Author Topic: ALL CIE PHYSICS DOUBTS HERE !!!  (Read 174074 times)

Offline ashish

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Re: ALL CIE PHYSICS DOUBTS HERE !!!
« Reply #15 on: October 10, 2010, 08:12:05 am »
i like to help friends  ;D

Offline Ghost Of Highbury

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Re: ALL CIE PHYSICS DOUBTS HERE !!!
« Reply #16 on: October 10, 2010, 10:53:08 am »
Can u explain the steps too please? Thank you
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Offline Deadly_king

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Re: ALL CIE PHYSICS DOUBTS HERE !!!
« Reply #17 on: October 10, 2010, 11:07:53 am »
Can u explain the steps too please? Thank you
20.

You should first know that points at the same level/height experience same pressure and that pressure due to a liquid is heg.

Back to the diagram, let's take it at the level x from below.

Since in both limbs, points at same level experience same pressure :
Patm + xeqg = Patm + 2xepg

Therefore simplify this equation to obtain ;
ep/eq = 1/2

Answer is A

12.
Since the forces are in equilibrium, the resultant force should be zero. For this, the forces should make a complete cycle back to the same original point.

Only triangles R and S show this possiblity. Hence answer is C

Hope it helps :)
« Last Edit: October 10, 2010, 11:10:52 am by Deadly_king »

Offline Ghost Of Highbury

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Re: ALL CIE PHYSICS DOUBTS HERE !!!
« Reply #18 on: October 10, 2010, 12:01:31 pm »
20.

You should first know that points at the same level/height experience same pressure and that pressure due to a liquid is heg.

Back to the diagram, let's take it at the level x from below.

Since in both limbs, points at same level experience same pressure :
Patm + xeqg = Patm + 2xepg

Therefore simplify this equation to obtain ;
ep/eq = 1/2

Answer is A

12.
Since the forces are in equilibrium, the resultant force should be zero. For this, the forces should make a complete cycle back to the same original point.

Only triangles R and S show this possiblity. Hence answer is C

Hope it helps :)

Thank you. +rep
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Offline Deadly_king

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Re: ALL CIE PHYSICS DOUBTS HERE !!!
« Reply #19 on: October 10, 2010, 12:08:47 pm »

Offline Deadly_king

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Re: ALL CIE PHYSICS DOUBTS HERE !!!
« Reply #20 on: October 10, 2010, 12:15:38 pm »
One doubt, which point have u considered?
Height x above the ground.

It will be 2x for the left limb but x in the right one. Got me?

Offline ruby92

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Re: ALL CIE PHYSICS DOUBTS HERE !!!
« Reply #21 on: October 10, 2010, 04:19:25 pm »
Keplers law T^2 is direcly proportional to R^3

therefore T^2/R^3 = constant

(27.4*24*60*60)^2 / (3.84*10^8)^3 =9.8977*10^-14

let the period be T

T^2 / R^3 = 9.8977*10^-14
T^2 = (9.8977*10^-14)* (6.4*10^6)^3
T = 5093.74 sec
T=1h24min

is that right?
Thanks:)

Offline ruby92

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Re: ALL CIE PHYSICS DOUBTS HERE !!!
« Reply #22 on: October 10, 2010, 04:38:23 pm »
Physics A2 oct/nov 2008 paper 4 q 1aiii
why is it that the normal reaction force=force of gravity - centerpital force.
shouldnt it be plus?

Also m/j 2007 paper 4 q1 bi
« Last Edit: October 10, 2010, 04:49:50 pm by ruby92 »

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Re: ALL CIE PHYSICS DOUBTS HERE !!!
« Reply #23 on: October 10, 2010, 04:54:51 pm »
i need help in this question

The load on each of the arrangements is L.
For each arrangement , complete the table by determining
(i) the total extension in terms of e,
(ii) the spring constant in terms of k.
« Last Edit: October 10, 2010, 05:01:07 pm by Ghost Of Highbury~ »
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Offline Deadly_king

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Re: ALL CIE PHYSICS DOUBTS HERE !!!
« Reply #24 on: October 11, 2010, 05:15:34 am »
i need help in this question

The load on each of the arrangements is L.
For each arrangement , complete the table by determining
(i) the total extension in terms of e,
(ii) the spring constant in terms of k.

