0.1 ;D
Post by: Saladin on May 11, 2010, 10:40:52 am
the precision is the least value that can be measured. Example: for these values 12.0,12.5,13.0 the precision is 0.5
but in our case 14.2 14.7 15.1 the precision is 0.1 ;D
O ok, but the question asked something different. O well.
Post by: vanibharutham on May 11, 2010, 10:42:23 am
On the May/June 2009 Paper
Question 4,
the one with the potential difference across the diode,
the mark scheme seems to have a mistake, because it misses out question 3b)
The answer to question 4 needs to be A
It can't be D because the diode is reverse biased.
Post by: T.Q on May 11, 2010, 10:49:01 am
Just a note,
On the May/June 2009 Paper
Question 4,
the one with the potential difference across the diode,
the mark scheme seems to have a mistake, because it misses out question 3b)
The answer to question 4 needs to be A
It can't be D because the diode is reverse biased.
yes thats true, THE ans should be A
Post by: Saladin on May 11, 2010, 10:52:58 am
Just a note,
On the May/June 2009 Paper
Question 4,
the one with the potential difference across the diode,
the mark scheme seems to have a mistake, because it misses out question 3b)
The answer to question 4 needs to be A
It can't be D because the diode is reverse biased.
I knew it! I wrote down A as well. It just had to be!
Post by: Ascer Zayan on May 11, 2010, 12:24:19 pm
This is what you have to do.This is to find Percentage of Uncertainty right??????
1. Find the mean
2. Then find the difference between the mean and the highest value
3.
cuz i am losstttttt in physics Unit 3 :(
Post by: Saladin on May 11, 2010, 01:08:24 pm
This is to find Percentage of Uncertainty right??????
cuz i am losstttttt in physics Unit 3 :(
No, in the markscheme, it says that you have to find it in GPa. I know this paper had a bad mark scheme.
Post by: 7ooD on May 11, 2010, 01:56:45 pm
Post by: vanibharutham on May 11, 2010, 02:17:13 pm
When a diode is forward biased it will allow current to pass through, as in 4 A
However, when a diode is reverse biased it will not allow current to pass through, as in 4 D
to completely understand how diodes work would go past the A level syllabus as you would need to understand P and N type semi conductor doping and then valence and conduction bands. With that, you will be able to explain why the diode conducts in one direction.
But for A level just know that if the anode of a diode is connected to the positive terminal of a power supply it is considered Forward Biased and will allow conduction.
Post by: Darth Austrakast on May 11, 2010, 03:32:45 pm
here's a question i need help solving:
A student observes the diffraction of white light by holding a diffraction grating close to her eyes
a)draw a simple diagram to illustrate what the student observes
b) a tube of KMn04 solution is placed in front of white light source.Then the student views the filament through the solution with the grating close to her eyes. she notices that the blue, yellow and orange colors have vanished. Give an explanation.
Can you please help me asap? Thanks ! :)
can't anyone solve this question? ???
it would be really helpful...............exams are less than 24 hours away!
so please help
Post by: Saladin on May 11, 2010, 03:39:50 pm
here's a question i need help solving:
A student observes the diffraction of white light by holding a diffraction grating close to her eyes
a)draw a simple diagram to illustrate what the student observes
b) a tube of KMn04 solution is placed in front of white light source.Then the student views the filament through the solution with the grating close to her eyes. she notices that the blue, yellow and orange colors have vanished. Give an explanation.
Can you please help me asap? Thanks ! :)
(a) A variety of colours, like that of a rainbow, all mixed up in one place.
(b) The light will be polarised. This means that the vibrations that have caused the other different colours of light no longer vibrate due to this polarisation.
Post by: Darth Austrakast on May 11, 2010, 04:17:47 pm
but i wrote exactly same for the last answer but still got wrong
but thanks a lot for confirming it :)
Post by: vanibharutham on May 11, 2010, 06:13:07 pm
Post by: vanibharutham on May 12, 2010, 07:03:50 pm
absorption spectra
the KMnO4 solution must have electrons that get excited whenever there is blue, yellow or orange light.
The excited electrons return to ground state and reemit the light equally in all directions. To an observer with a diffraction grating it seems like there is no orange, yellow or blue because the intensity is soo less
Post by: Saladin on May 12, 2010, 07:07:10 pm
i got it!
absorption spectra
the KMnO4 solution must have electrons that get excited whenever there is blue, yellow or orange light.
The excited electrons return to ground state and reemit the light equally in all directions. To an observer with a diffraction grating it seems like there is no orange, yellow or blue because the intensity is soo less
It has to be optically active then, which it is not. I believe your first point was correct.
Post by: wakemeup on May 15, 2010, 07:39:17 pm
Q16 please.
Post by: vanibharutham on May 16, 2010, 08:59:40 pm
Ill write down a model answer if you must have one:
"The debate about the true nature of light was "resolved" when Loius de Broglie proposed the wave-particle duality theory. The photoelectric effect demonstrates the particle like properties of light.
If we assumed a wave model, the observations seen could not be explained. We know that the energy of a wave is proportional to the intensity. Therefore, increasing the intensity should increase the kinetic energy of the particles released. But this is not seen. What is seen is that MORE photoelectrons of the SAME kinetic energy are released.
Also, waves transfer energy continuously. If a light wave continuously transmitted energy to a metal, eventually the electrons would gain enough energy to escape the metal. Thus the frequency would not matter. However, this is not the case and electrons are not emitted below a certain threshold frequency.
A particle model of light can explain these observations. If we assume light is made up of particles known as photons (quanta of EM radiation) that have a discrete amount of energy denoted by E = hf. We can see that by increasing the frequency we increase the energy of a photon. And thus by increasing the frequency, the SAME number of photoelectrons are released but of a higher kinetic energy. Because one photon interacts with only one electron, the energy is transmitted and a higher KE of the photoelectrons is seen.
Every metal has a work function - the minimum amount of energy required to JUST cause photoemission. If the energy of photons is not greater than the work function, no photoemission occurs. Because E is proportional to f, below a certain frequency the photons dont have enough energy to cause photoemission."
Haha that was good revision :)
Post by: Darth Austrakast on May 19, 2010, 04:28:36 pm
Post by: Darkstar3000 on May 25, 2010, 11:50:13 am
Energy stored per unit mass J/kg
Modern spring steel 130
Tendon in leg 2500
Rubber chord 8000
show that a car of mass 1200 kg would need steel springs of total mass approximately equal to 3kg to store energy when it encounters pot-holes of depth 3cm
how do i solve this
and
The sum of the mass of leg tendons of a skier might be of the order of 04.kg. Estimate the size 'bump' that a skier of mass 75kg could theoretically negotiate
Post by: T.Q on May 25, 2010, 12:26:37 pm
Post by: Darkstar3000 on May 25, 2010, 12:41:16 pm
from where did u got this question
It says jan 02 at the top, my teacher collected some papers with stuff that is in the syllabus for us to solve and this is one of the questions
Post by: ABF on May 25, 2010, 02:13:41 pm
Post by: wakemeup on May 25, 2010, 03:56:02 pm
Post by: Igcseboy on May 26, 2010, 08:44:16 am
2morrow is the 6PH01 paper
so i have some questions
in free body diagrams from where we should draw the contact force??
and anybody have notes on projection.....
any important points on the speed time graphs and all ???
i can understand wat to do as the portion for unit 1 is too small i guess
any suggestions on wat i shud do
Post by: Darth Austrakast on May 26, 2010, 06:38:07 pm
Hey
2morrow is the 6PH01 paper
so i have some questions
in free body diagrams from where we should draw the contact force??
and anybody have notes on projection.....
any important points on the speed time graphs and all ???
i can understand wat to do as the portion for unit 1 is too small i guess
any suggestions on wat i shud do
the normal reaction force should always be a normal to the/perpenicular to center of gravity
here r some stuffs on projectiles
http://www.astarmathsandphysics.com/a_level_physics_notes/forces_and_motion/a_level_physics_notes_equations_for_projectiles.html (http://www.astarmathsandphysics.com/a_level_physics_notes/forces_and_motion/a_level_physics_notes_equations_for_projectiles.html)
if u still have problem with free body force diagram
take a look 2 this http://www.astarmathsandphysics.com/a_level_physics_notes/forces_and_motion/a_level_physics_notes_free_body_diagrams.html (http://www.astarmathsandphysics.com/a_level_physics_notes/forces_and_motion/a_level_physics_notes_free_body_diagrams.html)
Post by: Igcseboy on May 26, 2010, 06:44:03 pm
the normal reaction force should always be a normal to the/perpenicular to center of gravity
here r some stuffs on projectiles
http://www.astarmathsandphysics.com/a_level_physics_notes/forces_and_motion/a_level_physics_notes_equations_for_projectiles.html (http://www.astarmathsandphysics.com/a_level_physics_notes/forces_and_motion/a_level_physics_notes_equations_for_projectiles.html)
if u still have problem with free body force diagram
take a look 2 this http://www.astarmathsandphysics.com/a_level_physics_notes/forces_and_motion/a_level_physics_notes_free_body_diagrams.html (http://www.astarmathsandphysics.com/a_level_physics_notes/forces_and_motion/a_level_physics_notes_free_body_diagrams.html)
thanxx
hey so wat do u expect of 2morrows paper is it gonna be easy like chemistry or ????
Post by: Darkstar3000 on May 26, 2010, 08:35:48 pm
A particle is projected from level ground in such a way that its horizontal and vertical components of velocity are 20m/s and 10 m/s respectively. Find
a. the max height of the ball (i can do this)
b. its horizontal distance from the point of projection when it returns to the ground
c. the magnitude and direction of its velocity on landing
....................................
A pencil is accidentally knocked off the edge of a (horizontal) desk top. The height of the desk is 64.8 cm and the pencil hits the floor a horizontal distance of 32.4cm from the edge of the desk. What was the speed of the pencil as it left the desk?
Post by: T.Q on May 26, 2010, 09:37:44 pm
How are these supposed to be done
A particle is projected from level ground in such a way that its horizontal and vertical components of velocity are 20m/s and 10 m/s respectively. Find
a. the max height of the ball (i can do this)
b. its horizontal distance from the point of projection when it returns to the ground
c. the magnitude and direction of its velocity on landing
....................................
A pencil is accidentally knocked off the edge of a (horizontal) desk top. The height of the desk is 64.8 cm and the pencil hits the floor a horizontal distance of 32.4cm from the edge of the desk. What was the speed of the pencil as it left the desk?
a)
b)
c)
Post by: T.Q on May 26, 2010, 09:45:40 pm
A pencil is accidentally knocked off the edge of a (horizontal) desk top. The height of the desk is 64.8 cm and the pencil hits the floor a horizontal distance of 32.4cm from the edge of the desk. What was the speed of the pencil as it left the desk?
Post by: Darth Austrakast on May 27, 2010, 02:31:42 am
thanxx
hey so wat do u expect of 2morrows paper is it gonna be easy like chemistry or ????
dude 1 thing i can tell u from my experience Physics is never easy :P
but i found jan10 okay so i m afraid this may they might decide to change the pattern
just hoping no freaky questions!
Best of luck !
Post by: Darkstar3000 on May 27, 2010, 07:11:13 am
a)
b)
c)
thank you
Post by: Igcseboy on May 27, 2010, 10:55:46 am
Post by: Saladin on May 27, 2010, 02:02:25 pm
how was unit 1 paper????
NO DISCUSSION UNTIL 24 HRS!
Post by: wakemeup on May 28, 2010, 04:43:37 pm
Well, it started out pretty simple but became completely mind boggling towards the end, particularly that kite vector question. I understand it now but I messed it up in the exam so it doesn't matter. Plus, there were 20 questions. Most questions required quite a bit of thinking and I had much trouble handling the time.
Man, this is not good at all. I didn't well in the Physics practical and now this? It really sucks because I prepared pretty hard for this one. My only potential saving grace is paper 2 now and that is no walk in the park.
Post by: Saladin on May 28, 2010, 05:34:27 pm
I'm assuming we can discuss the paper now.
Well, it started out pretty simple but became completely mind boggling towards the end, particularly that kite vector question. I understand it now but I messed it up in the exam so it doesn't matter. Plus, there were 20 questions. Most questions required quite a bit of thinking and I had much trouble handling the time.
Man, this is not good at all. I didn't well in the Physics practical and now this? It really sucks because I prepared pretty hard for this one. My only potential saving grace is paper 2 now and that is no walk in the park.
Yes for the kite thing you needed to use higher trig to answer it. But I got it in the end.
Post by: Igcseboy on May 28, 2010, 06:57:17 pm
Yes for the kite thing you needed to use higher trig to answer it. But I got it in the end.
wat was the tension and angle and all the questions tht followed????
Post by: wakemeup on May 28, 2010, 07:25:02 pm
wat was the tension and angle and all the questions tht followed????
I couldn't solve it myself during the exam but here's what you had to do. You had to resolve the Tension vector into horizontal and vertical components. Then, the vertical comp of T + weight = Upthrust and the horizontal comp of T = drag force. Now you can find the Tension and the angle.
Post by: T.Q on May 28, 2010, 11:39:58 pm
wat was the tension and angle and all the questions tht followed????
T=7.14
angel = 32
Post by: Saladin on May 29, 2010, 07:06:03 am
T=7.14
angel = 32
I got the same hi5!
Post by: Igcseboy on May 29, 2010, 08:02:41 am
I got the same hi5!
i got the same :D:D:D:D:D:D:D:D:D:D:D:D:D:D:D:D
Alhamdulillah ... abit releifed
Post by: T.Q on May 29, 2010, 09:06:31 am
i used this equ.
Post by: Saladin on May 29, 2010, 09:47:38 am
how much was the work done in the tension question
i used this equ.i dont remember the distance
Dunno about that. I equated like finding the value of T through sin and cos values, to find a tan value.
And used that tan value to calculate tension.
Post by: T.Q on May 29, 2010, 09:51:17 am
Dunno about that. I equated like finding the value of T through sin and cos values, to find a tan value.
And used that tan value to calculate tension.
ya i know , but im talking about the workdone question (next part)
after the tension and angel part
Post by: Saladin on May 29, 2010, 10:30:07 am
ya i know , but im talking about the workdone question (next part)
after the tension and angel part
thats easy, you just simply multiply the tension by the distance moved.
because the girl is the source of tension.
Post by: T.Q on May 29, 2010, 10:48:45 am
thats easy, you just simply multiply the tension by the distance moved.
because the girl is the source of tension.
ya , but u had to multiply force with cos theta
because the force is not in same direction of motion
Post by: Saladin on May 29, 2010, 12:04:20 pm
ya , but u had to multiply force with cos theta
because the force is not in same direction of motion
yes, but it is force moved into the direction of force, the only force she is applying is the tension.
And she is acting in the opposite direction to it.
Post by: student on May 31, 2010, 10:02:33 am
(http://img709.imageshack.us/img709/5949/p1100041ss.th.jpg) (http://img709.imageshack.us/f/p1100041ss.jpg/)
(A note on the graph from the book.) (http://img24.imageshack.us/f/eda2physabp184.jpg/)
If this were a model of an automatic hand dryer circuit which requires 4.0V to operate, use the graph to work out for how long it will remain on.
The spreadsheet says the time constant is 20s, and thats what I could read off the graph.
But aint time constant = RC? And RC is 200*0.0001=0.02s, not 20s..
So.. As the minimum required voltage is 4V it won't work below that? And initial is 6V:
V = V0e-t/RC
V/V0 = e-t/20
20 ln(4/6) = -t
t = 8.109s
Wrong..
And using a lower RC would only give a shorter time, but it's 50s+.
Post by: Saladin on May 31, 2010, 10:43:28 am
A question from A2 ActiveBook.
(http://img709.imageshack.us/img709/5949/p1100041ss.th.jpg) (http://img709.imageshack.us/f/p1100041ss.jpg/)
(A note on the graph from the book.) (http://img24.imageshack.us/f/eda2physabp184.jpg/)
If this were a model of an automatic hand dryer circuit which requires 4.0V to operate, use the graph to work out for how long it will remain on.
The spreadsheet says the time constant is 20s, and thats what I could read off the graph.
But aint time constant = RC? And RC is 200*0.0001=0.02s, not 20s..
So.. As the minimum required voltage is 4V it won't work below that? And initial is 6V:
V = V0e-t/RC
V/V0 = e-t/20
20 ln(4/6) = -t
t = 8.109s
Wrong..
And using a lower RC would only give a shorter time, but it's 50s+.
I regret not having the ability to answer that question. I have not finished my AS yet. I am deeply sorry.
Post by: seth99 on May 31, 2010, 01:31:03 pm
I have A levels physics paper 5 on 9th , i have problem answering first question. if you could tell me where do i add the additional details? shall i make it as another subheading and write all additional details(as answerd in mark scheme) or shall i keep adding in the subheadings i give while answering the question.
Answer to the question number 1 is in this format given in mark scheme.
Defining the problems
Methods of data collection
Annalysing the data
safety precautions
ADDITIONAL DETAILS
Post by: Saladin on May 31, 2010, 01:32:20 pm
hey this is my frirst question on this forum.
I have A levels physics paper 5 on 9th , i have problem answering first question. if you could tell me where do i add the additional details? shall i make it as another subheading and write all additional details(as answerd in mark scheme) or shall i keep adding in the subheadings i give while answering the question.
Answer to the question number 1 is in this format given in mark scheme.
Defining the problems
Methods of data collection
Annalysing the data
safety precautions
ADDITIONAL DETAILS
Just post your doubt here!
Post by: nid404 on May 31, 2010, 01:34:26 pm
Just post your doubt here!
That sounds out of place :P
Please post your doubts here ;)
Post by: Saladin on May 31, 2010, 01:43:42 pm
That sounds out of place :P
Please post your doubts here ;)
Wait let me add some effects.
Post by: Saladin on May 31, 2010, 01:44:36 pm
You may simply post them here.
Post by: Saladin on May 31, 2010, 01:47:30 pm
hey this is my frirst question on this forum.
I have A levels physics paper 5 on 9th , i have problem answering first question. if you could tell me where do i add the additional details? shall i make it as another subheading and write all additional details(as answerd in mark scheme) or shall i keep adding in the subheadings i give while answering the question.
Answer to the question number 1 is in this format given in mark scheme.
Defining the problems
Methods of data collection
Annalysing the data
safety precautions
ADDITIONAL DETAILS
Just one problem, what is the question?
Post by: seth99 on May 31, 2010, 02:01:54 pm
http://www.xtremepapers.net/CIE/International%20A%20And%20AS%20Level/9702%20-%20Physics/9702_s07_qp_5.pdf
Post by: Saladin on May 31, 2010, 02:03:55 pm
well in a levels physics paper 5(planning,Annalysis and evaluation) we have only 2 questions to answer. format for question number can be seen on any of the previous years question paper
http://www.xtremepapers.net/CIE/International%20A%20And%20AS%20Level/9702%20-%20Physics/9702_s07_qp_5.pdf
I humbly apologize, I don't do A2 Physics.
Post by: seth99 on May 31, 2010, 02:05:40 pm
Post by: ikh on May 31, 2010, 03:53:46 pm
HOw is the amplitude and maximum displacment calculated ???
Post by: Meticulous on May 31, 2010, 04:00:29 pm
Post by: ikh on May 31, 2010, 04:10:13 pm
The dotted vertical lines show the positions of two coils which at this moment are undisplaced.
Mark on the lower diagram a compression C and a rarefaction R
Measure the wavelength of the wave
Wavelength ...
Mark on the lower diagram a coil with maximum displacement, M.
Measure the amplitude of the wave, i.e. the displacement of coil M.
Amplitude ...
(Total 4 marks)
couldnt get the pic here so attached file
Post by: Meticulous on May 31, 2010, 04:12:58 pm
Post by: ikh on May 31, 2010, 04:20:59 pm
Amplitude=max displacement
i do reliase amplitude = displacment but how do you measure the amplitude in this graph (this is not a transverse graph but a longitudinal one displayed as a sinisoudial one) please explain how exactly to measure.
Thanks in advance
Post by: ikh on May 31, 2010, 07:28:29 pm
Post by: vanibharutham on May 31, 2010, 09:12:49 pm
Longitudinal waves have compressions and rarefactions and im sure you know that the distance between the centres of two compression or two rarefactions is the wavelength.
However, to work out amplitude you must know that at the centre of the rarefaction or at the centre of a compression the displacement is zero. Using that and the slinky given at the rest position, you can work out the distance between the maximum displacement of the slinky.
Post by: vanibharutham on May 31, 2010, 09:14:15 pm
When Supernova 1987A was detected on Earth, neutrinos emitted in the explosion arrived about 20 hours earlier than the light, even though both were created at the same instant. How is this consistent with the idea that nothing can travel faster than the speed of light? (Hint: space is not empty)
What I'm guessing, though not sure, is that because "space is not empty" the light rays must be refracting in the somewhat denser medium and must be slowing down. However, the neutrinos, being particles, must have sped up and thus reached the earth before the light reached us. However, since nothing is faster than the speed of light, neutrinos only reached the earth because light slowed down, the consistency of the idea that nothing can travel faster than light is true.
Post by: ikh on June 01, 2010, 02:29:25 pm
@ ikh
Longitudinal waves have compressions and rarefactions and im sure you know that the distance between the centres of two compression or two rarefactions is the wavelength.
However, to work out amplitude you must know that at the centre of the rarefaction or at the centre of a compression the displacement is zero. Using that and the slinky given at the rest position, you can work out the distance between the maximum displacement of the slinky.
could u plz elaborate a bit more if i assume the centre of the rarefaction or at the centre of a compression the displacement is zero then what i am supposed to measure using the slinky at rest position....
i would appreciate if u could use the uploaded file and put marks wht exactly to measure as that would be really helpful
Thanks
Post by: Saladin on June 01, 2010, 07:27:42 pm
could u plz elaborate a bit more if i assume the centre of the rarefaction or at the centre of a compression the displacement is zero then what i am supposed to measure using the slinky at rest position....
i would appreciate if u could use the uploaded file and put marks wht exactly to measure as that would be really helpful
Thanks
You can't actually. You have to measure from compression to compression or rarefaction to rarefacton.
Post by: student on June 01, 2010, 07:43:17 pm
A 4MeV particle collides head-on with a stationary proton, and its kinetic energy is reduced to 1MeV. What is the mass of the particle? Mass of a proton = 1.6726*10-27kg.
But I can't get it.
Post by: ABF on June 01, 2010, 09:38:20 pm
Post by: Meticulous on June 01, 2010, 09:45:54 pm
does anyone have any notes on the nature of light?please i need it urgently..
http://www.s-cool.co.uk/alevel/physics/quantum-physics.html
http://www.s-cool.co.uk/alevel/physics/wave-particle-duality-and-electron-energy-levels.html
Post by: ABF on June 01, 2010, 10:27:54 pm
Post by: Darth Austrakast on June 04, 2010, 06:30:25 am
In an experiment with photoelectric emitter if frequency of incident electromagnetic radiation is increased
wat will happen to the stopping potential and saturation current? can n e 1 plz explain y?
thanx !
and another question what do we have to know from absorption spectra?
Post by: Igcseboy on June 04, 2010, 10:59:27 am
does anybody has some good notes on Waves....... i need them urgently...
some notes on electricity and nature of light will appreciated...
Post by: wakemeup on June 04, 2010, 03:32:07 pm
I have a doubt from nature of light topic
In an experiment with photoelectric emitter if frequency of incident electromagnetic radiation is increased
wat will happen to the stopping potential and saturation current? can n e 1 plz explain y?
thanx !
and another question what do we have to know from absorption spectra?
I'm not sure what 'stopping potential' and 'saturation current' mean so I assume you want to know what will happen to the current and the potential difference. The more you increase the frequency above the threshold value, the more kinetic energy (1/2mv^2) electrons gain. So, both the current(I=nave) and potential difference(V=W/Q) will increase.
As for absorption spectra, we're basically supposed to know that different objects absorb light differently. They all absorb specific frequencies of light. That's why you see dark parts on the absorption spectra. These are the frequencies of light that have been absorbed by the object in question. This and the photoelectric effect both provide evidence for the particle nature of light.
Hope I helped.
Post by: wakemeup on June 04, 2010, 04:16:34 pm
hello
does anybody has some good notes on Waves....... i need them urgently...
some notes on electricity and nature of light will appreciated...
You might find this useful. Some stuff is not in the course but it's good revision nevertheless
http://www.slideshare.net/pburgess/edexcel-as-physics-waves-presentation
Post by: Darth Austrakast on June 04, 2010, 06:19:18 pm
I'm not sure what 'stopping potential' and 'saturation current' mean so I assume you want to know what will happen to the current and the potential difference. The more you increase the frequency above the threshold value, the more kinetic energy (1/2mv^2) electrons gain. So, both the current(I=nave) and potential difference(V=W/Q) will increase.
As for absorption spectra, we're basically supposed to know that different objects absorb light differently. They all absorb specific frequencies of light. That's why you see dark parts on the absorption spectra. These are the frequencies of light that have been absorbed by the object in question. This and the photoelectric effect both provide evidence for the particle nature of light.
Hope I helped.
hey thanx that helped a lot !
yeah stopping potential is voltage and saturation current is the maximum current after which the photocurrent does not increase
Post by: Sweet_03 on June 05, 2010, 03:29:52 pm
from June 2009 !
Post by: Saladin on June 05, 2010, 03:51:09 pm
can anyone plz explain how to do Question 21 (c)(ii)
from June 2009 !
On it now.
Post by: Saladin on June 05, 2010, 04:02:43 pm
Then use V=IR as the current stays the same!
Post by: Ibiza279 on June 05, 2010, 04:03:05 pm
Post by: Meticulous on June 05, 2010, 04:05:04 pm
http://www.s-cool.co.uk/alevel/physics/wave-particle-duality-and-electron-energy-levels.html
Post by: abuelzouz on June 05, 2010, 04:06:41 pm
Just use V=lA
Then use V=IR as the current stays the same!
whts V=IA ??
Post by: Ibiza279 on June 05, 2010, 04:11:25 pm
http://www.s-cool.co.uk/alevel/physics/quantum-physics.html
http://www.s-cool.co.uk/alevel/physics/wave-particle-duality-and-electron-energy-levels.html
Thanks :)
Post by: Saladin on June 05, 2010, 04:12:35 pm
whts V=IA ??
its the constant in the question, V= length x current.
Post by: Ibiza279 on June 05, 2010, 05:21:07 pm
Post by: Ibiza279 on June 05, 2010, 06:51:14 pm
Post by: studz on June 05, 2010, 07:50:27 pm
those were real nice websites but i need summarised notes on the particle wave theory cause i didnt get the question in the jan 2010 paper U2 physics based of the particle wave theory :(
just learn that table given in the notes..that differentiates particle theory with the wave theory, BIB :P :P
Post by: Ibiza279 on June 05, 2010, 08:05:50 pm
Post by: studz on June 05, 2010, 08:07:35 pm
Post by: Ibiza279 on June 05, 2010, 08:17:33 pm
Post by: studz on June 05, 2010, 08:20:26 pm
and i'm talking about that table that was given in the end..about particle and wave theories! study that..thats important
Post by: Ibiza279 on June 05, 2010, 08:21:57 pm
yeah u do lol! thats why i called u bib dude!
and i'm talking about that table that was given in the end..about particle and wave theories! study that..thats important
abai kanjar, kon sa ??? ??? jo rafi ne likhwaya hai ??? if so then i don't hav that :( n who is this exactly ??? Danny ???
Post by: studz on June 05, 2010, 08:25:07 pm
abai kanjar, kon sa ??? ??? jo rafi ne likhwaya hai ??? if so then i don't hav that :( n who is this exactly ??? Danny ???
yeap that is the one!
well try to get it from somewhere because thats important...
and who the hell is danny ???
Post by: Ibiza279 on June 05, 2010, 08:28:32 pm
n crap, jus tell me who u are n ill try to get the notes from u :O
Post by: Ibiza279 on June 05, 2010, 08:46:33 pm
Post by: nid404 on June 06, 2010, 06:50:32 am
Hope this helps :)
Post by: Ibiza279 on June 06, 2010, 10:50:09 am
i wanted stuff like this, my teacher gave these notes but i don't hav um cause i was sick or soemthing so couldnt make it to school, all the links were extremely helpful, thank u all :) but wat i require are the comparison notes :)
Post by: Saladin on June 06, 2010, 01:11:27 pm
well since tomorrow, im getting the whole story behind de broglie and his experiment, wat i require is that for example, in the particle theory, one photon interacts with one electron on a metal surface to release one photo electron where as the wave theory states that the more the intensity of the light falling on the metal surface, more the energy is build up in the electron.
i wanted stuff like this, my teacher gave these notes but i don't hav um cause i was sick or soemthing so couldnt make it to school, all the links were extremely helpful, thank u all :) but wat i require are the comparison notes :)
A wave model would predict that weak radiation would eventually release large numbers of electrons but this is not what happens.
Considering light as photons means that weak radiation can cause instantaneous release of some electrons if the frequency is greater than the threshold frequency. Hence one photon is responsible for the release of one photo-electron.
Post by: Ibiza279 on June 06, 2010, 05:19:20 pm
A wave model would predict that weak radiation would eventually release large numbers of electrons but this is not what happens.
Considering light as photons means that weak radiation can cause instantaneous release of some electrons if the frequency is greater than the threshold frequency. Hence one photon is responsible for the release of one photo-electron.
ok i got it till a point, ill clear the rest with the book :) Thanks dude ! JazakAllah
Post by: Saladin on June 06, 2010, 05:26:46 pm
ok i got it till a point, ill clear the rest with the book :) Thanks dude ! JazakAllah
What are you unclear about?
Post by: Ibiza279 on June 08, 2010, 01:37:20 pm
Post by: Darth Austrakast on June 08, 2010, 03:54:23 pm
5. Two points on a progressive wave differ in phase by radian. The distance between them is 0.50 m. The frequency of the oscillations is 10 Hz. The maximum speed of the wave is
A 2.50 m s–1
B 5.00 m s–1
C 12.5 m s–1
D 40.0 m s–1
Post by: Darth Austrakast on June 08, 2010, 05:49:29 pm
By the way wat r u'r predictions for unit 2???
hope it isnt a killer xam like unit 1 :P
Post by: Ibiza279 on June 08, 2010, 06:15:56 pm
i need the ryt answer wid explanation for this question
5. Two points on a progressive wave differ in phase by radian. The distance between them is 0.50 m. The frequency of the oscillations is 10 Hz. The maximum speed of the wave is
A 2.50 m s–1
B 5.00 m s–1
C 12.5 m s–1
D 40.0 m s–1
how many radian is the phase difference of ?
Post by: Darth Austrakast on June 08, 2010, 06:28:07 pm
Post by: Ibiza279 on June 08, 2010, 06:31:57 pm
how many radian is the phase difference of ?
dude could u tell me exactly which year's paper this question is from ?
Post by: Darth Austrakast on June 08, 2010, 06:34:00 pm
In ms they say the answer is D
Post by: Saladin on June 08, 2010, 06:36:38 pm
well i got this from exam wizard questions
In ms they say the answer is D
It is not pi it is pi/4. That should explain a lot more.
Post by: Darth Austrakast on June 08, 2010, 06:38:39 pm
srry for the inconvenience
any tips fr the xam? any predictions?
Post by: Saladin on June 08, 2010, 06:42:38 pm
eah thats wat i was thinking too
srry for the inconvenience
any tips fr the xam? any predictions?
Yes, there are going to be 10 multiple choice questions.! :P
Post by: Ibiza279 on June 08, 2010, 06:47:59 pm
eah thats wat i was thinking too
srry for the inconvenience
any tips fr the xam? any predictions?
lol good one engraved :P
well i was thinking so hard on the question that i remembered i was hungry, so im gonna go n eat something, b bak in half an hour ! :D
Post by: Darth Austrakast on June 08, 2010, 06:49:13 pm
Post by: Fenomenoe9 on June 08, 2010, 07:20:46 pm
Post by: Ascer Zayan on June 08, 2010, 07:40:18 pm
Post by: Ibiza279 on June 08, 2010, 07:54:49 pm
How can u find the Number of Photons ( IN General ) ??? assuming tht he gave u the energy??
jus divide the energy by charge, that is for example, i hav 2 joules of energy, so my photons will b 2/1.6x10^-19 which is gonna b a huge number, so don't worry,cause the energy given out by each photon is really low.
jus remember this, if u want to find the photon, then its energy divided by charge, if u want to find the energy, then its photon x charge
Post by: Ibiza279 on June 08, 2010, 07:56:25 pm
i need the ryt answer wid explanation for this question
5. Two points on a progressive wave differ in phase by radian. The distance between them is 0.50 m. The frequency of the oscillations is 10 Hz. The maximum speed of the wave is
A 2.50 m s–1
B 5.00 m s–1
C 12.5 m s–1
D 40.0 m s–1
By the way the way, how do u solve this question after knowing that the value for radian is pi/4 :o lol :D
Post by: GodOfWar on June 08, 2010, 08:21:02 pm
By the way the way, how do u solve this question after knowing that the value for radian is pi/4 :o lol :D
Well
wavelength is 2 pi(360o).
