Physics: Post your doubts here!

[anastasia grey113]

Can you please explain why are you multiplying 0.018 by e^(-48*10-2)2 i
I understand e^ part but why is 0.018 being multiplied to it.
shouldn't it just be this:
0.012=e^((-48*10-2)2)
i just dont understand why you're using the coefficient alpha thats been calculated in the first part of the question
Because itn't the formula:
I=Io e^-αx
the Intensity cancel out, but what i dont understand is that why are we using the α and the µ at the same time in this question.
i'm so confused.

okay imagine the beam entering
There will be three phases.
1- It will move through the layer of fat. As it moves, it will attenuate and intensity will decrease. (e^48 x*10-2)
2- It will be reflected at the tissue boundary. But some of it will be transmitted thru the muscle tissue. (0.018 of the total is reflected).
3- It will move across the layer of fat and will be equally attenuated once more (e^48 x 2*102)
So you just have to multiply all these three phases in order to find the intensity detected.
The detected intensity is already given (0.012) so put it's value in the equation which is like.
Intensity detected = not attenuated x reflected x not attenuated

In short, some of the beam is also 'transmitted and NOT REFLECTED' at the tissue boundary. In this case only 0.018 of it is reflected. The rest is not so obvio it is not detected.
To get an accurate value, this should also be considered.

 

[Hamnah Zahoor]

Can anyone help me in the following questions from oct/nov/2017/13 Q36 Q37

 

[MShaheerUddin]

http://maxpapers.com/wp-content/uploads/2012/11/9702_s12_qp_21.pdf
Can anybody explain last part of Q6?

 

[kity way]

can someone plz help in b) the ans is 0.85 and 0.51

 

[dumbledore.]

One possible nuclear fission reaction is
235 1 U + 092 n 141 1
56 0 Ba + 92
36 Kr + 3 n + energy.
Barium-141 (141
56 Ba) and krypton-92 (92
36 Kr) are both β-emitters.
Barium-141 has a half-life of 18 minutes and a decay constant of 6.4 × 10–4 s–1.
The half-life of krypton-92 is 3.0 seconds.
(b) A mass of 1.2 g of uranium-235 undergoes this nuclear reaction in a very short time
(a few nanoseconds).
(i) Calculate the number of barium-141 nuclei that are present immediately after the
reaction has been completed.
Ans is 3.1×10^21
(ii) Using your answer in (b)(i), calculate the total activity of the barium-141 and the
krypton-92 a time of 1.0 hours after the fission reaction has taken place.
activity
Ms says krypton activity is negligible . Why?
0c13/43

 

[Hamnah Zahoor]

http://maxpapers.com/wp-content/uploads/2012/11/9702_s12_qp_21.pdf
Can anybody explain last part of Q6?

I consider when the amplitude increases so does the resultant amplitude/displacement thus more or greater constructive interference occurs resulting in more brighter fringes produced. In other case A^2 is directly proportional to Intensity thus increase in A will ultimately inc. the Intensity of the bright fringes no effect on any other results.

 

[hamadh]

http://pastpapers.papacambridge.com...el/Physics (9702)/2017 Nov/9702_w17_qp_22.pdf
could somebody please explain me Q3)d) Where is the force 2.7 on the spring acting and how the spring can not have extension when weight is acting on it. Answers to previous parts are a)500 b)1.3 c)2.7

 

[MShaheerUddin]

Can Anybody solve b i and ii?

 

[MShaheerUddin]

Part B

 

[hamadh]

View attachment 63296
Can Anybody solve b i and ii?

For the first question, the wave that will be drawn on the graph is going to be a simple wave like you see in the textbook with crests and troughs . Stationary waves only move up and down. There is no propagation of wave. At t =0 , the stationary wave is in the middle . It takes 20 ms for it to complete one oscillation. From mean position to move upwards, it takes 5ms, then back to mean position another 5ms then from mean position to lowest position(downwards) it take 5ms then back to mean position it take another 5ms completing 20ms or one oscillation. Now they asked, what will be the position at 5 ms ? At 5ms it is either up or down, you can draw either of these positions.
For ii) the phase difference between any two loops is 180 degrees or 2 pie radian

 

[MShaheerUddin]

For the first question, the wave that will be drawn on the graph is going to be a simple wave like you see in the textbook with crests and troughs . Stationary waves only move up and down. There is no propagation of wave. At t =0 , the stationary wave is in the middle . It takes 20 ms for it to complete one oscillation. From mean position to move upwards, it takes 5ms, then back to mean position another 5ms then from mean position to lowest position(downwards) it take 5ms then back to mean position it take another 5ms completing 20ms or one oscillation. Now they asked, what will be the position at 5 ms ? At 5ms it is either up or down, you can draw either of these positions.
For ii) the phase difference between any two loops is 180 degrees or 2 pie radian

But doesn't it say to draw a line?
And you are saying to draw wave...

 

[hamadh]

But doesn't it say to draw a line?
And you are saying to draw wave...

If that's the answer why would they ask the question
It means to draw a single line . It need not to be horizontal. A curved path is a line too.

 

[MShaheerUddin]

Thanks

 

[Pabosto12]

could somebody please explain me Q3)d) Where is the force 2.7 on the spring acting and how the spring can not have extension when weight is acting on it. Answers to previous parts are a)500 b)1.3 c)2.7

F = ke
e = f/k = 2.7/30 = 0.09m = 9 cm
height = 9 - 7 = 2 cm

 

[hamadh]

F = ke
e = f/k = 2.7/30 = 0.09m = 9 cm
height = 9 - 7 = 2 cm

Thank you but why do we use 2.7?

 

[MIK3Y.]

hey I need help with 11b(ii).

 

[anastasia grey113]

View attachment 63304 hey I need help with 11b(ii).

Well the answer for b(i) is 1.68 so draw arrows between energy levels above which have a difference greater than 1.68 but lesser than 2.96eV on subtraction.
Remember the arrows should point upwards because energy from light is being 'absorbed' and this results in dark lines.

 

[Wâlèé Atèéq]

Can anyone help me with this ques.? Explanation needed..☺

 

[GarryTheGhost]

can someone explain part A and part B i'm stuck

 

[anastasia grey113]

can someone explain part A and part B i'm stuck

Well for the first part, it all lies on how well you understand both the terms
Activity is the number of Nuclei decayed per unit time...just like velocity so yh now you get how the answer came up
Okay probability are the chances of a single nucleus to decay out of all the nuclei out there
So for the time t obviously, the probability would be the no. of nuclei decayed/total nuclei

Part b demands relating uncertainty to readings
You may know very well that as the reading on any apparatus decreases, its uncertainty increases.
So if u want the uncertainty to increase, you will have to make sure the reading is smaller.
So if the sample keeps decaying and fewer nuclei are left, obviously the reading of activity recorded will also decrease and automatically uncertainty will increase
So if you want it to increase to 10% from 2% i.e if you want an increase of 8%, you will have to decrease the reading by 8% too.
So the percentage uncertainty will increase only when 100% - 8% = 92% of the 3.5x105 is being recorded.
Put in the values in the equeation A = Aoe^-λt