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Physics: Post your doubts here!

$$AK$$

$$AK$$

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GUYS HOW ARE U ALL?
PLZ I NEED UR HELP IN PHYSICS...I NEED TO KNOW THE PERFECT TEXT BOOK TO STUDY FROM........ :p ............AND ALSO WAT SUBJECTS DO U ADVICE ME TO TAKE IF I AM GOING INTO MEDICINE FIELD. :D

PLZ NAOTE THAT I NEED 4 AS SUBJECTS AND 3 A LEVELS :)
 
Muhammad TAimoor

Muhammad TAimoor

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Mushroom Juu said:
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s10_qp_23.pdf

PLS HELP! >_< does anyone know how to calculate the area under the graph for question 3? :(
Click to expand...

Make an appropriate line just under the curve that covers most of the boxes. The Shape formed would be a Trapezium. Calculate the are for trapezium by 1/2 x height x sum of parallel sides.
Now remains the boxes above the line. Count the boxes and add them to the trapezium area. That's it. (y)
 
W

wanderer100

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2mrrw is my physics paper2 and im not quite sure about this vector diagram thing, if u can help? thankyou so much .. also can figure out when to use cos sin stuff...waiting for the reply...
 
Muhammad TAimoor

Muhammad TAimoor

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100 said:
how to find the uncertainity of an area of a circle?
Click to expand...

Area of circle has the formula Pie Radius square. the uncertainty would definitely be in Radius value, so use the formula to calculate it and multiply it by 2 because of the square on Radius. Thats it (y)
 
Saad Mughal

Saad Mughal

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imack94 said:
ASLM guys.. explanation for nov 2012 pp 21 no. 2(b) 3rd part?? cheers!! :)
Click to expand...
There is a constant error present in all the readings (zero error, a systematic error). Now, when there is 0 voltage, the current should also be 0, right? But, the graph shows that when the voltage is 0, the current is 0.05 A which tells us that all reading are incorrect by +0.05 A. To account for this error, you'll subtract 0.05 from the reading across any voltage. This is the explanation for getting the correct answer. You need to describe/show this in your answer.

Here's the solution:
Resistance = V/I
when V = 2 V, I = 0.14 A,
correcting error, I = 0.14 - 0.05 = 0.09 A
Resistance = 2/0.09 = 22.2 ohms (correct to 3 s.f.)
 
imack94

imack94

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Saad Mughal said:
There is a constant error present in all the readings (zero error, a systematic error). Now, when there is 0 voltage, the current should also be 0, right? But, the graph shows that when the voltage is 0, the current is 0.05 A which tells us that all reading are incorrect by +0.05 A. To account for this error, you'll subtract 0.05 from the reading across any voltage. This is the explanation for getting the correct answer. You need to describe/show this in your answer.

Here's the solution:
Resistance = V/I
when V = 2 V, I = 0.14 A,
correcting error, I = 0.14 - 0.05 = 0.09 A
Resistance = 2/0.09 = 22.2 ohms (correct to 3 s.f.)
Click to expand...

Thanks man :)
 
fjmskt

fjmskt

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Can anybody help with the question 16.7.?image.jpg
It's a question from Robert Hutching's book.
 
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