Physics: Post your doubts here!

[RoyalPurple]

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

 

[aleezay]

http://papers.xtremepapers.com/CIE/...nd AS Level/Physics (9702)/9702_w10_qp_43.pdf

Can someone please try Q3b? Our teacher says it's not in the syllabus. I personally think there's a misprint in parts (i) and (ii) but I'm not sure.. if not, why is displacement equal to 2.0cm in part (iii) (and not 3.0, as calculated in parts (i) and (ii))?

 

[itallion stallion]

Yarh u r rite.I have paper 3 left for maths!!!

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

 

[thats_me]

http://papers.xtremepapers.com/CIE/...nd AS Level/Physics (9702)/9702_w13_qp_43.pdf
9bii ?

 

[aleezay]

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

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

 

[aleezay]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w13_qp_43.pdf
9bii ?

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.

 

[aleezay]

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

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]

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?

 

[thats_me]

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.

5small squares or 10?

 

[itallion stallion]

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..

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!

 

[aleezay]

5small squares or 10? View attachment 42728

10. You're right. I misread the scale

 

[thats_me]

10. You're right. I misread the scale

thankyou

 

[aleezay]

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!

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

 

[itallion stallion]

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

Hmmm
It's definitely your logic

 

[princeali97]

http://papers.xtremepapers.com/CIE/...nd AS Level/Physics (9702)/9702_w10_qp_11.pdf....Q5,.Pls explain.Thanks in ADVANCE!

 

[itallion stallion]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w07_qp_4.pdf
can anyone help me in Q5(b) part 2[/quote
Just take the area under the curve.for accuracy take the area of the small block any multiply by their total no.a bit tiring though but that's the it has to be solved!

 

[kitkat <3 :P]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w10_qp_11.pdf....Q5,.Pls explain.Thanks in ADVANCE!

Is It B._.?

 

[RoyalPurple]

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

ohh k i understand but i am confused in the end why did u multiplied .39*.33 again with .39?

 

[princeali97]

Is It B._.?

Yeah kaise?

 

[kitkat <3 :P]

Yeah kaise?

Look they said 3000 revolutions per minute
So I divided 3000 by 60*10 cuz it said 10cm wide I got 5 rounded it to 10 ._. Not sure if its correct