So wavelength of this wave = 8 x0.5 = 4m
V=f*wavelength which is 10 x4 = 40 m/s
Post by: Ibiza279 on June 08, 2010, 08:37:47 pm
Post by: GodOfWar on June 08, 2010, 08:59:22 pm
Post by: 7ooD on June 08, 2010, 10:14:45 pm
last thing is a question from exam wizard the one t.q gave us it says
one way to study the energy levels of an atom is to scatter elecrons from it and measure their kinetic energies before and after the collision. if an electron of k.e 0.92 ev is scattered from a lithium atom which is initially in the -5.02 ev level the scattered elecron can have only two possible k.e s
state these two k.e values and explain what has happened to the lithium atom in each case ( u should assum that the lithium atom was at rest both before and after the collision) -5.02 is the ground level in the diagram
Post by: 7ooD on June 08, 2010, 10:30:25 pm
Post by: Fenomenoe9 on June 08, 2010, 10:33:48 pm
Post by: Fenomenoe9 on June 08, 2010, 10:42:42 pm
The first kinetic energy 1 is 0.92 V ... now the explanation is that nothing happens to the atom it stays at -5.02 (eV) level and nothing happens to it.
The second kinetic energy is i dont know sorry but im trying to think about it.
Hope it helped a bit :)
Post by: 7ooD on June 08, 2010, 11:52:50 pm
Post by: S.M.A.T on July 14, 2010, 02:58:10 pm
#A ceiling fan is turning at a rate of 100 revolutions per minute.A spider is clinging to a blade of the fan.If the spider experiences a centripetal acceleration greater than 0.3g,it will lose its grip on the blade and be flung off. how far from the centre of the fan can the spider safely go?
Please give the complete solution.
The answer in the book is 1.05m
(g means acceleration due to gravity)
Post by: Saladin on July 14, 2010, 10:22:50 pm
Please i need help in a question.
#A ceiling fan is turning at a rate of 100 revolutions per minute.A spider is clinging to a blade of the fan.If the spider experiences a centripetal acceleration greater than 0.3g,it will lose its grip on the blade and be flung off. how far from the centre of the fan can the spider safely go?
Please give the complete solution.
The answer in the book is 1.05m
(g means acceleration due to gravity)
Is this AS or A2?
If it is A2, I am afraid I can not answer the question....yet...
Post by: S.M.A.T on July 14, 2010, 11:22:23 pm
Post by: Saladin on July 14, 2010, 11:38:28 pm
Post by: Saladin on July 30, 2010, 11:05:22 pm
I have attatched the questions in a .pdf.
For 1(b): Do we add both the weights of the cart and the human? I did so, and I got an answer of 4750 Kgms-1
2(a): I wrote that the ball does work against air resistance, and thus, it reduces the speed of the ball and therefore the momentum.
2(b): I said, that the wind might be against the batsman, and thus this may increase the ball's momentum. And this will require more energy to work against.
4(a): Is the answer 1025 Kgms-1?
4(b): Some of the momentum is absorbed by the dash-board? I am not to sure about what the question is asking.
5(b): I got 5.736 Kgms-1
5(c): I got 478 N
5(e): The rubber will bounce back with more speed, meaning that it had a greater impulse. Thus this means, that the force will be greater, and thus more advantageous.
Please can anyone help me check my answers?
Post by: astarmathsandphysics on July 31, 2010, 09:28:56 am
Post by: Saladin on July 31, 2010, 11:39:05 am
Post by: Saladin on August 02, 2010, 12:18:55 am
Post by: astarmathsandphysics on August 02, 2010, 01:30:08 pm
b)Impulse=m(v-u)=m(-u--u)=-2mu becuase ball is hit right back to batsman
5c)rubber is better at absorbing energy than metal or wood.
Post by: Saladin on August 02, 2010, 06:30:06 pm
for 2a)if the ball hardly changes direction becuase the batman hots a glancing blow then then the momentum will hardly change. Inpulse =change in momentum so is small
b)Impulse=m(v-u)=m(-u--u)=-2mu becuase ball is hit right back to batsman
5c)rubber is better at absorbing energy than metal or wood.
Thanks again! :D
Post by: Ghost Of Highbury on August 04, 2010, 03:59:54 pm
Cud someone explain with a diagram?
Post by: Saladin on August 04, 2010, 04:39:40 pm
Y is it B?
Cud someone explain with a diagram?
This is because the car is moving, and the reaction is opposite to the combined force of the car. Which is a diagonal force.
Post by: Ghost Of Highbury on August 05, 2010, 04:21:34 pm
http://img94.imageshack.us/img94/7417/blahfu.png
Answer : 1.4m/s^2
Post by: Saladin on August 05, 2010, 06:08:56 pm
Thanks. This one too.
http://img94.imageshack.us/img94/7417/blahfu.png
Answer : 1.4m/s^2
LOL at the image name "Blahfu" :P
There are two forces acting here, the force of gravity acting on the block, and the force of friction, which is 6N.
The force pulling the weight down is,
Therefore, the net force is,
Now, when doing these calculations, there is uniform acceleration, because the acceleration of the 8 KG box is the same as the acceleration of the 2 KG box.
So,
Post by: nid404 on August 06, 2010, 05:48:34 am
Thanks. This one too.
http://img94.imageshack.us/img94/7417/blahfu.png
Answer : 1.4m/s^2
2g-T=2a
T-6=8a
solve simultaneously
14=10a
a=1.4m/s
Post by: Ghost Of Highbury on August 06, 2010, 04:27:37 pm
@Nid - Cud u plz rite the steps abts solving the equations simultaneously with 3 variables. Thanks
Post by: cooldude on August 06, 2010, 04:30:48 pm
Thanks dude, that is what i wanted.
@Nid - Cud u plz rite the steps abts solving the equations simultaneously with 3 variables. Thanks
dude, there arent three variables g is a constant, neway if u still want the working here it is-->
2g-T=2a
T-6=8a
T=2g-2a, and T=8a+6
8a+6=2g-2a
10a=2g-6
g=10
10a=20-6
10a=14
a=14/10=1.4
Post by: Ghost Of Highbury on August 06, 2010, 04:37:42 pm
dude, there arent three variables g is a constant, neway if u still want the working here it is-->
2g-T=2a
T-6=8a
T=2g-2a, and T=8a+6
8a+6=2g-2a
10a=2g-6
g=10
10a=20-6
10a=14
a=14/10=1.4
ahh, terrible error.
Thanks mate and thanks nid.
Post by: nid404 on August 07, 2010, 05:22:56 am
Post by: Saladin on August 12, 2010, 08:12:05 pm
The questions are attached.
1) I got velocity as
2) a) I got velocity as
b) I got force as 100N
3) I explained that The boy goes to the right of the boat, and according to Newton's third law, the boat exerts an equal and opposite force moving away from the boy.
4) Confused about the calculations involved.
5) Please help on this as well, I will check my answers when someone has posted.
Post by: M-H on August 13, 2010, 01:33:20 pm
Post by: elemis on August 13, 2010, 02:16:21 pm
i'll post soon
Its been nearly an hour ! ::)
Post by: nid404 on August 13, 2010, 02:37:26 pm
Guys I need help with this Physics question. Please assist!
The questions are attached.
1) I got velocity asms-1
2) a) I got velocity asms-1
b) I got force as 100N
3) I explained that The boy goes to the right of the boat, and according to Newton's third law, the boat exerts an equal and opposite force moving away from the boy.
4) Confused about the calculations involved.
5) Please help on this as well, I will check my answers when someone has posted.
1) 8/1000X 358= 91 X v
v=0.03m/s
2) a)5X4= 100Xv
v=0.2m/s
b) F=ma
a=v/t
If you could tell me what you did?
3) that's right
4) astar has done this somewhere...I'll look for it
5) a) linear momentum of a body is the product of mass and velocity, and for a system equal to the vector sum of the products of mass and velocity of each particle in the system.
b) IIIrd statement is the second law
F=ma
F=m(v-u)/t
c) IInd
d) Not sure but i think he assumes it's a head-on collision.
e) should be using photogates. Will look for the entire setup.
Post by: M-H on August 13, 2010, 05:43:12 pm
Guys I need help with this Physics question. Please assist!
The questions are attached.
1) I got velocity as
2) a) I got velocity as
b) I got force as 100N
3) I explained that The boy goes to the right of the boat, and according to Newton's third law, the boat exerts an equal and opposite force moving away from the boy.
4) Confused about the calculations involved.
5) Please help on this as well, I will check my answers when someone has posted.
1st 2nd and 3rd are right..
4th and 5th, il b trying!! so sorry for the delay!..i know the diagram for the 4th one, but i don't know how to work it out cz we haven't yet started those stuff. anyway, ill try to get it asap!
Post by: Saladin on August 17, 2010, 05:57:22 pm
The hero must be pretty fat to be 100 KG....
Post by: M-H on August 17, 2010, 07:35:30 pm
I only have one thing to add.
The hero must be pretty fat to be 100 KG....
yeah fatso :P
Post by: Saladin on August 23, 2010, 08:24:35 pm
This is what I did:
Before collision
Horizontal
After Collision
Horizontal
Cue=
Black=
Therefore resolving horizontally to find the speed of the Cue ball
Vertical
Black=
Black=
At this point, it is safe to presume that as it is an elastic collision, they will move at equal and opposite angles. This means that the cue ball will also move at 45o to the right.
Thus now finding the velocity:
Thus, this means that it will end up in the bottom right pocket as it is 45
Post by: Saladin on August 23, 2010, 09:27:36 pm
There are 6283 milliradians in a complete circle. The army’s rounding of this to 6400 mils causes an error.
a How far sideways from the target could this rounding cause an artillery shell to be when aimed at a target 20km away?
I know this is most probably so simple that I just cant see it infront of me. But I just can't seem to be able to solve this... :(
The answer to this by the way is 36.6 cm.
Post by: Saladin on August 24, 2010, 01:06:07 am
This is another problem....
There are 6283 milliradians in a complete circle. The army’s rounding of this to 6400 mils causes an error.
a How far sideways from the target could this rounding cause an artillery shell to be when aimed at a target 20km away?
I know this is most probably so simple that I just cant see it infront of me. But I just can't seem to be able to solve this... :(
The answer to this by the way is 36.6 cm.
I already got this answered. Thanks everyone that had a look.
Post by: Saladin on August 24, 2010, 01:15:41 am
I want to be sure about question 2.
You see there are two answers for part 2(a)
1. They act on the same body or do not act on different bodies
2. They are different types of, or they are not the same type of, force
I agree with the first one, but I believe that the second answer is incorrect, can anyone please tell me as to why the second answer is correct or if I am the one who is correct.
Because Newton's third law only talks about two separate forces on two separate bodies. The reaction force is a different kind of force in comparison to gravity....
Can anyone help?
Post by: Saladin on August 24, 2010, 01:40:00 am
Hey guys, can you please take a look at this for me?
I want to be sure about question 2.
You see there are two answers for part 2(a)
1. They act on the same body or do not act on different bodies
2. They are different types of, or they are not the same type of, force
I agree with the first one, but I believe that the second answer is incorrect, can anyone please tell me as to why the second answer is correct or if I am the one who is correct.
Because Newton's third law only talks about two separate forces on two separate bodies. The reaction force is a different kind of force in comparison to gravity....
Can anyone help?
In fact, the whole question is wrong because in circular motion, there is acceleration! So, acceleration is caused by a resultant force!
How can this be?
Post by: nid404 on August 24, 2010, 06:14:45 am
As far as your question is concerned, lol, I don't quite get the first one itself :-[ I feel dumb :-X
Post by: nid404 on August 24, 2010, 06:36:52 am
Post by: S.M.A.T on August 24, 2010, 11:07:33 am
Hey guys, can you please take a look at this for me?Newton's third law,
I want to be sure about question 2.
You see there are two answers for part 2(a)
1. They act on the same body or do not act on different bodies
2. They are different types of, or they are not the same type of, force
I agree with the first one, but I believe that the second answer is incorrect, can anyone please tell me as to why the second answer is correct or if I am the one who is correct.
Because Newton's third law only talks about two separate forces on two separate bodies. The reaction force is a different kind of force in comparison to gravity....
Can anyone help?
'When a body 'A' exerts a force on body ''B',then body 'B' exerts a force of the same type on body 'A' which is equal in magnitude but opposite in direction.'
From the above statement it could be understand that
1)The two forces are equal in magnitude but opposite in direction
2)Two forces must be acting on two separate body.(in the above statement 'A' &'B'.)
3)The two forces must be same type of force.
4)The two forces must be acting for the same period of time i.e if body 'A' exerts a force for 't' second than on body 'B' then body 'B' exerts a force for 't' second than on body 'A'.
In the question,
1) 'F-1' &'F-2' are acting on the same body.So it is violating the point 2.
2) 'F-1' &'F-2' are not same type of force.'F-1' is normal reaction force whereas 'F-2' is weight.So it is violating point 3.
Post by: Saladin on August 24, 2010, 11:27:44 am
Newton's third law,
'When a body 'A' exerts a force on body ''B',then body 'B' exerts a force of the same type on body 'A' which is equal in magnitude but opposite in direction.'
From the above statement it could be understand that
1)The two forces are equal in magnitude but opposite in direction
2)Two forces must be acting on two separate body.(in the above statement 'A' &'B'.)
3)The two forces must be same type of force.
4)The two forces must be acting for the same period of time i.e if body 'A' exerts a force for 't' second than on body 'B' then body 'B' exerts a force for 't' second than on body 'A'.
In the question,
1) 'F-1' &'F-2' are acting on the same body.So it is violating the point 2.
2) 'F-1' &'F-2' are not same type of force.'F-1' is normal reaction force whereas 'F-2' is weight.So it is violating point 3.
Thanks man! :)
Post by: Saladin on August 24, 2010, 11:31:24 am
Post by: S.M.A.T on August 24, 2010, 11:52:19 am
Just to clarify, the two are different forces because one is not a cause of the other...
Well :-\ you can say that.
The thing is that here F-1 is contact force and F-2 is gravitational force.
Post by: nid404 on August 24, 2010, 11:53:11 am
Just to clarify, the two are different forces because one is not a cause of the other...
Yup that's how it is i suppose.
Thanks asif :)
Post by: haris94 on September 12, 2010, 12:07:42 pm
Determine the base units of 'G'.
pls explain how do i go about this question
Thanks
Post by: ashish on September 12, 2010, 02:40:22 pm
M=kg
r=radius =M
F= GM1M2/r^2
G=F*r^2/M1M2
=NM^2/Kg^2
or
M^3/(Kg*s^2)
Post by: haris94 on September 12, 2010, 04:51:51 pm
u hv to make G the subject of the formula
i wasnt aware of that
Thanks again
Post by: ashish on September 14, 2010, 04:42:04 pm
Post by: The Golden Girl =D on September 28, 2010, 05:37:32 pm
case 1 :
acceleration is directly proportional to resultant force when Mass is CONSTANT
case 2 :
acceleration is Inversely proportional to Mass when Force is Kept CONSTANT
and he gave an experiment for both cases ,can some one give me good notes/site which explains it in detail (cuz i need it to FULLY understand it )
Thx :)
Post by: The Golden Girl =D on September 28, 2010, 05:57:03 pm
in newton's Second Law of Motion , the teacher gave us two cases ;
case 1 :
acceleration is directly proportional to resultant force when Mass is CONSTANT
case 2 :
acceleration is Inversely proportional to Mass when Force is Kept CONSTANT
and he gave an experiment for both cases ,can some one give me good notes/site which explains it in detail (cuz i need it to FULLY understand it )
Thx :)
Anyone :-\
Post by: The Golden Girl =D on September 28, 2010, 06:04:43 pm
Anyone :-\
I have to go for now but i'll came perhaps tmrrw iA :)
Post by: Arthur Bon Zavi on September 28, 2010, 06:07:22 pm
http://www.physicsclassroom.com/class/newtlaws/u2l3a.cfm
Then this:
http://en.wikipedia.org/wiki/Newton%27s_laws_of_motion
Post by: Arthur Bon Zavi on September 28, 2010, 06:09:38 pm
A good presentation:
http://www.wisc-online.com/Objects/ViewObject.aspx?ID=tp1302
seee this too:
http://cnx.org/content/m14042/latest/
Post by: The Golden Girl =D on September 28, 2010, 06:16:40 pm
This is a good physics ref site.
http://www.physicsclassroom.com/class/newtlaws/u2l3a.cfm
Then this:
http://en.wikipedia.org/wiki/Newton%27s_laws_of_motion
http://zonalandeducation.com/mstm/physics/mechanics/forces/newton/newtonLaw2.html
A good presentation:
http://www.wisc-online.com/Objects/ViewObject.aspx?ID=tp1302
seee this too:
http://cnx.org/content/m14042/latest/
Thx loads man it really helped :D ..I'd really appreciate it if u had notes/site that emphasize on the experiments as well :)
Post by: astarmathsandphysics on September 28, 2010, 10:18:25 pm
Post by: The Golden Girl =D on September 29, 2010, 12:00:54 pm
I think I can help you there. If you email paul.smith@astarmathsandphysics.com I will email you back a set of 50 or so experiments with notes
ya sure will send the email :)
Post by: The Golden Girl =D on October 26, 2010, 04:20:19 pm
Thank You In Advance :)
Post by: Deadly_king on October 26, 2010, 04:43:54 pm
Hey Everyone ;D ...Well I have a Doubt .....Can someone give me notes or a Website that has notes on Fluid Flows ...The teacher just taught us some of the basics And the whole time I was -__- cuz i didn't get it :S :S
Thank You In Advance :)
Am not sure about it myself. So i'll ask you to consult your syllabus before proceeding to this link. ;)
http://www.engineersedge.com/fluid_flow/fluid_flow_table_content.htm
Post by: The Golden Girl =D on October 26, 2010, 04:53:52 pm
Thanks but the thing obviously freaked me out :S
Post by: blackberry on November 26, 2010, 10:42:40 pm
so i assumed that k = E, considering that both measure the stiffness of the material and both are found using the gradient of F/x graph in the case of k and Stress/Strain graph in the case of E.
BUT.....i think my teacher told me that they are both different, so im kinda confused as to which is what so can anyone tell me the difference between spring constant and young modulus please??
Post by: elemis on November 27, 2010, 09:43:08 am
The YM is true for a PARTICULAR material no matter what dimensions it takes in a particular spring.
Post by: blackberry on November 27, 2010, 11:16:30 am
Post by: The Golden Girl =D on November 27, 2010, 11:41:03 am
so i was just trying to study physics today and then WHAAAM!!! i got stuck at this point where my textbook mentioned that in hooke's law(F=kx) the spring constant(k)represents the 'stiffness' of the spring and then in another textbook it mentions that the young modulus(E)of a material is a measure of how stiff it is!! ??? ??? ???
so i assumed that k = E, considering that both measure the stiffness of the material and both are found using the gradient of F/x graph in the case of k and Stress/Strain graph in the case of E.
BUT.....i think my teacher told me that they are both different, so im kinda confused as to which is what so can anyone tell me the difference between spring constant and young modulus please??
Look when you draw a force-extension graph ,the gradient represents the stiffness of the material . the STEEPER the gradient is the STIFFER it is ;) and since stiffness is the gradient it's calculated as follows :
Stiffness/Gradient = Force/Extension
Young Modulus is the MEASUREMENT of how stiff the material is and
YM = stress/ strain
YM= (f/a) divided by (x/l)
hence , YM = (f/x) * (l/A) -> Pascals ;)
If you don't get it , let me know ;)
Post by: elemis on November 27, 2010, 11:41:47 am
urmm????
If i have a 10 cm long steel wire of 1mm2 cross sectional area the way it behaves under a load will differ from a 20 cm long steel wire of 0.5 mm2
This difference in behavior i.e. extension will affect the Spring Constant of the springs. That is, the first wire will have a different SK from the second one.
However, with the Young's Modulus we compare the stress and strain of the steel wire. Incidentally the YM of both wires will be the SAME.
Post by: The Golden Girl =D on November 27, 2010, 12:01:04 pm
Post by: blackberry on November 27, 2010, 12:46:48 pm
By the way golden girl im a she not a he!! :P :P :P
nyywayyy thanxxx so much guys!!!!gosh physics is reallly mind-boggling!!! :-[
Post by: The Golden Girl =D on November 27, 2010, 01:31:59 pm
oooooooooooooooooooooooooooooohhhhh nowww i get it!!!! :o :o
By the way golden girl im a she not a he!! :P :P :P
nyywayyy thanxxx so much guys!!!!gosh physics is reallly mind-boggling!!! :-[
glad u do
my bad :-X
It's NOT once you get it ;)
Post by: blackberry on November 27, 2010, 03:46:11 pm
oo hey i got nother question!! :P :P
could u plzz temme bout acceleration of free fall with and without air resistance??
i keep getting confused bout the acceleration part! ???
Post by: elemis on November 27, 2010, 04:29:05 pm
Post by: blackberry on November 27, 2010, 04:45:56 pm
and while it is falling is it accelerating or decelerating??
what is the situation with air resistance??
and how exactly would the graph of velocity against time for both cases come out?? ???
(Sorry for the 1000000000 questions i just get confused when i try to understand and concentrate on things alot plus i have a baad case of amnesia(not literally)!!!! :P :P)
Post by: The Golden Girl =D on November 27, 2010, 05:38:54 pm
so just at the point the object is dropped will g=9.8 or 0
and while it is falling is it accelerating or decelerating??
what is the situation with air resistance??
and how exactly would the graph of velocity against time for both cases come out?? ???
(Sorry for the 1000000000 questions i just get confused when i try to understand and concentrate on things alot plus i have a baad case of amnesia(not literally)!!!! :P :P)
through out the whole free fall the Acceleration is 9.81
what changes is the SPEED ;)
speed decreases going down due to air resistance due to the formula :
Air resistance is directly proportional to Square Speed *meaning speed to the power of two *
I'm sorry but if u are doing AS physics Edexcel Unit One , You don't need to know that cuz the air resistance is ALWAYS there . Unless it is stated in the Q that it is negligible .
I'm sorry but I don't know it or perhaps I'm too sleepy to focus -.-
Post by: The Golden Girl =D on November 27, 2010, 05:47:08 pm
http://www.physicsclassroom.com/class/newtlaws/u2l3e.cfm
Post by: ashish on November 28, 2010, 06:04:29 am
through out the whole free fall the Acceleration is 9.81
what changes is the SPEED ;)
speed decreases going down due to air resistance due to the formula :
Air resistance is directly proportional to Square Speed *meaning speed to the power of two *
I'm sorry but if u are doing AS physics Edexcel Unit One , You don't need to know that cuz the air resistance is ALWAYS there . Unless it is stated in the Q that it is negligible .
I'm sorry but I don't know it or perhaps I'm too sleepy to focus -.-
gg i think you got something wrong
when falling the speed of the object increases, that is the body accelerate until it reaches terminal velocity.check the graph
and while throwing the object upward, the object suffers a deceleration and after some time it's velocity will become 0
on the other hand if there is no air resistance the object will fall with constant acceleration (g) !
Post by: The Golden Girl =D on November 28, 2010, 12:18:38 pm
gg i think you got something wrong
when falling the speed of the object increases, that is the body accelerate until it reaches terminal velocity.check the graph
and while throwing the object upward, the object suffers a deceleration and after some time it's velocity will become 0
on the other hand if there is no air resistance the object will fall with constant acceleration (g) !
My Bad :-X
Thanks for clearing this out for us :)
Post by: blackberry on November 28, 2010, 02:00:50 pm
see thats the part where i get confused when the object falls it accelerates and decelerates and the speed/velocity
increases and decreases at the same time so i dont get it!! what is going on?? (ahhhhhhhhhhhhhh!!!!)
Post by: astarmathsandphysics on November 28, 2010, 09:02:35 pm
The speed is increasing but the rate at which it is decreasing - the acceleration - is slowing.
Post by: The Golden Girl =D on December 03, 2010, 05:46:36 pm
And Q 16]b)ii) ...I just go blank when I read the Q , what does the Q want ? -.-
as well as Q19] b)ii) ... how do I Answer it +i don't get how will the drawing will look like ? ???
Thanks In Advance :)
Post by: elemis on December 04, 2010, 03:44:23 am
Basically they're asking if I create one small bubble and another large bubble and let them go, explain why the big bubble catches up with the smaller one ?
Question 16
How suitable are the Diesel Generators for use in this project ? Remember, standby means if the other power sources fail Diesel generators will be switched on.
So in effect they must provide the same power output or even more than the power output of the other power sources.
Question 19
In such a case where you are unsure draw a small diagram at the bottom of the page and on the lines explain your working fully. This avoids misunderstandings.
Post by: elemis on December 04, 2010, 03:48:55 am
Using tan :
tan x = 4.8/0.485
x = 84.2 degrees.
Post by: The Golden Girl =D on December 04, 2010, 10:36:52 am
QUestion 17
Basically they're asking if I create one small bubble and another large bubble and let them go, explain why the big bubble catches up with the smaller one ?
Q17] catches up means that it'll be at the same level after a while *that's what i understood*
Answer says : if r increases ,speed increases too
1. I don't get why increases ? 2.if it does increases why does it catch up in the end ?
Thanks mate
Post by: elemis on December 04, 2010, 11:39:34 am
Q17] catches up means that it'll be at the same level after a while *that's what i understood*
Answer says : if r increases ,speed increases too
1. I don't get why increases ? 2.if it does increases why does it catch up in the end ?
Thanks mate
If you look at the equation you'll notice radius is on the numerator. If the numerator of a fraction increases in size then the fraction becomes LARGER in value.
Therefore, as radius increases the value of the fraction, in this case speed, will increase too.
The smaller bubble has a lower speed compared to the larger bubble.
Post by: The Golden Girl =D on December 04, 2010, 12:50:45 pm
If you look at the equation you'll notice radius is on the numerator. If the numerator of a fraction increases in size then the fraction becomes LARGER in value.
Therefore, as radius increases the value of the fraction, in this case speed, will increase too.
The smaller bubble has a lower speed compared to the larger bubble.
Aha Okay , Now I get it ;D ..Thanks mate :)
Post by: zbzbzb25 on December 04, 2010, 05:10:52 pm
Lawrence's second cyclotron, built with the assistance of Stanley Livingstone,was 25 cm in diameter.(continued?
it accelerated protons to 1 MeV.
(a) calculate the speed of these protons.
(b)calculate the momentum of these protons.
(c)Calculate the magnetic field strength Lawrence used in this cyclotron, assuming that the protons move in a circle around the very edge of the machine.
(d)Calculate the frequency of the voltage that Lawrence and Stanley had to apply to achieve this proton acceleration.
and another question
The Large Hadron Collider is a giant circular accelerator near Geneva in Switzerland.?
The LHC website povides some facts about it :
1- the circular ring has a circumference of 27km
2-each proton goes around over 11000 times a second
3-each proton has 7 tera eV of Kinetic energy ( tera is ten to the power of 12)
(a) calculate the speed of these protons.
(b)calculate 1/2 m (v squared) for these protons using mass of proton equal to 1.67* ten to the power of (-27) kg.
(c) how does your answer to (b) compare with the 7 T(tera) eV of kinetic energy the
protons are given through the p.d. acceleration?
(d) Explain this difference.
Post by: astarmathsandphysics on December 06, 2010, 10:24:05 pm
10^6*1.6*10^-19=1/2*1.67*10^-27v^2 so v=1.38*10^7m/s
b)p=mv=1.67*10^-27*1.38*10^7=2.3*10^-20 kgm/s
c)B=mv/er=1.67*10^-27*1.38*10^7/1.6*10^-19*0.125=1.152T
d)f=v/2%pir=1.38*10^7/6.28*0.125=17.6MHz
2a)v=2pi r f=27000*11000=2.97*10^8m/s
b)1/2mv^2=1/2*1.67*10^-27*(2.97*10^8)^2=7.4*10^-11J
c)7TeV=7*10^12*1.6*10^-19=1.12810^-6J
d)Much of this energy is given off as electromagnetic radiation through the sides of the accelerator as the protons accelerate in a circle
Post by: astarmathsandphysics on December 06, 2010, 10:25:24 pm
Post by: meeeeeee on December 08, 2010, 05:29:51 pm
Post by: MKh on December 08, 2010, 08:00:39 pm
Post by: MKh on December 08, 2010, 08:11:29 pm
Post by: Deadly_king on December 09, 2010, 11:44:57 am
For A2 physics, do we need to know the details of milikan's oil drop experiment?
Yes, you need to have a brief overview about it. Rather you should know about the principle. The small details are not important. ;)
n cud any1 suggest me hw i am supposed 2 prepare 4 d edexcel A2 unit 6B? Thanks...
Please mention which topic unit 6B is... :-[
Post by: MKh on December 09, 2010, 08:31:57 pm
Post by: Arthur Bon Zavi on December 10, 2010, 01:05:42 pm
unit 6B iz d A2 Alternative to Practical paper (for international candidates)...iz any1 preparing 4 it???
May be - ~ Amatu Allah ~ Aka GG.
Post by: The Golden Girl =D on December 10, 2010, 01:08:09 pm
May be - ~ Amatu Allah ~ Aka GG.
Nah I'm doing AS , which means 3B *in june tho * ,Sorry mate :-\
Post by: Arthur Bon Zavi on December 10, 2010, 01:33:57 pm
Nah I'm doing AS , which means 3B *in june tho * ,Sorry mate :-\
Sorry, I misjudged ! :D
Post by: Assi1993 on December 10, 2010, 04:38:53 pm
Q:A bungee jumper of mass 75 kg uses a bungee cord of length 25 m when jumping from a high bridge.Calculate the momentum of the jumper at the instant when the cord begins to stretch.
answer has 2 b=1660Ns
Q:An ice-hockey puch has a momentum of 1.2Ns just before making impact with the boards at the edge of the rink.It rebounds with a momentum of 1Ns after making contact for 0.15s.Calculate the average force acting on the puck during the collision.Ans=15N
Post by: elemis on December 10, 2010, 05:32:17 pm
Momentum = mass * velocity
To determine velocity when the rope is fully uncoiled we use :
v2 = u2 + 2as
Hence, v2 = 0 + 2*9.8*25
V = 22.14
75*22.14 = 1660 Ns
Post by: elemis on December 10, 2010, 05:35:18 pm
When the puck collides with the wall it will experience an IMPULSE.
Impulse = Force * time
We have time but need to find Impulse.
Hence, 1--1.2 = 2.2 Ns = impulse exerted in puck
Substituting into Impulse = Force * time
We get : 2.2 / 0.15 = force
Force = 14.666666 N i.e 15 N (2 s.f.)
Post by: Assi1993 on December 10, 2010, 07:02:14 pm
Post by: Assi1993 on December 10, 2010, 07:40:02 pm
Q:A circular space station of radius 1000m is designed to rotate about an axis through its centre to stimulate gravity.How many revolutions per minute must it perform in order to produce an artificial gravity of 9.8m/s
Q:A ceiling fan is turned at a rate of 100 revolutions per minute. A spider is clinging to a blade of the fan.If the spider experiences a centripetal acceleration greater than 0.3g, it will lose its grip on the blade and be flung off.How far from the centre of the fan can the spider safely go?
Q:A car of mass 1600kg takes a bend of radius of curvature 30m on a horizontal road at a constant speed.If the maximum frictional force from the road is 65% of the car's weight,what is the greatest speed at which the car can take the bend without skidding?
Post by: The Golden Girl =D on December 24, 2010, 08:18:13 am
The third Question in the June 2008 Paper *unit 1* . I s it a part of the Syllabus ?
this Q's Basics are of Mechanics , so I got a bit Confused , You know :-\ :-X
By the way it's the first link in the following Page :
http://www.freeexampapers.com/past_papers.php?l=Past_Papers/A+Level/Physics/Edexcel/2008+Jun/
Post by: elemis on December 24, 2010, 08:45:22 am
Post by: elemis on December 24, 2010, 08:53:07 am
(a) To find the angle between the acceleration and the i vector imagine that 6 and 8 were x and y coordinates respectively.
Plot them if you want on a sketch graph. Then do tan-1 8/6 = 53.1 degrees.
(b) F = ma <----- you must know this simple formula.
Since we are dealing with magnitudes simply find the modulus of the acceleration vector. 6^2 + 8^2 = 100 Hence, magnitude of acceleration is 10 m/s^2
10*0.4 = 4 N = magnitude of F.
(c) Let x be the unknown velocity vector.
Solving for x you should get : 39i + 30j
Post by: The Golden Girl =D on December 24, 2010, 09:31:38 am
it's this one :
http://www.freeexampapers.com/get_past_papers.php?l=Past_Papers/A+Level/Physics/Edexcel/2008+Jun/6731-01+Physics.pdf
third Q.
Post by: The Golden Girl =D on December 24, 2010, 09:42:57 am
Post by: elemis on December 24, 2010, 09:56:35 am
Post by: elemis on December 24, 2010, 09:58:32 am
I think the syllabus for Edexcel sciences changed after 2008. I'm not sure though since I dont do Edexcel science subjects.
Post by: The Golden Girl =D on December 24, 2010, 10:01:33 am
Whats your problem ? You cant get the answers ?
I think the syllabus for Edexcel sciences changed after 2008. I'm not sure though since I dont do Edexcel science subjects.
1.I didn't Know how to do it , And I wanted to Know whether it's a Part of the Syllabus or Not ,That's All .
Post by: elemis on December 24, 2010, 10:03:02 am
1.I didn't Know how to do it , And I wanted to Know whether it's a Part of the Syllabus or Not ,That's All .
I'll answer them.
Post by: The Golden Girl =D on December 24, 2010, 10:04:18 am
Post by: elemis on December 24, 2010, 10:14:59 am
b) i)To determine the velocity you could have a metre rule placed on the track. As Trolley A moves you use a stopwatch to determine the time it takes to travel from its origin to the point of collision.
Then you would use s = d/t to determine its speed. Repeat to gain an average and increase accuracy.
ii) Mass of the trolleys is the other quantity required as momentum = mass*velocity
iii) The velocities we are using to show the conservation of momentum are actually the velocities JUST before collision. We must rely on constant velocities as we are considering a system with no resultant force and hence no acceleration of the trolleys.