Hmm......Ashish......your spring constants are ok. But am afraid the extensions are not  :-[

Anyway dude, you should know that for spring constant,
When springs are in parallel, their net spring constants are added together. ( Same as for a capacitor and opposite to that of resistors)

When they are in series they are calculated as shown KR = [1/k + 1/k]-1 or 1/KR = 1/k + 1/k

For extension,
When in series, net extension is the sum of the extension of each spring.
When in parallel, it is calculated as shown eR = [1/e + 1/e]-1 or 1/eR = 1/e + 1/e

Let's back to the question now :
1. Total extension will be the sum of extension by each springs when they are connected in series. Each spring extends by e
Hence total extension will be 2e
Spring constant will be [1/k + 1/k]-1 = k/2

2. Since the two springs are connected in parallel, total extension will be e/2 and net spring constant will be 2k
eR = [1/e + 1/e]-1 = e/2
KR = k + k = 2k

3. Total extension : 3e/2
Net spring constant : 2k/3
Use the values you obtained in part(ii)
Total extension = e/2 + e = 3e/2
For the two springs in parallel, net spring constant will be 2k. Hence spring constant of the whole system will be [1/2k + 1/k]-1 = 2k/3

NOTE : kR and eR are the resultant spring constants and resultant extensions respectively.
« Last Edit: October 11, 2010, 05:19:07 am by Deadly_king »

Offline Deadly_king

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Re: ALL CIE PHYSICS DOUBTS HERE !!!
« Reply #25 on: October 11, 2010, 05:44:47 am »
Physics A2 oct/nov 2008 paper 4 q 1aiii
why is it that the normal reaction force=force of gravity - centerpital force.
shouldnt it be plus?

Also m/j 2007 paper 4 q1 bi

Nov 08 p4 No 1

a)(iii) The normal reaction exerted by a planet on a mass is equal to the weight of the mass.

Normally for an object undergoing circular motion,
Resultant towards centre = Centripetal force

Resultant force towards centre of the planet will be : Gravitational force - Weight

Hence
Gravitational force - Weight = Centripetal force ---> Weight = GMm/R2 - mRw2

Since Weight = Normal reaction = GMm/R2 - mRw2

Jun 07 p4 No 1

b)(i) Gravitational potential energy is given by the formula = -GMm/R

In the question you are asked to find the change in gravitational potential.
Let ^Ep = Change in gravitational potential energy
Uf = Final gravitational potential energy = -GMm/3R
Ui = Initial gravitational potential energy = -GMm/2R

Hence ^Ep = Uf - Ui
         ^Ep = (-GMm/3R) - (-GMm/2R)

Simplify this equation and you'll get your required solution which is ^Ep = GMm/6R
« Last Edit: October 11, 2010, 05:46:54 am by Deadly_king »

Offline ashish

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Re: ALL CIE PHYSICS DOUBTS HERE !!!
« Reply #26 on: October 11, 2010, 11:26:10 am »
Hmm......Ashish......your spring constants are ok. But am afraid the extensions are not  :-[

Anyway dude, you should know that for spring constant,
When springs are in parallel, their net spring constants are added together. ( Same as for a capacitor and opposite to that of resistors)

When they are in series they are calculated as shown KR = [1/k + 1/k]-1 or 1/KR = 1/k + 1/k

For extension,
When in series, net extension is the sum of the extension of each spring.
When in parallel, it is calculated as shown eR = [1/e + 1/e]-1 or 1/eR = 1/e + 1/e

Let's back to the question now :
1. Total extension will be the sum of extension by each springs when they are connected in series. Each spring extends by e
Hence total extension will be 2e
Spring constant will be [1/k + 1/k]-1 = k/2

2. Since the two springs are connected in parallel, total extension will be e/2 and net spring constant will be 2k
eR = [1/e + 1/e]-1 = e/2
KR = k + k = 2k

3. Total extension : 3e/2
Net spring constant : 2k/3
Use the values you obtained in part(ii)
Total extension = e/2 + e = 3e/2
For the two springs in parallel, net spring constant will be 2k. Hence spring constant of the whole system will be [1/2k + 1/k]-1 = 2k/3

NOTE : kR and eR are the resultant spring constants and resultant extensions respectively.

the answer i gave were in terms of k an L :(
« Last Edit: October 11, 2010, 03:50:51 pm by ashish »

Offline Deadly_king

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Re: ALL CIE PHYSICS DOUBTS HERE !!!
« Reply #27 on: October 11, 2010, 05:10:51 pm »
the answer i gave were in terms of k an L :(

Yeah.......the question asked in terms of e ;)

Offline ashish

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Re: ALL CIE PHYSICS DOUBTS HERE !!!
« Reply #28 on: October 12, 2010, 02:34:28 pm »
lol i i gave him these as i did a similar question in class in which it was asked in terms of k and l ::)

Offline WARRIOR

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Re: ALL CIE PHYSICS DOUBTS HERE !!!
« Reply #29 on: October 12, 2010, 03:07:47 pm »
a simple test if anyone has the time to do it , good for basics practise

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