- Messages
- 195
- Reaction score
- 495
- Points
- 73
http://papers.xtremepapers.com/CIE/...nd AS Level/Physics (9702)/9702_w10_qp_11.pdf....Q5,.Pls explain.Thanks in ADVANCE!
We are currently struggling to cover the operational costs of Xtremepapers, as a result we might have to shut this website down. Please donate if we have helped you and help make a difference in other students' lives!
Click here to Donate Now (View Announcement)
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!
Is It B._.?http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w10_qp_11.pdf....Q5,.Pls explain.Thanks in ADVANCE!
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
Yeah kaise?Is It B._.?
Look they said 3000 revolutions per minuteYeah kaise?
my ans was 3.6.. but my frnd got 6.67... now we r nt sure that which one is correct :/does anyone remember what was the answer to the question about the circuit with a variable and constant resistor in parallel and the other six ohm resistor in series , the one with the 24 volt battery ? in the part where he asked to calculate the resistance of the constant resistor X ? this ques could be the difference between my getting an A or a B .. please help .... Tense!
p.s im talking about v 22
at the nth maxima angle is 90. sin90=1A parallel beam of light of wavelength 450 nm falls normally on a diffraction grating which has
300 lines / mm.
What is the total number of transmitted maxima?
mohammad salik
Thought blocker
Suchal Riaz
sin(90) = 1 is the largest real value that sin(theta) can have. In other words 90 degree is the limit for possible diffraction angle magnitude. Rearrange the diffraction grating formula and you haveA parallel beam of light of wavelength 450 nm falls normally on a diffraction grating which has
300 lines / mm.
What is the total number of transmitted maxima?
Mohammed salik
Thought blocker
Suchal Riaz
the ans is 15
i did the same n i got 7 number of maximas... while it is 15 in ms :/sin(90) = 1 is the largest real value that sin(theta) can have. In other words 90 degree is the limit for possible diffraction angle magnitude. Rearrange the diffraction grating formula and you have
sin(theta)=lambda/d
Clearly n can only take values that result in the R.H.S. being less than or equal to one.
Note that the system is symmetric about the center line from the diffraction grating to the projected image. That means maxima will occur on both sides of the center line with the same angular spacing (and there is also a central maximum on the center line).
GOT IT ?
ohhh..... i didnt knw that... i just calculated the max no. of orders...at the nth maxima angle is 90. sin90=1
so in formula we get
d=nλ ==> n=d/λ
d=1*10^-3/300=3.33X10^6
so n=3.33X10^6/450X10^-9 = 7.4
so maximum order = 7
there are 7 maxima above middle maxima and 7 below
to total 7+7+1=15
physics paper 1 n 3 r leftbut the physics paper is over why is everyone still doing physics paper?
i still have p4. 1. 3 and 5but the physics paper is over why is everyone still doing physics paper?
Look this is what i could do.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
Look this is what i could do.
60secs-3000 revolutions
1sec-3000/60=50revolutions
T=1/f
=1/50=20ms
So if its 10ms the wave can be easily be shown on the c.r.o .LOL i asked for an answer and solved it myself
Thanks brotherKindly 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
For almost 10 years, the site XtremePapers has been trying very hard to serve its users.
However, we are now struggling to cover its operational costs due to unforeseen circumstances. If we helped you in any way, kindly contribute and be the part of this effort. No act of kindness, no matter how small, is ever wasted.
Click here to Donate Now