Author Topic: ALL CIE PHYSICS DOUBTS HERE !!!  (Read 171970 times)

Offline TJ-56

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Re: ALL CIE PHYSICS DOUBTS HERE !!!
« Reply #360 on: November 14, 2010, 11:33:15 am »
Why is the answer A?

Q=It (charge=current*time)
There are for charges so 4Q
I=Q\T
 =4Q\T
Frequency=1/T
so replacing f=1/T   into I= 4Q \ T
will give I=4Qf

Offline Dania

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Re: ALL CIE PHYSICS DOUBTS HERE !!!
« Reply #361 on: November 14, 2010, 11:40:27 am »
Q=It (charge=current*time)
There are for charges so 4Q
I=Q\T
 =4Q\T
Frequency=1/T
so replacing f=1/T   into I= 4Q \ T
will give I=4Qf


Thanks :)
:)

Offline thecandydoll

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Re: ALL CIE PHYSICS DOUBTS HERE !!!
« Reply #362 on: November 14, 2010, 11:55:50 am »
S08 P1 Q34!

Offline thecandydoll

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Re: ALL CIE PHYSICS DOUBTS HERE !!!
« Reply #363 on: November 14, 2010, 12:12:08 pm »
2009 JUNE -Q10/21

When there is force/extension curve not a straight line,how to find strain energy

Offline Deadly_king

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Re: ALL CIE PHYSICS DOUBTS HERE !!!
« Reply #364 on: November 14, 2010, 12:13:52 pm »
S08 P1 Q34!

Resistance of a wire is given by the formula R = eL/A where
e : Resistivity of wire
L : length of wire
A : Area of cross-section of wire.

Since both wires are made of copper, they have the same resistivity and they will cancel out.

It is said that both wires have same volume which implies that Vx = Vy

Vx = AxL and Vy = 4AyL

Now we can find Ax in terms of Ay and L ----> Ax = 4Ay

Now we replace in the formula.
Rx = L/Ax or L/4Ay and Ry = 4L/Ay

Now Ry/Rx = (4L/Ay) / (L/4Ay)

Simplify and you'll get it to be 16.

Answer is C

Hope it helps :)

Offline cs

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Re: ALL CIE PHYSICS DOUBTS HERE !!!
« Reply #365 on: November 14, 2010, 12:17:22 pm »
Deadly king to the rescue.. glad that you are here, i don't dare to see June 2010 paper 1, i can't do most of them cos my teacher didn't go through, can you help me Q5 and 6 ( any variant cos they are the same) before i go to sleep? =)
edited: whoops, its paper 11..
« Last Edit: November 14, 2010, 12:25:11 pm by cs »

Offline Deadly_king

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Re: ALL CIE PHYSICS DOUBTS HERE !!!
« Reply #366 on: November 14, 2010, 12:32:01 pm »
Deadly king to the rescue.. glad that you are here, i don't dare to see June 2010 paper 1, i can't do most of them cos my teacher didn't go through, can you help me Q5 and 6 ( any variant cos they are the same) before i go to sleep? =)

Hehe.........indeed they are all the same though the numbers are not in the same order. :-[

Anyway i'll do No 5 and 6 from variance 1 but if they are not the numbers you need, let me know. ;)

Jun 10 p1

5. let theta be x.
This is a right-angled triangle. If you resolve the resultant V in the X direction, it should be in terms of cos x while in the direction of Y it will be in terms of sin x.

As the angle is increased, we can note the sin x will rise as well. But cos x will decrease.

Hence X is decreasing and Y is increasing. :)

Answer is C.

6. The equation is Density = Mass / Volume ----> e = m/v

Therefore % uncertainty in e (^e) will be % (^m) + %(^v)

^e/e x 100 = [2(1)/(70-20) x 100] + [0.6/10 x 100]

Now you need to find ^e = 0.5

NOTE : The uncertainty in mass is multiplied by two since two values are measured.
(initial : empty beaker and final : beaker + liquid).

Answer is B

« Last Edit: November 14, 2010, 12:34:54 pm by Deadly_king »

Offline $!$RatJumper$!$

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Re: ALL CIE PHYSICS DOUBTS HERE !!!
« Reply #367 on: November 14, 2010, 01:20:53 pm »
S07 Q 3, 7, 8, 23, 31, 40
Thank you
« Last Edit: November 14, 2010, 02:02:22 pm by $!$RatJumper$!$ »

Offline Deadly_king

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Re: ALL CIE PHYSICS DOUBTS HERE !!!
« Reply #368 on: November 14, 2010, 01:59:35 pm »
2009 JUNE -Q10/21

When there is force/extension curve not a straight line,how to find strain energy

You need to find the area under graph. Most of the time you need to estimate since it's not easy to calculate the area under graph when a curve is involved.

