Edexcel PHYSICS Doubts

[guMnam]

jan 2010 unit 1 q6 can someone explain ?

[mustafa09]

can someone please explain how can i use Fleming left hand rule to find the charge in a hydrogen bubble chamber??

[MKh]

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.

[astarmathsandphysics]

you can only use the left hand rule to find the sign of a charge, and only if you know the direction it is moving in.

[xainer]

Hi! Can I get the answer for this question, ASAP, please?
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?

[astarmathsandphysics]

sqrt(22^2+11^2)=24.6m/s

[xainer]

Thankss a lot!! realised later that it was so easy...

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!

[astarmathsandphysics]

Thankss a lot!! realised later that it was so easy...

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!

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
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

[astarmathsandphysics]

And for 5cii constant deceleration so use suvat
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

[xainer]

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
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

why is the workdone against friction taken as 60F?
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)

[astarmathsandphysics]

Thought it said distance is 60m but is 80
80F=750*9.8*50-1/2*750*27^2 =94125 so F=94125/80=1177N

[xainer]

Thought it said distance is 60m but is 80
80F=750*9.8*50-1/2*750*27^2 =94125 so F=94125/80=1177N

ohh..okay..Thanks a lot
i have another question:
http://www.edexcel.com/migrationdocuments/GCE%20New%20GCE/sam-gce-physics.pdf
q 22) c)ii)

[Saladin]

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.

[astarmathsandphysics]

22cii

[xainer]

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!