Post by: The Golden Girl =D on December 24, 2010, 10:21:36 am
Thanks Again .
Post by: The Golden Girl =D on December 24, 2010, 04:17:03 pm
3 ques regarding circular motion
Q:A circular space station of radius 1000m is designed to rotate about an axis through its centre to stimulate gravity.How many revolutions per minute must it perform in order to produce an artificial gravity of 9.8m/s
Q:A ceiling fan is turned at a rate of 100 revolutions per minute. A spider is clinging to a blade of the fan.If the spider experiences a centripetal acceleration greater than 0.3g, it will lose its grip on the blade and be flung off.How far from the centre of the fan can the spider safely go?
Q:A car of mass 1600kg takes a bend of radius of curvature 30m on a horizontal road at a constant speed.If the maximum frictional force from the road is 65% of the car's weight,what is the greatest speed at which the car can take the bend without skidding?
Can't help ya sorry :-X
Anyone ?! :-\
Post by: Assi1993 on December 25, 2010, 09:22:54 am
Post by: The Golden Girl =D on December 27, 2010, 11:52:58 am
1.I didn't Know how to do it , And I wanted to Know whether it's a Part of the Syllabus or Not ,That's All .
It Ain't A Part of the Syllabus *I'll ask my tutor soon iA About it for those who want to know*
Post by: Vampire-Love4ever on December 27, 2010, 11:57:50 am
I have some doubts. Please help me out here
Q1- Explain the experimental procedure to find centre of gravity of an irregular object.
Q2- If in a collision, the loss of kinetic energy is 0.08J.
Will it be elastic or inelastic collision?
Q3- How to use strobe photography or video camera to analyse motion?
Q4- What are coplanar vectors?
Q5- Application of mechanics for safety or to sports.
Q6- Explain the term of tensile/ compressive stress and breaking stress.
Post by: Vampire-Love4ever on December 27, 2010, 12:38:20 pm
Q7- A school treasure hunt uses the following set of directions:
A, 80m at 45 degree
B, 40m at 150 degree
C, 50m at 180 degree
D, 60m at 270 degree.
a) Team 1 follows the direction correctly. Draw a vector diagram to find their final displacement.
b) Team 2 gets the directions mixed up and follows them in the order C-B-D-A. What effect will this have on their final position compared to team 1?
Q8- A block is sliding down a plank at a constant speed. The weight of the block is 5N and the plank is 35 degree to the horizontal. Sketch a diagram for the three forces and use it to calculate the friction.
Q9- A ball is kicked off a flat with a horizontal velocity of 12ms^-1. How would you :
a) Calculate the horizontal distance travelled after it has fallen 15m?
b) Calculate the resultant velocity after 1.9s?
Q10- During a school play, a dagger is kicked and slides off the stage with a horizontal velocity of 1.5ms^-1. It lands on the floor a horizontal distance of 80cm away. Calculate the height of the stage.
Q11- What is the volume of a cylinder, of length 26cm and diameter 56mm.
Q12- Calculate the force applied to a surface of 220mm^2 to create a pressure of 10MPa.
Give your answer in kN.
Q13- Estimate average acceleration of a car travelling at 6mph in 8.0s.
Q14- A boat of mass 800kg is pulled along the sand at constant velocity by a cable attached to a winch. The cable is 15 degree to the horizontal.
Draw a free body force diagram for the boat. Label the forces: Tension, Friction, Normal reaction and weight.
Q15- If volume of water displaced for a person having weight 600N is 0.05m^3 in the Dead sea. How is this compared with the volume that needs to be displaced for the person to float in a swimming pool?
Q16- A spring of length 10cm extends to 12.5cm when a 8.0N weight is suspended from it. Calculate the spring constant of the spring.
Post by: Vampire-Love4ever on December 27, 2010, 12:42:57 pm
Post by: Malak on December 30, 2010, 02:32:53 pm
Thx
Post by: The Golden Girl =D on December 30, 2010, 03:27:35 pm
Hey can anyone Help with these Question?
Thx
The Teacher said its Something Completely New but we need to know it -.-
I hope the Pictures are Not Blurry =]
Attached .
Post by: The Golden Girl =D on December 30, 2010, 03:29:33 pm
Lift Weight = F
4.3 - 0.44 = 3.86 N
Don't Rep Me just Pray for me =]
Post by: Malak on December 30, 2010, 04:03:44 pm
Ohh I skipped One part :S
Lift Weight = F
4.3 - 0.44 = 3.86 N
Don't Rep Me just Pray for me =]
Thx alot :)
Inshallah we all will do good...! Ameen
I hav another Q By the way :P
Post by: The Golden Girl =D on December 30, 2010, 06:49:36 pm
Thx alot :)
Inshallah we all will do good...! Ameen
I hav another Q By the way :P
So Since they said HALF the time is spent Accelerating and HALF the Time is spent accelerating .
Acceleration .....Deceleration Means that HALF the Journey the Train was Accelerating and the Other HALF it was decelerating Hence Distance is HALVED too ;)
1. 5000 Km splits to :
2500,000 and 2500,000
2.3600 Seconds Splits to :
1,800 and 1,800
Okay Now we Start With the Calculation ;
S = u + 0.5at
2500,000 = 0 + a * (1,800)2
a = 2*2500,000/(1,800)2
a = 1.54 ms-2
Which is About 2 ms-2
Ameen =]
Post by: Vampire-Love4ever on January 02, 2011, 12:10:12 am
(i) Curved line passing through the origin.
(ii) Straight line not passing through origin.
Is Hooke's law obeyed in each case?
Post by: Vampire-Love4ever on January 02, 2011, 12:13:48 am
Post by: Vampire-Love4ever on January 02, 2011, 12:15:02 am
Post by: The Golden Girl =D on January 02, 2011, 08:32:10 am
Q3 :We are NOT going to Consider the Weight in our calculation since it's NOT in the direction of the force .
So , work done = force * distance in the direction of the force
= 20 N * (8*1000) m
= 160,000 J
Hence Answer is C .
Post by: The Golden Girl =D on January 02, 2011, 08:34:59 am
If an force-extension graph shows:
(i) Curved line passing through the origin.
(ii) Straight line not passing through origin.
Is Hooke's law obeyed in each case?
it MUST pass through the Origin and hence ii) is NOT correct.
it should be a straight line to show the fact that they are directly proportional and Hence i) is NOT correct *I'll let someone confirm this though*
Post by: Malak on January 02, 2011, 12:22:43 pm
it MUST pass through the Origin and hence ii) is NOT correct.
it should be a straight line to show the fact that they are directly proportional and Hence i) is NOT correct *I'll let someone confirm this though*
About passing through the origin..I thinkt the same too but then this question confuses me
Q19 2010 Jun
The question asks if hookes law is obeyed and in the Ms it says that it does obey hooke's law
Post by: elemis on January 02, 2011, 12:37:00 pm
About passing through the origin..I thinkt the same too but then this question confuses me
Q19 2010 Jun
The question asks if hookes law is obeyed and in the Ms it says that it does obey hooke's law
Thats not a force EXTENSION graph in the diagram. It shows data for force plotted against TOTAL length.
Hence, hookes is law is obeyed.
Post by: Vampire-Love4ever on January 02, 2011, 04:54:55 pm
Thats not a force EXTENSION graph in the diagram. It shows data for force plotted against TOTAL length.
Hence, hookes is law is obeyed.
So, they just showed the total length in the graph. Not the extension?
Post by: The Golden Girl =D on January 02, 2011, 07:34:07 pm
Post by: Vampire-Love4ever on January 02, 2011, 07:37:16 pm
first time they do that actually ,but yes.
Thanks for your help!
I have one more doubt please :$
Q- What is the relationship between the average velocity of the cyclist and the average velocity of the car for the time interval covered by the graph?
...
Post by: The Golden Girl =D on January 02, 2011, 07:45:14 pm
Thanks for your help!
I have one more doubt please :$
Q- What is the relationship between the average velocity of the cyclist and the average velocity of the car for the time interval covered by the graph?
...
My Answer is : They are the Same .
Why : it's because of the fact that they have the Same distance and same time .=]
Post by: Vampire-Love4ever on January 02, 2011, 08:00:21 pm
My Answer is : They are the Same .
Why : it's because of the fact that they have the Same distance and same time .=]
thanks so much!
May Allah bless you :D
Post by: The Golden Girl =D on January 02, 2011, 08:02:44 pm
And you too :)
Post by: Vampire-Love4ever on January 02, 2011, 08:34:53 pm
Post by: Vampire-Love4ever on January 02, 2011, 08:40:08 pm
My answer is different from the markscheme answer.
I get 2.95 but the mark scheme answer is 3.1
Does it make such a big difference?
Post by: Vampire-Love4ever on January 02, 2011, 09:05:02 pm
Post by: Vampire-Love4ever on January 02, 2011, 11:41:11 pm
it MUST pass through the Origin and hence ii) is NOT correct.
it should be a straight line to show the fact that they are directly proportional and Hence i) is NOT correct *I'll let someone confirm this though*
So, is hooke's law obeyed here? :S
Post by: The Golden Girl =D on January 03, 2011, 03:41:16 am
If you have noticed ; they said ALLOW BOTH marks for no , cuz they aren't proportional .
Why ; it's a Curve , it's not proportional , the K is not constant - as stated by the MS =]
I'll confirm this with someone though :)
Post by: The Golden Girl =D on January 03, 2011, 07:41:23 am
Please explain this, it's really confusing :S
Since it's moving Forward , this means F1 is the friction , and friction is caused by the road ....
Hence i) F1 is the force of the Road on the Wheel
Since it's moving forward ,the wheels go forward , and hence they are exerting a force on the Road in order to move forward.
Hence ii) F2 is the force of the Wheel on the Road
Let me know if you don't get it ;)
Post by: The Golden Girl =D on January 03, 2011, 07:47:53 am
Another question:
My answer is different from the markscheme answer.
I get 2.95 but the mark scheme answer is 3.1
Does it make such a big difference?
can you get me the range Please from the MS .
usually if they said Estimate , the Question will have a WIDE range of answers and hence it'll be fine I guess. =]
Post by: The Golden Girl =D on January 03, 2011, 07:58:25 am
How do we know that the contact force will be downwards?
The First part of the Question before this One asks you to draw a fully labelled free-boy diagram for the magnet M1 .
which is you draw a dot an Arrow comes out UPWARDS , this represents the Repulsion due to M2
another Arrow comes out DOWNWARDS , this represents the Weight *Pull of Earth*
This Means that ; When you draw the free-body Diagram of M2 you will do as follows* I draw it so that I don't get confused , I suggest u do it too ;) * :
Dot representing M2
an Arrow UPWARDS which represents the force exerted by the Wooden Base
another Arrow DOWNWARDS which represents the Repulsion from M1
And Hence Since it's Newton's third law Pair , the forced has to be reversed *kind of*
Since the Wooden Base acted UPWARDS , it's Pair will act DOWNWARDS
If you don't get it Let Me know ;)
Post by: Vampire-Love4ever on January 03, 2011, 08:30:42 am
Since it's moving Forward , this means F1 is the friction , and friction is caused by the road ....
Hence i) F1 is the force of the Road on the Wheel
Since it's moving forward ,the wheels go forward , and hence they are exerting a force on the Road in order to move forward.
Hence ii) F2 is the force of the Wheel on the Road
Let me know if you don't get it ;)
I did the same as you but got it wrong.
This is the markscheme answer.
Post by: Vampire-Love4ever on January 03, 2011, 08:34:28 am
The First part of the Question before this One asks you to draw a fully labelled free-boy diagram for the magnet M1 .
which is you draw a dot an Arrow comes out UPWARDS , this represents the Repulsion due to M2
another Arrow comes out DOWNWARDS , this represents the Weight *Pull of Earth*
This Means that ; When you draw the free-body Diagram of M2 you will do as follows* I draw it so that I don't get confused , I suggest u do it too ;) * :
Dot representing M2
an Arrow UPWARDS which represents the force exerted by the Wooden Base
another Arrow DOWNWARDS which represents the Repulsion from M1
And Hence Since it's Newton's third law Pair , the forced has to be reversed *kind of*
Since the Wooden Base acted UPWARDS , it's Pair will act DOWNWARDS
If you don't get it Let Me know ;)
JazakAllah Khair.
I didn't see the wooden base :$ I was thinking of the stand the whole time :P
Thanks alot!
Post by: Vampire-Love4ever on January 03, 2011, 08:37:25 am
can you get me the range Please from the MS .
usually if they said Estimate , the Question will have a WIDE range of answers and hence it'll be fine I guess. =]
Post by: aloha32 on January 03, 2011, 12:10:12 pm
I'm confused..What is the difference between the celcius scale and centigrade scale?
Thanks :)
Post by: The Golden Girl =D on January 03, 2011, 01:27:36 pm
(a)
(i) As the car is moving forward, there should be a force applied on the road so that the car can move froth, that force is applied by the wheel of the car, power (energy) is given by the engine.
(ii) There is a backward arrow, showing that there is some force acting backwards. This might be due to the friction of the road. This force is faced by the wheel, so it gets the speed.
Otherwise the car will NOT BE SEEN.
by fidato - my friend =]
Post by: Vampire-Love4ever on January 03, 2011, 02:24:07 pm
Hello!
I'm confused..What is the difference between the celcius scale and centigrade scale?
Thanks :)
Celsius and centrigrade are two names for essentially the same temperature scale (with slight differences). The centrigrade scale is divided into degrees based on dividing the temperature between which water freezes and boils into 100 equal gradients or degrees. The word centigrade comes from "centi-" for 100 and "grade" for gradients. The centigrade scale was introduced in 1744 and remained the primary scale of temperature until 1948. In 1948 the CGPM (Conference General des Poids et Measures) decided to standardize several units of measurement, including the temperature scale. Since the "grade" was in use as a unit (including the "centigrade"), a new name was chosen for the temperature scale: Celsius.
The Celsius scale remains a centigrade scale in which there are 100 degrees from the freezing point (0°C) and boiling point (100°C) of water, though the size of the degree has been more precisely defined. A degree Celsius (or a Kelvin) is what you get when divide the thermodynamic range between absolute zero and the triple point of a specific type of water into 273.16 equal parts. There is a 0.01°C difference between the triple point of water and the freezing point of water at standard pressure.
Source: http://chemistry.about.com/b/2010/09/20/difference-between-celsius-and-centigrade.htm
Please do correct me if I've posted something wrong. :)
Post by: Vampire-Love4ever on January 04, 2011, 08:21:14 am
the car thing ;
(a)
(i) As the car is moving forward, there should be a force applied on the road so that the car can move froth, that force is applied by the wheel of the car, power (energy) is given by the engine.
(ii) There is a backward arrow, showing that there is some force acting backwards. This might be due to the friction of the road. This force is faced by the wheel, so it gets the speed.
Otherwise the car will NOT BE SEEN.
by fidato - my friend =]
Thx. I get it now :)
Post by: Vampire-Love4ever on January 04, 2011, 08:25:01 am
Post by: Vampire-Love4ever on January 04, 2011, 08:38:06 am
Please!!!!
Post by: Vampire-Love4ever on January 04, 2011, 08:59:48 am
Post by: Vampire-Love4ever on January 04, 2011, 09:00:33 am
Post by: Vampire-Love4ever on January 04, 2011, 09:03:28 am
The Teacher said its Something Completely New but we need to know it -.-
I hope the Pictures are Not Blurry =]
Attached .
I still didn't get a part :(
Post by: Vampire-Love4ever on January 04, 2011, 09:05:25 am
Post by: elemis on January 04, 2011, 09:16:45 am
Jm-3 = Pa
kgm2s-2m-3 = kgm-1s-2
kgm-1s-2 = kgm-1s-2
Hence, it is homogenous.
Note that Strain has no units since it is a ratio. Additionally, 0.5 was ignored since it is a constant.
Know your base units.
Post by: The Golden Girl =D on January 04, 2011, 10:07:58 am
I still didn't get a part :(
You can't see the picture or you don't understand it ?
Post by: The Golden Girl =D on January 04, 2011, 10:24:50 am
Will my answer be enough for the 4 marks?
her is an Answer that'll get you FULL 4 marks ;)
ELASTIC ;
Elastic Deformation : when a copper is stretched upto elastic limit and then the stretching force is removed , the material will regain it's Original Shape and Size .
Elastic Region : Stress is Directly Proportional to Strain
PLASTIC ;
Plastic Deformation : When the Copper wire is stretched Beyond Elastic limit and if the force is removed , the material will have Permanent extension
Plastic Region : Stress is NOT Directly Proportional to Strain
Post by: The Golden Girl =D on January 04, 2011, 10:33:05 am
Explain please?
the Question is Wrong , I know it sounds Weird but it's supposed to be at Point B instead of C.
Explanation ;
For a Circular Path :
You Have to move AWAY from the center . You will be Pushed to the left side (direction of motion)
Let me know if you don't get it ?
Post by: The Golden Girl =D on January 04, 2011, 10:35:49 am
How to do the bii part????!!! =/
Graph : the Lines are STRAIGHT , but I couldn't manage to do that ,Sorry about that.
distance = about 100 -> based on my calculations , so I assume it's 150 ,i'll make it easier for us to draw. ;)
Attached .
Post by: Vampire-Love4ever on January 04, 2011, 10:53:39 am
You can't see the picture or you don't understand it ?
Don't understand :(
Post by: The Golden Girl =D on January 04, 2011, 10:56:35 am
And this too! Please! :$
Please!!!!
I think the Question is Incomplete , but I'll answer from my copy ;)
Weight is 750 N -> value from the Graph in the Question
Attached to show you how I calculated it.
Here's the Drawing:
An Arrow going UPWARDS showing R -> 250 how => 750 N - 300 N = 250 N
An Arrow DOWNWARDS which is LONGER *because she is crouching- i.e exerting MORE force # -> 750 N
# => Weight does NOT change unless he's in another planet.
Post by: The Golden Girl =D on January 04, 2011, 11:02:03 am
The Teacher said its Something Completely New but we need to know it -.-
I hope the Pictures are Not Blurry =]
Attached .
I've got FOUR forces ,that I want to reduce to THREE to be able to draw a vector diagram *in my words - triangle showing the forces*
Since the Lift's Arrow is LONGER than the Weight it means it's GREATER in terms of Value.
LIFT : Hence we can reduce it into ONE force which is the Lift, which is UPWARDS
the Drag to the RIGHT
the tension slightly to the LEFT
third Question : is another example so never mind.
Post by: Vampire-Love4ever on January 04, 2011, 11:03:32 am
the Question is Wrong , I know it sounds Weird but it's supposed to be at Point B instead of C.
Explanation ;
For a Circular Path :
You Have to move AWAY from the center . You will be Pushed to the left side (direction of motion)
Let me know if you don't get it ?
Answer is A.
Post by: The Golden Girl =D on January 04, 2011, 11:04:31 am
Answer is A.
Yes ;D
Post by: Vampire-Love4ever on January 04, 2011, 11:19:02 am
I think the Question is Incomplete , but I'll answer from my copy ;)
Weight is 750 N -> value from the Graph in the Question
Attached to show you how I calculated it.
Here's the Drawing:
An Arrow going UPWARDS showing R -> 250 how => 750 N - 300 N = 250 N
An Arrow DOWNWARDS which is LONGER *because she is crouching- i.e exerting MORE force # -> 750 N
# => Weight does NOT change unless he's in another planet.
700-300= 450N
Plus, I get R at 1.2s as 340N!
Post by: Vampire-Love4ever on January 04, 2011, 11:23:27 am
I've got FOUR forces ,that I want to reduce to THREE to be able to draw a vector diagram *in my words - triangle showing the forces*
Since the Lift's Arrow is LONGER than the Weight it means it's GREATER in terms of Value.
LIFT : Hence we can reduce it into ONE force which is the Lift, which is UPWARDS
the Drag to the RIGHT
the tension slightly to the LEFT
third Question : is another example so never mind.
Thanks! Got it now! :D
Post by: Vampire-Love4ever on January 04, 2011, 11:29:08 am
Graph : the Lines are STRAIGHT , but I couldn't manage to do that ,Sorry about that.
distance = about 100 -> based on my calculations , so I assume it's 150 ,i'll make it easier for us to draw. ;)
Attached .
Why is the speed maximum at 1.5s? :S
Post by: The Golden Girl =D on January 04, 2011, 11:36:37 am
Why is the speed maximum at 1.5s? :S
Where in the Graph is the Gradient STEEPEST , at about 1.5 , that's why.
when you solved a]i) I did it as follows ;
Vmax= delta distance/ elta time
= (80 - 20) / (1.6 -1.0)
= 100
l.6 -1.0 => 1.5 cm/s
Post by: The Golden Girl =D on January 04, 2011, 11:42:17 am
700-300= 450N
Plus, I get R at 1.2s as 340N!
me too :o
I'll let Someone look into this.
Post by: Vampire-Love4ever on January 04, 2011, 11:42:35 am
Where in the Graph is the Gradient STEEPEST , at about 1.5 , that's why.
when you solved a]i) I did it as follows ;
Vmax= delta distance/ elta time
= (80 - 20) / (1.6 -1.0)
= 100
l.6 -1.0 => 1.5 cm/s
I did: 120/(1.9-0.9)= 120cms^-1
So for ii part I use the maximum speed of120 at 1s. Will it be right?
Post by: Vampire-Love4ever on January 04, 2011, 11:44:11 am
me too :o
I'll let Someone look into this.
I'll be waiting :)
Post by: Vampire-Love4ever on January 04, 2011, 11:46:38 am
can you get me the range Please from the MS .
usually if they said Estimate , the Question will have a WIDE range of answers and hence it'll be fine I guess. =]
Did you get how to do this? =s
Post by: The Golden Girl =D on January 04, 2011, 12:03:46 pm
Did you get how to do this? =s
Attached .
This is weird 0.o -> well let someone look into it.
Post by: The Golden Girl =D on January 04, 2011, 12:12:26 pm
I did: 120/(1.9-0.9)= 120cms^-1
So for ii part I use the maximum speed of120 at 1s. Will it be right?
MS says :
Sketch graph
Shows:
Speed increasing from 0 and then decreases
Max speed at correct time (accept between 1.0 and 1.5 s) OR
correct magnitude (must be indicated)
Speed decreasing to 0 at between 3.4 and 4.0 s
so I guess it's right then :D
Post by: The Golden Girl =D on January 04, 2011, 12:36:25 pm
Post by: Vampire-Love4ever on January 04, 2011, 01:05:49 pm
It might take time ,it seems they're Not Online at the moment.
It's ok. I'll be waiting.
Thanks alot for your time and help. I'll pray for your exams ;)
Post by: Vin on January 04, 2011, 01:29:13 pm
Post by: Vampire-Love4ever on January 04, 2011, 01:33:59 pm
I'm confused. :/ Which questions have not been answered? Plus, please post the whole paper because I dont do edexcel.
Post by: The Golden Girl =D on January 04, 2011, 02:09:01 pm
Attached.
Post by: Vin on January 04, 2011, 02:26:08 pm
While crouching down, weight does not change. therefore, W = 750N
and, from graph, R = 340 N
Clearly there has been a printing error. You cant really rely on old past papers I think.
Arrow W is longer cuz the magnitude of the vector is greater than that of R.
Post by: The Golden Girl =D on January 04, 2011, 02:29:59 pm
Post by: Vin on January 04, 2011, 02:45:41 pm
For the second one, take two points, for instance check the attatchment.
It is advised that you select points from over more than half of the line.
I took, (1.1, 3) & (2.8, 8.4)
m = y2 - y1 / x2 - x1
= 8.4 - 3 / 2.8 - 1.1
= 3.18 x10^3
I believe CIE always gives and a range of values, though I'm pretty much not sure about edexcel.
Post by: Vampire-Love4ever on January 04, 2011, 09:23:27 pm
Weight of Person (W) = 750 N cuz the initial reading states the weight.
While crouching down, weight does not change. therefore, W = 750N
and, from graph, R = 340 N
Clearly there has been a printing error. You cant really rely on old past papers I think.
Arrow W is longer cuz the magnitude of the vector is greater than that of R.
Thanks alot for your help.
Post by: Vampire-Love4ever on January 04, 2011, 11:35:28 pm
Please! Please! Help me out if you can!
I am a week away from my exam :'(
Post by: Vampire-Love4ever on January 04, 2011, 11:40:07 pm
Answer from markscheme: Use of mgh 1
Vertical drop per second = (8.4 m) sin (3°) 1
3.9 × 102 J/Js–1/W
Post by: Vampire-Love4ever on January 04, 2011, 11:46:46 pm
Post by: Vampire-Love4ever on January 04, 2011, 11:48:26 pm
Please explain this :)
Post by: Vampire-Love4ever on January 04, 2011, 11:50:15 pm
Post by: Vampire-Love4ever on January 04, 2011, 11:53:13 pm
Post by: Vampire-Love4ever on January 04, 2011, 11:54:51 pm
Post by: Vampire-Love4ever on January 04, 2011, 11:57:05 pm
ii-How to calculate the tension? why do we use sin here instead of cos?
Post by: Vampire-Love4ever on January 05, 2011, 12:02:40 am
Post by: Vampire-Love4ever on January 05, 2011, 12:05:06 am
Post by: Vampire-Love4ever on January 05, 2011, 12:06:41 am
Post by: Vampire-Love4ever on January 05, 2011, 12:08:14 am
Please explain!
Post by: Vampire-Love4ever on January 05, 2011, 12:11:08 am
But, I really need help here!!!!
Much thanks in advance!
Pleaseeee Help!!!
Post by: elemis on January 05, 2011, 05:08:00 am
The ii part please?
You have calculated a resultant force of 5600 N for each truck in the first part, right ?
Now truck B exerts a force of 11200 N on A. Thus, A will exert an equal force in the opposite direction.
Since the resultant is 5600 N a force of 16800 N must act to the left. (16800 - 11200 = 5600)
The weight of the truck must act vertically downwards and hence a normal force equal to the weight will act upwards.
Post by: The Golden Girl =D on January 05, 2011, 05:33:46 am
I know I have posted alot of doubts! I am sorry about it!
But, I really need help here!!!!
Much thanks in advance!
Pleaseeee Help!!!
I'll try My best =]
first Drawing is Attached , let me know if you don't get it ;)
as for the second , You calculated the Distance and you got it 1.75 m
Attached as well.
Ohh and the displacement is - 1.27 m
Why : the whole drawing is under the line and hence Nothing Cancels out and so it's -1.27 m
Post by: The Golden Girl =D on January 05, 2011, 05:44:44 am
I really don't know how to draw the graph here! =s
Please explain!
Attached.
Post by: The Golden Girl =D on January 05, 2011, 05:58:34 am
Explain this question! =)
Not sure about the first answer,is the Answer A ?
Attached.
Post by: The Golden Girl =D on January 05, 2011, 06:03:02 am
The ii part. please?
There is NOTHING given , Could you Upload the first Part of the Question!
And if you're solving those sheets for materials and mechanics (exam wizard*I'd really appreciate it if you told me the Number *it's takes me a life time to find it*
Post by: The Golden Girl =D on January 05, 2011, 06:06:51 am
How to do draw C?
Attached.
Post by: The Golden Girl =D on January 05, 2011, 06:27:51 am
How to draw the graph?
Attached.
Post by: The Golden Girl =D on January 05, 2011, 06:37:02 am
i-i know how to do it!
ii-How to calculate the tension? why do we use sin here instead of cos?
Attached.
Post by: The Golden Girl =D on January 05, 2011, 06:38:52 am
Please help!
Indeed it is :D
Post by: The Golden Girl =D on January 05, 2011, 06:53:50 am
How to do this, plz?
Attached.
I could don't Solve the Question you first Posted.
I'm Sorry But I really Have to go if you have Any further doubts when I'm Not ONLINE - send a mesg of a link of the page that has the Q , and ask for help from *fidato/Vin/Ari Ben Canaan*
Good Luck ;)
Post by: Vampire-Love4ever on January 05, 2011, 05:31:32 pm
You have calculated a resultant force of 5600 N for each truck in the first part, right ?
Now truck B exerts a force of 11200 N on A. Thus, A will exert an equal force in the opposite direction.
Since the resultant is 5600 N a force of 16800 N must act to the left. (16800 - 11200 = 5600)
The weight of the truck must act vertically downwards and hence a normal force equal to the weight will act upwards.
Thank you.
If B exerts 11200N on A which summed with the resultant force 5600N = 16800N. Right?
Why is the weight showed as 840000N? =S
Post by: Vampire-Love4ever on January 05, 2011, 05:39:57 pm
I'll try My best =]
first Drawing is Attached , let me know if you don't get it ;)
as for the second , You calculated the Distance and you got it 1.75 m
Attached as well.
Ohh and the displacement is - 1.27 m
Why : the whole drawing is under the line and hence Nothing Cancels out and so it's -1.27 m
I didnt get how to draw the displacement time graph!! :S
Post by: Vampire-Love4ever on January 05, 2011, 05:52:12 pm
Attached.
How do we know that it must be drawn like this? :S
Post by: Vampire-Love4ever on January 05, 2011, 05:56:33 pm
Attached.
Shouldn't it be less steep because it has a smaller young modulus? :S
Post by: Vampire-Love4ever on January 05, 2011, 05:58:40 pm
There is NOTHING given , Could you Upload the first Part of the Question!
And if you're solving those sheets for materials and mechanics (exam wizard*I'd really appreciate it if you told me the Number *it's takes me a life time to find it*
No, I took them from a solomon user site :S
I am so sorry to waste your time like this :(
Post by: Vampire-Love4ever on January 05, 2011, 06:09:16 pm
Not sure about the first answer,is the Answer A ?
Attached.
I am not sure about the markscheme. I guessed it as A too. because it has greatest stress.
Post by: Vampire-Love4ever on January 05, 2011, 06:51:21 pm
There is NOTHING given , Could you Upload the first Part of the Question!
And if you're solving those sheets for materials and mechanics (exam wizard*I'd really appreciate it if you told me the Number *it's takes me a life time to find it*
Post by: The Golden Girl =D on January 05, 2011, 06:52:22 pm
How do we know that it must be drawn like this? :S
Here you go :
Attached.
Post by: The Golden Girl =D on January 05, 2011, 07:08:06 pm
https://studentforums.biz/revison-notes/important-graphs-you-must-know-for-physics-edexcel-unit-1-!!!!/
Post by: Vampire-Love4ever on January 05, 2011, 07:14:35 pm
check this :
https://studentforums.biz/revison-notes/important-graphs-you-must-know-for-physics-edexcel-unit-1-!!!!/
JazakAllah Khair!!!!
THANK YOU SOOOOOOO MUCH!! =D
ps- I am sorry but from my doubts Q108, Q120, Q121 AND Q158 haven't been attempted. Please help me out?
Post by: The Golden Girl =D on January 05, 2011, 07:25:53 pm
I'll try My best =]
first Drawing is Attached , let me know if you don't get it ;)
as for the second , You calculated the Distance and you got it 1.75 m
Attached as well.
Ohh and the displacement is - 1.27 m
Why : the whole drawing is under the line and hence Nothing Cancels out and so it's -1.27 m
Attached.
girl - got easy on me , since I will sleep late tonight I'll try my best to be done with them for you =]
Post by: The Golden Girl =D on January 05, 2011, 07:35:24 pm
Attached.
Post by: The Golden Girl =D on January 05, 2011, 07:47:36 pm
Attached.
Let me know if you don't get it. *I'm not sure about this One ,to be honest*
I'm sorry It's COMPLETELY Wrong .
Attached a Picture.
Post by: The Golden Girl =D on January 05, 2011, 07:51:58 pm
Shouldn't it be less steep because it has a smaller young modulus? :S
I'll contact someone About the ones I am Not sure of including this one.
Post by: The Golden Girl =D on January 05, 2011, 07:57:21 pm
I have many doubts and I am totally relying on this thread.
Please! Please! Help me out if you can!
I am a week away from my exam :'(
Attached.
Post by: The Golden Girl =D on January 05, 2011, 08:22:14 pm
Please help!
Attached.
Post by: The Golden Girl =D on January 05, 2011, 08:46:23 pm
How to do this, plz?
I didn't Notice this one.
Attached.
Post by: The Golden Girl =D on January 05, 2011, 08:54:40 pm
Attached.
I could don't Solve the Question you first Posted.
I'm Sorry But I really Have to go if you have Any further doubts when I'm Not ONLINE - send a mesg of a link of the page that has the Q , and ask for help from *fidato/Vin/Ari Ben Canaan*
Good Luck ;)
check this one - I modified it.
Post by: The Golden Girl =D on January 05, 2011, 08:57:55 pm
So can you Upload it for the Last time. -> I'm going to go ask for help ,Plus I've got to Sleep or I might as well wake up with a HUGE headache.
Salam , Pray for me =]
And let me know if you don't get anything, alright ;)
Post by: Vampire-Love4ever on January 05, 2011, 09:04:27 pm
for Question 142]ii)
Attached.
How do we get the weight here? :S
Post by: The Golden Girl =D on January 05, 2011, 09:08:40 pm
How do we get the weight here? :S
Weight = Mass * gravity
= 84,000 Kg * 9.81 in this case We took it as 10 m/s^2 it's the same
= 840,000 N
if we used 9.81 Answer is 824,040 N almost the same
Post by: Vampire-Love4ever on January 05, 2011, 09:14:23 pm
Post by: Vampire-Love4ever on January 05, 2011, 09:15:30 pm
Weight = Mass * gravity
= 84,000 Kg * 9.81 in this case We took it as 10 m/s^2 it's the same
= 840,000 N
if we used 9.81 Answer is 824,040 N almost the same
THANKS SO MUCH! =D
Post by: Vampire-Love4ever on January 05, 2011, 09:32:32 pm
Attached.
