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Physics: Post your doubts here!

B

bomberblitz

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Oliveme said:
We subtract the two to get the resultant force. Frictional force opposes the driving force and so the total force becomes 2N.
Click to expand...

The driving force is a result of the gravitational force of the mass(2kg), connected to the box by the string, thus it should be 2g, the resultant force (friction taken into account) should be (2g-6)N shouldn't it... the weight of the boxed being pulled shouldn't be taken into account as the whole movement is horizontal.
 
H

hela

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Mishaa

Mishaa

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hela said:
please help
http://papers.xtremepapers.com/CIE/...and AS Level/Physics (9702)/9702_w08_qp_1.pdf Q32 B
http://papers.xtremepapers.com/CIE/...nd AS Level/Physics (9702)/9702_w10_qp_12.pdf Q7 D Q13 D Q25 D Q31 A
http://papers.xtremepapers.com/CIE/...nd AS Level/Physics (9702)/9702_w12_qp_11.pdf Q23 A

http://papers.xtremepapers.com/CIE/...nd AS Level/Physics (9702)/9702_s12_qp_11.pdf Q12 A
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w09_qp_11.pdf Q12 B
Click to expand...

1. Take the 7 wires to be parallel to each other, so 1/R=6(1/10)+1/100. Answer is 1. 63.
 
Mishaa

Mishaa

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hela said:
please help
http://papers.xtremepapers.com/CIE/...and AS Level/Physics (9702)/9702_w08_qp_1.pdf Q32 B
http://papers.xtremepapers.com/CIE/...nd AS Level/Physics (9702)/9702_w10_qp_12.pdf Q7 D Q13 D Q25 D Q31 A
http://papers.xtremepapers.com/CIE/...nd AS Level/Physics (9702)/9702_w12_qp_11.pdf Q23 A

http://papers.xtremepapers.com/CIE/...nd AS Level/Physics (9702)/9702_s12_qp_11.pdf Q12 A
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w09_qp_11.pdf Q12 B
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2. Q7 Acceleration upwards is always -9.81 and downwards 9.81. As upward motion is positive, + multiplied ny - gives you -9.81. Vice versa with downward motion. - multiplied with +9.81. The magnitude stays the same as long as air resistance is neglected, and the sign depends on which direction you take as positive.
Q13 Once the cube is fully immersed, no change in the reading should be seen. The reading corresponds to the WEIGHT - UPTHRUST, not the upthrust. And the pressure/force on vertical sides is not at all accountable/considerable, it's always the pressure on the bottom and top. Thus, the last is the correct option.
 
mushoo

mushoo

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down the column in question no 2, the sig fig shud remain same or the decimal place??
Rep asap.. please..
 
Mishaa

Mishaa

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hela said:
please help
http://papers.xtremepapers.com/CIE/...and AS Level/Physics (9702)/9702_w08_qp_1.pdf Q32 B
http://papers.xtremepapers.com/CIE/...nd AS Level/Physics (9702)/9702_w10_qp_12.pdf Q7 D Q13 D Q25 D Q31 A
http://papers.xtremepapers.com/CIE/...nd AS Level/Physics (9702)/9702_w12_qp_11.pdf Q23 A

http://papers.xtremepapers.com/CIE/...nd AS Level/Physics (9702)/9702_s12_qp_11.pdf Q12 A
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w09_qp_11.pdf Q12 B
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Q25) Drawing the waveforms, l=1/4wavelength in X and l=1/2 wavelength in Y. As the speed of sound is constant, the ratios of the frequencies should be reciprocal to the ratio of wavelengths, that would be frequency of X:Y = 1:2. So frequency of Y is double of the frequency of X.
Q31) Conduction wires conduct due to mobile electrons, so C and D are incorrect. The movement of electrons random, so B is eliminated. A is correct.

3. Q23) I'm not entirely sure of this but the extension is denoted by W/k. Now P and Q have identical spring constants so total extenion = 2. Now PQ and R are taken in series so product over sum gives us the answer.

4. Q12) SOMEONE PLEASE SOLVE THIS FOR ME TOO.

5. Q12) R1 and R2 both act against the weight of the frame so their magnitude would be equal. T1 has a greater horizontal component than T2, the vertical one is in the same line of force as the weight and R for both and is not considered, so wire T1 is more tensed upp. More like how picture frames fall when you suspend them with too short a wire and they snap. Real life scenarios. ;)

And I had a powercut scheduled in the middle so the 3 seperate posts. Sorry. :p
 
TaffsAsLevel

TaffsAsLevel

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hassanhijazi1995 said:
u need to MULTIPLY THE PERCENTAGE UNCERTINITY OF L BY 2
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How come, then it wil be +- 0.8 for all then? What about this?
Oliveme said:
Let's assume l is 10.2 (+- 0.4). When l is squared, we square the two extreme points and subtract them to get uncertainty in the l squared value.
10.2^2 = 104.04 and uncertainty = (10.2+0.4)^2 - (10.2-0.4)^2
= 10.6^2 - 9.8^2
= 112.36 - 96.04 = 16.32
So, we have 104.04 (+- 16.32).
Hope you get it.
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TaffsAsLevel

TaffsAsLevel

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Soldier313 said:
Aoa wr wb
Can someone please upload a diagram for question 1 of this paper?

http://papers.xtremepapers.com/CIE/...and AS Level/Physics (9702)/9702_w07_qp_5.pdf
http://papers.xtremepapers.com/CIE/...and AS Level/Physics (9702)/9702_w07_ms_5.pdf

Thank you so much!
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A loudspeaker incident and the sound received by a microphone or source detector. We put the detector as close to the glass to avoid reflection of sound and the detector behaves in such a way we want to know the details about the sound wave at the glass window, my thought though, I am doing it right now btw, what about the second question where they ask to draw best and worst fit line I don't get it
 
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