Physics: Post your doubts here!

[omg]

path difference = 28 cm
frequency is increased from 1.0 kHz to 4.0 kHz
that means (33o/1000)m to (330/4000)m
= 0.33 m to 0.0825 m
in cm,
= 33 cm to 8.25 cm
Now, lets check wavelengths for minima (n + ½)λ
0.5λ = 28 --> λ = 28/0.5 = 56 cm
1.5λ = 28 --> λ = 28/1.5 = 18.7 cm
2.5λ = 28 --> λ = 28/2.5 = 11.2 cm
3.5λ = 28 --> λ = 28/3.5 = 8 cm

In the range, 33 cm to 8.25 cm
There are only two wavelenghts, which are:
18.7 cm and 11.2 cm

Therefore two minima

thanks a bunch =D

 

[XPFMember]

Any Genious Here Giving A2
PLZ Help Me

Assalamoalaikum wr wb!
i'm giving A2.....

 

[A.ELWY 7]

- Question 11
Driving force = m x a = 2 x 9.81 = 19.62
Driving force - frictional force = m x a
19.62 - 6 = (8+2) x a
a = 1.362. Approximately = 1.4 m/s^2
Now the main doubt here is why are we doing 8+2. It would help if you've studied P4 mathematics, but if you haven't, then this is why:
T = tension, F = frictional force
Forces of box: T-F = 8a
Forces on 2kg mass = 2g - T = 2a (Note: we're taking -T here because in the previous equation we took T as positive when it was away from the box, not T is towards the box and away from the 2kg mass)
Substitute the first equation in the second one. 2x9.8 - 8a - F = 2a
10a = 13.6
a = 13.6/10 = 1.36 m/s^2. Approximately = 1.4 m/s^2

- Question 28 was already explained above.

- Question 32
All the wires are connected in parallel.
So 1/R = (1/10)X6 + 1/100
Therefore, 1/R = 61/100
R = 100/61 = 1.6 ohms.

thanx very much jaf....for question 11 i did it the same way i just wanted to check if my way was correct....and for question 32 how did u no they where in parrallel ?

 

[Jaf]

thanx very much jaf....for question 11 i did it the same way i just wanted to check if my way was correct....and for question 32 how did u no they where in parrallel ?

Think about it... how else could they have been connected? There are clearly spaces between the individual wires. This shows that they are infact 7 distinct wires running alongside each other. The only way they could have been in series would be if they were connected end to end.
(PS: Real Madrid FTW!! )

 

[A.ELWY 7]

well thanx

 

[A.ELWY 7]

so how does everyone do in paper 1....i wish the curve become low this year!!

 

[A.ELWY 7]

hey guys these are short AS physics notes...all the credit goes to KEITAK from coloumbia, he posted them as with A2 notes in another thread, so i thought of putting them here as this thread have more views.

 

[A.ELWY 7]

and these are for the A2 guys....all my thanx to Keitak from coloumbia

 

[Mustehssun Iqbal]

Assalamu alaikum,
Can you please explain how to solve Q.9?? thanks!
http://www.xtremepapers.com/papers/...and AS Level/Physics (9702)/9702_s09_qp_4.pdf
here's the marking scheme;
http://www.xtremepapers.com/papers/...and AS Level/Physics (9702)/9702_s09_ms_4.pdf

 

[unique840]

Assalamu alaikum,
Can you please explain how to solve Q.9?? thanks!
http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s09_qp_4.pdf
here's the marking scheme;
http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s09_ms_4.pdf

the activity A0 is stated with 2% error. we hav to get the time for 10% ka error. means original se 8% error. if original is A0, then after 8% error, it will be 0.92A0. we will consider the expression: A = A0* e^(-lambda*time)
lambda will be 0.693/half life
lambda = 0.693/(5.27*365*24*3600)
A = 0.92 A0
so the equation will be:
0.92A0 = A0 * e^(-lambda*time)
A0 will be cancelled
0.92 = e^ (-lambda*time)
ln (0.92) = -lambda*time

 

[Mustehssun Iqbal]

the activity A0 is stated with 2% error. we hav to get the time for 10% ka error. means original se 8% error. if original is A0, then after 8% error, it will be 0.92A0. we will consider the expression: A = A0* e^(-lambda*time)
lambda will be 0.693/half life
lambda = 0.693/(5.27*365*24*3600)
A = 0.92 A0
so the equation will be:
0.92A0 = A0 * e^(-lambda*time)
A0 will be cancelled
0.92 = e^ (-lambda*time)
ln (0.92) = -lambda*time

thanks! btw I was asking about Q.6 and it got mistyped. Can you solve that one too, about electromagnetism??
And thanks!

 

[unique840]

which part of q6?

 

[Mustehssun Iqbal]

which part of q6?

part b and c??

 

[unique840]

part b and c??

bi) reading on balance is increasing which means force on balance is acting downwards. according to newton's third law, every action has an equal and opposite reaction. so the wire will have force upwards with the same magnitude. the direction of current in from X to Y and the force is upwards. according to fleming's rule, magnetic field is from ryt to left so pole P is north cox magnetic field is directed from north to south

 

[unique840]

part b and c??

bii) B = F/(IL)
I and L is given. force is mg
m is given.
B = mg/IL

 

[Mustehssun Iqbal]

Thanks!
And how do we know that the force due to current-carrying wire is acting upwards??

 

[ABDSyed]

Any one know how to Take Uncertainty when dealing with Log

 

[XPFMember]

Assalamoaalikum wr wb!
some doubts in Nov 2005 Paper 4 Physics...
· Q:3 I’m sorta confused with the calculation...part b ...and I don’t understand why they give the ans to C to 2s.f when they’re penalising 2s.f in part b?
· Q:4 c Why speed max when displacement is zero? :s
· Q:5b(ii) why when speed is reduced, deflection is larger? :s
· Q:6 b I don’t understand

 

[XPFMember]

Thanks A lot Bro
JazzakAllah

waeyyakum..
and sorry..forgot to say salam..assalamoalaikum

 

[hassam]

http://www.xtremepapers.com/papers/...and AS Level/Physics (9702)/9702_s09_qp_4.pdf
q5 part c
ms says plastic is an insulator / not a conductor / has no free electrons B1
charges do not move (on an insulator) B1
either so no single value for the potential
or charge cannot be considered to be at centre
wats the purpose of this point that charges do not move (on an insulator)