How to do this graph?
Post by: The Golden Girl =D on January 05, 2011, 09:34:43 pm
Decreasing gradient ;)
Post by: The Golden Girl =D on January 05, 2011, 09:35:46 pm
Post by: Vampire-Love4ever on January 05, 2011, 09:37:02 pm
Salam For Now =]
Thanks ! :)
Good night! :D
Sleep well!
God bless.
xx
Post by: Deadly_king on January 06, 2011, 05:18:55 am
Can you specify your doubts, please?
I'll try to help as much as I can. :)
Post by: The Golden Girl =D on January 06, 2011, 05:35:14 am
Am sorry but am being confused here :-[
Can you specify your doubts, please?
I'll try to help as much as I can. :)
Click on this Quote and you'll see the Qs we need help in :)
Posted them again.
Thanks =]
Post by: The Golden Girl =D on January 06, 2011, 07:51:24 am
Attached.
Post by: The Golden Girl =D on January 06, 2011, 07:59:50 am
Post by: Vin on January 06, 2011, 02:16:45 pm
How to do draw C?
[Attached]
Remember,
Young Modulus = stress / strain
therefore, the gradient of the graph determines the young modulus.
The question says :
C has smaller modulus than A & B, so the gradient is the least/
C is stronger than A & B, so the maximum point of the graph (stress and strain) should be greater than A&B
C is a brittle substance, that means when crossed the limit of proportionality, the substance breaks, so the line C will have an abrupt end. (< This is the property of the graphs for brittle substances - remember it, it'll come in handy with other ques as well.
Post by: Vampire-Love4ever on January 06, 2011, 03:01:20 pm
[Attached]
Remember,
Young Modulus = stress / strain
therefore, the gradient of the graph determines the young modulus.
The question says :
C has smaller modulus than A & B, so the gradient is the least/
C is stronger than A & B, so the maximum point of the graph (stress and strain) should be greater than A&B
C is a brittle substance, that means when crossed the limit of proportionality, the substance breaks, so the line C will have an abrupt end. (< This is the property of the graphs for brittle substances - remember it, it'll come in handy with other ques as well.
Thank you.
Post by: Vampire-Love4ever on January 08, 2011, 06:08:23 pm
The horse is attached to a point on the barge 2m from the bank. As the barge starts to move, the
tension in the rope is 500N. Calculate the barge's initial acceleration parallel to the bank.
Post by: astarmathsandphysics on January 08, 2011, 07:52:56 pm
Post by: Vampire-Love4ever on January 08, 2011, 09:28:17 pm
Is that right?
Please correct me if I am wrong.
Post by: astarmathsandphysics on January 08, 2011, 09:29:42 pm
Post by: Vampire-Love4ever on January 08, 2011, 09:52:17 pm
Yes gradient is Young's modulus
Thank you for your help.
Post by: astarmathsandphysics on January 08, 2011, 10:16:00 pm
Blogs dont work on freeexampapers/courseworkbak/astarmathsandphysics
theeducationchannel is down
astarmathsandphysics was hacked over xmas
On the other hand my server is fixed
So thats all right then
Post by: Vampire-Love4ever on January 08, 2011, 10:56:30 pm
No probs. Always something going wrong.
Blogs dont work on freeexampapers/courseworkbak/astarmathsandphysics
theeducationchannel is down
astarmathsandphysics was hacked over xmas
On the other hand my server is fixed
So thats all right then
Hope everything gets better :)
Post by: Vampire-Love4ever on January 08, 2011, 11:08:37 pm
acceleration of 2ms^-2.
a-With what force must the car pull the caravan? What motive force must the
engine provide? [caravan has a drag force of 400N and the car has a drag force of 200N]
b- The engine of the car can provide a maximum motive force of 6000N. Assuming that the drag
forces are unchanged. Calculate the maximum acceleration which can be achieved?
c- The force which the car exerts on the caravan at this acceleration.
answers: a- 4200N, b- 3ms^-2 and c- 2800N.
Please explain how to do this question.
Post by: zetsuboushita on January 09, 2011, 03:38:43 am
a=2m/s^2, (caravan 800kg)---(car 1000kg)---->Engine force
400N<--- 200N<---
Resolve vertically,
Engine force -200-400=(1800)(2)
Engine force=4200N
b)Given engine force = 6000N
Sub into above equation, 6000-200-400=(1800)(a)
a=3 m/s^2
c) Consider caravan,
a=3 m/s^2
400N<---(caravan 800kg)--->Tension of tow bar(i.e. force car acting on caravan)
Resolve vertically,
T-400=(800)(3)
T=2800N
Post by: Vampire-Love4ever on January 09, 2011, 12:15:12 pm
a) consider this:-
a=2m/s^2, (caravan 800kg)---(car 1000kg)---->Engine force
400N<--- 200N<---
Resolve vertically,
Engine force -200-400=(1800)(2)
Engine force=4200N
b)Given engine force = 6000N
Sub into above equation, 6000-200-400=(1800)(a)
a=3 m/s^2
c) Consider caravan,
a=3 m/s^2
400N<---(caravan 800kg)--->Tension of tow bar(i.e. force car acting on caravan)
Resolve vertically,
T-400=(800)(3)
T=2800N
For part b- why dont we minus 6000 from 4200N?
Post by: Vampire-Love4ever on January 09, 2011, 08:15:56 pm
Use the graph to estimate the maximum height reached by the rocket.
[please ignore the numbers in the image attached].
Post by: Vampire-Love4ever on January 09, 2011, 08:20:20 pm
Please answer part c of the question.
Post by: Vampire-Love4ever on January 09, 2011, 08:24:10 pm
(i)- If lap takes 3mins, show that the length of the track is 5km. I know this part :)
(ii)- What is the magnitude of displacement of the car after 1.5mins? <--- Explain how to do this, please!
Post by: Vampire-Love4ever on January 09, 2011, 09:21:51 pm
Post by: astarmathsandphysics on January 09, 2011, 10:34:59 pm
Post by: astarmathsandphysics on January 09, 2011, 10:37:07 pm
5000=pi*diamter so diameter=5000/pi=1591m
Post by: Vampire-Love4ever on January 10, 2011, 05:25:42 pm
after 1.5 minc car has travelled halfway around the track so dsplacement =diammeter
5000=pi*diamter so diameter=5000/pi=1591m
How do we know that we must use this?
Post by: blackberry on January 10, 2011, 06:00:44 pm
How do we know that we must use this?
you cant use 2.5km cuz that wud be the distance travelled not displacement.
they also mention that the track is circular so 5km is basically the circumference of the track
circumference formula--> C=2(pi)r
wen you solve the equation you get the radius as about 796 m. Double this to get diameter.
this will be the displacement.
Post by: blackberry on January 10, 2011, 06:01:56 pm
what is the difference between air resistance and viscous drag???
Post by: The Golden Girl =D on January 10, 2011, 06:05:49 pm
if there is a balloon going UPWARDS , then the force opposing the motion is either viscous drag *the balloon is in a MEDIUM which is AIR* or air resistance.
I think they're the same. CORRECT me if I'm wrong though.
Post by: blackberry on January 10, 2011, 06:13:53 pm
it's almost the same. they both are OPPOSING forces.
if there is a balloon going UPWARDS , then the force opposing the motion is either viscous drag *the balloon is in a MEDIUM which is AIR* or air resistance.
I think they're the same. CORRECT me if I'm wrong though.
hey so can air resistance act under water since there's air under water as well or will it just be called viscouse drag???
oooo do we have to learn the formula for terminal velocity of stuff in fluids or is it given in the paper??
Post by: The Golden Girl =D on January 10, 2011, 06:22:01 pm
hey so can air resistance act under water since there's air under water as well or will it just be called viscouse drag???
oooo do we have to learn the formula for terminal velocity of stuff in fluids or is it given in the paper??
1. Air Resistance has the Word AIR in it , which means it can't be AIR resistance if the object in in a liquid.
2. to be honest my teacher is ALWAYS Emphasizing on learning formulas.
I just checked June 2009 and the formula for the viscous drag*the one that has terminal velocity in it * is THERE ! but in my case ,I have Know it just like I know my name.
there is one that you might Not know though you MUST know it which is :
Terminal Velocity = {(2*r2*g) / (9 *eita) } * (Density of Solid - Density of Liquid)
Unit -> m/s ;)
so ya :)
Post by: blackberry on January 10, 2011, 06:25:02 pm
Post by: The Golden Girl =D on January 10, 2011, 06:28:52 pm
your welcome =]
Post by: Vampire-Love4ever on January 10, 2011, 08:38:41 pm
you cant use 2.5km cuz that wud be the distance travelled not displacement.
they also mention that the track is circular so 5km is basically the circumference of the track
circumference formula--> C=2(pi)r
wen you solve the equation you get the radius as about 796 m. Double this to get diameter.
this will be the displacement.
The answer is 1.59km.
Method given:
Circumference= pi*D
5km= pi*D
D= 5km/pi = 1.59km
But I dont know why we must use this. :/
Post by: Vampire-Love4ever on January 10, 2011, 08:43:34 pm
I read from one of my photocopied notes that : the diameter of a wire should be measured using a micrometer screw gauge and calculate area using formula = 1/4*pi*d^2.
Is it right if we use formula: pi*r^2 instead? :S
Post by: Vampire-Love4ever on January 10, 2011, 11:30:20 pm
Please help!
Post by: Deadly_king on January 11, 2011, 04:54:14 am
I have a doubt.
I read from one of my photocopied notes that : the diameter of a wire should be measured using a micrometer screw gauge and calculate area using formula = 1/4*pi*d^2.
Is it right if we use formula: pi*r^2 instead? :S
Sure it,s the same formula. You just changed the values.
d = 2r ------> That's all you need to take care, otherwise the formula remains the same.
Good luck for the exams. :)
Post by: Vampire-Love4ever on January 11, 2011, 09:58:57 am
Sure it,s the same formula. You just changed the values.
d = 2r ------> That's all you need to take care, otherwise the formula remains the same.
Good luck for the exams. :)
=O Ok.
Thank you so much for replying! :D
Post by: Malak on January 11, 2011, 10:36:17 am
thx
Post by: The Golden Girl =D on January 11, 2011, 10:57:06 am
Can anyone help solve this Question..
thx
since the rate of flow = 0.13 ms-1 .... it means the Volume = 0.13 m
Mass/time = 0.13 *1000 = 130 Kg
Gravitational Potential Energy = mgh
= 130 * 9.81 * 30 ...blue is the Delta h *change in height*
= 38,259 J
Power = Work done or engery / Time
= 38,259 / 1
= 38,259 J
Post by: Assi1993 on January 11, 2011, 12:24:17 pm
can some1 plz tell me that how will we draw the graph for this???
thanks in advance.....!
Post by: Deadly_king on January 11, 2011, 12:41:36 pm
hello
can some1 plz tell me that how will we draw the graph for this???
thanks in advance.....!
You just need to know this formula ----> F = k
From the second equation,
In the second case, we can note that e is inversely proportional to cross-sectional area. So by multiplying the area by 3, we need to reduce the extension by 3 (
In both cases same type of curve applies. :)
Hope it helps :D
Post by: Assi1993 on January 11, 2011, 02:32:08 pm
thankYou soooo much!!!!! :)
Post by: Vampire-Love4ever on January 11, 2011, 02:50:55 pm
Post by: Vampire-Love4ever on January 11, 2011, 02:54:27 pm
(i) Determine the object's final displacement from it's point of release from the balloon. [2]
Mark scheme answer:
rise 5m and fell 5m [1]
displacement 40m before point of release [1]
Post by: Malak on January 11, 2011, 02:57:02 pm
why do we add the drag and not minus it???
i think thats because u need to find the Total FORCE from which when u minus the resultant, you will get the drag=430 N..
Hope u get it?
Post by: Vampire-Love4ever on January 11, 2011, 02:57:32 pm
Post by: Vampire-Love4ever on January 11, 2011, 03:03:59 pm
Isn't stoke's law applicable to object in laminar flow? ? ?
Post by: The Golden Girl =D on January 11, 2011, 03:08:05 pm
why isnt the answer B ?
Isn't stoke's law applicable to object in laminar flow? ? ?
To be honest I never heard of stoke's law applying to a certain thing ONLY. Which is Why I think it shouldn't be B *I'll ask for back up :) *
Post by: Vampire-Love4ever on January 11, 2011, 03:10:10 pm
To be honest I never heard of stoke's law applying to a certain thing ONLY. Which is Why I think it shouldn't be B *I'll ask for back up :) *
In my Edexcel revision guide it's written: Stoke's law appiled to small spherical object moving at low speed in laminar flow.
I agree with A but I am lil confused cos I am stuck with both A and B. :/
Post by: Malak on January 11, 2011, 03:14:40 pm
In my Edexcel revision guide it's written: Stoke's law appiled to small spherical object moving at low speed in laminar flow.
I agree with A but I am lil confused cos I am stuck with both A and B. :/
By laminar flow i think it means for the velocity to be terminal..
And as it says SMALL speherical objects...the answer is possibly A
i dnt knw..Mayb A is more reasonable thts y
Post by: blackberry on January 11, 2011, 03:15:34 pm
i tried finidng the answers to these questions online but i cudnt find them.
Post by: astarmathsandphysics on January 11, 2011, 03:15:47 pm
Working vertically, v^2 =u^2 +2as
s=-21
u=0
v=?
a=-9.8
t=?
s=1/2at^2 so t=sqrt(2s/g) =sqrt(2*21/9.7)=2.07s
Post by: astarmathsandphysics on January 11, 2011, 03:17:08 pm
Post by: blackberry on January 11, 2011, 04:20:46 pm
how to do part c???
consider the 2nd image here, so the vertical distance from which it is launched is 16+4+1=21m
s=ut + .5at^2
21=(0xt) + (.5x9.81xt^2)
t=2.07s
Post by: Vampire-Love4ever on January 11, 2011, 04:42:50 pm
evn i selected B, vampire are you sure the answer is A??
i tried finidng the answers to these questions online but i cudnt find them.
Yupe, it is A!
Post by: Vampire-Love4ever on January 11, 2011, 04:44:29 pm
AND OTHERS, THANKS FOR UR HELP =]
Post by: The Golden Girl =D on January 11, 2011, 06:16:07 pm
Q-A hot air ballon is rising vertically at a speed of 10ms^-1. An object is released from the balloon. Graph shows how the velocity if the object varies with time from when it leaves the balloon to when it reaches the ground 4s later. It's assumed that the air resistance is neglected.
(i) Determine the object's final displacement from it's point of release from the balloon. [2]
Mark scheme answer:
rise 5m and fell 5m [1]
displacement 40m before point of release [1]
i] area/distance = 0.5*b*h
= 0.5 * 1*10
= 5m
ii] total distance = A1 + A2 = 5 + (0.5*3*30*)
= 5 + 45
= 50 m
b] -45 + 5 = -40 m
hence 40 m below the level of release.
Post by: Vampire-Love4ever on January 11, 2011, 07:29:21 pm
i] area/distance = 0.5*b*h
= 0.5 * 1*10
= 5m
ii] total distance = A1 + A2 = 5 + (0.5*3*30*)
= 5 + 45
= 50 m
b] -45 + 5 = -40 m
hence 40 m below the level of release.
Thanks. I didnt get this point: rise 5m and fell 5m [1]
How can it be rising and falling 5m? :S
Plus, please help with this question:
Why do we add the drag force here? Shouldn't we minus it?
Post by: Vampire-Love4ever on January 11, 2011, 11:43:32 pm
The projectile is fired from a height of 21mwhere is s= 21m? =S
Working vertically, v^2 =u^2 +2as
s=-21
u=0
v=?
a=-9.8
t=?
s=1/2at^2 so t=sqrt(2s/g) =sqrt(2*21/9.7)=2.07s
Post by: The Golden Girl =D on January 12, 2011, 04:40:41 am
drag Q : I think we should minus it *since drag ALWAYS opposes the motion ,that's weird. :-X *
I think.
Post by: blackberry on January 12, 2011, 01:48:22 pm
a) F= ma so find out F. Now F here is actually the resultant force (RF) NOT the driving force, since the general formula for finding RF=ma. So, RF= Driving Force - Drag. Rearrange the equation to find driving force: Driving Force= RF + Drag
b) Find F using F= ma. Here we consider F as the Forward Force but NOT the Resultant Driving Force. To find the Resultant Driving Force use RF= Forward Force - Drag. Notice that Drag is acting in the opposite direction and so drag will be taken as a negative value. In other words RF= Forward Force - ( - Drag) which is equal to RF= Forward Force + Drag.
However, i am not too sure if the 2nd method is acceptable so just stick to the 1st method. But if you get stuck you could use the 2nd method to help you out!!! :) :)
Post by: elemis on January 12, 2011, 01:58:39 pm
Thanks. I didnt get this point: rise 5m and fell 5m [1]
How can it be rising and falling 5m? :S
Plus, please help with this question:
Why do we add the drag force here? Shouldn't we minus it?
Using F = ma
Driving force - Drag = mass * acceleration
X - 430 = 520*3.5
X = 520*35 + 430
X = 2250 N
Post by: Vampire-Love4ever on January 12, 2011, 04:06:01 pm
hey vampire i found out why drag is added instead of substracted. There are 2 ways to do this:
a) F= ma so find out F. Now F here is actually the resultant force (RF) NOT the driving force, since the general formula for finding RF=ma. So, RF= Driving Force - Drag. Rearrange the equation to find driving force: Driving Force= RF + Drag
b) Find F using F= ma. Here we consider F as the Forward Force but NOT the Resultant Driving Force. To find the Resultant Driving Force use RF= Forward Force - Drag. Notice that Drag is acting in the opposite direction and so drag will be taken as a negative value. In other words RF= Forward Force - ( - Drag) which is equal to RF= Forward Force + Drag.
However, i am not too sure if the 2nd method is acceptable so just stick to the 1st method. But if you get stuck you could use the 2nd method to help you out!!! :) :)
Thanks. Though I exam got over today already. ;)
Post by: blackberry on January 13, 2011, 05:23:16 pm
Post by: 3bood on January 15, 2011, 06:25:43 pm
the diameter of wire A is twice that of B. calculate the ratio of the drift velocity in wire A to the drift velocity in wire B and explain your answer.
urgent plz because phy unit 2 is on monday :( :(
thx in advance ;D ;D ;D ;D
Post by: GodOfWar on January 23, 2011, 01:42:48 pm
http://www.scribd.com/doc/46124612/Physics-Jun-2010-Actual-Exam-Paper-Unit-4
Question 4
The answer's D, but how do you get there?
It's June 2010 Unit 4 By the way
Post by: elemis on January 23, 2011, 02:09:16 pm
I have a question :
http://www.scribd.com/doc/46124612/Physics-Jun-2010-Actual-Exam-Paper-Unit-4
Question 4
The answer's D, but how do you get there?
It's June 2010 Unit 4 By the way
Ek = 0.5mv2
P2/2m = 0.5mv2
Moving the 0.5 to the other side :
P2/2m ÷ 0.5 = mv^2
Hence, simplifying to the answer in D.
Post by: GodOfWar on January 23, 2011, 04:09:22 pm
What does the word derivation mean here? I thought we had use derivatives ???
Post by: elemis on January 23, 2011, 04:26:31 pm
What does the word derivation mean here? I thought we had use derivatives ???
You can derive a formula from an already known relationship.
Derive = Prove in this case.
Post by: GodOfWar on January 26, 2011, 06:15:37 pm
Post by: The Golden Girl =D on February 14, 2011, 02:42:35 pm
Attached.
than you in advance =]
Post by: elemis on February 14, 2011, 03:42:01 pm
Hey can someone explain to me what is in the Photo but in a bit more detail.
Attached.
than you in advance =]
On it.
Post by: elemis on February 14, 2011, 03:47:12 pm
The path difference is the difference in distance covered by the waves travelling from the source to the screen.
If Wave A from speaker A travels 10 m to get to a point on a screen and Wave B from speaker B travels 20 m to get to the SAME POINT, the path difference is 10 m. P.D. is usually expressed in terms of wavelengths. If the wavelengths of A and B were 5m then the P.D. can be said to be 2 lambda.
When the P.D = 0 phase difference = 0 hence constructive interference occurs i.e. the two waves superimpose such that the vector sum of their amplitudes leads to a doubling of amplitude.
When P.D. = whole no. of wavelengths constructive interference occurs.
When P.D. = non-integer, half wavelengths you have destructive interference.
Post by: The Golden Girl =D on February 14, 2011, 03:52:12 pm
Post by: The Golden Girl =D on February 19, 2011, 08:00:32 am
Well there is this little thing I don't get :S Which is the following sentence ;
the glass prism offers LESS refractive index for RED color which has a GREATER wavelength and offers LARGER refractive index for VIOLET color having SMALLER wavelength.
it's is the SAME prism *GLASS prism* so how can it give slightly differenct refractive indices for different colors o.0 ?!..... I know the idea of refractive index being INVERSELY proportional to wavelength but I just don't get the refractive index being different for different colors part.
Thank you in advance.
Post by: elemis on February 19, 2011, 08:51:40 am
Dispersion of light :
Well there is this little thing I don't get :S Which is the following sentence ;
the glass prism offers LESS refractive index for RED color which has a GREATER wavelength and offers LARGER refractive index for VIOLET color having SMALLER wavelength.
it's is the SAME prism *GLASS prism* so how can it give slightly differenct refractive indices for different colors o.0 ?!..... I know the idea of refractive index being INVERSELY proportional to wavelength but I just don't get the refractive index being different for different colors part.
Thank you in advance.
Check your source again. I think you've misread/misunderstood the given statement.
Post by: Amelia on February 19, 2011, 09:43:48 am
Dispersion of light :
Well there is this little thing I don't get :S Which is the following sentence ;
the glass prism offers LESS refractive index for RED color which has a GREATER wavelength and offers LARGER refractive index for VIOLET color having SMALLER wavelength.
it's is the SAME prism *GLASS prism* so how can it give slightly differenct refractive indices for different colors o.0 ?!..... I know the idea of refractive index being INVERSELY proportional to wavelength but I just don't get the refractive index being different for different colors part.
Thank you in advance.
I found something. :P I do hope this explains it.
The index of refraction values are dependent upon the frequency of light. For visible light, the 'n' (refractive index) value does not show a large variation with frequency, but nonetheless it shows a variation. For instance, the 'n' value for frequencies of violet light is 1.53; and the 'n' value for frequencies of red light is 1.51. The absorption and reemission process causes the higher frequency (lower wavelength) violet light to travel slower through crown glass than the lower frequency (higher wavelength) red light. It is this difference in 'n' value for the varying frequencies (and wavelengths) that causes the dispersion of light by a triangular prism. Violet light, being slowed down to a greater extent by the absorption and reemission process, refracts more than red light. Upon entry of white light at the first boundary of a triangular prism, there will be a slight separation of the white light into the component colors of the spectrum. Upon exiting the triangular prism at the second boundary, the separation becomes even greater and ROYGBIV is observed in its splendor.
http://www.physicsclassroom.com/class/refrn/u14l4a.cfm
:)
Post by: The Golden Girl =D on February 19, 2011, 11:38:27 am
I found something. :P I do hope this explains it.
The index of refraction values are dependent upon the frequency of light. For visible light, the 'n' (refractive index) value does not show a large variation with frequency, but nonetheless it shows a variation. For instance, the 'n' value for frequencies of violet light is 1.53; and the 'n' value for frequencies of red light is 1.51. The absorption and reemission process causes the higher frequency (lower wavelength) violet light to travel slower through crown glass than the lower frequency (higher wavelength) red light. It is this difference in 'n' value for the varying frequencies (and wavelengths) that causes the dispersion of light by a triangular prism. Violet light, being slowed down to a greater extent by the absorption and reemission process, refracts more than red light. Upon entry of white light at the first boundary of a triangular prism, there will be a slight separation of the white light into the component colors of the spectrum. Upon exiting the triangular prism at the second boundary, the separation becomes even greater and ROYGBIV is observed in its splendor.
http://www.physicsclassroom.com/class/refrn/u14l4a.cfm
:)
lia -Thanks it helped me link my ideas =D :-*
Ari - nope it's correct but it seems I forgot that there is the equation ;
lamda 1/lamda 2 = refractive index ..meaning wavelength is what varies the refractive in this case.
lamda 1 -> in air/vaccum
lamda 2 -> in medium
Post by: mayisha on March 27, 2011, 10:06:57 pm
Post by: red_911 on April 23, 2011, 05:37:52 pm
thanksss alot in advance
Post by: Aadeez || Zafar on April 24, 2011, 06:18:00 am
hey can anyone tell me the approximate wavelength of all the electromagnetic radiation ? there are different values written on different websites ...see this it is the stansard one
thanksss alot in advance
(http://www.kollewin.com/EX/09-15-03/electromagnetic-spectrum.jpg)
here
http://www.kollewin.com/EX/09-15-03/electromagnetic-spectrum.jpg (http://www.kollewin.com/EX/09-15-03/electromagnetic-spectrum.jpg)
Post by: Vampire-Love4ever on April 25, 2011, 06:22:46 am
Please help :)
Post by: Vampire-Love4ever on April 25, 2011, 07:29:02 am
Post by: Mike Benn on April 25, 2011, 10:52:32 am
Much Appreciated if u Helped/..
Post by: astarmathsandphysics on April 25, 2011, 03:25:08 pm
could be to do with emmission spectra
radon decays by alpha emmision. The alpha particles pass through the glass and collect at the top of the tube where they are neutralised and become helium.
. It is subjected to high voltage and becomess ionised. When neutralised the electrons emit characteristic radiation as the drop to the lowest energy level, which is detected.
Post by: Vampire-Love4ever on April 25, 2011, 07:28:10 pm
Vampire love4ever
could be to do with emmission spectra
radon decays by alpha emmision. The alpha particles pass through the glass and collect at the top of the tube where they are neutralised and become helium.
. It is subjected to high voltage and becomess ionised. When neutralised the electrons emit characteristic radiation as the drop to the lowest energy level, which is detected.
Thanks alot!
What about my second post? Please help me with that as well!
[For the 2nd Question: How do we know that the electrons are pushed towards the top-right (PQ)? I tried using Fleming's Left Hand Rule & I still couldn't get it :S]
Please help!!!!!!!!!
Post by: Vampire-Love4ever on April 25, 2011, 07:42:14 pm
Vampire love4ever
could be to do with emmission spectra
radon decays by alpha emmision. The alpha particles pass through the glass and collect at the top of the tube where they are neutralised and become helium.
. It is subjected to high voltage and becomess ionised. When neutralised the electrons emit characteristic radiation as the drop to the lowest energy level, which is detected.
How does it get neutralized and change to Helium? :S
Post by: astarmathsandphysics on April 25, 2011, 09:35:47 pm
How does it get neutralized and change to Helium? :S
cos it attracts electrons from the -ve electrode
Post by: astarmathsandphysics on April 25, 2011, 09:37:39 pm
Post by: Vampire-Love4ever on April 26, 2011, 09:59:50 am
As for the field question, you have to realise that electric current is positve and electrons negative, so electrons travel in the opposite direction to the electric current
I realised that but I still couldn't get it- even by using the Left Hand Rule =/
Post by: astarmathsandphysics on April 26, 2011, 03:14:14 pm
Is this what you get?
Current to right
field into paper
Post by: EMO123 on April 26, 2011, 03:39:40 pm
Hey Guys ..Hi.....acutally i needed something kinda file ( which includes different past papers ) so that i could practisce....any one got it..have watch to this link you would get it all if there :)
Much Appreciated if u Helped/..
https://studentforums.biz/pastpapers/
Post by: Vampire-Love4ever on April 26, 2011, 04:51:16 pm
I get force up page
Is this what you get?
Current to right
field into paper
I am not sure, how do we know that the field is into the paper? :S
Post by: astarmathsandphysics on April 26, 2011, 05:48:14 pm
dots represent tips and the field is out of the paper
Post by: Mike Benn on April 26, 2011, 09:28:14 pm
Much Appreciated if Helped
Post by: Vampire-Love4ever on April 26, 2011, 10:06:52 pm
the crosses represent the tails of arrows
dots represent tips and the field is out of the paper
but the field is down the page :s so i get force out of the page
im really confused :/
Post by: Vampire-Love4ever on April 26, 2011, 10:15:14 pm
its said that the magnetic field is perpendicular to the electron's spiral & that the left hand rule says that the field is directed out of the plane
how do we know that the field is oyt of the plane????? :/ :s
Post by: astarmathsandphysics on April 26, 2011, 10:34:21 pm
Cant tell from picture.
Is electron spiralling in or out?
Post by: Vampire-Love4ever on April 26, 2011, 10:49:16 pm
Left hand rule.
Cant tell from picture.
Is electron spiralling in or out?
i think inwards!
Post by: Vampire-Love4ever on April 26, 2011, 11:05:14 pm
how do we know that A&B is positive while C&D and E&F is negative??? :/
Post by: astarmathsandphysics on April 27, 2011, 06:44:17 am
Post by: Mike Benn on April 27, 2011, 04:25:03 pm
Post by: elemis on April 27, 2011, 04:34:48 pm
R all waves Progressive Waves
No. Progressive waves are defined as waves that transmit energy.
Stationary waves on the other hand do not transmit energy and do not fall under the heading of Progressive waves.
Post by: Vampire-Love4ever on April 28, 2011, 01:20:07 am
If yes-what for??
Post by: astarmathsandphysics on April 28, 2011, 09:09:49 am
Post by: The Golden Girl =D on April 30, 2011, 06:39:06 pm
http://www.scribd.com/doc/32843706/June-9th-Edexcel-Physics-Unit-2
Qs ;3] , 10] , 14] , 15]b)ii) , 20]c)ii)
It's really Urgent !!!!
Post by: elemis on April 30, 2011, 06:57:25 pm
Current into junction must equal current exiting junction.
Look for a junction where I1 and I2 enter it whilst I3 and I4 exit it.
Question 10
The TOTAL current across the two must add up to 9V. The potential drop across the resistor and lamp respecticvely must be either :
9 and 0
8 and 1
7 and 2
6 and 3
And there are more combintions.
Determine the current of both the resistor and lamp at the given voltages and add the curents up to see if they match any of the options.
Question 14
Ignore the fact that this about oil and water coatings and simply write the same points you would have written if you had simply been asked what is the reason for an interference pattern forming.
Post by: Mike Benn on May 01, 2011, 08:12:19 pm
Post by: Vampire-Love4ever on May 02, 2011, 11:18:31 pm
Post by: Vampire-Love4ever on May 02, 2011, 11:28:30 pm
I need help with a(ii) & c!
Please!!!!!!!!!!!!!!
Hope the images are clear enough.
Post by: Mike Benn on May 05, 2011, 09:23:31 am
Post by: The Golden Girl =D on May 05, 2011, 12:46:20 pm
hi guys...Can someone tell me the Key experiments for Unit 3 ....as me got exam on this 18th....Will be greatful
Finding :
Young-Modulus
Free fall
Viscosity
Those are the ones I've noticed they come a lot but I'll let you know if there's more that you should know iA ;)
Post by: s.shamo on May 05, 2011, 07:04:53 pm
5. A student is doing an experiment to find the resistivity of constantan. His apparatus inclides a length of constantan wire, an ammeter, a voltmeter, a variable resistor and a micrometer.
(b)The student has been told to use a range of current valures to plot a graph of p.d against current.
(ii)Explain how plotting a graph should improve his results?
Help :(
Post by: EMO123 on May 05, 2011, 07:15:33 pm
Unit 3B January 2010sry but plz can i get the full paper link
5. A student is doing an experiment to find the resistivity of constantan. His apparatus inclides a length of constantan wire, an ammeter, a voltmeter, a variable resistor and a micrometer.
(b)The student has been told to use a range of current valures to plot a graph of p.d against current.
(ii)Explain how plotting a graph should improve his results?
Help :(
Post by: s.shamo on May 05, 2011, 08:00:02 pm
sry but plz can i get the full paper link
:/ I dont have one, but that's all that was written on the question paper regarding that question, no diagrans other than the cirut you had to draw.
Post by: The Golden Girl =D on May 06, 2011, 07:23:07 am
Unit 3B January 2010
5. A student is doing an experiment to find the resistivity of constantan. His apparatus inclides a length of constantan wire, an ammeter, a voltmeter, a variable resistor and a micrometer.
(b)The student has been told to use a range of current valures to plot a graph of p.d against current.
(ii)Explain how plotting a graph should improve his results?
Help :(
The answer could be ;
That it would smooth out the readings*i.e. it would cancel the wrong readings*
I hope I helped :)
Post by: narnia on May 06, 2011, 09:49:49 pm
occurs ?
Historically, physicists found that electrons of low energy could not be used to find
out information about the nucleus of neutral atoms. Suggest why ?
Post by: Master_Key on May 07, 2011, 07:47:11 am
Atoms which emit alpha or beta-particles usually emit gamma-rays as well. Explain why thisThey emit Gamma-rays in order to get stable. When they emit Alpha/beta they are in excited state so to return to its ground state the atoms emit radiation.
occurs ?
Historically, physicists found that electrons of low energy could not be used to find
out information about the nucleus of neutral atoms. Suggest why ?
Low energy electrons are nearer to nucleus and thus is difficult to find information about nucleus. Hint:- Compare charges, velocity, mass, distance from nucleus and difficulty to trace them.
