Physics: Post your doubts here!

[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

 

[bassamkhan35]

Hey everyone out there. I was wondering if someone has the same paper variant as me in other countries so that due to time difference we can help with the questions coming. My paper variants are listed below.
Physics 22, physics 12 physics 33
Chemistry 22, Chemistry 12 and Chemistry 33
Mathematics 12 and Mathematics 62
My time zone Pakistan Standard time GMT+5

 

[GarryTheGhost]

You and other people doing the same variant would be doing the exam at the same time that's why there are variants so students can't use other peoples time zones to see the questions coming. However the 22 and 21 usually have one or 2 similar questions but I doubt they can tell you since they probably don't know or remember the questions coming. Plus the 21 variant and 22, 23 all happen at different dates. So lol I don't know how people are gonna help eachother.

Hey everyone out there. I was wondering if someone has the same paper variant as me in other countries so that due to time difference we can help with the questions coming. My paper variants are listed below.
Physics 22, physics 12 physics 33
Chemistry 22, Chemistry 12 and Chemistry 33
Mathematics 12 and Mathematics 62
My time zone Pakistan Standard time GMT+5

 

[KernelPanic]

Hi guys, this is kinda unrelated to physics but I wanted some help with further maths question since further thread is pretty much dead. If anyone taking further maths can help me out it would be very helpful. I can do (a) but I need help with (b). It's S07P1Q11 btw if you need Mark Scheme which didn't help me enough to understand.

 

[nageena]

winter 2016 p22
ques 2 (b) and (c) (i)
explanation plz
thankyou.

 

[Ebrahim12]

oct nov 2017 13
any help would be appreciated

 

[Ebrahim12]

Can anyone help me in the following questions from oct/nov/2017/13 Q36 Q37View attachment 63284

View attachment 63285

36)
Applying KVL around the circuit gives
2E = IR + Ir1 + Ir2

Since the voltmeter reading zero this means there is no voltage difference between its contacts points which gives
Vrise=Vdrop
E = Ir1

Substituting this into the first equation gives
2Ir1 = IR + Ir1+ Ir2
R = r1 - r2 (B)

37)
Applying KVL around the circuit gives
2E - E = 3IR
E = 3IR

The PD between P and Q is due to the battery and resistor between them
PD = E - IR
= E - 1/3 E (substituiting first equation)
= 2/3 E (C)

I would recommend practicing how to use KCL and KVL using YouTube videos.