Physics: Post your doubts here!

[sumaiyarox:)]

Question 2b plzzz...

 

[smzimran]

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

AoA,
Q3:
We know that when a liquid turns into gas, work is done by the liquid
In this case, they are asking about work done on the system, so this must mean - (work done by the system)
(i)
W = - p(deltaV)
W = - 1.03 × 10^5 [(2.96 × 10^-2) - (1.87 × 10^-5 )]
W = - 3050 J

(ii)
Heating we know it is given Q = 4.05 × 10^4 J

(iii) U = Q + W
U = 4.05 × 10^4 + (- 3050)
U = 3750 J
I dont think the calculation is much tricky
And, I dont get their approach of penalising 2 s.f in (b) but not in (c)
What i think is because in (b) they have given all data values to 3 s.f so they want the answer to atleast 3 s.f
In (c) they have not given any values, so they accept s.f below 3,
This is what i think altough im not sure

 

[smzimran]

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

AoA,
Q4:
Speed is max when displacement is zero,
Consider the ball thrown in the air and down again case (which you must have studied)
At max. height the ball has max. gravitational potential energy (P.E) and zero kinetic energy (K.E) meaning zero speed
Similarly, when the ball falls and touches the ground,
P.E is zero as the height is zero and by law of conservation of energy the whole P.E at top is converted into max. K.E at bottom, so speed is max at ground (at zero displacement)

Same applies in S.H.M:
At amplitude, speed is zero (K.E is zero) because the object has max P.E as it is at max height and at momentary rest.
At zero displacement, speed has increased to maximum value as whole of P.E at amplitude has been converted into K.E
That is why speed is taken max at displacement = 0

 

[sumaiyarox:)]

Q2? AND
how to find the magnitude in Q 6A 1?

 

[confused123]

Q2? AND
how to find the magnitude in Q 6A 1?

Electric field stregth = Voltage / Distance between the plates
350 / 2.5 into 10^-2 = 1.4 into 10^4 N/C

 

[smzimran]

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

AoA,
Q5
When an electron is moving fast in a magnetic field, it experiences the magnetic force for a shorter duration of time thus is deflected less!
When it is passing slowly (with smaller value of speed), it experiences the magnetic force for a longer duration of time thus is deflected more!

Imagine you have to pass a patch of quicksand;
If you run through you will go through easily and will not be much drenched in the sand
If you walk slowly, you will be going down and down in the sand

^ Just came across my mind, hope it helps

Q6(b)
You just have to derive the expression for magnetic flux linkage (Φ) :
We know that
Φ = BAsinθ
For N number of lines,
Φ = NBAsinθ
Done
Small derivation so worth 2 marks only

 

[sumaiyarox:)]

Electric field stregth = Voltage / Distance between the plates
350 / 2.5 into 10^-2 = 1.4 into 10^4 N/C

umm..soo magnitude is equal to electric field strength?

 

[confused123]

umm..soo magnitude is equal to electric field strength?

the question asked for the magnitude of electric field which means the same as to find the electric field strength.

 

[smzimran]

http://www.xtremepapers.com/papers/CIE/Cambridge International A 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)

Just defining the term 'insulator'
Thats all, otherwise i agree there is no need

 

[XPFMember]

AoA,
Q5
When an electron is moving fast in a magnetic field, it experiences the magnetic force for a shorter duration of time thus is deflected less!
When it is passing slowly (with smaller value of speed), it experiences the magnetic force for a longer duration of time thus is deflected more!

Imagine you have to pass a patch of quicksand;
If you run through you will go through easily and will not be much drenched in the sand
If you walk slowly, you will be going down and down in the sand

^ Just came across my mind, hope it helps

Q6(b)
You just have to derive the expression for magnetic flux linkage (Φ) :
We know that
Φ = BAsinθ
For N number of lines,
Φ = NBAsinθ
Done
Small derivation so worth 2 marks only

waalaikumassalam wr wb
haha...nice one

ok and for ques 6..i wanna know what's the difference b/w flux and flux linkage..see ms...it says smthng like that...

and jazakAlllahhhhhhhhhhhhh for the help...
many duassss 4 u

 

[XPFMember]

...some more doubts in June 2006
· Q:2 b (ii) how to do?? :s
· Q:4 c can you show, plz?

 

[sumaiyarox:)]

the question asked for the magnitude of electric field which means the same as to find the electric field strength.

thnx

 

[unique840]

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

if the mass in inc the force on balance is downwards and on wire is upwards. if mass is decreasing thn the force on balance is upwards and on wire is downwards

 

[A.ELWY 7]

http://www.xtremepapers.com/papers/...and AS Level/Physics (9702)/9702_w07_qp_1.pdf
Q 23, need some 1 to explain...thanx

 

[Jaf]

http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w07_qp_1.pdf
Q 23, need some 1 to explain...thanx

Wavelength of the waves [from the graph] = 50m
Velocity of the wave = 8m/s
[f]Frequency = Velocity/Wavelength = 8/50 = 0.16 Hz
[a]Amplitude (or max displacement) of the waves [from the graph] = 2m
Maximum velocity= 2(pi)af = 2 x pi x 2 x 0.16 = (16/25)pi
Maximum KE of the object = (1/2)m(v^2) = (1/2) x (2x(10^-3)) x ((16/25)pi)^2 = 0.00404 J. Approximately 4mJ.

 

[A.ELWY 7]

tha

Wavelength of the waves [from the graph] = 50m
Velocity of the wave = 8m/s
[f]Frequency = Velocity/Wavelength = 8/50 = 0.16 Hz
[a]Amplitude (or max displacement) of the waves [from the graph] = 2m
Maximum velocity= 2(pi)af = 2 x pi x 2 x 0.16 = (16/25)pi
Maximum KE of the object = (1/2)m(v^2) = (1/2) x (2x(10^-3)) x ((16/25)pi)^2 = 0.00404 J. Approximately 4mJ.

thanx very much...if u may jaf im spending more than 1 hour to solve q1 b in http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w10_qp_22.pdf, we all know that cosx=adj/hyp....so it must be 25/cos(35) to get R but the mark scheme says 25cos(35)... ill blow pls say how???

 

[Jaf]

^I don't get why they're complicating things and finding 'R' in the marking scheme lol. This is how I got the answer (sorry my handwriting is kinda messy ) :

 

[A.ELWY 7]

^I don't get why they're complicating things and finding 'R' in the marking scheme lol. This is how I got the answer (sorry my handwriting is kinda messy ) :
View attachment 6625

thank you but, ALL of this i do know jaf...but i cant understand why 25cos(55)??

 

[ullahabd]

http://www.xtremepapers.com/papers/...nd AS Level/Physics (9702)/9702_s11_qp_11.pdf
help in q13, q15, q25, q27, q37

 

[XPFMember]

...some more doubts in June 2006


· Q:2 b (ii) how to do?? :s
· Q:4 c can you show, plz?
· Q:6 a what I understand is the field should be circle..right? I don’t get this :s
· Q:6 c what values do I use? :s
· Q:7