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Physics: Post your doubts here!

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Why is q negative for the first row?
no thermal energy is supplied to the system since it's constant temperature right so then why zero??
 
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Can someone please solve this

any help is much appreciated.
 

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Can someone please solve this

any help is much appreciated.
When it's on the floor horizontally:
Fr = uR
u = Fr/R = 10N/12N = 0.833
When it's titled:
R = 12cos@
12sin@ > Fr
12sin@ > uR
12sin@ > 0.833*12cos@
tan@ > 0.833
@ > 39.8°
 
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I need help with finding the gradient of the lines in P5
These are the two papers
http://papers.xtremepapers.com/CIE/...nd AS Level/Physics (9702)/9702_s12_qp_51.pdf
http://papers.xtremepapers.com/CIE/...nd AS Level/Physics (9702)/9702_s10_qp_51.pdf
Please look at question 2 of both papers.

In the first link (s12 paper) to find the gradient I did,
(0.82-0.36)/(1.95-1.67)x10ˉ²
This gives 164
But the answer is 1.64 so my teacher said don't multiply by 10ˉ² since the graph says I/10ˉ²

In the second paper (s10) to find the gradient I did what my teacher said above
(334-152)/(4.5-2)
This gives 68.8
Now in part d, the answer should be 10^-6 whereas I get 10^-3
So in this case I had to multiply the
(334-152)/(4.5-2)x10^-3
to get the correct answer

So I am confused when to multiple by the power given on the axis and when not to? :/
Please help!
 
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I need help with finding the gradient of the lines in P5
These are the two papers
http://papers.xtremepapers.com/CIE/...nd AS Level/Physics (9702)/9702_s12_qp_51.pdf
http://papers.xtremepapers.com/CIE/...nd AS Level/Physics (9702)/9702_s10_qp_51.pdf
Please look at question 2 of both papers.

In the first link (s12 paper) to find the gradient I did,
(0.82-0.36)/(1.95-1.67)x10ˉ²
This gives 164
But the answer is 1.64 so my teacher said don't multiply by 10ˉ² since the graph says I/10ˉ²

In the second paper (s10) to find the gradient I did what my teacher said above
(334-152)/(4.5-2)
This gives 68.8
Now in part d, the answer should be 10^-6 whereas I get 10^-3
So in this case I had to multiply the
(334-152)/(4.5-2)x10^-3
to get the correct answer

So I am confused when to multiple by the power given on the axis and when not to? :/
Please help!
When making calculations based on values of gradients, make sure you use the REAL values (ie the one with powers of ten multiplied)
When asked to calculate the gradient of a graph, I'd prefer to use the real value still, but it seems they want without power of ten adjustments.
 
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When making calculations based on values of gradients, make sure you use the REAL values (ie the one with powers of ten multiplied)
When asked to calculate the gradient of a graph, I'd prefer to use the real value still, but it seems they want without power of ten adjustments.
That's what i thought as well like always take the powers
but then in the first paper they're not using the power... is it something to do with logs?
 
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That's what i thought as well like always take the powers
but then in the first paper they're not using the power... is it something to do with logs?
Unlikely to do with logs. It's simple scales. Powers of ten simply prevent having to like long decimals like 0.0035
 
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Re: Physics Help here! Stuck somewhere?? Ask here! :)

Can anyone tell me what should i learn for pressure in lquids for As level and please give me the detailed explanation of viscosity in liquids asap !!! thanks
viscosity is not in the syllabus, download the syllabus from the cie guide and read what u want
 
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View attachment 57071
Why is q negative for the first row?
no thermal energy is supplied to the system since it's constant temperature right so then why zero??

Sorry for late reply you might have already got the answer but just in case I'll explain it.
For an ideal gas, Internal energy is proportional to Temperatuer.
Since temp= constant, Change in U=0
According to first law of thermodynamics, Q-W=0
Therefore, Q=W
Meaning all the heat added to the system is used to do work( thats why the -ve sign at q and +ve sign at w)
 
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Sorry for late reply you might have already got the answer but just in case I'll explain it.
For an ideal gas, Internal energy is proportional to Temperatuer.
Since temp= constant, Change in U=0
According to first law of thermodynamics, Q-W=0
Therefore, Q=W
Meaning all the heat added to the system is used to do work( thats why the -ve sign at q and +ve sign at w)
Thankyou! And better late than never :) I had actually almost forgotten about this question and hadn't found an answer :p
Thanks!
 
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A whole hearted thanks to smzimran the notes were very organized, hope it will be helpful for my upcoming exam..........
 
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how many s.f should my answers be in paper 2? i always see 2 s.f in final answer and i use 3

Either 2 or 3. They only penalise if you write to 1sf or to more than 3sf( unless specified in the question)

This statement does not apply for uncertainty questions!
 
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