Physics: Post your doubts here!

[Thought blocker]

9702/41/O/N/13
qtn 11 (a) & (c)
can someone plzz explain

part(a)

(c) No Idea. This chapter, I sucked at it xD

 

[DragonCub]

View attachment 61794
(iii)

Shouldn't it be 9000 Hz? The signal bandwidth is 18 kHz; the audio bandwidth is half of that, which means 9 kHz or 9000 Hz.

Audio (or message) bandwidth is the same as maximum frequency of the audio (or message) signal.

 

[Thought blocker]

Shouldn't it be 9000 Hz? The signal bandwidth is 18 kHz; the audio bandwidth is half of that, which means 9 kHz or 9000 Hz.

Audio (or message) bandwidth is the same as maximum frequency of the audio (or message) signal.

Yes. xD I answered part (ii) xD

M sorry.

 

[AnonymousX9]

Hey guys could someone please help me with M/J 2009 P4 q9 part (b). Really confused. Thanks a lot.

 

[areeba240]

qtn 5(b)

 

[i_try9621]

9702/41/O/N/13
qtn 11 (a) & (c)
can someone plzz explain

For this if you look at the samples taken you see that some peaks get lost (between 0.25ms and 0.50ms & between 1.00ms and 1.25ms) which we don't want. So there is a significant change in voltage measured and the sampling frequency(or sampling rate) is too low.

 

[student2]

Can any1 pls xplain (b)?
y do we use A= -λN?
cuz N gives us undecayed nuclei nt amnt in atmosphere?

 

[areeba240]

s14/42
plzz explain the complete qtn...

 

[areeba240]

w13/43
plzz explain the complete qtn
(i) no of uranium nuclei = no of barium nuclei why is that so why arent including krypton
(ii) why neglecting krypton

 

[i_try9621]

View attachment 61806

Can any1 pls xplain (b)?
y do we use A= -λN?
cuz N gives us undecayed nuclei nt amnt in atmosphere?

We use A= -λN because we have Activity, Decay constant and N( the no of particles that are present).
Using that formula we get N= 9.52 x 10^7 of Radon-222 atoms
So the ratio will be (2.52 ×10^25)/(9.52 x 10^7) giving 2.6 x 10^17. I believe this is the correct answer.

 

[i_try9621]

View attachment 61807
s14/42
plzz explain the complete qtn...

 

[student2]

We use A= -λN because we have Activity, Decay constant and N( the no of particles that are present).
Using that formula we get N= 9.52 x 10^7 of Radon-222 atoms
So the ratio will be (2.52 ×10^25)/(9.52 x 10^7) giving 2.6 x 10^17. I believe this is the correct answer.

bt isn't N the no. of particles present in the sample n not in the air? Lyk the undecayed amnt?

 

[i_try9621]

bt isn't N the no. of particles present in the sample n not in the air? Lyk the undecayed amnt?

In 1m^3 of atmospheric air 9.52 x 10^7 atoms of Radon-222 are present

 

[Djlucifer]

Specifically looking for a candidate who appeared in A level exams and received an A and had Physics Chemistry and Biology just wanted few advices regarding how to approach the examintions and how to revise having 40 days in hand

 

[Thought blocker]

Specifically looking for a candidate who appeared in A level exams and received an A and had Physics Chemistry and Biology just wanted few advices regarding how to approach the examintions and how to revise having 40 days in hand

I am an Engineering student. Got AAB in Phy, Math, and Chem respectively.
Only forty days remaining. I would strongly encourage you to focus on past papers right now.
Solve as many papers as you can, find in which concepts you are loosing marks.
Target your weaknesses, work on them.
Plan your studies properly. Two papers per subject in a day, if you can manage it. Else, just one paper per subject in a day will be also great.
No one can stop you from an A.

Good luck.

P.S. This is the time. Work on your weaknesses. Strengthen it. A is yours.

 

[krishnapatelzz]

yes but how?

(918 - 900) x 1000 = 18000

 

[dreamerbeliever]

9702_w15_qp_21



Q5 (i) On Fig. 5.1, draw a line to represent the position of the string at t = 5.0 ms.
How??
I don't understand the 5 ms

 

[techgeek]

Hey guys could someone please help me with M/J 2009 P4 q9 part (b). Really confused. Thanks a lot.

 

[Aishayasin]

Can anyone explain these questions
Anwer is D

 

[Aishayasin]

Answer is C how?