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Physics: Post your doubts here!

RoyalPurple

RoyalPurple

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papers.xtremepapers.com/CIE/Cambridge%20International%20A%20and%20AS%20Level/Physics%20(9702)/9702_s09_qp_4.pdf

question 9 b and question 11 b (iii)..
can someone help me plz ?
thank you :)
 
itallion stallion

itallion stallion

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periyasamy said:
Yarh u r rite.I have paper 3 left for maths!!!
Click to expand...
Kindly make correction in the answer which I posted for your correction!
increase number of bits in digital number at each sampling ......................................M1
so that step height is reduced ................................................................................... A1 increase sampling frequency / reduce time between samples ..................................M1
so that depth / width of step is reduced ..................................................................... A1 [4] (do not allow ‘smoother output’)
I wrote it upside down,just saw it now,sorry
 
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aleezay

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RoyalPurple said:
papers.xtremepapers.com/CIE/Cambridge%20International%20A%20and%20AS%20Level/Physics%20(9702)/9702_s09_qp_4.pdf

question 9 b and question 11 b (iii)..
can someone help me plz ?
thank you :)
Click to expand...


I'm assuming that you got the answers for the previous two parts.. for part (iii),
Intensity received back= (intensity reflected at the muscle-bone boundary)e^-(23 * 4.1 * 10^-2)

[ e^-(23 * 4.1 * 10^-2) = 0.39]


intensity of the wave incident on the boundary was 0.39I, as calculated in part (ii)
0.33 of this was reflected at the boundary.
So intensity reflected at the muscle-bone boundary= .39 * .33I

answer= .39 * .33I * .39 = .050I

I hope this helps
 
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aleezay

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thats_me said:
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w13_qp_43.pdf
9bii ?
Click to expand...
Gain = 10

maximum value of output voltage = 9V since the op-amp becomes saturated at that point

G= Vout/Vin

for Vout = 9V, Vin is 0.9 which is approximately equal to 1V.

So you start your graph at the origin. Draw a line such that Vout= 9V when Vin= 1V (that happens after 5 small squares on the x-axis).. draw a horizontal line at y= 9V until Vin falls to -2. At that point, you draw a vertical line from y= +9V to -9V.
Draw a horizontal line at -9V there on. Then, for the last 5 small squares, Vout changes from -9V to 0.

I'm sorry I couldn't send a picture of the graph. I hope this explanation makes up for that. :)
 
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aleezay

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itallion stallion said:
The equation that is given showing 4 this is the amplitude from the mean position which will be the center of ab and cd.
To move from halfway to cd is 4 so complete will be B.
Next part is just put 2pie(f)=W
Next part u should know that the place where displacement is zero acceleration will be max and vice versa.so displacement zero at mean which is 4 cm midway.draw it and the next part just put values in v=rw
Click to expand...

I get why they've written 4. I just think that Y and Z should be positioned 2.0cm above AB. I don't get why it's 3.0..
 
MissBellum

MissBellum

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Relation between electric field strength E, potential difference (deltaV) and separation (delta r)
Workdone = F * delta r cos180
W = F * delta r (-1)
W = -qE delta r
W/q = -E delta r
delta V = -E delta r
E = -delta V/ delta r

Im not sure where the cos180 came from? Could someone explain? :confused:
 
thats_me

thats_me

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aleezay said:
Gain = 10

maximum value of output voltage = 9V since the op-amp becomes saturated at that point

G= Vout/Vin

for Vout = 9V, Vin is 0.9 which is approximately equal to 1V.

So you start your graph at the origin. Draw a line such that Vout= 9V when Vin= 1V (that happens after 5 small squares on the x-axis).. draw a horizontal line at y= 9V until Vin falls to -2. At that point, you draw a vertical line from y= +9V to -9V.
Draw a horizontal line at -9V there on. Then, for the last 5 small squares, Vout changes from -9V to 0.

I'm sorry I couldn't send a picture of the graph. I hope this explanation makes up for that. :)
Click to expand...
5small squares or 10? image.jpg
 
itallion stallion

itallion stallion

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aleezay said:
I get why they've written 4. I just think that Y and Z should be positioned 2.0cm above AB. I don't get why it's 3.0..
Click to expand...
Sorry I just believe I could do the question so i am half way there! I can't just understand the rest!
By the how do u think it is 2 by guess or some formula!
 
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aleezay

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itallion stallion said:
Sorry I just believe I could do the question so i am half way there! I can't just understand the rest!
By the how do u think it is 2 by guess or some formula!
Click to expand...
well.. you see X's motion can be represented using a cosine wave. For Y, a phase angle of 2pi/3 means cos(2pi/3) = -.5

So Y is 0.5* 4= 2.0 cm above the ean position, moving upwards (2pi/3 is less than 2pi indicating that the wave is not complete.. so Y moves upwards)

=> My logic :p
 
itallion stallion

itallion stallion

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aleezay said:
well.. you see X's motion can be represented using a cosine wave. For Y, a phase angle of 2pi/3 means cos(2pi/3) = -.5

So Y is 0.5* 4= 2.0 cm above the ean position, moving upwards (2pi/3 is less than 2pi indicating that the wave is not complete.. so Y moves upwards)

=> My logic :p
Click to expand...
Hmmm
It's definitely your logic:)
 
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