Physics: Post your doubts here!

[kishen1001]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s15_qp_13.pdf
number (19) please

 

[kishen1001]

How do I solve this? It's from http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s05_qp_1.pdf
View attachment 59108

When you throw something upwards, gravity always acts downwards, so the acceleration is always -9.81ms^-1. It doesn't matter if it falling down again after that, if your initial direction is upwards, the acceleration will always be negative. This a very confusing concept to understand though.

 

[kishen1001]

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can someone tell me why the ans isnt B???! :'(

In this case, the initial direction of motion is upwards, as shown in the graph. When it first reaches its highest point, its velocity is 0, which is at point B, after that it starts to move downwards, and when it reaches its lowest point, its velocity will be 0 again, thus the answer D

 

[The Sarcastic Retard]

Q1.b
http://papers.gceguide.com/A Levels/Physics (9702)/9702_y16_sp_4.pdf
ms http://papers.gceguide.com/A Levels/Physics (9702)/9702_y16_sm_4.pdf
How is resultant force the difference in gravitational force of Earth and centripetal force of an object on the earth?

As u did in above part, Fg - Fc = 9.77. So We know Fg = 9.77 + Fc therefore let Fc be any number, the resultant will always be 9.77.

 

[The Sarcastic Retard]

??

http://studyguide.pk/Past Papers/CIE/International A And AS Level/9702 - Physics/9702_s08_qp_4.pdf
3b(ii)

http://studyguide.pk/Past Papers/CIE/International A And AS Level/9702 - Physics/9702_s06_qp_4.pdf
4(a(i)) (c)

http://studyguide.pk/Past Papers/CIE/International A And AS Level/9702 - Physics/9702_w08_qp_4.pdf
3(b)(ii)1 (c(ii))

 

[a_wiserME!!]

shouldn't it be 3.8-1.9??

 

[kishen1001]

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shouldn't it be 3.8-1.9??

Work done in this case is area under the graph. Since they want the work done from 10mm to 20mm, the shape of the area under the graph is a trapezium, hence we use to formula for the area of a trapezium which is A=0.5*(a+b)*h.
First we convert all the data into SI units since the answer is in Joules. The load on the y-axis should be converted to its weight so we multiply it by 9.81. The extension on the y axis is multiplied by 10^-3 to change it to metres. Now for the calculation.
A=0.5 * ((1.9*9.81)+(3.8*9.81)) * ((20x10^-3)-(10x10^-3))
A=0.28J

 

[a_wiserME!!]

Work done in this case is area under the graph. Since they want the work done from 10mm to 20mm, the shape of the area under the graph is a trapezium, hence we use to formula for the area of a trapezium which is A=0.5*(a+b)*h.
First we convert all the data into SI units since the answer is in Joules. The load on the y-axis should be converted to its weight so we multiply it by 9.81. The extension on the y axis is multiplied by 10^-3 to change it to metres. Now for the calculation.
A=0.5 * ((1.9*9.81)+(3.8*9.81)) * ((20x10^-3)-(10x10^-3))
A=0.28J

yup got it.. thnx

 

[a_wiserME!!]

anyone with an easier explanation pls? TIA

 

[kishen1001]

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anyone with an easier explanation pls? TIA

If you have learnt the chapter Superposition, you should know that S2X-S1X is the path difference. In this case, the signal decreases until X, so we assume at X, the amplitude is 0. This is called destructive interference. For this to happen, S2X-S1X must equal to n(lambda/2), where n=odd numbers=1,3,5,7.... This is because the maxima of one wave will meet the minima of the other wave when the the wavelength difference between them is a multiple of lambda/2, resulting in total cancellation.

 

[princess Anu]

Please help me with Q9bii

http://papers.xtremepapers.com/CIE/...nd AS Level/Physics (9702)/9702_w13_qp_43.pdf
Marking scheme;
http://papers.xtremepapers.com/CIE/...nd AS Level/Physics (9702)/9702_w13_ms_43.pdf

 

[a_wiserME!!]

the (ii) part pls

 

[a_wiserME!!]

how is the phase difference between P and Q= 0??

 

[Konstantino Nikolas]

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the (ii) part pls

coz there is no work done against friction nor kinetic energy change? right?

 

[Rizwan Javed]

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how is the phase difference between P and Q= 0??

In stationary waves, all the points in between the two adjacent nodes are in phase. Since they are in phase the phase difference between them will be zero.

 

[Rizwan Javed]

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the (ii) part pls

First reason would be that there is no work done against air resistance.
Second could be that since the speed is constant, there is no gain or loss of Kinetic Energy.
Therefore the rate of increase of P.E is equal to power calculated.

 

[amal sharkawi]

 

[amal sharkawi]

View attachment 59282

 

[amal sharkawi]

 

[Salonee B]

QUESTION: An optic fibre has a signal attenuation of 0.20 dB km–1. The input signal to the optic fibre has a power of 26 mW. The receiver at the output of the fibre has a noise power of 6.5 µW. Calculate the maximum uninterrupted length of optic fibre given that the signal-to-noise ratio at the receiver must not be less than 30 dB

ANSWER SCHEME : if P is power at receiver, 30 = 10lg(P / (6.5 × 10–6) C1 P = 6.5 × 10–3 W
loss along cable = 10lg({26 × 10–3} / {6.5 × 10-3})
= 6.0 dB
length = 6.0 / 0.2 = 30 km

I dont understand how to approach this whole problem :/ its from s08 qp 4 of cie a level physics, help will be very very much appreciated.