Physics: Post your doubts here!

[donewithtime80]

http://papers.gceguide.com/A Levels/Physics (9702)/9702_w15_qp_53.pdf

qs 2(a)...pls show how to find the gradiesnt

 

[Konstantino Nikolas]

http://papers.gceguide.com/A Levels/Physics (9702)/9702_w15_qp_53.pdf

qs 2(a)...pls show how to find the gradiesnt

v^2 is plotted against r ... so make v^2 the subject of your eqn -> v^2 = Pgr/m
So from that we know the graph is a straight line as it is in the form y=mx+c .... where y = v^2 ; x = r ; c = o (that's why there's no c term) and gradient = Pg/m (the coefficient of x)

 

[donewithtime80]

v^2 is plotted against r ... so make v^2 the subject of your eqn -> v^2 = Pgr/m
So from that we know the graph is a straight line as it is in the form y=mx+c .... where y = v^2 ; x = r ; c = o (that's why there's no c term) and gradient = Pg/m (the coefficient of x)

ohhh got it
thank you

 

[Konstantino Nikolas]

ohhh got it
thank you

no problem mate

 

[Fahad Afzal]

http://papers.xtremepapers.com/CIE/...nd AS Level/Physics (9702)/9702_s12_qp_41.pdf
Q6 (c) Anyone? How can we obtain the average p.d.?

Thanks in advance for your help

 

[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))

 

[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]

View attachment 58723
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]

View attachment 59260

View attachment 59261

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]

View attachment 59267

View attachment 59268

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]

View attachment 59280

the (ii) part pls

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