Post by: red_911 on May 07, 2011, 09:16:40 pm
Post by: Arthur Bon Zavi on May 08, 2011, 04:52:55 am
0.20 0.19 0.18 0.08 .. what is the best mean value ? and what is the uncertinity ? will we consider 0.08 when writing uncertinity ?
Mean - 0.19
Uncertainity - 0.08 (42 %)
(42 % cannot be uncertainty)
Neglect 0.08 then. It will be alright. Take uncertainity of 0.01 (5%).
Post by: Ramprakash on May 08, 2011, 06:54:10 am
thanks
Post by: The Golden Girl =D on May 08, 2011, 06:55:55 pm
were there any unit 3 physics papers before the new syllabus started?if there were,can u give the paper code number and the source from u can get it?
thanks
I hope these help :
http://www.freeexampapers.com/past_papers.php?l=Past_Papers%2FA+Level%2FPhysics%2FEdexcel%2FResources%2FPHY3/
http://www.freeexampapers.com/past_papers.php?l=Past_Papers%2FA+Level%2FPhysics%2FEdexcel%2FResources%2FPHY3_-_Practical/
Post by: Vampire-Love4ever on May 08, 2011, 07:21:14 pm
Post by: Ramprakash on May 09, 2011, 10:06:27 am
I hope these help :thank u very much.do know where i can get bio 3 papers(old syllabus)?
http://www.freeexampapers.com/past_papers.php?l=Past_Papers%2FA+Level%2FPhysics%2FEdexcel%2FResources%2FPHY3/
http://www.freeexampapers.com/past_papers.php?l=Past_Papers%2FA+Level%2FPhysics%2FEdexcel%2FResources%2FPHY3_-_Practical/
Post by: Ramprakash on May 09, 2011, 10:07:22 am
For those doing Unit 3B. Hope this will be of help!thank u very much
Post by: narnia on May 11, 2011, 09:39:24 pm
question 2 b (i)
how do we know that we have to subtract 40 from 220 while calculating anti-clockwise moments??
i was just goin the normal way : 220*10^-3 W=228 *10^-3 (9)
but the answer does not end up to 11 N?!
please explain asap!
thanks!
Post by: The Golden Girl =D on May 12, 2011, 08:55:18 pm
Thx in advance :)
Post by: Malak on May 12, 2011, 09:16:04 pm
State what happens to the following properties as the light goes from the air into the glass.
Frequency
Wavelength
Speed
The ms says that wavelength will decrease while the frequency stays the same..Confused at this part, y would the wavelength change?
I know the speed increases.
Thanks
Post by: Hanooush on May 12, 2011, 10:14:08 pm
could you plz tell me the safety precautions for any experiment in physics.
Thx in advance :)
well, there arent plenty of safety precautions. however, these are the ones that I know.
for experiments that invlove hanging masses e.g. testing the relationship between force and acceleration OR acceleration due to gravity experiments.
You would place a piece of sponge beneath the hanging masses to avoid any injury and to reduce their impact on the ground.
If you are to measure the mass of heavy objects, measure them using scales placed on the floor.
Post by: Hanooush on May 12, 2011, 10:24:20 pm
Make sure that the trolley and light gates are positioned away from the edge so that the trolley will stop on the bench (that one doesnt make too much sense to me, I copied it off a book).
Force-extention experiments:
Wear safety goggles in case of the wire breaks. Also sponge/soft material is placed beneath the masses.
Polarised light exp.:
Do not look directly into the beam and wear eye protection =/
That's it ,, there may be a couple more but it depends on the experiment given
Post by: Hanooush on May 12, 2011, 11:13:49 pm
The ms says that wavelength will decrease while the frequency stays the same..Confused at this part, y would the wavelength change?First of all, the speed doesn't increase. it DECREASES. This is because light has moved from a less dense medium (air) to more dense one (glass). The denser the object, the harder something can pass through, that's why the speed decreases and so the wavelength decreases as well.
I know the speed increases.
The reason why the wavelength decreases can be explained using the equation:
speed=wavelength x frequency. Frequency remains the same and we know that speed decreases, so wavelength must decrease as well.
Post by: Hanooush on May 12, 2011, 11:21:23 pm
Here's a link to a website which includes an image of the graph.
http://www.pfk.ff.vu.lt/lectures/funkc_dariniai/diod/p-n_devices.htm (http://www.pfk.ff.vu.lt/lectures/funkc_dariniai/diod/p-n_devices.htm)
Thanks in advance
:)
Post by: The Golden Girl =D on May 13, 2011, 10:41:05 am
Could someone please explain me the graph of current against voltage (I against V) for a semiconductor diode? Why does it behave that way? Current starts off at 0 for some time, then it increases.
Here's a link to a website which includes an image of the graph.
http://www.pfk.ff.vu.lt/lectures/funkc_dariniai/diod/p-n_devices.htm (http://www.pfk.ff.vu.lt/lectures/funkc_dariniai/diod/p-n_devices.htm)
Thanks in advance
:)
Well it's because the semiconductor diode doesn't work unless it's 0.6v if it's made of Silicon and it won't work if it's made from germanium unless it's 0.4 v
Post by: The Golden Girl =D on May 13, 2011, 10:41:47 am
If an experiment involved using light gates/trolley
Make sure that the trolley and light gates are positioned away from the edge so that the trolley will stop on the bench (that one doesnt make too much sense to me, I copied it off a book).
Force-extention experiments:
Wear safety goggles in case of the wire breaks. Also sponge/soft material is placed beneath the masses.
Polarised light exp.:
Do not look directly into the beam and wear eye protection =/
That's it ,, there may be a couple more but it depends on the experiment given
Thanks a lot mate , if you know more plz do let me know.
I have one though :) , wear goggles if you're working with wires cuz it might snap.
Post by: red_911 on May 13, 2011, 05:14:41 pm
and we plot a graph of Distance/time with time .. how can we prove that the graph will be a straight line ? ???
it is question 9 C of physics specimen paper
http://www.xtremepapers.me/Edexcel/Advanced%20Level/Physics/Specimen/Physics-Practical-Alternative-SAMs.pdf
Post by: Malak on May 13, 2011, 05:36:46 pm
Scientists have discovered recently that the dung beetle can navigate using polarised moonlight. The beetles hunt for fresh dung. When they find some each beetle makes a small ball. To keep this ball for itself it needs to remove it quickly. The beetle pushes the ball along with its back legs while moving with its front legs and keeping its head down. Using the plane of the polarised moonlight as a guide lets the beetle run away in a straight line.
The beetles have sensors in their eyes which act as polarising filters.
Describe and explain the effect of rotating a polarising filter in front of a source of plane polarised light.
Thx
Post by: Ramprakash on May 13, 2011, 06:10:31 pm
You are talking abt mj 2009 3b rite?do u consider the sign when calculating max difference?thanks
Ok,
The mean:
Now find the maximum difference:
So, the maximum uncertainty is
Post by: elemis on May 13, 2011, 06:21:28 pm
if an equation is given : s=ut+0.5at^2
and we plot a graph of Distance/time with time .. how can we prove that the graph will be a straight line ? ???
it is question 9 C of physics specimen paper
http://www.xtremepapers.me/Edexcel/Advanced%20Level/Physics/Specimen/Physics-Practical-Alternative-SAMs.pdf
Re-arranging the equation :
s/t = u + 0.5at2
This conforms with the general equation of a STRAIGHT LINE :
y = mx +c
Post by: The Golden Girl =D on May 13, 2011, 06:22:52 pm
if an equation is given : s=ut+0.5at^2
and we plot a graph of Distance/time with time .. how can we prove that the graph will be a straight line ? ???
it is question 9 C of physics specimen paper
http://www.xtremepapers.me/Edexcel/Advanced%20Level/Physics/Specimen/Physics-Practical-Alternative-SAMs.pdf
S/t = u + 0.5at
u and a are constants
so s/t is directly proportional to t
:)
Post by: The Golden Girl =D on May 13, 2011, 06:24:43 pm
Q:
Scientists have discovered recently that the dung beetle can navigate using polarised moonlight. The beetles hunt for fresh dung. When they find some each beetle makes a small ball. To keep this ball for itself it needs to remove it quickly. The beetle pushes the ball along with its back legs while moving with its front legs and keeping its head down. Using the plane of the polarised moonlight as a guide lets the beetle run away in a straight line.
The beetles have sensors in their eyes which act as polarising filters.
Describe and explain the effect of rotating a polarising filter in front of a source of plane polarised light.
Thx
Which paper is this ? :-\
Post by: Malak on May 13, 2011, 06:33:23 pm
Which paper is this ? :-\
Its a question from exam wizard
Post by: mustafa09 on May 13, 2011, 11:39:14 pm
Post by: Malak on May 14, 2011, 12:01:59 am
By reading from the frequency spectrum (figure 2) state the fundamental frequency of the note
Thx
Post by: elemis on May 14, 2011, 03:51:02 am
Post by: The Golden Girl =D on May 14, 2011, 05:41:13 am
I need the whole question.
It's Question 3 , here ;
http://www.xtremepapers.me/Edexcel/Advanced%20Level/Physics/Specimen/Physics-Practical-Alternative-SAMs.pdf
Post by: elemis on May 14, 2011, 07:30:31 am
It's Question 3 , here ;
http://www.xtremepapers.me/Edexcel/Advanced%20Level/Physics/Specimen/Physics-Practical-Alternative-SAMs.pdf
Is it 440 Hz ?
Post by: The Golden Girl =D on May 14, 2011, 11:14:17 am
Post by: red_911 on May 14, 2011, 03:49:44 pm
if there is a question like : using sonometer show that frequency of sound in inversely proportional to the length. what are the points that we will write ? :o ...
Post by: mustafa09 on May 14, 2011, 06:05:18 pm
i need notes or something
Post by: The Golden Girl =D on May 15, 2011, 08:06:26 am
physics unit 3 help needed
i need notes or something
Check the last page for notes.
https://studentforums.biz/revison-notes/useful-websites/new/#new
Post by: The Golden Girl =D on May 15, 2011, 08:22:46 am
hey i ve a important question
if there is a question like : using sonometer show that frequency of sound in inversely proportional to the length. what are the points that we will write ? :o ...
Check out Page 7 , here ;
http://physics.slss.ie/resources/downloads/ph_pr_soundexperiments.pdf
Check Notes attached , page 6.
Post by: deadlyelder on May 15, 2011, 02:43:56 pm
I need to know how to get the answers for the Physics 3B International Question Number 3 and 9 (c) and (d) of the following document :
http://www.edexcel.com/migrationdocuments/GCE%20New%20GCE/Physics-Practical-Alternative-SAMs.pdf
I need to know the method not the answer. Hope Someone helps
Post by: Malak on May 15, 2011, 03:36:35 pm
Hey,
I need to know how to get the answers for the Physics 3B International Question Number 3 and 9 (c) and (d) of the following document :
http://www.edexcel.com/migrationdocuments/GCE%20New%20GCE/Physics-Practical-Alternative-SAMs.pdf
I need to know the method not the answer. Hope Someone helps
For Q3:
Each small line is 40 Hz and the lowest frequency is the the first line on the graph therefore if you count them until the first frequency line, you will get 440 Hz
Q9 c:
Eq. of a straight line is the form y = mx + c
You know that the graph is of s/t for y-axis and t for x-axis
Therefore your y = s/t and you know that s/t is speed therefore y = v
So eq. will be like:
s = ut + 1/2at2
Dividing by t will give : s/t = u + 1/2at
Substituting for s/t :
v = u + 1/2at ------> y = c + mx
And for part d..m not so sure
Post by: MasterMath on May 15, 2011, 03:50:41 pm
and for question 3 a) .. Is it D or C :S? is the MS all wrong? cause Golden girl says C while the MS says D
thanks in advance!
Post by: Malak on May 15, 2011, 03:56:13 pm
Question 4 2009.. Why is the answer not A instead of D(which is totally wrong cause the diode is reverse biased.. the MS says D :S) Anyone could help please?
and for question 3 a) .. Is it D or C :S? is the MS all wrong? cause Golden girl says C while the MS says D
thanks in advance!
For Q3 a its D
You always give the answer for your average to same d.p. or sf as the values so its D
Post by: MasterMath on May 15, 2011, 04:04:54 pm
Post by: Malak on May 15, 2011, 04:17:58 pm
Yeah just read through the forums.. seems that q4 on the MS is wrongMy teacher said D too but i cant remember how is it D.
as what you are saying makes sense, i guess too the ms is wrong unless we are missing out something :S
I have a question
Q1 a for 2010 may
I dont understand the question itself? can anyone explain
Thx
Post by: The Golden Girl =D on May 15, 2011, 04:41:04 pm
My teacher said D too but i cant remember how is it D.
as what you are saying makes sense, i guess too the ms is wrong unless we are missing out something :S
I have a question
Q1 a for 2010 may
I dont understand the question itself? can anyone explain
Thx
I believe because it is conducting since the circuit has a fixed battery
Post by: MasterMath on May 15, 2011, 04:55:16 pm
Post by: deadlyelder on May 15, 2011, 05:01:10 pm
For Q3:
Each small line is 40 Hz and the lowest frequency is the the first line on the graph therefore if you count them until the first frequency line, you will get 440 Hz
Q9 c:
Eq. of a straight line is the form y = mx + c
You know that the graph is of s/t for y-axis and t for x-axis
Therefore your y = s/t and you know that s/t is speed therefore y = v
So eq. will be like:
s = ut + 1/2at2
Dividing by t will give : s/t = u + 1/2at
Substituting for s/t :
v = u + 1/2at ------> y = c + mx
And for part d..m not so sure
Thanx alot. But regrading the Q3. I just found one more way to solve, hopefully more easy and less error prone.
First, take the least frequncy, which in the graph will be 2200 Hz, and divde it by the total number of bar graps, i.e. 5. You get 440.
Which method is correct though? I merely tried to use Logic.
As for Question 9 (d) first calculate the gradient obtained from the graph, i.e 0.04,
Now according to the formula given : s=ut+1/2at^2
As u = 0; Initial Value
S=0+1/2at^2
s/t=1/2at
s/t/t=1/2a
a=s/t/t x 2
a=gradient x 2 [s/t/t = Gradient]
a=0.04x2=0.08 m/s
Now, s=ut+1/2at^2
s-1/2at^2=ut
u=(s-1/2at^2)/t
=s/t-1/2at
=0.86-1/2 x0.08 x2.32
=0.86-0.04 x2.32
=0.7672 m/sec
Hope it is not to confusing.
Post by: Malak on May 15, 2011, 05:09:55 pm
Thanx alot. But regrading the Q3. I just found one more way to solve, hopefully more easy and less error prone.
First, take the least frequncy, which in the graph will be 2200 Hz, and divde it by the total number of bar graps, i.e. 5. You get 440.
Which method is correct though? I merely tried to use Logic.
As for Question 9 (d) first calculate the gradient obtained from the graph, i.e 0.04,
Now according to the formula given : s=ut+1/2at^2
As u = 0; Initial Value
S=0+1/2at^2
s/t=1/2at
s/t/t=1/2a
a=s/t/t x 2
a=gradient x 2 [s/t/t = Gradient]
a=0.04x2=0.08 m/s
Now, s=ut+1/2at^2
s-1/2at^2=ut
u=(s-1/2at^2)/t
=s/t-1/2at
=0.86-1/2 x0.08 x2.32
=0.86-0.04 x2.32
=0.7672 m/sec
Hope it is not to confusing.
I think both methods are correct as 2200 Hz is a multiple of the fundamental frequency i.e. 440 Hz therefore using ur method gives the right answer too :D
Post by: Malak on May 15, 2011, 06:48:11 pm
My a comes out as 0.078
and for u i get a weird answer
Post by: DreamlessSea on May 15, 2011, 10:08:56 pm
Post by: deadlyelder on May 16, 2011, 06:40:16 am
^^ Why did u use 0.86 and 2.32 as the values?
My a comes out as 0.078
and for u i get a weird answer
Got the vaules from the graph check you graph and then substitute the values.
And regrading the uncertainly here is what I know :
Quote
what is uncertainty?
fault in the apparatus due to limitation of apparatus
limitation of apparatus = least count/ precision
reliable ---> less error
Precise ---> small value
less reliable ---> more precise >>> reason: because its difficult to take small measurement!
absolute uncertainty = least count = maximum value of uncertainty
greater than least count ---> personal error
percentage uncertainty = least count/measurement*100
This is posted on this forums by the users so credit to them, hope it helps
Post by: sean321 on May 16, 2011, 06:59:12 am
Any specific chapters? HELP!!
Post by: EMO123 on May 16, 2011, 09:08:03 am
What can you study for paper 3 AS level?ya i suggest read the syllabus and do as given for as
Any specific chapters? HELP!!
hope i helped
Post by: The Golden Girl =D on May 16, 2011, 09:39:24 am
What can you study for paper 3 AS level?
Any specific chapters? HELP!!
If you're taking about Physcis Unit 3 Edexcel AS level , then study ALL the experiments from Unit 1 and 2.
and check this ;
https://studentforums.biz/revison-notes/useful-websites/30/
Post by: The Golden Girl =D on May 16, 2011, 10:22:04 am
hey humans, who knows how to find the uncertancy of a value. I find a different equation and method everywhere and now im confused.
Attached.
Post by: Malak on May 16, 2011, 12:21:54 pm
Q1 from 2011 (attachment) and Q1 from 2010 Jun
Thx
Post by: NightHawkerr on May 16, 2011, 12:44:45 pm
Yeah just read through the forums.. seems that q4 on the MS is wrong
Yea i put C for that one, I just cant seem to find any logical reason as to why it's D. ???
Post by: NightHawkerr on May 16, 2011, 12:46:55 pm
The one where they ask about the intensity of light and distance, and you change the distance from a lamp.
Which one of the following is the measured independent variable? Ans is DISTANCE.
Arent we the ones changing it? So it should be dependent :s
& the next question which of the following is dependent, Ans is Resistance of LDR.. Why?
Post by: Malak on May 16, 2011, 01:00:25 pm
May 2010 Q 2,
The one where they ask about the intensity of light and distance, and you change the distance from a lamp.
Which one of the following is the measured independent variable? Ans is DISTANCE.
Arent we the ones changing it? So it should be dependent :s
& the next question which of the following is dependent, Ans is Resistance of LDR.. Why?
Independent is the one WE change..and we are chaning the distance so its independent
Dependent is the one which depends on the variable we change
So it resistance of LDR
Post by: Malak on May 16, 2011, 01:07:03 pm
Q1 a from 2010 May
Q1 fro 2011
Q4 from Jan 2010
and
Attachment..
Justify the choice of meter rule?
thx
Post by: MasterMath on May 16, 2011, 02:03:33 pm
Post by: The Golden Girl =D on May 16, 2011, 02:48:45 pm
Anyone could help finding the uncertainty? I dont seem to understand how :'( .. You have 4 values ..0.20, 0.18, 0.19 and 0.08 s. you ignore the 0.08 cause its really off.. so you get a mean value of 0.19.. How do you find the uncertainty from here ??? ?
can you tell me which paper is that ?
Post by: The Golden Girl =D on May 16, 2011, 02:57:49 pm
Questions:
Q1 a from 2010 May
Q1 fro 2011
Q4 from Jan 2010
and
Justify the choice of meter rule?
thx
May '10 Q.1]
first of all find total resistance = 100 + 10 = 110 ohms
then calculate the current using formula I = V/R = 3v / 110 ohms = 0.0272
multiply it by 1000 , you'll get 200 mA hence answer is C
Jan '11 Q.1]
same as above .. total resistance = 110 ohms , V = 3 , I = 3/110 = 0.0273 A
then find V across R {100 ohms } ....V = 0.027 * 100 = 2.7v
hence the one that BEST fits is B that is the 20v
y a meter rule is used?
longer length so measurement to nearest mm appropriate ....
I believe the answer or the meaning is that because it's more accurate.
Post by: MasterMath on May 16, 2011, 02:58:24 pm
can you tell me which paper is that ?2010 MAY 5c)
Post by: The Golden Girl =D on May 16, 2011, 03:02:19 pm
2010 MAY 5c)
uncertainty = (0.20 - 0.18)/2 = +- 0.01
hence answer is 0.01
to calculate any uncertainty you should do as follows ;
[Max - Min]/2
they didn't ask for percentage which is why i DIDN'T MULTIPLY IT BY 100
Post by: The Golden Girl =D on May 16, 2011, 03:13:55 pm
Q4 from Jan 2010
My QP is pretty mess so I'll just describe the experiment.
place a meter rule and a stand right next to it with a pointer
and make sure the meter rule is vertical using set squares
Measure the mass of the ball using top pan balance
drop the ball from a known height say 1.5 m it's just an example By the way
then when the ball bounces back since the reading is not accurate much *difficult to get the exact reading of it's height* we drop the ball about 10 times and take average height of when the ball bounces back [h2].
initial K.E = G.P.E
% = (mgh1 -mgh2 ) / 100
Precaution ; make measurements of height at eye level.
Post by: NightHawkerr on May 16, 2011, 03:46:11 pm
Post by: Malak on May 16, 2011, 03:52:59 pm
Can anyone give me the link for AS Physics 6pH07 2011 Jan past paper? Thankyou :D
here
https://studentforums.biz/pastpapers/edexcel-january-2011-examiners-reports-%28physics-chemistry-biology-maths%29/ (https://studentforums.biz/pastpapers/edexcel-january-2011-examiners-reports-%28physics-chemistry-biology-maths%29/)
^^ GG
why do you multiply by 1000 for 2010 Q1:S
Post by: The Golden Girl =D on May 16, 2011, 05:55:49 pm
^^ GG
why do you multiply by 1000 for 2010 Q1:S
Because I want to know the answer in terms of Milli-Ammeter to find out whether any of the answers in millimeter actually are suitable for it or not.
GG
Post by: mustafa09 on May 17, 2011, 10:20:16 am
Post by: Malak on May 17, 2011, 11:40:37 am
2009 May 3B ... 6 c) anyone :) ? thanks in advance!
If you have done part b then ur answer should come out as 14.7 Gpa
1 G = 109
So therefore yes it is consistent as 14.7 GPa = 1.47 x 10 10 Pa
Post by: MasterMath on May 17, 2011, 11:48:26 am
Post by: The Golden Girl =D on May 17, 2011, 12:08:29 pm
Do we have to know how to calculate internal resistance in the unit 3B, if yes can someone tell me how
you just have to know the formulas cuz they might ask you to change it into the form y = mx +c
but the formuls for Internal resistance it e = IR +Ir
e => emf
I => current
R=> load /external resistance
r => internal resistance
that's it.
Post by: The Golden Girl =D on May 17, 2011, 12:26:41 pm
2009 June 7 a) I get the question but in the MS they said "comment on liquid level" what do they mean :O ?
is for you to comment on how you would find the distance of the liquid from the ground.
what I wrote ; find the liquid level from the top of the bench and subtract readings obtained "SA" and "SB"
I hope I helped.
Post by: sean321 on May 17, 2011, 07:24:55 pm
Post by: KJ1995 on May 17, 2011, 09:32:10 pm
Post by: Malak on May 17, 2011, 10:26:55 pm
Does anyone know how to choose the best setting for a multimeter?? I saw a question in 2010 paper 3b and one in 2011 jan paper 3b...unfortunately i didnt know how to solve them both, they were the first question in multiple choiceI had the same question..its on the last page..
But anywayz
Quote
May '10 Q.1
first of all find total resistance = 100 + 10 = 110 ohms
then calculate the current using formula I = V/R = 3v / 110 ohms = 0.0272
multiply it by 1000 , you'll get 200 mA hence answer is C
Jan '11 Q.1
same as above .. total resistance = 110 ohms , V = 3 , I = 3/110 = 0.0273 A
then find V across R {100 ohms } ....V = 0.027 * 100 = 2.7v
hence the one that BEST fits is B that is the 20v
Thx to Golden Girl
Post by: KJ1995 on May 18, 2011, 09:30:16 am
Post by: Malak on May 18, 2011, 10:25:21 am
guys can u plz tell me how to find the uncertainty between 2 values and 3 values
Check pm
Post by: KJ1995 on May 18, 2011, 10:35:15 am
Check pm
Thanks man uve helped me alot :D its (max-min)/2...right?
Post by: Malak on May 18, 2011, 10:40:40 am
Thanks man uve helped me alot :D its (max-min)/2...right?I think u can do it that way too..It works
But i do it the other way as my teacher told me..
For eg if u have these values 0.20, 0.18, 0.19
I would do it like:
Find the avg. which i get as 0.19
and then the greatest difference is 0.20 - 0.19 = 0.01
So i get my uncertainty that is +/- 0.01
But u can do it that way too i guess...Jst do the way u find it easy.. :D
And Good Luck Everyone :D
Post by: NightHawkerr on May 20, 2011, 03:48:50 pm
Post by: The Golden Girl =D on May 20, 2011, 03:51:12 pm
Post by: Malak on May 20, 2011, 04:08:21 pm
It was good except for the last Question.lol..same here everything was good except the last one..The uncertainty and angle thing was weird huh
Post by: The Golden Girl =D on May 21, 2011, 09:14:58 am
Post by: MasterMath on May 21, 2011, 10:00:26 am
The apparatus not in the diagram.. I had everything on my diagram :S
Post by: NightHawkerr on May 21, 2011, 02:32:23 pm
Lol what apparatus did you use for the thermistor question..?
Post by: The Golden Girl =D on May 21, 2011, 04:01:09 pm
https://studentforums.biz/gce-as-a2-level-(cie-edexcel)/edexcel-2011-mayjune-examination-discussion-!/ (https://studentforums.biz/gce-as-a2-level-(cie-edexcel)/edexcel-2011-mayjune-examination-discussion-!/)
GG :D
Post by: studz on May 21, 2011, 09:32:51 pm
i have no idea what that paper's like ??? :( >:(
and i want to know the important things to remember for that paper
Post by: The Golden Girl =D on May 22, 2011, 05:55:49 am
anyone doing unit 6b for physics?
i have no idea what that paper's like ??? :( >:(
and i want to know the important things to remember for that paper
check this :
https://studentforums.biz/pastpapers/edexcel-unit-6b-(6ph08)-practicals/ (https://studentforums.biz/pastpapers/edexcel-unit-6b-(6ph08)-practicals/)
Post by: MasterMath on May 22, 2011, 12:20:30 pm
Post by: astarmathsandphysics on May 22, 2011, 11:28:49 pm
Can you post the paper
Post by: Malak on May 29, 2011, 01:33:10 pm
But does that alter the wavelength?
Thanks
Post by: Vampire-Love4ever on May 29, 2011, 01:41:07 pm
anyone doing unit 6b for physics?
i have no idea what that paper's like ??? :( >:(
and i want to know the important things to remember for that paper
Check this out: https://studentforums.biz/reference-material-83/edexcel-chemistry-and-physics-unit-6b/
Post by: The Golden Girl =D on May 29, 2011, 02:28:40 pm
Can anyone clarify, waves travel slower is shallow water and faster in deep right?going from deep to shallow the wavelength decreases and the speed also decreases.
But does that alter the wavelength?
Thanks
I hope I cleared your doubt :)
Post by: Malak on May 29, 2011, 02:38:38 pm
And also sometimes in paper in the diagram they show small gaps between wave lines, does that mean that the wavelength is smaller? :-\
What i mean is this (attachment the 2nd lines)
Post by: The Golden Girl =D on May 29, 2011, 02:39:45 pm
Post by: NidZ- Hero on May 30, 2011, 05:23:26 pm
Measurements don't agree 0.86 s ± 0.02 s and 0.98 s ± 0.02 s
Measurements agree 0.86 s ± 0.08 s and 0.98 s ± 0.08 s
If the ranges of two measured values don't overlap, the measurements are discrepant (the two numbers don't agree). If the rangesoverlap, the measurements are said to be consistent.
CAn someone explain this part :S
Post by: Malak on May 30, 2011, 05:26:56 pm
Post by: Malak on May 30, 2011, 08:09:58 pm
thx
Post by: NidZ- Hero on May 30, 2011, 09:03:52 pm
^ Is this a past paper or what? :-\ can you tell which year then? *confused* :P
Dude ppl in cie physcis wont reaply MAybe cuz they r busy So.. i posted in here
Well its in practical Nvermind :)
Post by: wakemeup on May 30, 2011, 09:38:57 pm
Uncertainty estimates are crucial for comparing experimental numbers. Are the measurements 0.86 s and 0.98 s the same or different? The answer depends on how exact these two numbers are. If the uncertainty too large, it is impossible to say whether the difference between the two numbers is real or just due to sloppy measurements. That's why estimating uncertainty is so important!
Measurements don't agree 0.86 s ± 0.02 s and 0.98 s ± 0.02 s
Measurements agree 0.86 s ± 0.08 s and 0.98 s ± 0.08 s
If the ranges of two measured values don't overlap, the measurements are discrepant (the two numbers don't agree). If the rangesoverlap, the measurements are said to be consistent.
CAn someone explain this part :S
We actually kind of have this in Edexcel practical as well (mainly for A2). It's pretty simple. In the first case, think about the max value the 0.86 measurement could have. 0.86 + 0.02 = 0.88. Now, think about the minimum value the 0.98 value could have. 0.98 - 0.02 = 0.96. So the 2 values don't overlap. In the second case repeat the same process. 0.86 + 0.08 = 0.94. And 0.96 - 0.08 = 0.88. The values of the range do overlap each other so the measurements are consistent. Basically, if the ranges intersect, the values are consistent. If not, then they aren't.
Post by: deadlyelder on May 31, 2011, 06:57:44 am
(b) Using the correct circuit the student obtains the following results.
Current in the cell
I/A Terminal potential difference
across the cell V/V
0.5 1.2
0.9 1.0
1.5 0.8
1.9 0.6
2.5 0.4
2.9 0.2
?
(i) On the grid below, plot these results and draw the line of best fit through your points.
(ii) Use your graph to determine the e.m.f. of the cell.
e.m.f. = .........................................
Post by: Malak on May 31, 2011, 10:49:09 am
And if u have plotted the graph then Emf is the y intercept :)
Post by: deadlyelder on May 31, 2011, 12:20:08 pm
^ Which paper is that, cuz i cant plot a grap here :P so maybe i cant post my paper
And if u have plotted the graph then Emf is the y intercept :)
Thanx. Yes I did ploted the graph and you are right the y-intercept from graph does match the limits given in the answers. It is from a bunch of papers of Physics (each chapter) like Examzone with answers. I downloaded them from the forums. And its answer is stated as : e.m.f. = [1.36 – 1.44 V]
Post by: Malak on May 31, 2011, 12:21:54 pm
Thanx. Yes I did ploted the graph and you are right the y-intercept from graph does match the limits given in the answers. It is from a bunch of papers of Physics (each chapter) like Examzone with answers. I downloaded them from the forums. And its answer is stated as : e.m.f. = [1.36 – 1.44 V]Oh okay..Good luck with those :)
Post by: Malak on June 01, 2011, 01:34:23 pm
The ms says this:
See 2 resistors in parallel with supply (1)
Supply across ends of variable resistor (10 ?) (1)
Fixed resistor across one end and slider (consequent mark) (1)
Thanks
Post by: The Golden Girl =D on June 01, 2011, 05:31:04 pm
Question in attachment..Can anyone draw the circuit
The ms says this:
See 2 resistors in parallel with supply (1)
Supply across ends of variable resistor (10 ?) (1)
Fixed resistor across one end and slider (consequent mark) (1)
Thanks
I have it somewhere so I'll get back to you in a few minutes iA.
EDIT: Attached.
Post by: Malak on June 01, 2011, 09:37:54 pm
I have few more doubts ::)
In first Q..Why is the answer D? ???
The last two Qs are of the same sort, just need a basic explanation.
Thanks Alot :)
Post by: High-R on June 02, 2011, 06:49:12 am
or where r u prctising from??
Post by: The Golden Girl =D on June 02, 2011, 07:48:27 am
GG...Ur awesome :D
I have few more doubts ::)
In first Q..Why is the answer D? ???
The last two Qs are of the same sort, just need a basic explanation.
Thanks Alot :)
It'll take me time to type all my answers so I hope you have patience.
EDIT : I don't get the first one and the last one :S
my answer for the second Question is as follows ;
Lamp
As the temperature increases , the atoms vibrate violently .
flow of charge is blocked by the vibrating atoms.
the flow of charge decreases and ,since the flow of charges is directly proportional to the current , the current decreases hence the resistance increases
Thermistor
As temperature increases , the electrons gain Kinetic energy and they hit the lattice atoms and the flow of charges increases , and since the flow of charges is directly proportional to the current , the current increases and hence the resistance decreases
EDIT :
for the First Question
Thanks to Master_Key ;
Divide the current into two parts. Like if 3 A Max.
then 1A for R
and 2A for R1+R2.
P = I2R
For P of R
P = 12R
For P of R1+R2
P = 22R1 or R2
P in R1 and R2 are same
Ratio is 1 : 4.
or Conside R and R1+R2 in parallel.
Take V as a fixed V. In parallel circuits, V is always same.
P = V2/R
For P of R
P = V2/R
For P of R1+R2
P = V2/(R1+R2)
and find the ratio.
EDIT again :P
Question 3.
I = nAqv where
I = current
n = charge carrier density
q = charge of an electron
v = drift speed
Increase in temperature decreases the resistance of Thermistor.
The increase in temperature causes an increase in drift velocity, so current increases if voltage is to remain same. That explains I against V graphs that defy Ohm's law (which would predict a straight line) for thermistor.
hope i helped.