Jun 09 p01

21.
You need to find are over the curve since it is rubber and the question ask for the energy stored.

Try to divide it into small triangles and trapeziums.

Answer is A

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Re: ALL CIE PHYSICS DOUBTS HERE !!!
« Reply #369 on: November 14, 2010, 02:13:51 pm »
S07 Q 3, 7, 8, 23, 31, 40
Thank you

Jun 07 p1

3. Do it in terms of units. The first one itself is the answer.

V has unit ms-1.
g has unit ms-2 while lambda has unit m. Hence g(lamnda) will have unit m2s-2. Its square root will come out to be ms-1.

Answer is A.

7. You should use the equation v = u + at to describe the motion of this object.

From vector diagram we can note that V = u + X

Hence X comes out to be at. ;)

Answer is C

I'll complete the rest asap

Offline TJ-56

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Re: ALL CIE PHYSICS DOUBTS HERE !!!
« Reply #370 on: November 14, 2010, 02:14:53 pm »
S07 Q 3, 7, 8, 23, 31, 40
Thank you

3
Its not about deriving a formula here, its about using SI units to get the result:
Balancing SI units on each side of the 4 choices, only A is balanced,
m/s = root(m s-2 * m )  sqr both sides
m2 s-2 = m2 s-2

8
I found the time taken to fall 40m - the time taken to fall 30 m , which is equivalent to the required time
s=40 t=? g=a=9.81 u=0
s=ut + 0.5at2 is used to find t, with using 40m once as s and 30m the other, and finding the difference, which is the answer A

31
Ohm’s Law: if a conductor obeys Ohm’s law, then the current I through it is directly proportional to the potential difference V across its ends provided that its temperature remains constant. That is current  I ? V if temperature is constant.

40
You need to calculate the ratios proton/nucleons of each to find the speed, with the lowest being 0.428 of lithium


Offline Dania

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Re: ALL CIE PHYSICS DOUBTS HERE !!!
« Reply #371 on: November 14, 2010, 02:16:10 pm »
Beats me why the answer to this is B.
:)

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Re: ALL CIE PHYSICS DOUBTS HERE !!!
« Reply #372 on: November 14, 2010, 02:17:55 pm »
3
Its not about deriving a formula here, its about using SI units to get the result:
Balancing SI units on each side of the 4 choices, only A is balanced,
m/s = root(m s-2 * m )  sqr both sides
m2 s-2 = m2 s-2

8
I found the time taken to fall 40m - the time taken to fall 30 m , which is equivalent to the required time
s=40 t=? g=a=9.81 u=0
s=ut + 0.5at2 is used to find t, with using 40m once as s and 30m the other, and finding the difference, which is the answer A

31
Ohm’s Law: if a conductor obeys Ohm’s law, then the current I through it is directly proportional to the potential difference V across its ends provided that its temperature remains constant. That is current  I ? V if temperature is constant.

40
You need to calculate the ratios proton/nucleons of each to find the speed, with the lowest being 0.428 of lithium



Thank you so much. +rep

Now Dead King can rest a lil bit.

Offline $!$RatJumper$!$

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Re: ALL CIE PHYSICS DOUBTS HERE !!!
« Reply #373 on: November 14, 2010, 02:28:11 pm »
3
Its not about deriving a formula here, its about using SI units to get the result:
Balancing SI units on each side of the 4 choices, only A is balanced,
m/s = root(m s-2 * m )  sqr both sides
m2 s-2 = m2 s-2

8
I found the time taken to fall 40m - the time taken to fall 30 m , which is equivalent to the required time
s=40 t=? g=a=9.81 u=0
s=ut + 0.5at2 is used to find t, with using 40m once as s and 30m the other, and finding the difference, which is the answer A

31
Ohm’s Law: if a conductor obeys Ohm’s law, then the current I through it is directly proportional to the potential difference V across its ends provided that its temperature remains constant. That is current  I ? V if temperature is constant.

40
You need to calculate the ratios proton/nucleons of each to find the speed, with the lowest being 0.428 of lithium



Thank you :)
so why cant we do this: 10 = 0.5 * 9.81 * t2?
and so the ratio of the two gives the speed of them?

Offline TJ-56

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Re: ALL CIE PHYSICS DOUBTS HERE !!!
« Reply #374 on: November 14, 2010, 02:51:17 pm »
Thank you :)
so why cant we do this: 10 = 0.5 * 9.81 * t2?
and so the ratio of the two gives the speed of them?

Because at 30m, it has an initial velocity, whereas ur equation assumes initial velocity is 0  (ut=0)

i remember reading it off a book, so i think yeah thats true.