A2 Physics | Post your doubts here

[Jonathan127]

jut draw a AM signal graph with modulating frequency of 5kHz(200us) and carrier frequency of 50KHz(20us)

I still don't get it. Like, do you mean, draw the carrier frequency separate?

 

[alphabravocharlie]

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!!

This question please, help needed

 

[smzimran]

This question please, help needed

SkyPilotage has answered this question posted by me in this very thread!
Please search

 

[Mobeen]

http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s08_qp_4.pdf
http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s08_ms_4.pdf
please can anyone explain question 6 part (a) part (b) (i) and (b) (ii)

(a) when it would be parallel(horizontal) the magnetic field strenght would pass through it for a longer distance (2.8cm) .when it is perpendicular magnetic field strength passes through for a very short distance which is thickness of the coil.

(b)(i) torque = force x distance
2.1 x 10^3 = F x 0.028m
F=0.075N

bii there are no field lines cutting the side BC so there wont be a force there .

 

[alphabravocharlie]

SkyPilotage has answered this question posted by me in this very thread!
Please search

like finding a needle in a haystack,
Okay will do, thanks though

 

[Mobeen]

I still don't get it. Like, do you mean, draw the carrier frequency separate?

no . have you seen an AM signal graph ?

 

[Jonathan127]

no . have you seen an AM signal graph ?

Yeah, isn't it the one where the frequency stays the same, but the amplitude varies with the displacement?

 

[Mobeen]

http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s08_qp_4.pdf
anyone for 9 (b).
thanks in advance!

just draw two LEDs one with forward direction and one reversed connected from the input to the earth . the forward biased LED is of red light and the reversed biased LED is green

 

[Mobeen]

Yeah, isn't it the one where the frequency stays the same, but the amplitude varies with the displacement?

open this link .AM graph is like this
https://ccrma.stanford.edu/~jos/st/img494.png
the time for each individual peak is of the carrier frequency
and the time for one large peak to the other large peak is of the information signal

 

[smartangel]

SkyPilotage has answered this question posted by me in this very thread!
Please search

i dont get it from his solution..could you explain the working? pleeease bro?

 

[smzimran]

like finding a needle in a haystack,
Okay will do, thanks though

Here it is:
From Part i) The Intensity that is reflected back from the muscle bone boundary is 0.331 i.e 0.331 x 0.389 I

The key to this part is that you have to deduce the Intersity reflection coefficient from part ii).
When the ultrasound reaches the surface of the muscle, 0.661 will be reflected and 0.389 will pass through ----> 0.331 x 0.389 x 0.389 I = 0.05 I

 

[hassam]

calculate current in 6 kiloohm resistor

 

[alphabravocharlie]

Here it is:
From Part i) The Intensity that is reflected back from the muscle bone boundary is 0.331 i.e 0.331 x 0.389 I

The key to this part is that you have to deduce the Intersity reflection coefficient from part ii).
When the ultrasound reaches the surface of the muscle, 0.661 will be reflected and 0.389 will pass through ----> 0.331 x 0.389 x 0.389 I = 0.05 I

Jazakallah brother

 

[Jonathan127]

open this link .AM graph is like this
https://ccrma.stanford.edu/~jos/st/img494.png
the time for each individual peak is of the carrier frequency
and the time for one large peak to the other large peak is of the information signal

So basically, you just draw a graph similar to that one, but each individual peak is every 20µs and the lwhole waveform takes 200µs to complete?

 

[Mobeen]

So basically, you just draw a graph similar to that one, but each individual peak is every 20µs and the lwhole waveform takes 200µs to complete?

exactly! the amplitude is not given so dont worry about the amplitude !

 

[Jonathan127]

exactly! the amplitude is not given so dont worry about the amplitude !

Thanks so much!

 

[Mobeen]

Thanks so much!

no problem!

 

[angelgirl:)]

ok every1 best of luck for your examsss...May Allah Subhana Wa Taa'llah helps all of us...Ameen

 

[Abdulrab]

thank

For Q9 b) the difference in the uncertainties is 10%-2% =8%

That's an uncertainty of 0.08. So to calculate the activity for when the source is not decaying 0.08, it would be decaying 1-0.08= 0.92.
so the activity= 0.092 decay constant=0.693/half life .. and the rest is just
A
= A0 exp(–ln2 t / T½
)

thanku angelic

 

[angelgirl:)]

ok every1 best of luck for your examsss...May Allah Subhana Wa Taa'llah helps all of us...Ameen

and thanks alot for all the help i recieved from every 1...