MORE HELP FOR PHYSICS FROM HERE.
resources.r9paul.org/ASA2/Physics/AS_RevisionBook.pdf
Post by: Vampire-Love4ever on June 02, 2011, 08:00:45 am
6B how many pastpapers r u people doing?
or where r u prctising from??
Sample paper, June 2010 and Jan 2011 papers.
Revise notes.
https://studentforums.biz/reference-material-83/edexcel-chemistry-and-physics-unit-6b/
Post by: Malak on June 02, 2011, 12:03:01 pm
Post by: Banana on June 02, 2011, 12:12:56 pm
GG...Ur awesome :D
I have few more doubts ::)
In first Q..Why is the answer D? ???
The last two Qs are of the same sort, just need a basic explanation.
Thanks Alot :)
Lol the 1st question is the 1st question of the specimen paper :) See My Diagram Attached
Suppose the resistance of each resistor is 10 ohm......the total R through line 1 will be 10 ohm and the total R through line 2 will be 20 ohm......
So the current through line 2 will be halved( use the equation V=IR if R doubles, I is halved)
So using the Eq: P= I2 x R for line 2 if the current is halved, the P is quartered....D is the answer
lol....i know its tricky but go in sequence and use the right equatons ;)
http://imageshack.us/photo/my-images/29/physicsl.jpg (http://imageshack.us/photo/my-images/29/physicsl.jpg)
Post by: Malak on June 02, 2011, 12:19:02 pm
Lol..I hate when we have to do all this just for one mark >:(
And i still didnt do the past papers so I didnt knew this Q is there :P
Post by: Malak on June 02, 2011, 04:05:39 pm
2009 Jan Q16 (b)
I understand that the voltage is same, and I wrote that A is brighter 'cause more current flows through it as it has less resistance than B.
But I dont get why the MS is relating it to power equations ???
Is my answer considered wrong or what? :-\
Thx
Hate electricity chapter >:(
Post by: The Golden Girl =D on June 02, 2011, 04:38:49 pm
One more doubt..
2009 Jan Q16 (b)
I understand that the voltage is same, and I wrote that A is brighter 'cause more current flows through it as it has less resistance than B.
But I dont get why the MS is relating it to power equations ???
Is my answer considered wrong or what? :-\
Thx
Hate electricity chapter >:(
Usually when a lamp is brighter that's cuz of the Power it dissipates ;) that's why ,Which is why I suggest you write your answer according to the power equations if you don't get it ,let me know :)
Post by: Banana on June 02, 2011, 04:42:26 pm
Usually when a lamp is brighter that's cuz of the Power it dissipates ;) that's why ,Which is why I suggest you write your answer according to the power equations if you don't get it ,let me know :)
thats because power relates to brightness....the greater the power, the brighter the lamp! ;)
Post by: High-R on June 02, 2011, 04:43:48 pm
;)
Post by: The Golden Girl =D on June 02, 2011, 04:44:28 pm
Welcome :)
Post by: Malak on June 02, 2011, 06:16:48 pm
I fail at this chapter.. :-[
Nywayz, thx you both :)
So whenever its about brightness I should talk about power? And the more power it dissipates the less brighter it is ? right ? ???
Post by: The Golden Girl =D on June 02, 2011, 06:37:16 pm
Oh god..Was is that simple :o
I fail at this chapter.. :-[
Nywayz, thx you both :)
So whenever its about brightness I should talk about power? And the more power it dissipates the less brighter it is ? right ? ???
Nope , the more power it dissipates ,the brighter it is ;)
Post by: Banana on June 02, 2011, 06:49:34 pm
Oh god..Was is that simple :o
I fail at this chapter.. :-[
Nywayz, thx you both :)
So whenever its about brightness I should talk about power? And the more power it dissipates the less brighter it is ? right ? ???
Lol.....ur confused!! :o....study harder, this papers gonna be tough....*gut feeling* :-[
Post by: The Golden Girl =D on June 02, 2011, 06:51:16 pm
Lol.....ur confused!! :o....study harder, this papers gonna be tough....*gut feeling* :-[
Let's ALL Study harder ..who knows it might turn out to be easy ;)
Post by: Malak on June 02, 2011, 07:14:43 pm
I Know M studying hard, M okay with waves and particle phy but this makes me go.... >:(
Ahh..I should go through it again :-[
thx guys :)
Post by: The Golden Girl =D on June 02, 2011, 07:28:26 pm
Post by: Banana on June 03, 2011, 04:21:33 am
Let's ALL Study harder ..who knows it might turn out to be easy ;)
Lol.....Inshallah it'll be alright! ;)
Post by: The Golden Girl =D on June 03, 2011, 08:36:59 am
Post by: High-R on June 03, 2011, 01:46:39 pm
can any1 plz help me with 2011 phys unit6B (6PH08)
i dont get the Q2 part c ???
and even in Q2 part d how exactly do we do that?
all i know is u find the uncertainity by first finding the mean of the values given and then subtracting the mean from the lowest value(from the set of values given) thats how uncertainity is found right? so its +/- 0.2!!
and what do i do next? ???
Post by: High-R on June 03, 2011, 01:55:10 pm
percentage difference
uncertainity
etc
can u please tell how do u find ALL THESE?
:S
eg: in 2011 unit6b jan paper
Q3(ii) they ask percentage difference so (2.01-1.93)/ 1.97 = 4.1% is the working given in the markscheme, but why did we use 1.97(the mean value)? ???
Post by: Master_Key on June 03, 2011, 02:03:41 pm
Lol the 1st question is the 1st question of the specimen paper :) See My Diagram Attached
Suppose the resistance of each resistor is 10 ohm......the total R through line 1 will be 10 ohm and the total R through line 2 will be 20 ohm......
So the current through line 2 will be halved( use the equation V=IR if R doubles, I is halved)
So using the Eq: P= I2 x R for line 2 if the current is halved, the P is quartered....D is the answer
lol....i know its tricky but go in sequence and use the right equatons ;)
http://imageshack.us/photo/my-images/29/physicsl.jpg (http://imageshack.us/photo/my-images/29/physicsl.jpg)
From kirchoff's laws. Current is never used up. Current entering a circuit equals the current leaving it.
IN SERIES:-
Current is the same across all components and voltage is different across all components.
IN PARALLEL:-
Current is different across different components and voltage is same across all of them. ALL HOUSES AND INDUSTRIES ARE THUS CONNECTED IN PARALLEL TO THE NATIONAL GRID IN ORDER TO GET FULL P.D.
Post by: The Golden Girl =D on June 03, 2011, 03:13:03 pm
I've got loads of Qs :-X
hmm can someone explain in section A Qs 4 ,5 , 6
Section B
Qs , 12]b) , 13]a)iii) , 17]b) , 19]a)ii)
Thanks in Advance :D
Post by: elemis on June 03, 2011, 03:25:34 pm
Once the lamp in parallel breaks down the total resistance of the circuit will increase, hence, the total current flow will decrease. The PD across L increases whilst the PD across N falls.
Thus, the brightness of L increases and N falls.
Question 6
Power = I2R
W/A2 = R i.e. WA-2
Post by: The Golden Girl =D on June 03, 2011, 05:53:19 pm
Paper Attached.
I've got loads of Qs :-X
hmm can someone explain in section A Qs 4 ,5 , 6
Section B
Qs , 12]b) , 13]a)iii) , 17]b) , 19]a)ii)
Thanks in Advance :D
Qs 5 and 6 in Section A have been answered but I'd really appreciate it if someone answers the rest :)
And I got some more Qs :-\
Attached QP.
Section A
Qs .4 , 6 .
Ohh when we change from electron volts to Joules do we Multiply or divide by (1.6 * 10-19)
Section B
Qs 19]c)i) and ii) , 20]a)iii)
Q 18]c) I wrote the answer as : Because ALL the particles of the food will be exposed to micrwaves and get heated up.
is my answer considered correct :-[
Thanks Again :)
Post by: elemis on June 03, 2011, 06:00:20 pm
Q=It
The e.m.f of a battery is defined as the work done per unit charge. What is the unit of work ? What other quantity has this unit ?
Got it ?
Question 6
If I tell you that I gained 20 J of useful energy out of a total input of 100 J and I lost 80 J in the process, what would the efficiency be ?
Post by: Banana on June 04, 2011, 05:18:13 am
19]c)i) :
Well they've said that the spacing between the bands is inversely
proportional to the distance between the two slits. And so use the proportion method (Inverse Variation)
Slit Spacing Distance between Bands
1.2mm-------------->0.6mm
0.4mm--------------> X
So by finding x, you find the separation distance.....So to find x, dont cross multiply (its not directly proportional, its incersely proportional)
=> 1.2 x 0.6 = 0.4 x X................so X is 1.8mm
19]c)ii)
Okay now they say the slit spacing is increased....using the relationship given (inverse proprtion) an increase in split spacing will decrease the band spacing to the point that they donot overlap at all......thats all you have to say ;D
20]a)iii)
Lol....I think this question was easy :); they are asking for the power and power is given by P= I x V
So the power in the load is equal to the current in the load multiplied by the pd across the load so multiply the values of D7 and E7.......>
2.14 x 4.29 = 9.18 W
So GG do you get it? If you dont ask again!.....Glad to help.... :D
Post by: The Golden Girl =D on June 04, 2011, 09:44:16 am
Ya M&M I get it :)
Post by: MasterMath on June 04, 2011, 02:36:48 pm
Look at 2009 June 20b.. >_>
The distance moved by the microphone between two adjacent maxima is
0.050 m.
Calculate the wavelength of the sound wave.
They say the answer is 0.1 ???
Post by: elemis on June 04, 2011, 02:50:47 pm
Post by: Malak on June 04, 2011, 02:58:03 pm
Q11 (c)
Q16 (b) (never understood polarisation) >:(
Thankyou very much
Post by: The Golden Girl =D on June 04, 2011, 03:03:51 pm
Few doubts:
Q11 (c)
Q16 (b) (never understood polarisation) >:(
Thankyou very much
Q.11]c)
you measure the distance between the compressions let's say it's 4 cm , you move the compressions backwards by 1 cm , I hope you got it ...if not let me know so that I scan my paper :)
16 (b)
My answer is ;
I would place a Polaroid film and observe .If the intensity decreased it means it might be polarized but I would confirm it by using another Polaroid fiml as an Analyser and rotate it by 90 and observe the intensity of the light ..it would be darker but not 100 % dark .
I hope you got my idea and excuse my poor representation :P
:D
Post by: Malak on June 04, 2011, 03:07:38 pm
But for Q11, why would you move it backwards by 1 cm ???
Thx
Post by: The Golden Girl =D on June 04, 2011, 03:09:33 pm
I get he second one,
But for Q11, why would you move it backwards by 1 cm ???
Thx
I'll ask my friend and get back to you :-X
EDIT : I copied it from her and she drew it before the compression cuz she though it was a continious wave but well that's not mentioned so it's actually after the first compression by 1 cm :)
Post by: Malak on June 04, 2011, 03:23:21 pm
And thx alot :)
Post by: The Golden Girl =D on June 04, 2011, 03:27:45 pm
Sorry for being dumb..But why 1 cm ? Is it 'cause the line is t + 3/4 T ???
And thx alot :)
have the four is 2 and then you take the Quarter of that which I guess ain't 1 :S
srry for confusing u :(
Post by: Malak on June 04, 2011, 03:34:27 pm
Thanks for helping though :D
Post by: The Golden Girl =D on June 04, 2011, 03:35:08 pm
Lol..Its okay...
Thanks for helping though :D
Welcome :)
Post by: Banana on June 04, 2011, 03:43:30 pm
Sorry for being dumb..But why 1 cm ? Is it 'cause the line is t + 3/4 T ???
And thx alot :)
Look....u measure the distance between 2 compressions and multiply by three quarters....thats wat GG meant ;)....then u measure out that distance from the compression points and just PLOT it :P
Post by: Malak on June 04, 2011, 03:49:13 pm
Post by: Banana on June 04, 2011, 04:20:54 pm
Okay k Thanks ::)
No problemo ;)
Post by: Banana on June 04, 2011, 04:40:37 pm
Why does the Kinetic E of emitted electrons in the photoelectric effect depend only on the frequency and not on the intensity of the light????
Post by: Malak on June 04, 2011, 05:23:52 pm
More energy means more K.E
Post by: The Golden Girl =D on June 04, 2011, 05:25:42 pm
Somebody pls help me out with this question.....im totally confused :-\:
Why does the Kinetic E of emitted electrons in the photoelectric effect depend only on the frequency and not on the intensity of the light????
It is due to the equation E = hf ...
h is Planck's constant ,and since it's constant this means there is a relation between Energy and frequency .
And from this equation we got to know that the Energy of a Photon depends on frequency
I don't think we should know WHY exactly it is this way ...I'm not sure if you we need to know it or not , but I hope you got my point :)
Post by: Banana on June 04, 2011, 05:56:36 pm
Post by: The Golden Girl =D on June 04, 2011, 06:12:31 pm
Post by: Romeesa-Chan on June 04, 2011, 07:46:43 pm
*Q1bii- where did 8 come from ? :S
*Q2ciii- No idea what the marking scheme says!
*Q3 a & b
*Q4 a & dii
Link: http://www.edexcel.com/migrationdocuments/GCE%20New%20GCE/Physics-Practical-Alternative-SAMs.pdf
Please help me. Have exam the next day (Monday)!!!
Post by: elemis on June 05, 2011, 05:42:25 am
% Uncertainty = uncertainty of instrument / actual measurement.
Each foil is 15 micrometres thick. When you decide to measure the thickness of the foil will you measure the thickness of only one foil or several ?
You would do it for several so as to gain an average and eliminate random uncertainties.
If I were you I would measure the thickness of 8 sheets or more. That's where the eight comes from.
Q2
In a roundabout fashion they are asking you to calculate the percentage uncertainty whilst showing workings.
Post by: Romeesa-Chan on June 05, 2011, 11:21:51 am
Q1
% Uncertainty = uncertainty of instrument / actual measurement.
Each foil is 15 micrometres thick. When you decide to measure the thickness of the foil will you measure the thickness of only one foil or several ?
You would do it for several so as to gain an average and eliminate random uncertainties.
If I were you I would measure the thickness of 8 sheets or more. That's where the eight comes from.
Q2
In a roundabout fashion they are asking you to calculate the percentage uncertainty whilst showing workings.
For Q1. Why 8? It can be any number, right? Like. . . 10/12?
Can you please help with Q. 3 & 4?
Post by: Romeesa-Chan on June 05, 2011, 11:31:37 am
Describe means of stopping the riders gently at the far end of the track.
Post by: deadlyelder on June 05, 2011, 02:24:16 pm
Thanx
Post by: The Golden Girl =D on June 05, 2011, 02:27:40 pm
How to calclulate the Critical Angle? If anyone has example please post it for clarification I am confused.
Thanx
If a ray of light is moving from Denser *glass* to rarer *air* then we use this formula
sin c= na / na
so Critical angle equals to sin inverse the refractive index of Air over the refractive index of Glass
i.e.
c = sin-1 [ na/ na ]
Example
c = sin-1 [ 1.00 / 1.5 ]
= 42
I believe it's 42 :)
I hope you got it :D
Post by: MasterMath on June 05, 2011, 02:40:43 pm
In the photo electric effect .. when your shinning radiation on a metal's surface. If you increase the intensity you increase the number of electrons getting knocked out, but if you increase the frequency your just increasing the K.E the electrons gain right? So frequency has no effect on the current(number of electrons emitted)?
Post by: The Golden Girl =D on June 05, 2011, 02:42:18 pm
I just want to make sure of something correct me if im wrong
In the photo electric effect .. when your shinning radiation on a metal's surface. If you increase the intensity you increase the number of electrons getting knocked out, but if you increase the frequency your just increasing the K.E the electrons gain right? So frequency has no effect on the current(number of electrons emitted)?
Yes :)
Post by: Malak on June 05, 2011, 02:43:53 pm
Frequency effects KE
And intensity effects no of electrons emitted (if the frequency is equal to the threshold frequency)
Good Luck Everyone...! :D
Post by: MasterMath on June 05, 2011, 02:45:06 pm
Post by: Malak on June 05, 2011, 03:14:20 pm
Oh..I am going to do Biology next year this way..! :-\
Looks like its quite stressful..
Post by: The Golden Girl =D on June 05, 2011, 03:44:46 pm
Lets hope its mostly electricity.. i hate waves.. Got no time to study it properly..Shouldnt of done A2 and AS in the same year :'(
Not trying to discourage you or freak you out , but my teacher said that a very big portion of the exam questions' in the exam are usually Waves ,so I suggest you do take it seriously :)
Post by: Malak on June 05, 2011, 04:01:09 pm
And If you need help with any specific point/topic in wave, we all can help :)
Post by: EMO123 on June 05, 2011, 05:14:39 pm
Can the edexcel students try this one too plz? :Dask this to master_key i know that he knows this
Post by: The Golden Girl =D on June 05, 2011, 06:11:21 pm
ask this to master_key i know that he knows this
Hmm where is the post :-X :-\
Post by: Ghost Of Highbury on June 05, 2011, 06:15:19 pm
Hmm where is the post :-X :-\
Ari answered it, I deleted it. Sorry :-)
Post by: The Golden Girl =D on June 05, 2011, 06:17:24 pm
Ari answered it, I deleted it. Sorry :-)
It's okay :)
Post by: The Golden Girl =D on June 05, 2011, 06:17:57 pm
My exam is tomorrow so please reply ASAP !
Hmm I need an explanation for :
in Section A in Q 5 , I just want to know why is it ODD , is it Always ODD ? *I thought it shouldn't be a whole number and that's it.* ,so let me know Please !
Section B
Q. 14]c) , 19]c) *I need a diagram of it :-\ * , 21]c)ii)
Thanks in advance :D
Post by: rameeziiii on June 05, 2011, 06:28:08 pm
Attached.
My exam is tomorrow so please reply ASAP !
Hmm I need an explanation for :
in Section A in Q 5 , I just want to know why is it ODD , is it Always ODD ? *I thought it shouldn't be a whole number and that's it.* ,so let me know Please !
Section B
Q. 14]c) , 19]c) *I need a diagram of it :-\ * , 21]c)ii)
Thanks in advance :D
it is an odd number of half wavelengths. the half wavelength gives a path difference of 1/2 lambda or 180 antiphase compared to the original wave. thus when superimposition occurs, destructive interference occurs. if the difference is of even number of half wavelengths or whole numbers of wavelengths, the phase difference is 0 and thus constructive interference will occur.
Post by: The Golden Girl =D on June 05, 2011, 06:30:41 pm
Post by: Malak on June 05, 2011, 06:37:06 pm
Post by: The Golden Girl =D on June 05, 2011, 06:41:36 pm
EDIT: Attached.
Post by: Malak on June 05, 2011, 07:05:25 pm
Post by: The Golden Girl =D on June 05, 2011, 07:09:35 pm
Post by: Romeesa-Chan on June 05, 2011, 07:12:34 pm
Post by: Malak on June 05, 2011, 07:21:59 pm
And for 19, yes its correct
By the way, for part a (the blue ray) it doesnt reflect 'cause its the cricital angle so it should pass along the prism ( 90 degrees)
Edit: the answer for R is 1.76 * 10^-3, just to confirm for you
And lol, thanks to your scan I realized that the angle is 'c', because I had refracted the blue light ::)
Post by: The Golden Girl =D on June 05, 2011, 07:57:32 pm
21 c attached (I find it easier typing in word 'cause the formulas are clear)
And for 19, yes its correct
By the way, for part a (the blue ray) it doesnt reflect 'cause its the cricital angle so it should pass along the prism ( 90 degrees)
Edit: the answer for R is 1.76 * 10^-3, just to confirm for you
And lol, thanks to your scan I realized that the angle is 'c', because I had refracted the blue light ::)
Thanks A LOT ..Jazaki Allah Kulla kari ... Good luck with tomorrow's exam ! ;D
Post by: Nobody on June 05, 2011, 08:39:51 pm
Physics 6B, Sample Paper:
*Q1bii- where did 8 come from ? :S
*Q2ciii- No idea what the marking scheme says!
*Q3 a & b
*Q4 a & dii
Link: http://www.edexcel.com/migrationdocuments/GCE%20New%20GCE/Physics-Practical-Alternative-SAMs.pdf
Please help me. Have exam the next day (Monday)!!!
okay, for Q1 b
assuming you've got the part 1
the uncertainty given is 0.01 mm.
and the thickness is 15 micrometre
so the uncertainty% is
uncertainty
the value
here in this case,
it will be
0.01 mm
DIVIDED BY
8 X 15 micrometre
now convert all of them into meters.
and the answer will be
0.00001
divided by
0.00012
that is 8.333%
Post by: Romeesa-Chan on June 05, 2011, 08:42:12 pm
okay, for Q1 b
assuming you've got the part 1
the uncertainty given is 0.01 mm.
and the thickness is 15 micrometre
so the uncertainty% is
uncertainty
the value
here in this case,
it will be
0.01 mm
DIVIDED BY
8 X 15 micrometre
now convert all of them into meters.
and the answer will be
0.00001
divided by
0.00012
that is 8.333%
Thank you, awesomeness! :D
I am sorry, but could you help me with the Q4? Pretty please with cherry on top? :-[
Post by: Nobody on June 05, 2011, 08:59:18 pm
Thank you, awesomeness! :D
I am sorry, but could you help me with the Q4? Pretty please with cherry on top? :-[
Q 4. okay..i'll answer that
hey, you know how you got 8 in Q 1 b (i) right?
don't hesitate. ;)
Post by: Romeesa-Chan on June 05, 2011, 09:06:06 pm
Q 4. okay..i'll answer that
hey, you know how you got 8 in Q 1 b (i) right?
don't hesitate. ;)
Hmmm . . . Nope. :$
I asked my friend and she told me that we must always use 8 for folding. :/
Post by: Nobody on June 05, 2011, 09:20:23 pm
Hmmm . . . Nope. :$
I asked my friend and she told me that we must always use 8 for folding. :/
no, i guess she's wrong here.
Quote
(b) An aluminium cooking foil manufacturer states that the thickness of the foil is 15 ?m.
You have a micrometer which can read to a precision of 0.01 mm.
(i) Describe how you would try to find an accurate value for the thickness of a sheet
of the foil.
Now, look the precision of micrometer is 0.01 mm. so that's the LEAST value it can read.
and, since the sheet is 15 MICROMETER, we first convert this into milimeter
15 micrometers = 0.015 millimeter
now, we need to fold the sheet so that the screw gauge(micrometer) can be used to read it.
so, try the following
fold the sheet into 2 > 0.015 x 2 > 0.03
fold the sheet into 5 > 0.015 x 5 > 0.075
fold the sheet into 7 > 0.015 x 7 > 0.105
fold the sheet into 8 > 0.015 x 8 > 0.2
since the one in bold is close to the precision of micrometer (0.01) we fold that sheet 8 times and read the value
why we don't take 7 folds > 8 is easier and taking 7 will be too close to the precision of micrometer (0.01).
therefore. 8 ;D
get it? ;)
Post by: Romeesa-Chan on June 05, 2011, 09:26:45 pm
no, i guess she's wrong here.
Now, look the precision of micrometer is 0.01 mm. so that's the LEAST value it can read.
and, since the sheet is 15 MICROMETER, we first convert this into milimeter
15 micrometers = 0.015 millimeter
now, we need to fold the sheet so that the screw gauge(micrometer) can be used to read it.
so, try the following
fold the sheet into 2 > 0.015 x 2 > 0.03
fold the sheet into 5 > 0.015 x 5 > 0.075
fold the sheet into 7 > 0.015 x 7 > 0.105
fold the sheet into 8 > 0.015 x 8 > 0.2
since the one in bold is close to the precision of micrometer (0.01) we fold that sheet 8 times and read the value
why we don't take 7 folds > 8 is easier and taking 7 will be too close to the precision of micrometer (0.01).
therefore. 8 ;D
get it? ;)
But. . . 0.015 x 8 = 0.12 :S
Post by: Romeesa-Chan on June 05, 2011, 09:29:22 pm
Why do we multiply them both with 2? :S
Post by: Romeesa-Chan on June 05, 2011, 09:32:30 pm
Post by: Nobody on June 05, 2011, 09:33:25 pm
But. . . 0.015 x 8 = 0.12 :S
haha! typo! :P
i wanted to type 0.12 but typed 0.2 instead :P
Post by: Romeesa-Chan on June 05, 2011, 10:18:42 pm
haha! typo! :P
i wanted to type 0.12 but typed 0.2 instead :P
;)
By the way, did you try the other question? :D
Post by: Dangerous99 on June 05, 2011, 10:30:58 pm
They give us 2 values 18.66 and 18.64 and ask us to find the percentage uncertainty.
Post by: Malak on June 05, 2011, 10:32:46 pm
How would you calculate % uncertainty. I know its the first thing u answered in the topic but its not working with this questionIs there anything else in the question? :S like original reading, instrument mentioned or something ???
They give us 2 values 18.66 and 18.64 and ask us to find the percentage uncertainty.
Post by: Nobody on June 05, 2011, 10:33:35 pm
Post by: Dangerous99 on June 05, 2011, 10:35:45 pm
Is there anything else in the question? :S like original reading, instrument mentioned or something ???
the values are the time taken for 20 oscillation per second, and they give us the mean time for one oscillation which is 0.932 seconds.
Post by: Dangerous99 on June 05, 2011, 10:39:04 pm
which question? the one i answered?
no not the exact same question. I meant this ones percentage uncertainty as well
Post by: Deadly_king on June 05, 2011, 10:43:21 pm
June 2010, Q2v
Why do we multiply them both with 2? :S
This is because you made two readings before calculating the answer. I've not read the question but am pretty sure there was something like initial and final readings.
Since two readings were made, same uncertainty was carried out twice. That's why we multiply it by two ;)
Hope it helps :D
Post by: Romeesa-Chan on June 05, 2011, 10:44:51 pm
How would you calculate % uncertainty. I know its the first thing u answered in the topic but its not working with this question
They give us 2 values 18.66 and 18.64 and ask us to find the percentage uncertainty.
Percentage uncertainty = uncertainty/mean value x 100%
Post by: Dangerous99 on June 05, 2011, 10:48:11 pm
Percentage uncertainty = uncertainty/mean value x 100%
yea they did something like that in the mark scheme, they did (0.5/18.64) x 100%
I really dont get where they got the 0.5 from tho
Post by: studz on June 05, 2011, 10:50:23 pm
yea they did something like that in the mark scheme, they did (0.5/18.64) x 100%
I really dont get where they got the 0.5 from tho
even i dint get where did they bring 0.5 from
anyway guys can you list out the uncertainties of measurement of the following apparatus?
1. metre rule
2. voltmeter (not digital)
3. ammeter (not digital)
4. voltmeter (digital)
5. ammeter (digital)
6. Digital stopwatch
by the way, is human error an uncertainty too? Is human error 0.01s?
If it is, then please consider this case:
when a ball is allowed to free fall for 2s, the distance is recorded. (different sizes of ball are used) the distance has an uncertainty, however the uncertainty is human error (0.01s+0.01s) or uncertainty of the metre rule (0.5mm+0.5mm)?
thanks :)
Post by: Nobody on June 05, 2011, 10:50:56 pm
sometimes it's what YOU HAVE to remember or figure it out yourself.
it's easy ;)
want me to help you out there? ::)
Post by: studz on June 05, 2011, 10:53:21 pm
0.5 is the uncertainty.
sometimes it's what YOU HAVE to remember or figure it out yourself.
it's easy ;)
want me to help you out there? ::)
yes please moderator help :P:$
Post by: Romeesa-Chan on June 05, 2011, 10:54:58 pm
This is because you made two readings before calculating the answer. I've not read the question but am pretty sure there was something like initial and final readings.
Since two readings were made, same uncertainty was carried out twice. That's why we multiply it by two ;)
Hope it helps :D
Thank you very much! XD
Post by: Dangerous99 on June 05, 2011, 10:55:25 pm
0.5 is the uncertainty.
sometimes it's what YOU HAVE to remember or figure it out yourself.
it's easy ;)
want me to help you out there? ::)
Yes please! how do u figure that out?
they also wrote "change in 20T= 0.5 - 1.0 s [reaction time] leading to correct %"
what reaction time?!
Post by: Nobody on June 05, 2011, 10:55:59 pm
even i dint get where did they bring 0.5 from
anyway guys can you list out the uncertainties of measurement of the following apparatus?
1. metre rule
2. voltmeter (not digital)
3. ammeter (not digital)
4. voltmeter (digital)
5. ammeter (digital)
6. Digital stopwatch
by the way, is human error an uncertainty too? Is human error 0.01s?
If it is, then please consider this case:
when a ball is allowed to free fall for 2s, the distance is recorded. (different sizes of ball are used) the distance has an uncertainty, however the uncertainty is human error (0.01s+0.01s) or uncertainty of the metre rule (0.5mm+0.5mm)?
thanks :)
okay, to all those who've got trouble with uncertainty.
Uncertainty of ANY instrument is the SMALLEST readable value of it.
in a metre rule
like this one (http://www.labscientificequipments.com/product/Chemistry%20Products/Chemistry%200016%20-%20Meter%20Rule%20Wooden%20cm%20cm.jpg)
the smallest readable value is 1 mm.
so the uncertainty is 1 mm for this ruler.
Post by: Dangerous99 on June 05, 2011, 11:02:18 pm
okay, to all those who've got trouble with uncertainty.
Uncertainty of ANY instrument is the SMALLEST readable value of it.
in a metre rule
like this one (http://www.labscientificequipments.com/product/Chemistry%20Products/Chemistry%200016%20-%20Meter%20Rule%20Wooden%20cm%20cm.jpg)
the smallest readable value is 1 mm.
so the uncertainty is 1 mm for this ruler.
then shouldnt the uncertainty be 0.01 because the values they give are 18.66 and 18.46 they can only read it upto 0.01?
sorry By the way im just panicking here thats y im pressing on it so much
Post by: studz on June 05, 2011, 11:03:23 pm
This is because you made two readings before calculating the answer. I've not read the question but am pretty sure there was something like initial and final readings.
Since two readings were made, same uncertainty was carried out twice. That's why we multiply it by two ;)
Hope it helps :D
i dont get it where did we take two readings ???
we just had a table of values for time in secs.
Post by: Nobody on June 05, 2011, 11:19:40 pm
then shouldnt the uncertainty be 0.01 because the values they give are 18.66 and 18.46 they can only read it upto 0.01?
sorry by the way im just panicking here thats y im pressing on it so much
look, let's take random values from a meter rule
ALL in CENTIMETRE
56.7
43.6
34.6
1.7
09.7
99.9
now, have a look. 56 CM & 7 MM
so, the smallest value is "*.X"
so, that .X will be 0.1 and not any less.
you can't read "26.77" cm on a meter rule. that's impossible value to get.
Post by: Deadly_king on June 05, 2011, 11:23:30 pm
!st reading : 0.00
2nd reading : 3.02
So we got the same uncertainty twice ;)
Same applies for the other measurement ;)
Post by: Dangerous99 on June 05, 2011, 11:25:30 pm
look, let's take random values from a meter rule
ALL in CENTIMETRE
56.7
43.6
34.6
1.7
09.7
99.9
now, have a look. 56 CM & 7 MM
so, the smallest value is "*.X"
so, that .X will be 0.1 and not any less.
you can't read "26.77" cm on a meter rule. that's impossible value to get.
I get what u mean. 0.1 is the uncertainty on the meter rule rite?
but the question I was doing they give the values 18.66 and 18.64 so obviously they can can only read it upto 0.01 sig figures, what will the uncertainty be in this case?
Post by: Nobody on June 05, 2011, 11:28:33 pm
I get what u mean. 0.1 is the uncertainty on the meter rule rite?
but the question I was doing they give the values 18.66 and 18.64 so obviously they can can only read it upto 2 DECIMAL PLACES, what will the uncertainty be in this case?
yes!
yes! It will be 0.01.
boy, you're smart! :D ;)
significant figures & decimal place are two different things ;)
Post by: Nobody on June 05, 2011, 11:31:44 pm
even i dint get where did they bring 0.5 from
anyway guys can you list out the uncertainties of measurement of the following apparatus?
1. metre rule
2. voltmeter (not digital)
3. ammeter (not digital)
4. voltmeter (digital)
5. ammeter (digital)
6. Digital stopwatch
by the way, is human error an uncertainty too? Is human error 0.01s?
If it is, then please consider this case:
when a ball is allowed to free fall for 2s, the distance is recorded. (different sizes of ball are used) the distance has an uncertainty, however the uncertainty is human error (0.01s+0.01s) or uncertainty of the metre rule (0.5mm+0.5mm)?
thanks :)
1. metre rule > 1 millimeter
2. voltmeter (not digital) >
it depends on the instrument that you use.
(http://www.hiwtc.com/photo/products/26/00/15/1507.jpg)here it will be 10V
(http://12volt.com.au/General%20Htmls/webcat2003/arrid%20large%2010160.gif)here it will be 2V
just measure the smallest value in the instrument.
that's the uncertainty ;)
3. ammeter (not digital) >
again depends..
(http://product-image.tradeindia.com/00157741/b/0/Analog-Ammeter.jpg) here it will be 2 A
4. voltmeter (digital) > if you read "5.034" uncertainty is "0.001 A" if you read "50.34" uncertainty is "0.01 A" ans so on
5. ammeter (digital) > same a above. ^^
6. Digital stopwatch > 0.1 second
Post by: Dangerous99 on June 05, 2011, 11:33:17 pm
yes!
yes! It will be 0.01.
boy, you're smart! :D ;)
significant figures & decimal place are two different things ;)
okai u know what? im feeling really dumb rite now cos according to what u just said the uncertainty is 0.01 not 0.5 like the mark scheme says.
