Physics: Post your doubts here!

[A*(a*)]

uncertainities of all instruments in phy are taken as their least counts eg. 0.1 mm in vernier caliper
as for in chem n bio uncertainities are taken to b 'half' of least count

and the least counts of the instruments are? sorry if it takes your time but I am in desperate need for them

 

[Dug]

But why (0.5+0.5)

Whenever we add or subtract 2 quantities, their uncertainty is added.

Change in Temperature = Final Temp. - Initial Temp.
Total uncertainty = ΔFinalT + ΔInitialT

 

[attitudehed]

Hi can someone help me with question 26 Nov/10/v2?
http://papers.xtremepapers.com/CIE/...nd AS Level/Physics (9702)/9702_w10_qp_12.pdf
A microwave transmitter emits waves towards a metal plate. The waves strike the plate and are
reflected back along their original path.

A microwave detector is moved along the line PT.
Points P, Q, R, S and T are the positions where minima of intensity are observed. These points
are found to be 15mm apart.
What is the frequency of the microwaves?
A 5.0GHz B 6.7GHz C 10GHz D 20GHz
The answer is C. Can someone please explain how to get it?

 

[snowbrood]

Hi can someone help me with question 26 Nov/10/v2?
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w10_qp_12.pdf
A microwave transmitter emits waves towards a metal plate. The waves strike the plate and are
reflected back along their original path.

A microwave detector is moved along the line PT.
Points P, Q, R, S and T are the positions where minima of intensity are observed. These points
are found to be 15mm apart.
What is the frequency of the microwaves?
A 5.0GHz B 6.7GHz C 10GHz D 20GHz
The answer is C. Can someone please explain how to get it?

distance between one minima to the other is 0.5 wavelength
0.5 wavelength=15*10^-3
wavelength =0.03m
we speed of the light is 3*10^8
frequency= velocity/wavelength
3*10^8/0.03 = we get 1*10^10
1*10^3 is kil0
1*10^6 is mega
1*10^9 is gega
so it is 10GHZ

 

[snowbrood]

velocity of all the electromagentic waves is the same i.e 3*10^8

 

[Shahed Shubeir]

I need urgent help in these questions below:
1) calculate the force needed to change the momentum of an object from 50 kgms-1 to 250 kgms-1 in 10 sec
2)Winter 2005 , paper 2 , question #4( I need Full explanation because I can't understand anything from the marking scheme )
3) winter 2002 , paper 2 , question #3 ( I need Full explanation because I can't understand anything from the marking scheme )

 

[mamoon123]

Just a thought, if in the structured paper the mark scheme says an answer is 4.2 and my answer is 4.19, will I lose the accuracy (A1) mark?
(This would be ironic, because I'd actually be more accurate than the answer itself e.e)

 

[Shahed Shubeir]

which question are you talking about?

 

[hino]

How do we solve questions 14, 19 and 20 in this paper:
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s09_qp_1.pdf

Q14
power=F*d/t
=F*v
substitute the value of F=kv^2 into the formula above
power(P)=kv^2*v
P =kv^3
k=P/v^3 (answer)

 

[Tabi Sheikh]

I need urgent help in these questions below:
1) calculate the force needed to change the momentum of an object from 50 kgms-1 to 250 kgms-1 in 10 sec
2)Winter 2005 , paper 2 , question #4( I need Full explanation because I can't understand anything from the marking scheme )
3) winter 2002 , paper 2 , question #3 ( I need Full explanation because I can't understand anything from the marking scheme )

Answer to ur first question is 20N using F*t=p

 

[attitudehed]

distance between one minima to the other is 0.5 wavelength
0.5 wavelength=15*10^-3
wavelength =0.03m
we speed of the light is 3*10^8
frequency= velocity/wavelength
3*10^8/0.03 = we get 1*10^10
1*10^3 is kil0
1*10^6 is mega
1*10^9 is gega
so it is 10GHZ

Thanks I got it now

 

[attitudehed]

How about this one?

Question 26.
http://papers.xtremepapers.com/CIE/...nd AS Level/Physics (9702)/9702_w11_qp_12.pdf
At time t = 0 the waves are in phase. At the dotted line, t = 18s.
At which time is the phase difference between the two oscillations ⅛ of a cycle?
A 4.0s B 4.5s C 8.0s D 9.0s

Answer is B

 

[snowbrood]

How about this one?

Question 26.
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w11_qp_12.pdf
At time t = 0 the waves are in phase. At the dotted line, t = 18s.
At which time is the phase difference between the two oscillations ⅛ of a cycle?
A 4.0s B 4.5s C 8.0s D 9.0s

Answer is B

in waveform p there are four crests and so there are four waves ok? (for the motion from 0 to 18)
in waveform q there four crest and and one trough that makes 4.5 waves
for waveform p

no of waves time period
4 18
x 1
it is directly proportional we get x =4/18=0.2222222hz this is f1(as frequency is the number of waves in one second)
now lets find T1
T1=1/f =1/0.222222=4.5s

now for waveform q
no of waves time period
4.5 18
x 1
x=0.25
T2=4s

now lets come to diagram
the phase difference of the two waves is a horizantal distance a similar part of one wave leads or lags the other wave

 

[snowbrood]

How about this one?

Question 26.
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w11_qp_12.pdf
At time t = 0 the waves are in phase. At the dotted line, t = 18s.
At which time is the phase difference between the two oscillations ⅛ of a cycle?
A 4.0s B 4.5s C 8.0s D 9.0s

Answer is B

now to find the phase difference draw two vertical lines from one wave form to the other remember it should be perpendicular to the x axis
4s means one wave of waveform Q
4.5 means one wave of waveformP
9s means two waves of waveform Q draw the lines and see what u get
at 4 s u will see 0 phase difference
at 4.5 s there is a difference of 1/8 of a wave
at 9 s there is a difference of 1/4 of wave

 

[mamoon123]

Q14
power=F*d/t
=F*v
substitute the value of F=kv^2 into the formula above
power(P)=kv^2*v
P =kv^3
k=P/v^3 (answer)

Cheers mate! I owe you one

 

[hino]

n

Cheers mate! I owe you one

no problem

 

[Kumkum]

pls help me with question 5(b)
http://papers.xtremepapers.com/CIE/...nd AS Level/Physics (9702)/9702_s10_qp_42.pdf

 

[hino]

pls help me with question 5(b)
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s10_qp_42.pdf

5b(i)first find the capacitance across the parallel combination of capacitors that is C+C=30+30=60
now using the capacitance above find the capacitance across the series combination that is 1/C+1/C=1/30+1/60
total capacitance=20
b(ii)voltage across each capacitor is 6V
find the voltage across the parallel combination that is 1/V+1/V =1/6+1/6=3V
now using the above voltage find the voltage across the series combination that is V+V=3+6
total voltage =9V

 

[Kumkum]

5b(i)first find the capacitance across the parallel combination of capacitors that is C+C=30+30=60
now using the capacitance above find the capacitance across the series combination that is 1/C+1/C=1/30+1/60
total capacitance=20
b(ii)voltage across each capacitor is 6V
find the voltage across the parallel combination that is 1/V+1/V =1/6+1/6=3V
now using the above voltage find the voltage across the series combination that is V+V=3+6
total voltage =9V

thnx

 

[white rose]

can anyone give me the link for oct/nov 2012 pastpapers for bio, chem & phy......thnx