Physics: Post your doubts here!

[geek101]

http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s04_qp_2.pdf
^Question 8please!! :'(

I really badly need help!! Please and thankyou!!

8) just use proportionality here...
a - temp at 0 degrees = 3900 ohms
the voltmeter reading on V = 1 Volts
and becuz the sum of the potential pds = the emf of the battery
therefore the pd across the thermistor is 1.5 - 1 = 0.5 V

if 3900 ohms >> 0.5 V
then x ohms >> 1 V
x = 3900 / 0.5 = 7800 ohms

b - resistance of the thermistor = 1250 ohms
and resistance of R = 7800 ohms

if the pd across R is x , then the pd across the thermistor is 1.5 - x
so 7800 >> x ohms
1250 >> 1.5 - x ohms
cross multiply, ull get x = 1.3

c - there are now gonna be two 7800 ohm resistors in parallel and in series with the thermistor (3900 ohms >> 0 degress)
the combined resistance of the the two parallel resistors is = 7800 x 7800 / 7800 + 7800 = 3900 ohms
so the voltage is equally shared
1.5 / 2 = 0.75 volts >> the reading

 

[bamteck]

http://www.xtremepapers.com/papers/...nd AS Level/Physics (9702)/9702_w09_qp_22.pdf

No. 6(c) & d(ii)

 

[Goku]

Omg. You just made this question so easy for me,THANK YOU SO MUCH!!! :') <3 <3 <3

8) just use proportionality here...
a - temp at 0 degrees = 3900 ohms
the voltmeter reading on V = 1 Volts
and becuz the sum of the potential pds = the emf of the battery
therefore the pd across the thermistor is 1.5 - 1 = 0.5 V

if 3900 ohms >> 0.5 V
then x ohms >> 1 V
x = 3900 / 0.5 = 7800 ohms

b - resistance of the thermistor = 1250 ohms
and resistance of R = 7800 ohms

if the pd across R is x , then the pd across the thermistor is 1.5 - x
so 7800 >> x ohms
1250 >> 1.5 - x ohms
cross multiply, ull get x = 1.3

c - there are now gonna be two 7800 ohm resistors in parallel and in series with the thermistor (3900 ohms >> 0 degress)
the combined resistance of the the two parallel resistors is = 7800 x 7800 / 7800 + 7800 = 3900 ohms
so the voltage is equally shared
1.5 / 2 = 0.75 volts >> the reading

 

[Goku]

P.S could you please help out with the other questions too?

 

[geek101]

8 a iii ....thnx and good luck

 

[deepum]

9702_w05_qp_2................number 8 b ii can someone explain how to draw the line please.....

 

[InnocentAngel]

Anyone has done the graph question june 2012 p 22 ??? Please if yes post the graph here. Im really lost

 

[Soldier313]

8 a iii ....thnx and good luck

Just connect two resistors in series, that will give you 200 ohms
next connect two resistors in parallel, that will give you 50 ohms
now connect that first one with second one in a parallel way
(1/200) + (1/50) = 1/R
thus, R = 40 ohms

@ symbol represents a resistor



am so sorry for the vague drawing:/

 

[hussamh10]

3b) ok here first you need to have a scale lets say : 1 ms-1 = 1 cm

and the vertical and horizontal velocities are perpendicular to each other
now draw 4.o cms >> v (horizontal)
6.2 cms >> v (vertical)
now complete the triangle and measure the length of the line >> the resultant velocity!
and measure the angle with the vertical

5b) phase difference = 360 ( x / lambda) where x is the distance between the points
now i cant figure out the 180 sorry

ci) speed = frequency x wavelength
wavelength = L and frequency = f
therefore speed = f . L

Particles in the same segment/ between 2 adjacent nodes, are in phase. Particles in adjacent segments are in anti-phase so 180..​

 

[geek101]

Just connect two resistors in series, that will give you 200 ohms
next connect two resistors in parallel, that will give you 50 ohms
now connect that first one with second one in a parallel way
(1/200) + (1/50) = 1/R
thus, R = 40 ohms

@ symbol represents a resistor

View attachment 17617

am so sorry for the vague drawing:/

omg thanks!