Post by: Assi1993 on June 05, 2011, 11:34:18 pm
* Stopwatch: 0.2 secs
* Metre rule: 1mm
* Vernier caliper: 0.1mm
* Micrometer: 0.01mm
0.1 for the voltmeter
Post by: Assi1993 on June 05, 2011, 11:37:45 pm
okai u know what? im feeling really dumb rite now cos according to what u just said the uncertainty is 0.01 not 0.5 like the mark scheme says.
which question,paper?? i want to check what it is
Post by: Romeesa-Chan on June 05, 2011, 11:43:19 pm
Post by: Nobody on June 05, 2011, 11:45:01 pm
Uncertainties in some instruments:
* Stopwatch: 0.2 secs
* Metre rule: 1mm
* Vernier caliper: 0.1mm
* Micrometer: 0.01mm
0.1 for the voltmeter
stop watch : 0.1 seconds. >>sure
Vernier caliper: 1mm
Micrometer: 0.1mm
Post by: Dangerous99 on June 05, 2011, 11:50:28 pm
which question,paper?? i want to check what it isquestion 2c page 29
Post by: Assi1993 on June 05, 2011, 11:51:01 pm
Please help me with the last question part d! :-\
u have 2 just find the gradient!!
rearranging the equation gives us,
lnA=-ux+lnAo
y =mx+c
-u is the gradient ...which u can find...
take a larger gradient triangle always...
so take the 1st plotted value and the last 1...
In this, u can take the coordinates(0,6.869) and (31.74, 5.257)
Post by: Romeesa-Chan on June 06, 2011, 12:00:02 am
u have 2 just find the gradient!!
rearranging the equation gives us,
lnA=-ux+lnAo
y =mx+c
-u is the gradient ...which u can find...
take a larger gradient triangle always...
so take the 1st plotted value and the last 1...
In this, u can take the coordinates(0,6.869) and (31.74, 5.257)
I got it. I am not sure whether my graph is right. Can you lemme know if it's right if I post a picture of it? Please? :D
Post by: Dangerous99 on June 06, 2011, 12:01:17 am
So its 0.5/18.64 x 100%
Post by: Nobody on June 06, 2011, 12:07:32 am
so, you're marring him now? ::) :P
reaction time is normally used only when you have to time something.
like calculating the time for a pendulum to complete 10 oscillation.
Post by: Assi1993 on June 06, 2011, 12:14:08 am
Please don't temme my graph is wrong! :P
well u should have used a different scale..with shorter intervals..like 6.4,6.6,6.8
and...go through the examiner reports..they have explained about how to do the line of best fit etc
Post by: Dangerous99 on June 06, 2011, 12:15:05 am
Thanks for ur help By the way, i mustve been eating ur head tonite :-[
Post by: Romeesa-Chan on June 06, 2011, 12:16:45 am
well u should have used a different scale..with shorter intervals..like 6.4,6.6,6.8
and...go through the examiner reports..they have explained about how to do the line of best fit etc
Sure! ;)
Thanks! :D
Post by: Nobody on June 06, 2011, 12:21:15 am
Please don't temme my graph is wrong! :P
the graph is good.
In this, u can take the coordinates(0,6.869) and (31.74, 5.257)
the graph goes a BIT away. makes sure you don't repeat it. ;)
and, like Assi1993 said use a shorter scale. ;)
lol im only 18 :P
Thanks for ur help by the way, i mustve been eating ur head tonite :-[
you said so. didn't you? ;) :P
nah, that's what we're here for.
Helping you guys. :)
Post by: Romeesa-Chan on June 06, 2011, 12:29:20 am
well u should have used a different scale..with shorter intervals..like 6.4,6.6,6.8
and...go through the examiner reports..they have explained about how to do the line of best fit etc
By the way, what if we get negative value for the gradient? We don't put the -ve sign, right?
Post by: Romeesa-Chan on June 06, 2011, 12:32:25 am
Post by: studz on June 06, 2011, 05:30:10 am
though it is still kinda confusing but anyway hope all goes well today!
good luck to all :D
and lmao at the marrying part hehe!! :P
Post by: Nobody on June 06, 2011, 04:10:12 pm
Post by: studz on June 06, 2011, 05:24:08 pm
Post by: wakemeup on June 20, 2011, 12:17:15 pm
Also, in Q13b, how are we supposed to estimate the radius if we don't have any reference in the diagram?
Post by: Romeesa-Chan on June 20, 2011, 12:17:31 pm
*A 47micro F capacitor is charged to a pd of 15V and then discharged through a 100k Ohm resistor. How long does it take for half of the energy stored in the capacitor to have dissipated in the circuit?
*A beam of electrons enters a region of uniform electric field, moving at right angles to the field and is deflected into a circular path.
Draw a diagram to show the path of electrons, including directions of the electron velocity 'v' and the magnetic field 'B'.
Post by: Romeesa-Chan on June 20, 2011, 12:31:15 pm
I'm pretty much completely lost on Q16 of Jun 10 paper 4. Would really appreciate it if someone could clearly explain parts bi, ci and cii.
Also, in Q13b, how are we supposed to estimate the radius if we don't have any reference in the diagram?
I am lost in 13b myself too. :(
Ok, Q16:
bi) Capacitor stores charge, as charge flows in during the charging process, pd is also formed.
If the voltage is constant, the capacitor will not discharge. The graph is during the charging process. Hence, pd is constant.
(remember: during discharging process, pd falls.)
ci) Use E= 1/2*C*V^2
(take voltage value from the graph)
cii) Compare the graph with the previous one and see the change in voltage value. (pd decreases)
Capacitor is initially charging from power supply then the capacitor is discharging exponentially.
(remember: the changing position of switches causes the discharge process, instead here. . . we added a power supply)
I hope that's clear. :)
Post by: viktress on June 20, 2011, 01:31:57 pm
A runner of mass 65kg is running a circular running track at a sped of 5.0m/s. The magnitude of the runner's change in momentum from starting to end position after half a revolution is
A. 0 N s
B. 325 N s
C. 460 N s
D. 625 N s
The correct answer according to the book is D, but I just can't seem to be able to get i. Could you please try to work it out? Thanks!
Post by: Romeesa-Chan on June 20, 2011, 02:43:09 pm
Hey guys could you help me out with this question I've trying to crack for almost a day now? It's from the Edexcel a2 revision guide (physics).
A runner of mass 65kg is running a circular running track at a sped of 5.0m/s. The magnitude of the runner's change in momentum from starting to end position after half a revolution is
A. 0 N s
B. 325 N s
C. 460 N s
D. 625 N s
The correct answer according to the book is D, but I just can't seem to be able to get i. Could you please try to work it out? Thanks!
The last option D is wrong- suppose to be 650N.
F*t= m*v => change in momentum
Think of half a circle.
at the "top" it is travelling ------->>> at p=mv = 325Ns
at the "bottom" it is travelling <<<------- at p=mv = -325Ns
Therefore the change in momentum is = 325 - (-325) = 650Ns
Post by: viktress on June 20, 2011, 03:19:02 pm
Post by: wakemeup on June 20, 2011, 03:39:46 pm
I am lost in 13b myself too. :(
Ok, Q16:
bi) Capacitor stores charge, as charge flows in during the charging process, pd is also formed.
If the voltage is constant, the capacitor will not discharge. The graph is during the charging process. Hence, pd is constant.
(remember: during discharging process, pd falls.)
ci) Use E= 1/2*C*V^2
(take voltage value from the graph)
cii) Compare the graph with the previous one and see the change in voltage value. (pd decreases)
Capacitor is initially charging from power supply then the capacitor is discharging exponentially.
(remember: the changing position of switches causes the discharge process, instead here. . . we added a power supply)
I hope that's clear. :)
Sorry, I'm still pretty lost. How does the capacitor keep the p.d constant? By discharging a constant pd? But the pd varies when the capacitor discharges. And if the capacitor is discharging exponentially why does the graph look like that? Could you please explain cii as well. (You explained bii where you wrote ci)
Post by: Romeesa-Chan on June 20, 2011, 03:48:59 pm
Sorry, I'm still pretty lost. How does the capacitor keep the p.d constant? By discharging a constant pd? But the pd varies when the capacitor discharges. And if the capacitor is discharging exponentially why does the graph look like that? Could you please explain cii as well. (You explained bii where you wrote ci)
The pd is constant during the CHARGING process. That's cos the pd is equal to the emf of the source.
During discharging, the pd value falls as the charge decreases. (the capacitor decreases exponentially, we must just say that pd falls)
Sorry, if I am not making sense here.^^ :(
cii) T= RC
R= T/C
(value for capacitance is given and T value is taken from the graph)
:)
Post by: wakemeup on June 20, 2011, 04:01:49 pm
The pd is constant during the CHARGING process. That's cos the pd is equal to the emf of the source.
During discharging, the pd value falls as the charge decreases. (the capacitor decreases exponentially, we must just say that pd falls)
Sorry, if I am not making sense here.^^ :(
cii) T= RC
R= T/C
(value for capacitance is given and T value is taken from the graph)
:)
Thanks, I understand a bit better now but as for cii what value of T are we supposed to take? Because the mark scheme has 15-20 ms in it and I don't know how that came about.
Post by: Romeesa-Chan on June 20, 2011, 04:05:11 pm
Thanks, I understand a bit better now but as for cii what value of T are we supposed to take? Because the mark scheme has 15-20 ms in it and I don't know how that came about.
That's good. :D
Lemme know if you don't.
Post by: wakemeup on June 20, 2011, 04:08:53 pm
That's good. :D
Lemme know if you don't.
For value of T, I took 16 ms. :)
Yes, but how? Are you supposed to take a period, because that's about 10 and nowhere near the range in the mark scheme.
Post by: Romeesa-Chan on June 20, 2011, 04:12:55 pm
Yes, but how? Are you supposed to take a period, because that's about 10 and nowhere near the range in the mark scheme.
Okay. This seems crazy but I tried it this way! (makes sense to me though)
See the previous graph: The time is constant from 8ms to 28 ms. The double peaks in the graph.
Time constant is always the same for every circuit. So we have 20ms for T value.
Does that make sense?
Post by: Romeesa-Chan on June 20, 2011, 04:36:01 pm
Post by: wakemeup on June 20, 2011, 04:43:08 pm
Any more doubts? I will be leaving now. (:
No, but thanks for all your help. That is the only chapter that's still not super clear to me. I'll just keep my fingers crossed and hope they don't ask any super complex question from that (or any other chapter) tomorrow.
Thanks again.
Post by: Dangerous99 on June 26, 2011, 07:58:08 pm
I dont get question 9 from june 2010 paper. It says For a black body radiator, the frequecy at which maximum radiation of energy occurs is proportional to?
a) T^-4
b) T^-1
c) T
d) T^4
Post by: Romeesa-Chan on June 26, 2011, 08:33:46 pm
Hey can somebody please help me out with physics unit 5?? ???
I dont get question 9 from june 2010 paper. It says For a black body radiator, the frequecy at which maximum radiation of energy occurs is proportional to?
a) T^-4
b) T^-1
c) T
d) T^4
It says 'max frequency'
Max wavelength is proportional to 1 / T (Wien's law)
Frequency is proportional to 1 / wavelength, so it's proportional to T.
:D
Post by: Dangerous99 on June 26, 2011, 08:43:52 pm
I was doing something like FA = stephen boltsman equation heh heh :-\
Post by: Romeesa-Chan on June 26, 2011, 08:49:57 pm
OH! Thanks it makes sense now ;D
I was doing something like FA = stephen boltsman equation heh heh :-\
No problem! :D
It's quite a tricky question which makes us jump to the answer immediately!
Any more doubts? (:
Post by: Dangerous99 on June 27, 2011, 12:25:52 am
No problem! :D
It's quite a tricky question which makes us jump to the answer immediately!
Any more doubts? (:
No thanks :) inshallah well ace the exam tom ;)
By the way i really like ur kuran kaname gif lol HAWT :P
Post by: Romeesa-Chan on June 27, 2011, 02:06:12 am
No thanks :) inshallah well ace the exam tom ;)
By the way i really like ur kuran kaname gif lol HAWT :P
Insha Allah!!!!!!!!!! :D
OMGGGGGGGGGGGGGGGGGGGGGG. You watched VK?!!!!!!!!!! AWESOME STUFF. :D
Haha. Good luck !!!!!!!!!!! :D
Post by: YU-GI-OH! on September 17, 2011, 04:54:39 pm
Post by: Malak on September 17, 2011, 05:00:22 pm
Why do aircraft take off and land into the wind ???Is that the question? :-\
Check this, might be helpful: http://answers.yahoo.com/question/index?qid=20080831193525AASbJXE (http://answers.yahoo.com/question/index?qid=20080831193525AASbJXE)
Post by: Banana on September 17, 2011, 06:07:17 pm
Why do aircraft take off and land into the wind ???
Thats to ensure the polishing of the plane isn't ruined ;D
Check out ang3l's link :)
Post by: YU-GI-OH! on September 28, 2011, 09:15:33 pm
Thanks in advance :)
Post by: astarmathsandphysics on September 28, 2011, 09:26:41 pm
Aircraft land into the wind becuase it is an aid to braking.
Post by: astarmathsandphysics on September 28, 2011, 09:33:27 pm
Don't understand this. Please expand.
Post by: The Golden Girl =D on September 29, 2011, 12:23:31 pm
If someone coasts from a greater speed than before, why would the deceleration be greater? ???
Thanks in advance :)
Can you give me a link to the paper or perhaps write down the Whole Questions , it'll make it easier for us to answer =]
Post by: YU-GI-OH! on September 29, 2011, 10:18:57 pm
Post by: astarmathsandphysics on September 29, 2011, 10:48:06 pm
Post by: astarmathsandphysics on September 29, 2011, 10:55:06 pm
Post by: The Golden Girl =D on October 01, 2011, 09:24:26 am
T = 600 revolutions per minute.
Omega at the end is supposed to be about 63 rads-1
I know this is Math , but I just can't seem to get the answer =,=
Thanks =]
Post by: Banana on October 01, 2011, 09:38:52 am
Omega (Angular velocity) = 2 pie / T
T = 600 revolutions per minute.
Omega at the end is supposed to be about 63 rads-1
I know this is Math , but I just can't seem to get the answer =,=
Thanks =]
w = 2pi / T
T is the period (time taken for one rotation) SO for 600 rpm each rotation takes 60/600 seconds = 0.1 sec
Substitute into the equation:
w= 2pi / (0.10) = 63 rad s-1 as the question says ;)
Post by: The Golden Girl =D on October 01, 2011, 09:46:53 am
w = 2pi / T
T is the period (rotations in one second) SO for 600 rpm there will be 600/60 revolitions per second( the period)
Substitute into the equation:
w= 2pi / (10) = 0.63 rad s-1
So basically it isn't 63 but 0.63 ;)
In the Question it says it's about 63 though. and isn't 600 supposed to be multiplied by 60 ? :-\
Another Question ,
Q. The average wind speed in the UK is 5.8 m s–1, which results in an actual average power output of 100W. Discuss whether it would be better for the environment to replace some filament light bulbs with low energy bulbs than to use this turbine. Assume each filament light bulb is rated at 100W and each low energy bulb is rated at 11 W.
Answer. The 100 W is an average over the whole day. Most households would use light bulbs for 6 hours a day in no more than 4 rooms, so this would mean no other energy was needed for lighting. 4 low energy bulbs would be 44 W for 6 each hours so would require energy from the National grid.
I don't get it.
Thanks =]
Post by: The Golden Girl =D on October 01, 2011, 11:36:35 am
w = 2pi / T
T is the period (time taken for one rotation) SO for 600 rpm each rotation takes 60/600 seconds = 0.1 sec
Substitute into the equation:
w= 2pi / (0.10) = 63 rad s-1 as the question says ;)
Today my brain is BLANK ::) so hmm why did you divide 60 by 600 ?
Explain it in details please :-[ :-\
Post by: Banana on October 01, 2011, 11:48:31 am
Today my brain is BLANK ::) so hmm why did you divide 60 by 600 ?
Explain it in details please :-[ :-\
Okies....
Period is the time taken for one rotation..
Using proportion:
Revolutions Time
600-------------------> 1 minute =60 seconds
1 --------------------> T seconds
Cross multiply: 600T = 60
Therefore, T = 60/600 = 0.1 seconds
This is the period!
All you have to do now is substitute into the equation: w= 2pi/ T
Voila!! 8)
Post by: astarmathsandphysics on October 01, 2011, 03:04:59 pm
Post by: The Golden Girl =D on October 01, 2011, 04:50:54 pm
Post by: astarmathsandphysics on October 01, 2011, 04:52:53 pm
Post by: The Golden Girl =D on October 01, 2011, 05:02:52 pm
Post by: astarmathsandphysics on October 01, 2011, 05:12:58 pm
Post by: The Golden Girl =D on October 08, 2011, 07:40:59 am
a) Calculate the size of the Force between the charges.
Substituing F = kQ1Q2/r2
F = [(9*109) * (30*10-9) * (30*10-9) ] / [0.25]2
F = 1.3*10-4 N
I don't get how they came up with the values in Red , they aren't in the Question (green) :-\
{ Unit 4 - Electric fields } I got this Question from a Book.
Thanks Guys =]
Post by: The Golden Girl =D on October 08, 2011, 08:18:37 am
That's the only Data given by the Question to be honest :(
Post by: Romeesa-Chan on October 08, 2011, 08:31:41 am
TWO Point charges of + 0.50 micro C and - 0.50 micro C ,from an electric dipole of length 0.12 m
a) Calculate the size of the Force between the charges.
Substituing F = kQ1Q2/r2
F = [(9*109) * (30*10-9) * (30*10-9) ] / [0.25]2
F = 1.3*10-4 N
I don't get how they came up with the values in Red , they aren't in the Question (green) :-\
{ Unit 4 - Electric fields } I got this Question from a Book.
Thanks Guys =]
Not sure ... it's all messed up. =X
Post by: The Golden Girl =D on October 08, 2011, 08:51:56 am
Post by: Romeesa-Chan on October 08, 2011, 09:42:14 am
^ Check your InboxI am pretty much useless.
I hope astar can help you out here. (:
Post by: The Golden Girl =D on October 08, 2011, 09:54:23 am
Can Anyone help me on this Question please , thanks again =D
Post by: astarmathsandphysics on October 09, 2011, 11:54:43 am
Post by: astarmathsandphysics on October 09, 2011, 12:02:56 pm
F=9*10^9*(0.5*10^-6)^2/0.12^2 =0.156 N
Post by: astarmathsandphysics on October 09, 2011, 12:06:31 pm
Post by: The Golden Girl =D on October 09, 2011, 02:13:39 pm
Post by: dk bose on October 26, 2011, 11:03:58 am
palaeontologists are able to deduce much about the behaviour of dinosaurs from the study of fossilised footprints.The tracks in the attachment show the path of a Tyrannosaurus rex as it attacks a stationary Triceratops.The time between the footprints is 0.62s.(a)Show that the maximum speed of the Tyrannosaurus rex is about 10m/s.(b)the tyrannasaurus rex has a mass of 7000kg.calculate the momentuemjust before it hits the triceratops.(c)Triceratopshas a mass of 5000kg.calculate their combined speed immediately after the collision.(d)The skull of tyrannosaurus rex is heavily reinforced to withstand the force produced in such a collision.Calculate the force exerted on the Tyrannosaurus rex if the time taken to reach their combined speed after the collision is 0.30s.
Post by: astarmathsandphysics on October 26, 2011, 10:08:44 pm
Post by: The Golden Girl =D on October 29, 2011, 02:48:24 pm
show the direction of induced e.m.f.
Diagram attached having answer. the pink arrow shows the direction of e.m.f.
how did we know the direction of the resistance :o
thx
Post by: The Golden Girl =D on October 29, 2011, 03:12:55 pm
Post by: iluvme on October 29, 2011, 04:18:08 pm
Post by: The Golden Girl =D on October 29, 2011, 04:20:47 pm
Post by: astarmathsandphysics on October 29, 2011, 07:01:09 pm
Use the left hand rule - it is upwards as you drew it in the diagram.
Post by: The Golden Girl =D on October 29, 2011, 07:08:30 pm
This topic sure does confuse me =[
can u explain in detail (perhaps a diagram could help) thanks sir =]
Post by: astarmathsandphysics on October 30, 2011, 08:01:00 am
Post by: The Golden Girl =D on October 30, 2011, 03:34:29 pm
Post by: astarmathsandphysics on October 30, 2011, 04:50:10 pm
http://astarmathsandphysics.com/a_level_physics_notes/electricity/a_level_physics_notes_flemings_left_hand_rule.html
Post by: The Golden Girl =D on October 31, 2011, 12:25:16 pm
Post by: astarmathsandphysics on October 31, 2011, 02:19:48 pm
Post by: SZM on November 02, 2011, 09:50:19 am
i know pastpaper questions, questions from revison guide, but i am teling u, apart from that, where can i do more questions based on mechanic in edexcel physics unit 1.
Under edexcel physics unit 1, under mechanic, the chapters are
1. motion equations and graphs
2. Combining and resolving vectors.
3. Force and acceleration
4. Gravity and free-body diagrams
5. Projectile motion
6. Work and power
7. Energy and energy conservation.
under mechanic 1, which is M1, the topics are-
1. Mathematical model in mechanics
2. Kinematics of a particle moving in a straight line
3. Dynamics of a particle moving in a straight line or plane
4. Statics of a particle
5. Moments
6. Vectors
under mechanic 2, which is M2, the topics are-
1. kinematics of a particle moving in a straight line
2.Centres of mass
3.Work, energy and power
4.Collisions
5.Statics of rigid bodies
so tel me wat should i do?? i need ur suggestion to this matter.
waitin for ur reply.
need ur reply urgenlty
Post by: astarmathsandphysics on November 02, 2011, 12:22:05 pm
Post by: dk bose on November 02, 2011, 08:40:29 pm
A ball on a snooker table is hit by another ball and travels at a distance of 50cm due west.it is then hit again and travels a distance of 30cm due north.Using a scale drawing,or by calculation,work out the snooker ball's displacement from its starting position.
A ship is travelling at 5m/s with a bearing of 20 deg east of north.there is a current of 1m/s flowing from the west.What is the resultant velocity of the ship?
Aristotle argued that a force was needed in order to keep an object moving.Describe some everyday situations that are consistent with this argument. Suggest a more scientific explanation for each case that you describe.
ANYONE PLZZZ TRY TO SOLVE THESE QUESTIONS AND GIVE THE SOLUTIONS AS FAST AS POSSSIBLE!!!!!! ITZZZZ URGENT!!!!!!!!
Post by: astarmathsandphysics on November 03, 2011, 08:55:51 pm
For the second question you have to draw a scale diagram for which you need an artist.
3rd question - pushing a car uphill.
See here http://www.astarmathsandphysics.com/ib_physics_notes/history_and_development_of_physics/ib_physics_notes_aristotles_idea_of_motion.html
The scientific view would be that air resistance acts against the direction of motion to slow the velocity.
Post by: EVIL DOCTOR on November 20, 2011, 03:39:35 pm
yard. He undergoes successive displacements 3.50 m south,
8.20 m northeast, and 15.0 m west. What is the resultant
displacement?
need asap
Post by: Malak on November 20, 2011, 06:58:12 pm
Your dog is running around the grass in your backDraw a diagram like this (attached)
yard. He undergoes successive displacements 3.50 m south,
8.20 m northeast, and 15.0 m west. What is the resultant
displacement?
need asap
You have to resolve the pink vector into its component.
As it is in the north-east direction the angle would be 45.
So the x-component will be 8.2cos45 and y will be 8.2sin45
Now add them up as we normally do in vectors.
Post by: astarmathsandphysics on November 20, 2011, 10:55:49 pm
east 8.2 sin 45 -15 =-9.2
displacement= 2.3 m north and -9.2 east
Post by: Romeesa-Chan on November 21, 2011, 04:46:19 am
dear in physics unit 1, i came across the spec. point, that is ==
demonstrate an understanding of how ICT can be used to collect data for, and display, displacement/time and velocity/time graphs for uniformly accelerated motion and compare this with traditional methods in terms of reliability and validity of data///
wat does this spec, point want and expect to know from us??? can u give me excellent notes for this.. Wat does ICT stand for???
Motions can be represented using graphs.
Traditionally, data were collected using rulers and stopwatches.
ICT methods are more accurate, efficient and fast.
An example of ICT method is a motion sensor.
Post by: EVIL DOCTOR on November 21, 2011, 12:00:06 pm
Draw a diagram like this (attached)the question didnot mention the angle of north east
You have to resolve the pink vector into its component.
As it is in the north-east direction the angle would be 45.
So the x-component will be 8.2cos45 and y will be 8.2sin45
Now add them up as we normally do in vectors.
who told u its 45 degree ??
Post by: astarmathsandphysics on November 21, 2011, 02:22:20 pm
north=0
east =90
(0+90)/2=45
Post by: EVIL DOCTOR on November 21, 2011, 02:50:29 pm
of aS
5 3.00j^ m/s2 and an initial velocity of vS
i 5 5.00i^ m/s.
Find (a) the vector position of the particle at any time t,
(b) the velocity of the particle at any time t, (c) the coordinates
of the particle at t 5 2.00 s, and (d) the speed of the
particle at t 5 2.00 s.
Post by: astarmathsandphysics on November 21, 2011, 03:09:43 pm
Post by: EVIL DOCTOR on November 21, 2011, 03:30:28 pm
of a
= 3.00j^ m/s2 and an initial velocity of v is
= 5.00i^ m/s.
Find (a) the vector position of the particle at any time t,
(b) the velocity of the particle at any time t, (c) the coordinates
of the particle at t = 2.00 s, and (d) the speed of the
particle at t= 2.00 s.
Post by: astarmathsandphysics on November 21, 2011, 10:41:57 pm
b)v=u+at=5i+3jt
c)s=5i*2+1.5j*2^2=10i+6j
d)v=5i+3j*2=5i+6j so speed =sqrt(5^2+6^2)=7.81m/s
Post by: dk bose on December 09, 2011, 02:20:04 pm
block of wood of mass 1.25 kg. Calculate the speed of the bullet if the
block moves forward with a speed of 3.9m/s after the impact.
plzz solve this question and give the solutions as fast as possible!!!!
Post by: astarmathsandphysics on December 09, 2011, 09:27:22 pm
to start, momentum is due to bullet only:m*v=0.05*v
after, momentum is due to bullet + block together m*v=(1.25+0.05)*3.9=5.07
momentum before=momentum after
0.05*v=5.07 so v=5.07/0.05=101.4 m/s
Post by: dk bose on December 10, 2011, 12:39:29 pm
The cars remain together after the collision. Calculate the velocity of the cars immediately after the collision, and the fraction of kinetic energy lost. Account for the loss of kinetic energy.
solve this question.its urgent!!!!
Post by: astarmathsandphysics on December 10, 2011, 07:36:27 pm
Post by: astarmathsandphysics on December 10, 2011, 10:28:36 pm
Post by: dk bose on December 11, 2011, 02:02:05 pm
Post by: dk bose on December 11, 2011, 03:01:32 pm
off. How far from the centre of the fan can the spider safely go?
one more tough questions,solve it if u can??
Post by: The Golden Girl =D on December 11, 2011, 05:37:22 pm
Whoever can help PLEASE HELP !
I have an exam tomorrow and I need someone to explain to me why I got my answers wrong.
I got 2/10 in the June 2011 Section A :o :o :-X :-X :-X :-X :'( :'(
ALL Question in Section A EXCLUDING Q.5 and 10.
so Please help !
Attached.
Post by: astarmathsandphysics on December 11, 2011, 06:15:51 pm
off. How far from the centre of the fan can the spider safely go?
need to know mass of s[ider
Post by: astarmathsandphysics on December 11, 2011, 06:39:25 pm
a=v^2/r ==(2 pi r f)^2/r =4 pi^2 f^2 r =0.3g so r=0.3g/(4 pi^2 f^2)=2.94/(4 pi^2 *(100/60)^2) = 0.026 m
Post by: astarmathsandphysics on December 11, 2011, 06:40:30 pm
I have an exam tomorrow and I need someone to explain to me why I got my answers wrong.
I got 2/10 in the June 2011 Section A
ALL Question in Section A EXCLUDING Q.5 and 10.
so Please help !
lokking now
Post by: astarmathsandphysics on December 11, 2011, 07:02:41 pm
1. D since f=ma=m dv/dt= d(mv)/dt= rate of change of momentum. F is equal to the weight.
2. A applying the left hand rule around the loop shows the force always inwards
3. C s since time in seconds
4.C flux linkage is BAN =0.002*0.0004*50
5. D. since gravity mg acts down and centrifugal force=mv^2/r acts up
6. B lenz's law - if the switch is opened the current is Y decreases so the current in X increases to maintain the current in Y
7.mv^2/r=Bev so v=Ber/m=0.02*1.6*10^-19*1.5/(1.67*10^-27)=2.9*10^6 B
Post by: The Golden Girl =D on December 11, 2011, 07:10:36 pm
Post by: The Golden Girl =D on December 11, 2011, 07:14:58 pm
Same paper :
Q.8 and Q.9... as well as Q]12)b) [ I never seem to understand these graphs ] .. Q.14 [ didn't get the MS at all :$ ] ..Q.17]c)iv)
I guess that's all :-[ :-[ :-\
Post by: astarmathsandphysics on December 11, 2011, 07:17:55 pm
9.1.8*10^-29 kg
1Mev/c^2=1*10^6*1.6*10^-19/(3*10^8)^2 =1.8*10^-30 kg
(1.8*10^-29)/1.8*10^-30=10 A
10 C since E=mc^2
Post by: The Golden Girl =D on December 11, 2011, 07:21:07 pm
and 9 the answer of nine isn't within the option of A to D :S
Post by: astarmathsandphysics on December 11, 2011, 07:24:52 pm
ln (R)=ln(R_0) - ut
-u is gradient = (3.26-4.28)/(24-8)=-1.02/16=-0.06375/cm
14. Between the plates the field E is constant so the force on the charged ball is F=Eq and acceleration is a=F/m=Eq/m also constant (so v is a straight line). When the ball its a plate is is charged with opposite sign so is repelled from plates but as before, acceleration constant.
Post by: astarmathsandphysics on December 11, 2011, 07:26:29 pm
Post by: astarmathsandphysics on December 11, 2011, 07:28:14 pm
Post by: astarmathsandphysics on December 11, 2011, 07:29:18 pm
Post by: The Golden Girl =D on December 11, 2011, 07:33:59 pm
Post by: The Golden Girl =D on December 11, 2011, 07:36:00 pm
Post by: astarmathsandphysics on December 11, 2011, 07:39:11 pm
Post by: The Golden Girl =D on December 11, 2011, 07:41:14 pm
Post by: astarmathsandphysics on December 11, 2011, 07:43:13 pm
Post by: astarmathsandphysics on December 11, 2011, 07:43:41 pm
Post by: The Golden Girl =D on December 11, 2011, 07:45:58 pm
Q.14 sir is about the electric fields and the carbon coated ball , it's in Section B.
Post by: The Golden Girl =D on December 11, 2011, 08:08:46 pm
Post by: The Golden Girl =D on December 11, 2011, 08:13:58 pm
Be right back
Post by: astarmathsandphysics on December 11, 2011, 08:14:47 pm
Post by: astarmathsandphysics on December 11, 2011, 08:19:33 pm
Q8 is B cos the mass increases with speed so if the mass is much larger than the mass it has when it is moving, it must be moving at relativistic speed 9close to the speed of light)
Post by: The Golden Girl =D on December 11, 2011, 08:32:29 pm
Post by: astarmathsandphysics on December 11, 2011, 08:35:24 pm
Post by: The Golden Girl =D on December 11, 2011, 08:36:56 pm
Post by: dk bose on December 12, 2011, 12:55:43 pm
Post by: astarmathsandphysics on December 12, 2011, 04:33:44 pm
Post by: dk bose on December 13, 2011, 01:10:57 pm
A ceiling fan is turning at a rate of 100 revolutions per minute. A spider
is clinging to a blade of the fan. If the spider experiences a centripetal
acceleration greater than 0.3g, it will lose its grip on the blade and be flung
off. How far from the centre of the fan can the spider safely go?
Plzz try it one more time and give the solutions as fast as possible.!!!!!
Post by: astarmathsandphysics on December 13, 2011, 01:35:42 pm
r<v^2/0.3g
v=100/60*circumference of circle=(100/60)*2 pi r msince 100/60 revs per second
so r <(100/60 *2 pi r)^2/0.3g=37.3r^2 so r<1/37.3=0.027m
Post by: dk bose on December 15, 2011, 03:13:15 pm
Post by: astarmathsandphysics on December 17, 2011, 05:56:07 pm
Like the retarded cowboy who thought Iraq could destroy the world?
Post by: dk bose on December 17, 2011, 09:41:18 pm
1)a racing car is moving at constant velocity along a track.explain how newton's first law is satisfied for this racing car.
2)Describe and explain how the resultant force on a skydiver varies from the moment they jump from a plane.
3)A person standing on a bus is thrown towards the rear of the bus as it starts to move forwards and to the front as it slows down.why?
Post by: astarmathsandphysics on December 18, 2011, 02:51:01 pm
2)Air reistance zero to start. It increases as the parachitist falls until it equals his weight and his velocity is constant. When he opens his chute air resistance increases suddenly so there is resultant upwards force and acceleration upwards until air resistance equals weight again and speed is constant.
