Physics: Post your doubts here!

[mehria]

MemoryMatrix 21

 

[MemoryMatrix 21]

MemoryMatrix 21

JazakAllah Khair !!!
i cnnt view the 1st file completely .....
May Allah Bless u immensly 4 this !!!

 

[Suchal Riaz]

Thank You can you help with the rest

did i do the winter 2008 papers?

 

[mehria]

JazakAllah Khair !!!
i cnnt view the 1st file completely .....
May Allah Bless u immensly 4 this !!!

U're welcum...
n if u need any help then feel free to ask...

 

[Mohammed salik]

MemoryMatrix 21

WOW!! Krichoff's Law Was Amazing!! Thnk uuuu (I always have problem with those)!

 

[MemoryMatrix 21]

U're welcum...
n if u need any help then feel free to ask...

thx a million ........ u always help me

 

[MissBellum]

u may b asked to find the *number* of moles ..... so we hv to ultiply with this avagadro
constant (6.02*10^23)
as, by definition : mole is the amount containing a no. of particles equal to avagadro constant

ok, so this equation is just the mean kinetic energy for one particle?

 

[Thought blocker]

mehria you are awesome.

 

[Suchal Riaz]

mahabaloch
w08
Q3(b)
the 12 and 2.5 N blocks exert clockwise moment. sack exerts clockwise moment. for them to be in equilibrium the sum of anticlockwise moment = sum of clockwise moment
12N * 84 cm + 2.5 N * 72 cm = W * 4.8 cm
4.8W=1188Ncm
W=1188/4.8=247.5N
the balance will be imprecise because the length of sliding of masses would be very small so there will be large uncertainty in the length.

 

[mehria]

mehria you are awesome.

i knw that

 

[Suchal Riaz]

mahabaloch
w08
q7(b)
the combined resistance of 5000 and thermister will decrease because at high temperature the resistance of thermistor decrease.
the p.d across the 5000 and thermistor will decrease.
by potential divider formula, the p.d across 2000 will increase.
volt meter shows the p.d across 2000 resistor.
so volt meter reading increase.
let R be resistance of 5000 resistor plus thermistor
so 2000/(2000+R) * 6=3.6 V [this is potential divider formula]
solve for R, R=1333 ohm
but R is combined resistance of thermistor and 500 resistor
so 1/R = 1/5000 + 1/Rt
Rt = 1820 ohm

 

[Suchal Riaz]

mahabaloch
s10_22
Q4(c)
at lowest frequency the wavelength = 4 L
speed = F λ
so F=speed/λ
F=330/4L = 330/4(0.45m)=183Hz
Q5(b)(ii)
no calculations are required but it is visible from graph that the strain after the limit of proportionality is 3.5 so the same will be final strain after unloading.

the markscheme says this mark was cancelled because the inconsistency of power of ten. i assume that they meant 2.5X10^8 not 6

 

[Thought blocker]

mahabaloch
s10_22
Q4(c)
at lowest frequency the wavelength = 4 L
speed = F λ
so F=speed/λ
F=330/4L = 330/4(0.45m)=183Hz
Q5(b)(ii)
no calculations are required but it is visible from graph that the strain after the limit of proportionality is 3.5 so the same will be final strain after unloading.
View attachment 40787
the markscheme says this mark was cancelled because the inconsistency of power of ten. i assume that they meant 2.5X10^8 not 6

Lets start waves

 

[♣♠ Magnanimous ♣♠]

Lets start waves

i wanna join as i said you in school.. dont forget.

 

[Thought blocker]

i wanna join as i said you in school.. dont forget.

Sure!

 

[Champ101]

http://papers.xtremepapers.com/CIE/...nd AS Level/Physics (9702)/9702_s10_qp_42.pdf

Q1(b)(ii), its the question on meteorite (Gravitation related sum )

im dont get it, from the graph, 2R corresponds to -3.1 J/g, but they are statin -2.1J/g .. please someone explain this!! :\

http://papers.xtremepapers.com/CIE/...nd AS Level/Physics (9702)/9702_s10_ms_42.pdf

 

[Suchal Riaz]

mahabaloch
s10 Q4(b)(ii)

each small square = 0.02mm * 2N = 0.04 mJ
count the small squares:
till 50 N there are 25*5=125 blocks
from 50-60 N there are 5*4=20 blocks
from 60-74 there are almost 15 blocks
total = 160 blocks
160 * 0.04 = 6.4 mJ
i know it is very tedious but this is how it has to be done. by counting blocks.

 

[***amd***]

View attachment 40788 Question 6a.. anyone please

***amd*** Suchal Riaz

i m not an A level student, but i guess the answer is NO.
becoz when a circuit gets closed, the electrons dont move from battery all around the circuit. but instead, the electrons present in the metallic wire start moving round the circuit at the same time. I mean, there is not a single wave of electrons running all round in the circuit. but, consider that instead of wires, we have a long round line of people. and when we close the switch, all the ppl just start walking round in the line.
so all the three bulbs would light at the same time.

hope you can understand it

 

[funky brat]

i m not an A level student, but i guess the answer is NO.
becoz when a circuit gets closed, the electrons dont move from battery all around the circuit. but instead, the electrons present in the metallic wire start moving round the circuit at the same time. I mean, there is not a single wave of electrons running all round in the circuit. but, consider that instead of wires, we have a long round line of people. and when we close the switch, all the ppl just start walking round in the line.
so all the three bulbs would light at the same time.

hope you can understand it

I think it's right. (Y).
I'm too not an A'level student.

 

[Suchal Riaz]

examiner can ask why both bulbs will light at same time.
as soon as the circuit is closed an electric fields will travel through out the circuit. as the speed of electric field is highest possible speed possible(speed of light) both bulbs light up instantly.