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umm..soo magnitude is equal to electric field strength?Electric field stregth = Voltage / Distance between the plates
350 / 2.5 into 10^-2 = 1.4 into 10^4 N/C
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umm..soo magnitude is equal to electric field strength?Electric field stregth = Voltage / Distance between the plates
350 / 2.5 into 10^-2 = 1.4 into 10^4 N/C
the question asked for the magnitude of electric field which means the same as to find the electric field strength.umm..soo magnitude is equal to electric field strength?
Just defining the term 'insulator'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)
waalaikumassalam wr wbAoA,
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
thnxthe question asked for the magnitude of electric field which means the same as to find the electric field strength.
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 downwardsThanks!
And how do we know that the force due to current-carrying wire is acting upwards??
Wavelength of the waves [from the graph] = 50mhttp://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
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???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.
thank you but, ALL of this i do know jaf...but i cant understand why 25cos(55)??^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
Whenever a force is given and you have to resolve it and you want to make a closed triangle make sure the given force is always the hypotenuse of your triangle. So to make the triangle in my diagram, drop a perpendicular from the arrowhead of the 25N on the dotted extension of T. This perpendicular will be the vertical. cos theta is adj/hyp. Adjacent side the dotted extension of T. So cos 55 X hyp = 25cos55.thank you but, ALL of this i do know jaf...but i cant understand why 25cos(55)??
jaf i really dont no how to thank you...and this note about the given force being the hypotinse it really save me thank youWhenever a force is given and you have to resolve it and you want to make a closed triangle make sure the given force is always the hypotenuse of your triangle. So to make the triangle in my diagram, drop a perpendicular from the arrowhead of the 25N on the dotted extension of T. This perpendicular will be the vertical. cos theta is adj/hyp. Adjacent side the dotted extension of T. So cos 55 X hyp = 25cos55.
Oh but I do... kind duas and prayers for me would suffice.jaf i really dont no how to thank you
Q13:CLOCK WISE moment = anticlockwise moment so FxD=FxD and the rule have a mass of 100g in the centre at 50cm mark so 10 cm from the 40 mark so it is (20g x 60)+(100g x 10)/50 = 44 which is Chttp://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s11_qp_11.pdf
help in q13, q15, q25, q27, q37
Q25)http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s11_qp_11.pdf
help in q13, q15, q25, q27, q37
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