A2 Physics | Post your doubts here

[ousamah112]

but the total distance (the raduis) b/e the planet and the ponit is 2R...so how come its 3R?

point at 2R distance from the SURFACE of planet has total distance 3R.
2R +radius of planet...

 

[RGBM211]

Anyone??

It said at a height o.5 above Earth's surface which means that r = R of Earth + 0.5 R(Height ABOVE Earth's Surface) = 1.5R
Hope you get it
shiningstar,

 

[arlery]

P=1\3 (NM<c^2>)\V
is the derivation of this equation important????????????

Yes it is, its in almost every year!

 

[Mobeen]

Anyone??

we have to find the gravitational field strength at 0.5r above the surface so total distance= 1.5r (because we add the total distance from the centre to the surface and the distance above the surface) so total distance = 1.5 x (6.4x10^6)
understand ?

 

[angelgirl:)]

Anyone??

radius of earth might be 1 and the other thing is 0.5 above the earths surface...so total distance b/w them is 1.5...i guess

 

[angelicsuccubus]

but in the syllabus it is menioned derive bt not rigorous!!!

it says "deduce".. so you should be able to explain it but you don't need to be able to derive it from the mu--mu=2mu stage.. go over it and understand what's happening though..

 

[miss irfan]

if there are two charged particles and the graph of e.p.e against distance between the particles show energy going towards zero as distance increases, and the epe is more negative as distance is decreased then how we know whether the particles have same charge or different?

 

[samwickz]

O

relax lad...for these sort of question u first need to calculate the area of one block like one small block inside the graph in this case it is 1cm and 2.5v/m
then count down the number of blocks under the graph if not complete do approximation and u come across 21 blocks then u multiply both to find the voltage simple

omg ur lyk d best thannnnnnnnnnnnnnnnnnxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx alot

 

[angelicsuccubus]

co

could you plz elobarte this mate!!!
espically didnt get the part when u said 1-0.08!!!

you know what.. .it doesnt make sense to me either anymore..

 

[miss irfan]

please answer my question

 

[Mobeen]

please answer my question

can you post the link or the image of the graph?

 

[prannoy]

and how the amount of smoothing is increased?

by increasing the capactaince Or the resistance. It leads to faster time constant CR so it will discharge faster and smoothing will be increased.

 

[NyTRiDr]

A sinusoidal carrier wave has frequency 500 kHz and amplitude 6.0 V. It is to be
frequency modulated by a sinusoidal wave of frequency 8 kHz and amplitude 1.5 V.
The frequency deviation of the carrier wave is 20 kHz V–1.
Describe, for the carrier wave, the variation (if any) of
A: amplitude
B: Frequency

can sum1 plz tell the answer of part B?? how to calculate the frequecy?
thanks!

 

[miss irfan]

its november 2011 paper 41 Q4. there is always some problem when uploading the question. i'll try again

 

[prannoy]

To find mean voltage, you need the sum of all the voltage under the sinusoidal graph, which is composed of peaks and valleys. Now if you add the areas under the (+) peak and the area above the (-) valley, the sum is zero. Through Mean=Sum of X/No. of X , whatever the number of X may be, the mean is zero is the sum of X is zero. So the mean voltage in AC is zero unless you remove the effect of direction by squaring.

thanx

 

[smartangel]

http://www.xtremepapers.com/papers/...and AS Level/Physics (9702)/9702_s09_qp_4.pdf
Q11 b iii. last part of the question..please explain working!!

 

[miss irfan]

can you post the link or the image of the graph?

there is a problem with my adobe, please can you see the question and answer?

 

[prannoy]

A sinusoidal carrier wave has frequency 500 kHz and amplitude 6.0 V. It is to be
frequency modulated by a sinusoidal wave of frequency 8 kHz and amplitude 1.5 V.
The frequency deviation of the carrier wave is 20 kHz V–1.
Describe, for the carrier wave, the variation (if any) of
A: amplitude
B: Frequency

can sum1 plz tell the answer of part B?? how to calculate the frequecy?
thanks!

frequency change= 20khz*1.5=+-30 so the frequncy fo the carrier wave will change by \+- 30 (470-530)Khz frequency

 

[angelgirl:)]

http://www.xtremepapers.com/papers/...and AS Level/Physics (9702)/9702_w08_qp_4.pdf
Q1bi....plz help?

 

[Mobeen]

there is a problem with my adobe, please can you see the question and answer?

ok let me see .