 

[TeKnOzOr]

http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w09_qp_22.pdf

No. 6(c) & d(ii)

(c)
when the lamps is connected in series, therefore the e.m.f of the battery will be divided into half....
means the p.d across each resistor will be (3.0/2) v , that is, 1.5 v .....
Now the current is obtained from the graph...look for the value of I when v=1.5--->> we get I=0.10A
you then can calculate the resistance of each lamp and the combined resistance.

when the lamps is connected in parallel,therefore the p.d across the 2 resistors will be the same as the e.m.f of the battery.that is 3.0 v itself....Now the current is obtained from the graph...look for the value of I when v=3.0--->> we get I=0.15A
from here you can calculate the resistance of each lamp and the combined resistance.

(d) (ii)
Answer: From a graph of I against V, the resistance is given by (1/gradient). From the graph, the gradient decreases when the p.d increases. This result in an increase in resistance of the lamp.
Note: the resistance is given by (1/gradient) rather than (gradient) itself because the graph is I against V.
If it would be V against I,then the resistance would be = the gradient.
Hope it helps

 

[alcoholic 111]

(c)
when the lamps is connected in series, therefore the e.m.f of the battery will be divided into half....
means the p.d across each resistor will be (3.0/2) v , that is, 1.5 v .....
Now the current is obtained from the graph...look for the value of I when v=1.5--->> we get I=0.10A
you then can calculate the resistance of each lamp and the combined resistance.

when the lamps is connected in parallel,therefore the p.d across the 2 resistors will be the same as the e.m.f of the battery.that is 3.0 v itself....Now the current is obtained from the graph...look for the value of I when v=3.0--->> we get I=0.15A
from here you can calculate the resistance of each lamp and the combined resistance.

(d) (ii)
Answer: From a graph of I against V, the resistance is given by (1/gradient). From the graph, the gradient decreases when the p.d increases. This result in an increase in resistance of the lamp.
Note: the resistance is given by (1/gradient) rather than (gradient) itself because the graph is I against V.
If it would be V against I,then the resistance would be = the gradient.
Hope it helps

For d (ii) you are wrong although it make sense but seriously you would be given 0 marks for your explaination
we don't consider gradient we consider point
Check Oct/Nov 2004 Question 6 a i read marking scheme and examiner report would would understand

 

[TeKnOzOr]

For d (ii) you are wrong although it make sense but seriously you would be given 0 marks for your explaination
we don't consider gradient we consider point
Check Oct/Nov 2004 Question 6 a i read marking scheme and examiner report would would understand

Hehe...dude....I don't think so...!! lol
and who would give me 0 marks man?!? because acording to the marking schemes,my answer is perfect,that is 3/3 !!
And this answer is my teacher's answer...so no doubts on that...!!!
Ohhhh....we don't consider gradient!?! lol....hmmm...then what points you considered you?!
Because normally,the question said 'with reference to graph'...and the p.d is given by 1/gradient.
Here is the ms:http://www.xtremepapers.com/papers/...nd AS Level/Physics (9702)/9702_w09_ms_22.pdf
Check it yourself and give your view dude...!!

 

[TeKnOzOr]

For d (ii) you are wrong although it make sense but seriously you would be given 0 marks for your explaination
we don't consider gradient we consider point
Check Oct/Nov 2004 Question 6 a i read marking scheme and examiner report would would understand

Hehe...but you may be correct man...!!! I read the reports....the answer of the ms doesn't talk about gradient at all !!
Anyway,thanks for your kind views...i'll talk this to my teacher..!!

 

[aiskw1]

http://xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w11_qp_23.pdf
HELP NEEDED

2.8x2.8 = intensity original

2 x intensity original = new intensity

square root new intensity = amplitude

amplitude is proportional to intensity

 

[alcoholic 111]

Hehe...but you may be correct man...!!! I read the reports....the answer of the ms doesn't talk about gradient at all !!
Anyway,thanks for your kind views...i'll talk this to my teacher..!!

It's Okay!!
I was too using the same gradient concept of yours
but still confused !!
can you talk to your teacher and explain this to me as well ?
thanks in advance

 

[deep mehta]

what will be the current time graph and voltage time graph after an ac current passes through a diode???? PLZ Reply fast I M really stuck on this question

 

[geek101]

emkay here ya go...

 

[emkay]

emkay here ya go...

variant 21 of 2012 please? you accidentally posted 2010

 

[geek101]

variant 21 of 2012 please? you accidentally posted 2010

oops sorry