3)Because the passenger has inertia - his body will resist any attempt to change its state of motion.
Post by: dk bose on December 18, 2011, 10:06:39 pm
1)When a driving force is removed,a moving object 's drag will slow it down to a stop.This means it loses all its kinetic energy.explain how this disappearance of energy can be in keeping with the law of conservation of energy.
2)one way of storing surplus electrical energy from a power station is to use it to raise water from a lower reservoir to a higher one.This water can then be released to generate electricity again later.explain how this system might be limited by the law of conservation of energy.
Post by: astarmathsandphysics on December 19, 2011, 11:51:43 am
2)Some of the energy input to raise the water is lost to friction, and some of the potential energy of the water is also lost to friction as it falls.
I like jokes, so you have to live with them.
Post by: dk bose on December 19, 2011, 10:04:30 pm
2)An aeroplane carrying out a parcel releases a parcel while travelling at a steady speed of 90m/s at an altitude of 200m.calculate:
a)th time between the parcel leaving the aeroplane and it striking the ground.
b)the horizontal distance travelled by the parcel in this time .
c)the speed at which the parcel strikes the ground.
Post by: astarmathsandphysics on December 19, 2011, 11:13:37 pm
Post by: dk bose on December 21, 2011, 02:24:23 pm
2)A penalty stroke is flicked from a distance of 6.40m from the goal line.The striker scoops it so that the ball leaves the ground at 45 degrees angle and a speed of 8m/s.How does the goalkeeper have to make a save before the ball crosses the goal line?
3)The study of mechanics in sport is a popular and often profitable new idea of scientific study. Descibe how a sport scientist could use ICT to collect data to study the movement of players and equipment overtime. Explain why technological developments have made the data collected more valid and reliable than with traditional methods of studying mechanics.
Post by: astarmathsandphysics on December 21, 2011, 09:40:52 pm
2)I presume you mean to find the height
The horizontal speed is 8cos 45=5.656m/s
Use x=u_x t to find the time:t=x/u_x =6.4/5.656=1.132 s
now work vertically
s=?
u=8sin45=5.656
v=? (not needed)
a=-9.8
t=1.132 s
s=ut+1/2at^2 =5.656*1.132-1/2*-9.8*1.132^2=0.12m the ball crosses at a height of 0.12m
3)Fit all the players with trackers and take their position every second. Computers are more reilable and can gather and analyse data methodically.
3)
Post by: dk bose on December 21, 2011, 11:04:12 pm
2)A 47*10^-6 F capacitor is charged to a potential difference of 15V and then discharged through a 100 k ohm resistor. How long does it take for half of the energy stored in the capacitor to have dissipated in the circuit?
3)Mains-operated power supplies have large-value capacitors to help keep the output voltage constant. In
a full wave rectified supply, without a capacitor the output falls to zero every 10 ms. If the output must be
maintained at a value at least 90% of its maximum value when a load of 1.0 k ohm is connected to the output
terminals, show that the minimum capacitance needed in the power supply circuit is about 100*10^-6F.
Post by: astarmathsandphysics on December 22, 2011, 08:34:28 pm
Post by: astarmathsandphysics on December 22, 2011, 08:38:08 pm
Post by: dk bose on December 24, 2011, 03:43:17 pm
2)A ball bearing of mass 180g is hung on a thread in oil of density 800kgm-3.Calculate the tension in the string ,if the density of the ball bearing is 8000kgm-3.
3) Give three examples of objects which are designed to reduce the amount of turbulent flow of air or water over them.
4)Explain these poetic observations of the flow of a lake district stream:In the gentle time of a late summer, a creek over boulder flowed smooth.As autumn fell,floating leaf after leaf skipped round the rock,chasing like giddy schoolgirls playing 'Follow the leader'. In winter's depth,all frozen stood,ice on stone,stone on ice.The bright thaw springs a maelstrom,water currents churning and swirling as drunken Maypolers.
5) How and why would holding a swimming competition in a warmer pool affect the times achieved by swimmers?
Post by: astarmathsandphysics on December 24, 2011, 07:15:56 pm
2)V=mass/density=0.180/8000=2.25*10^-5 m^3
Upthrus=weight of oil displaced =density*volume*g=800*2.25*10^-5=0.018g
tension=wieght-upthrust=0.180g-0.018g=1.5876N
3)fins/wings/shark skin
4)is this an english question?
5) would be quicker since water molecules move faster.
Post by: dk bose on December 26, 2011, 02:07:27 pm
1)cos fresh water is less dense than salt water so sho upthrust is less so ship is lower in the waer
2)V=mass/density=0.180/8000=2.25*10^-5 m^3
Upthrus=weight of oil displaced =density*volume*g=800*2.25*10^-5=0.018g
tension=wieght-upthrust=0.180g-0.018g=1.5876N
3)fins/wings/shark skin
4)is this an english question?
5) would be quicker since water molecules move faster.
The 4th question is not an english question,its a question from the Edexcel AS physics textbook from materials chapter.
Post by: dk bose on December 26, 2011, 02:09:25 pm
1)A capacitor charges through a 0.22 M ohm resistor. Calculate the capacitance if the time taken for it to reach full charge is 5 minutes.the first question is not done!! plzz solve it and post the solutions as fast as possible!!!!
2)A 47*10^-6 F capacitor is charged to a potential difference of 15V and then discharged through a 100 k ohm resistor. How long does it take for half of the energy stored in the capacitor to have dissipated in the circuit?
3)Mains-operated power supplies have large-value capacitors to help keep the output voltage constant. In
a full wave rectified supply, without a capacitor the output falls to zero every 10 ms. If the output must be
maintained at a value at least 90% of its maximum value when a load of 1.0 k ohm is connected to the output
terminals, show that the minimum capacitance needed in the power supply circuit is about 100*10^-6F.
Post by: astarmathsandphysics on December 26, 2011, 07:45:25 pm
Post by: dk bose on December 27, 2011, 08:29:09 am
Post by: astarmathsandphysics on December 28, 2011, 09:48:16 am
Post by: dk bose on December 29, 2011, 03:39:48 pm
85 rad s-1. The length of the propeller from tip to tip is 2.5 m. Calculate
the emf generated between the centre and the tip of the propeller. The
horizontal component of the Earth's magnetic field strength is 22*10^-6 T.
Post by: astarmathsandphysics on December 30, 2011, 07:36:16 pm
Post by: dk bose on December 31, 2011, 03:46:01 pm
Post by: astarmathsandphysics on January 01, 2012, 11:44:29 pm
do 22*10^-6 *3.14*1.25^2*13.5
Post by: dk bose on January 03, 2012, 05:16:24 am
Post by: astarmathsandphysics on January 04, 2012, 08:52:17 am
Post by: dk bose on January 04, 2012, 03:52:32 pm
A 2.2*10^-6F capacitor is charged to a potential difference of 15V, and a 3.3*10^-6F capacitor is charged to a potential difference of 30V.
a)Calculate - 1)the charge on each capacitor, 2)the energy stored by each capacitor.
b)Show that, once Charge has re-distributed, the ratio of charge on each capacitor is equal to the ratio of the two capacitances. Hence calculate the charge on each capacitor and the final potential difference.
c)Calculate the total energy stored by the capacitors and suggest why this is less
than the total energy that would be stored by the capacitors individually.
Post by: astarmathsandphysics on January 04, 2012, 11:37:06 pm
Post by: dk bose on January 05, 2012, 03:22:15 pm
Calculate the size of the resistor that she should connect in the circuit to achieve a 1 minute delay.
Post by: astarmathsandphysics on January 05, 2012, 09:43:31 pm
4.3=5e^-60/(R*2.2*10^-3))
4.3/5=0.86 =e^-60/(R*2.2*10^-3))
-60/(R*2.2*10^-3)) =ln 0.86=-0.151
R=180000 ohms
Post by: dk bose on January 06, 2012, 09:29:05 am
Post by: dk bose on January 06, 2012, 11:06:47 am
plzz solve this sum and give the solutions as fast as possible.!!!!
Post by: astarmathsandphysics on January 06, 2012, 11:43:14 pm
Checked the previous answer and get the same
Post by: dk bose on January 07, 2012, 09:26:56 am
Post by: The Golden Girl =D on January 07, 2012, 05:58:08 pm
It is demonstrated in this video => http://videosift.com/video/Magnet-Slowly-Falling-in-Copper-Pipe
Thanks in Advance =D
Post by: astarmathsandphysics on January 07, 2012, 11:15:20 pm
Post by: astarmathsandphysics on January 07, 2012, 11:16:27 pm
Post by: AS01 on January 08, 2012, 12:54:43 pm
Post by: astarmathsandphysics on January 08, 2012, 09:54:38 pm
Post by: The Golden Girl =D on January 09, 2012, 09:25:11 am
thanks again =D
Post by: astarmathsandphysics on January 10, 2012, 09:08:21 am
Post by: The Golden Girl =D on January 10, 2012, 03:58:00 pm
Post by: astarmathsandphysics on January 11, 2012, 11:45:52 pm
Post by: The Golden Girl =D on January 14, 2012, 06:24:11 pm
So please Answer the Question for the paper attached (Hopefully) ..in case it didn't get attached then it's January 2011 Unit 4 Physics (:
Questions : 3 , 7, 9, 10 , 11]c) , 13]a) , 14]e) , 15}a]ii)iii)and c), 16]b)iii)iv)
Thanks in Advance :)
Post by: Malak on January 14, 2012, 06:42:19 pm
I need someone to help me clear my doubts cuz I have loads of doubts my teacher will kill me if I ask him ALL of them :-X :-[ :-[ :-[Give me sometime (I'll edit)
So please Answer the Question for the paper attached (Hopefully) ..in case it didn't get attached then it's January 2011 Unit 4 Physics (:
Questions : 3 , 7, 9, 10 , 11]c) , 13]a) , 14]e) , 15}a]ii)iii)and c), 16]b)iii)iv)
Thanks in Advance :)
3 - B, as the distance increases the pd decreases
7 - C
When I can't decided on an answer, i try to cancel down what seems wrong and here energy, charge and momentum is always conserved so its C
9 - C, that formula is used for energy-mass conversion
10 - not sure :S
sorry, i'll answer the rest later
Post by: The Golden Girl =D on January 14, 2012, 07:17:22 pm
Will be waiting iA =D
Post by: Malak on January 15, 2012, 11:51:24 am
We know that a lambda particle and anti-proton in produced
so the equation will be:
(any letter, I used B) = (the symbol for lambda) + (the symbol for anti-proton)
We can use any letter 'cause meson is a group and therefore it doesnt have a specific symbol.
If you check the markscheme, the 2 marks are only for the RHS.
-------------------------------
13 a - I was never good with these motor questions so I hope someone else will explain better :-[
-----------------
14 e - (unit 2)
Emission spectra are porduced when electrons are exicted - jump to higher energy levels and then de-excited, reutring to their orignal state, releasing energy in the process.
Why is it X-ray ---> Not reallu sure about this part
---------------------------
15 - ii (attached) discharging curve of capacitor for current (you should learn these) :D
how to find the values:
Max current = V/R
10/5000 = 0.002 = 2 mA
then, Q says that typical values so we can find the the time constant
To find 37% of the original current, simply multiply by 0.37 and you will get 0.74
to find the time i takes to fall to this current use the equation I = I(max) x e^(-t/RC)
we know I(max), R and C so we can find t
t = 0.049 = 0.05 s
part iii - Same curve
c - use the equation V = V(max) x e^(-t/RC)
V will be 0.7 (0.07 of 10 = 0.7)
(If you can't get the answer then tell me and I will show you the full solution)
------------------------------------------------
16 b iii - Okay, we know that the smaller the wavelength the smaller objects can be investigated and these electrons have that wavelength
iv - use the knowledge of aplha scattering.
Passed through - empty space
scattered at large angles - there are other particles within a proton and those are quarks
Post by: KJ1995 on January 17, 2012, 06:56:11 pm
Thanks in advance!
Post by: The Golden Girl =D on January 18, 2012, 01:13:17 pm
if you have MeV then this is what you do ;
[2.2 * (1.6 * 10-13)]/[ (3*108)2] = _____ Kg
I think you should multiply by 2 since it's TWO gamma particles.
E = delta m * c2
you find the Energy ..I think that's how you do it :)
Mr.Paul can you please confirm this (:
Post by: The Golden Girl =D on January 18, 2012, 01:54:27 pm
Particle Lepton Baryon Hadron Meson
Neutron, n tick tick
Neutrino, ? tick
Muon, ? tick tick
(Total 3 marks)
they say Neutrino is considered a Lepton okay ..why is that ?
shouldn't it be either Muon neutrino , tau neutrino , or electron neutrino not just the word Neutrino ALONE or I'm misunderstanding something here?
Thanks in Advance (:
Post by: Malak on January 18, 2012, 05:16:42 pm
And By the way, Muon is not a baryon or hadron :-\
Post by: The Golden Girl =D on January 18, 2012, 05:35:13 pm
By just the word neutrino, it is understood that it can be any neutrino and all of them (electron neutrino etc.) are lepton anyway.
And By the way, Muon is not a baryon or hadron :-\
Are you sure about that ? :-\
I have a document that Vampire_Lover posted on TSR about a year ago and it really is Beneficial (:
(Attached Hopefully )
Post by: Malak on January 18, 2012, 06:20:19 pm
Are you sure about that ? :-\Thanks for the doc.
I have a document that Vampire_Lover posted on TSR about a year ago and it really is Beneficial (:
(Attached Hopefully )
Um, sure about what? The neutrino part? Yes.
and if the muon part, again yes.
Post by: MKh on January 18, 2012, 07:05:25 pm
Are you sure about that ? :-\
I have a document that Vampire_Lover posted on TSR about a year ago and it really is Beneficial (:
(Attached Hopefully )
Because i do not have the appropriate version of the software, can you please attach the doc in some other format like the MS Word 2003 format?
Thanks.
Post by: The Golden Girl =D on January 18, 2012, 07:30:55 pm
@ MKh : ALL my documents are 2007 version , I suggest you download a converter (Google :Word document 2007 to 2003 converter and iA you'll be able to find what you want (: )
Post by: Malak on January 19, 2012, 02:11:20 pm
@ Ang3l : Welcome. ... Muon is a Hadron and a meson but not a BARYON ..my teacher ALWAYS said a Mesons are NOT Baryons ..NEVER Mix them up. Thanks by the way (:
@ MKh : ALL my documents are 2007 version , I suggest you download a converter (Google :Word document 2007 to 2003 converter and iA you'll be able to find what you want (: )
Yeah, that is right mesons are not baryons
...but a muon is not a hadron, it is a lepton :o that is what my textbook says and also wikipedia
Here: http://en.wikipedia.org/wiki/Lepton (http://en.wikipedia.org/wiki/Lepton) (check the leptop table)
Post by: The Golden Girl =D on January 19, 2012, 02:50:49 pm
Post by: dk bose on January 21, 2012, 08:06:50 am
so there is a very real danger of skidding.
a)At a corner of radius 260m the track is banked at 20°. Calculate the maximum speed if no frictional force is required.
plzz anyone solve this question and give the solutions as fast as possible!!!!
Post by: dk bose on January 21, 2012, 08:26:58 am
(b)'Downforce' is used to keep a car in contact with the track. The motion through the air produces a force perpendicular to the direction of travel, pushing the car onto the track. A racing car of mass 720 kg takes an unbanked corner of radius 550 m at a speed of 50 m s-1. The maximum frictional force is 0.4 x (reaction from track).Calculate the downforce necessary to prevent the car from skidding outwards.
plzz solve this sum also and give the solutions as fast as possible!! itzzz urgent!!!!plzzz help me!!!
Post by: Malak on January 21, 2012, 01:34:31 pm
The answer is B but i dont get how :s
And Q9 of the paper attached
Post by: dk bose on January 21, 2012, 02:23:28 pm
Post by: Malak on January 21, 2012, 06:31:33 pm
Can anyone give me an answer to the above question and plus Q6 part a (i) to (iv) of the paper attached
Thanks a lot
Post by: The Golden Girl =D on January 21, 2012, 08:03:39 pm
Post by: dk bose on January 22, 2012, 05:48:30 am
Post by: The Golden Girl =D on January 22, 2012, 06:37:15 am
I hope Mr.Paul would answer you though :)
I hope I helped :D
Post by: Malak on January 22, 2012, 09:45:32 am
I'm confused ..Which ones do you have doubts in ?Q9 of unit 4 jan 2010
Q6 a (i - iv) of jun 2011
Post by: The Golden Girl =D on January 22, 2012, 01:11:11 pm
Q9 of unit 4 jan 2010
[ I'm studying Math currently so can only answer this part :-\ ]
My teacher's Answer :
When the circuit is closed (Switched ON) the current in the circuit grows from Minimum to Maximum and therefore the magnetic field around the coil also grows and this induces Emf in the coil. This is called Self-Induction.
When the circuit is Switched OFF ,the current in the coil falls RAPIDLY and causes the Magnetic field around the Coil to COLLAPSE SUDDENLY and this has Emf that is sufficient to LIGHT UP the Neon Lamp.
I hope I helped (:
Post by: Malak on January 22, 2012, 07:59:56 pm
Thanks a lot for replying even though you have an exam tomorrow =)
Post by: Malak on January 23, 2012, 05:57:41 pm
When t = 0, the markscheme says it is 1.6/16 = 100mA
Post by: MKh on January 23, 2012, 06:17:04 pm
How do you find the Max current?
When t = 0, the markscheme says it is 1.6/16 = 100mA
Io = 1.6 V/16 ohms = 10 A = 100mA ---> this is how it must have been worked out.
Io is actually initial current (yes, max. current). But what i don't understand over here is how do we get 1.6V - isn't it supposed to be 12 V as clearly stated in the question?
I think it is a misprint.
Post by: Malak on January 23, 2012, 06:20:56 pm
Io = 1.6 V/16 ohms = 10 A = 100mA ---> this is how it must have been worked out.Yea, exactly. I dont get why is it 1.6 :S
Io is actually initial current (yes, max. current). But what i don't understand over here is how do we get 1.6V - isn't it supposed to be 12 V as clearly stated in the question?
I think it is a misprint.
Hmm, i guess then it is a misprint.
Post by: Malak on January 23, 2012, 08:06:39 pm
I thought m about to fail physics :P
Good luck everyone!
Post by: Uchia on March 11, 2012, 01:48:52 am
can anyone help
A 200.0 m long thin wire carries a line charge density l = 361.0 nC/m. Find the absolute value of the
potential difference between points 3.0 m and 6.0 m on a perpendicular radius to the axis of the wire,
provided the perpendicular radius is not near either end of the wire. (The value of
eo = 8.85 x 10-12 C2/N oe m2.)
Post by: astarmathsandphysics on March 11, 2012, 07:49:09 am
Post by: SZM on March 11, 2012, 03:39:12 pm
can u pls tel me how can we indicate the amplitude, wavelength of a LONGITUDINAL and TRANSVERSE wave in a displacement-time graph???
Thanx in advance.
Waitn 4 ur reply asap.
Post by: Romeesa-Chan on March 11, 2012, 04:32:29 pm
Hello guys,
can u pls tel me how can we indicate the amplitude, wavelength of a LONGITUDINAL and TRANSVERSE wave in a displacement-time graph???
Thanx in advance.
Waitn 4 ur reply asap.
(http://sciencecity.oupchina.com.hk/npaw/student/supplementary/images/graph-1b_8.jpg)
Hope it's cleared. :)
Post by: Romeesa-Chan on March 11, 2012, 05:38:36 pm
\To identify the amplitude and wavelength in a displacement-time graph for a longitudinal wave is the same way as how we identify the amplitude and wavelength in a displacement-time graph for a transverse wave. Am i correct dear???
the diagram u r giving is the displacement versus distance and not time/// Are the graph of displacement-time graph and displacement-distance graph same?????
the amplitude of a longitudinal wave is a maximum displacement of any point on the wave from the equilibrium position .--- when they say any point, that means any point in rarefraction as well as compression region?? what do they mean by at any point??
The equilibrium position of a longitudinal wave is at the middle point of a rarefraction and compression area. AM i correct ??
Waitn 4 ur reply asap.
Post by: astarmathsandphysics on March 12, 2012, 07:31:29 pm
A 200.0 m long thin wire carries a line charge density l = 361.0 nC/m. Find the absolute value of the
potential difference between points 3.0 m and 6.0 m on a perpendicular radius to the axis of the wire,
provided the perpendicular radius is not near either end of the wire. (The value of
eo = 8.85 x 10-12 C2/N oe m2.)
really tricky question
Will try it in bed tomorrow
Post by: SZM on March 31, 2012, 08:39:18 pm
1. Is it not a correct answer in terms of PHYSICS, that newton's third law tells us that actions usually have a reaction?? if not pls tel me why along with their reason???
2. if air resistance is neglected, the horizontal velocity component of an arrow fired from a bow stays constant at a non-zero value. If air resistance is not neglected and is acting on the arrow, then the horizontal velocity component of an arrow fired from a bow will increase/decrease with distance travelled????
3. What happens to the velocity of a ball bearing falling from rest through syrup with distance fallen??? My answer is the velocity of the ball bearing will decrease withe distance fallen, becoz the syrup has higher viscosity. Therefore velocity will decrease. But my answer is wrong. the correct answer is it increases from zero to a maximum.How is dat?? explain me in details clearly.
4. If the height above the Moon's surface increases, what happens to the acceleration of a feather falling near to the Moon's surface??? If ur answer is constant, tel me how is dat ?? Why cant the acceleration increases with increasing the height???
5. steel can be classified as a strong materials, because it has a large young modulus value. But the markschemes says it is wrong. the correct answer is , it is because it has a large ultimate tensile stress value. Why my answer is wrong??
6. Modern cars include crumple zones to reduce the size of the impact force. Suggest how the crumpole zones do this????
7. what do u mean by the stiffness??? the type of materials//// am i correct??
8. when unloading the chest expander, it is found that at each extension the restoring force is always less than the loading force. Explain the significance of this, and describe what effect this would have on the rubber cords when performing a large number of repetitions with the expander??? pls explain me in details clearly.
Hope I will get clear explanation for the above doubtfuls. IN SHAA ALLAH.
Thanx in advance. Wasalaam
Post by: astarmathsandphysics on March 31, 2012, 11:56:04 pm
2. decreases. Resistance always acts to slow down
3. When ball starts to fall, fluid resistance is zero since it increases with speed, zero at the start. The only force acting is the weight of the ball bearing. As ball speeds up, resistance increases until it is equal to the weight. There is no net force on the ball bearing so acceleration is zero. The ball has reached terminal velocity.
4. Constant acceleration is only an approximation. In fact a=GM/r^2 so as r increases a decreases.
5. large young's modulus means stiff not strong. Strong means ability to withstand stress.
6. crumple zones absorb energy and reduce the magnitude of acceleration. Since F=ma, F is also reduced.
7.stiffness is resistance to stretching when a force is applied.
8.this is hysterisis. Work must be done to expand and compress the expander. With each cycle the expander heats up a little.
Post by: ifko on April 01, 2012, 05:32:27 pm
Have a question from A2 particle physics...Does photon have a mass?and how is photon considered as a particle?thank you in advance
Post by: Romeesa-Chan on April 01, 2012, 05:55:11 pm
Hey!im new here :)
Have a question from A2 particle physics...Does photon have a mass?and how is photon considered as a particle?thank you in advance
(i) No, photons do not have mass, but they do have momentum. The proper, general equation to use is E2 = m2c4 + p2c2 So in the case of a photon, m=0 so E = pc or p = E/c.
(ii) Photon is a discrete bundle (or quantum) of electromagnetic (or light) energy.
source ~ (http://physics.about.com/od/lightoptics/f/photon.htm)
Post by: astarmathsandphysics on April 01, 2012, 09:40:03 pm
Post by: YU-GI-OH! on April 18, 2012, 11:54:13 pm
Thanks in advance
Post by: SZM on April 22, 2012, 06:33:21 am
what is resistance/// In textbuk, it says the resistance is caused when the free-mobile delocalized electrons, which are mobiling to and fro, collide with the electrons that are orbiting around its orbitals of the metal lattice atoms, causes the resistance.
THAT MEANS WHEN they both collide, there is a repulsive forces between them due to like charges, so this causes them to repel each other and move apart with small/large distance. That depends on the way they collide and the amount of repulsive force they contain. The distance that they move apart is considered as a space, where the charged particles, which is the current, can flow easily without any difficulty. But this happens only when the temperature has increased not decreased / at room temperature.
Because when they collide, they both produce/feel repulsive forces, this repulsive forces is dependant. Because with higher temperature, more free-delocalized electrons and the electrons that are orbiting around the atoms, gain more kinetic energy and are excited state and unstable state and therefore they move at higher speed randomly and they both will collide with each other frequently with short period of time during collision, but they will feel very less repulsive forces, because both carry high amount of kinetic energy. So this gives more space for the flow of charged particles to move easily, hence this decreases the resistance.
Is my explanation correct?? May I know resistance is present in every electrical component? Is there any other electrical component, which does not have resistance?? if there is no, can u tel me why there is no resistance in any electrical component only few?? Can I relate the concept of friction to this concept?? when they collide, they cause resistance. But how??
may I know how charged particles produce magnetic field, when charged particles are moving through the conducting wire??
Normally a point charge either could be a positively charge or a negatively charge, they both will create their electric field// am I right??
so when a point charge creates an electric field, this point charge starts to accelerate and according to newton second law, this point charge produces the force. But i dont know how the charged particles produces the magnetic field and why do they form in a circular form, which is a magnetic field, around the conducting wire?
May I know when a point charge creates an electric field, which is a region of space, how can the point charge starts to accelerate through the field??? when a point charge creates the electric field and if we place another point charge on this field, then that charge can accelerate but NOT the point charge, the one who creates the electric field. AM i correct? So again here confusion, how a point charge is accelerating when this point charges creates an electric field??/
Sir what is electromotive force? It is the force, that pushes the electrons which carry negative charges inside the battery/power supply to move through the circuit. AM i correct??
Sir why in textbuks says, that the current flows from positive terminals to the negative terminal, which is called the conventional current?? Just image the proton carries positive charge and it is inside the nucleus of the atom. So how can the protons move through the circuit, when the electrons, which are too small in size, can move freely???
Sir when I asked myself what is the current, the textbuk says it is the flow of charged particles, that means the flow of electrons or the flow of protons??? If u say flow of electrons, why the protons cannot move, even though they carry charge// Maybe u would say that they are located in the nucleus , which seems to be far away when compare to the location of electron. Am I correct??
Sir electron and proton, they both carry charge// Right. So what does the charge contains?? I mean i know that proton carries positive charge and electron carries negative charge, but I want to know what does these charge contain? if there is no such description for that, its alright.
Need ur reply urgently//
If u dont know, pls can any one of the member inform this confusion to Sir astarmathandphysics immediately. Pls help me .. plssssssssssssssssssssssss
Post by: guMnam on May 12, 2012, 10:06:47 pm
Post by: mustafa09 on June 10, 2012, 09:17:08 pm
Post by: MKh on June 10, 2012, 10:10:03 pm
can someone please explain how can i use Fleming left hand rule to find the charge in a hydrogen bubble chamber????? ??? ??? ???
http://en.wikipedia.org/wiki/Fleming's_left-hand_rule_for_motors
Hope this helps.
Post by: astarmathsandphysics on July 06, 2012, 11:17:53 pm
Post by: xainer on August 20, 2012, 05:41:59 am
An athlete throws a javelin. Just as it hits the ground the javelin has a horizontal velocity component of 22.0 m/s and a vertical velocity component of 11.0 m/s. What is the magnitude of the javelin's velocity as it hits the ground?
Post by: astarmathsandphysics on August 20, 2012, 09:54:15 am
Post by: xainer on August 22, 2012, 09:08:28 am
I have a few more questions..
http://www.xtremepapers.com/papers/Edexcel/Advanced%20Level/Physics/2008%20Jan/6731_01_que_20080116.pdf
- Question no. 3) b)i) and 5) c)ii)
hope you can give the answers for these two questions....:)
thanks once again!
Post by: astarmathsandphysics on August 23, 2012, 07:00:31 pm
Thankss a lot!! :) realised later that it was so easy...Max speed will be at bottom. In travelling from P to Q the work done against friction is Force times distance =60F where F is the average force. This is the difference between the initial and final energy
I have a few more questions..
http://www.xtremepapers.com/papers/Edexcel/Advanced%20Level/Physics/2008%20Jan/6731_01_que_20080116.pdf
- Question no. 3) b)i) and 5) c)ii)
hope you can give the answers for these two questions....:)
thanks once again!
60F=mgh (at P) -1/2mv^2 at Q
60F=750*9.8*50-1/2*750*27^2 =94125 so F=94125/60=1568.75N
Post by: astarmathsandphysics on August 23, 2012, 07:03:28 pm
s
u=0.95
v=0
a
t=9.3*10^-2
s=(u+v)/2*t=0.95/2*9.3*10^-2=0.044m
Post by: xainer on August 25, 2012, 04:05:43 pm
Max speed will be at bottom. In travelling from P to Q the work done against friction is Force times distance =60F where F is the average force. This is the difference between the initial and final energywhy is the workdone against friction taken as 60F?
60F=mgh (at P) -1/2mv^2 at Q
60F=750*9.8*50-1/2*750*27^2 =94125 so F=94125/60=1568.75N
and, the answer isn't matching with the marking scheme. it says, answers in the range of 1100N-1300N
here's the link
http://www.xtremepapers.com/papers/Edexcel/Advanced%20Level/Physics/2008%20Jan/6731_01_rms_20080306.pdf
Q3) b)i)
Post by: astarmathsandphysics on August 25, 2012, 04:27:17 pm
80F=750*9.8*50-1/2*750*27^2 =94125 so F=94125/80=1177N
Post by: xainer on August 25, 2012, 04:31:59 pm
Thought it said distance is 60m but is 80ohh..okay..Thanks a lot :)
80F=750*9.8*50-1/2*750*27^2 =94125 so F=94125/80=1177N
i have another question:
http://www.edexcel.com/migrationdocuments/GCE%20New%20GCE/sam-gce-physics.pdf
q 22) c)ii)
Post by: Saladin on August 25, 2012, 06:17:02 pm
Post by: astarmathsandphysics on August 26, 2012, 11:15:17 am
Post by: xainer on August 26, 2012, 05:48:07 pm
What is particularly difficult about the question? They just ask you to calculate tension. Also, its not always a good idea to do questions from SAMs because they often have errors in them.I had been thinking whether the 2 forces are going to b equal or not. So got confused. And I have seen this question not only from SAMs but from another site as well, with some other questions. Our physics teacher gave this question. Anyways, thanks for the info and help! :)
Post by: xainer on August 26, 2012, 05:48:48 pm
22ciiThank you! It was very helpful! :)
Post by: astarmathsandphysics on August 26, 2012, 08:40:12 pm
Post by: xainer on August 27, 2012, 05:46:54 pm
a) A car of mass 1200kg is in a crash. The front bumper of the car deforms and the car is brought to rest from an initial speed of 10m/s in a distance of 0.12m. by considering the workdone on the car as it s brought to rest, calculate the average impact force that acts.
b) Modern cars crumple zones to reduce the size of the impact force. Suggest how the crumple zones do this.
Post by: astarmathsandphysics on August 27, 2012, 09:10:50 pm
Post by: xainer on August 28, 2012, 03:31:00 pm
xainer here you arethanks! but why does longer distance decrease the value of F?
Post by: xainer on August 28, 2012, 03:32:50 pm
thanks! but why does longer distance decrease the value of F?is it because force is inversely proportional to the distance according to the equation??
what's the concept behind it?
Post by: astarmathsandphysics on August 28, 2012, 05:12:50 pm
Post by: xainer on August 30, 2012, 12:01:13 pm
Post by: astarmathsandphysics on August 30, 2012, 05:34:23 pm
Post by: jenifer911 on April 02, 2013, 12:27:13 pm
Could any one solve the following problem for me??? it can be found on the edexcel a2 physics book p.41 Questions 4.
4. In the electron beam of a CRO, electrons are accelerated through a potential difference of 3000V which is set up between electrodes 3cm apart.
a)Calculate the electric field strength, assuming it is a uniform field
b) How fast will the electrons be moving when htey emerge from this field?
Im having problems with (b), the answer 1.62 x 10^7 m s-1 while my calculations are double it i.e. 3.25X10^7ms-1
Thx in advance!!
Post by: yo2646126 on April 04, 2013, 10:30:46 pm
Post by: jenifer911 on April 08, 2013, 07:23:36 am
Post by: astarmathsandphysics on November 08, 2013, 05:56:09 pm
Hi everyone,a)E=V/d=3000/0.03=100000 V/m
Could any one solve the following problem for me??? it can be found on the edexcel a2 physics book p.41 Questions 4.
4. In the electron beam of a CRO, electrons are accelerated through a potential difference of 3000V which is set up between electrodes 3cm apart.
a)Calculate the electric field strength, assuming it is a uniform field
b) How fast will the electrons be moving when htey emerge from this field?
Im having problems with (b), the answer 1.62 x 10^7 m s-1 while my calculations are double it i.e. 3.25X10^7ms-1
Thx in advance!!
b)
so
Post by: studx3m on April 19, 2014, 04:27:48 pm
also can anyone tell me why the answer is C to may 2010 unit 3b edexcel physics??
m a private student in physics! so i have no idea how to prepare for unit 6!!! any practicals that i have to know??
Post by: astarmathsandphysics on April 24, 2014, 11:43:47 am
Canyou post it?
Post by: conquerpro on May 07, 2014, 09:06:24 pm