ALL CIE PHYSICS DOUBTS HERE !!!

[WARRIOR]

thank you guys for answerin my question !!!!!!!!

[birchy33]

Hi I'm New to the forum and need some help with this question. Its from May/June 2009 paper 1st variant.
Question number 5a and 5b. Sorry i don't know how to link it but help would be appreciated because I'm just stuck for all money on how to approach it. Cheers guys.

[Deadly_king]

First the anchor is on the boat. It exerts a downward force on the boat, forcing the boat down, so displacing alot of water.

When the anchor is released, less of the ship is in the water therefore less water diplacement. The volume diplaced by the anchor is small.

The higher the water displacement, the higher the water level. Since water displacement decreased the water level should also decrease.

Imagine yourself in a bathtub. Imagine that the anchor is on you, you will go deeper into the water, the water will rise. when the anchor is released, less of you will be in the water, so the water level will decrease.

If the anchor doesnt reach the bottom the overall water level will not change cos the weight of water displaced equals weight of boat + weight of anchor.
If the anchor hits the bottom then archimedes principle only applies to the boat. The weight of water displaced equals the weight of the boat only, so less water is displaced and the water level falls.

That's right. Sorry for the confusion Kimo

[Deadly_king]

Hi I'm New to the forum and need some help with this question. Its from May/June 2009 paper 1st variant.
Question number 5a and 5b. Sorry i don't know how to link it but help would be appreciated because I'm just stuck for all money on how to approach it. Cheers guys.

Welcome to the forum.

I'll advice you to introduce yourself in the Introduction thread.

For your question, is it paper 2 or paper 4

[$!$RatJumper$!$]

I think he's talking about p2

[$!$RatJumper$!$]

W05 P2
Q3 (c)- what is the direction?
Q4 (a)(ii)- why doesnt it work when I find the area under of the triangle for the first 0.3 sec then times it by 2
Q5 (c)(ii)- please explain
 
Thank you

[birchy33]

yup spot on. Paper 2.

[astarmathsandphysics]

when I gwt bup shortly

[astarmathsandphysics]

Q3 (c)- what is the direction?
Q4 (a)(ii)- why doesnt it work when I find the area under of the triangle for the first 0.3 sec then times it by 2
Q5 (c)(ii)- please explain
3c)36 degrees below the horizontal, at right angles to the upper string
4aii) It should work the way you did it. I get about 2*1/2*0.3*0.10 =0.03 by APPROXIMATING THE slope by a line
5vii)amplitude of A=sqrt(1) and amplitude of B=sqrt(4/9)=2/3
1-2/3=1/3 then 1/3^2=1/9 . Subtract since there are 180 degrees out of phase

[$!$RatJumper$!$]

Q3 (c)- what is the direction?
Q4 (a)(ii)- why doesnt it work when I find the area under of the triangle for the first 0.3 sec then times it by 2
Q5 (c)(ii)- please explain

3c)36 degrees below the horizontal, at right angles to the upper string
4aii) It should work the way you did it. I get about 2*1/2*0.3*0.10 =0.03 by APPROXIMATING THE slope by a line
5vii)amplitude of A=sqrt(1) and amplitude of B=sqrt(4/9)=2/3
1-2/3=1/3 then 1/3^2=1/9 . Subtract since there are 180 degrees out of phase

i really dont understand 3 and 5.. please could you elaborate
and fir 4, the mark scheme says horizontal.. doesnt that mean 90 degrees? so why is it 36?

thankx a lot

[TJ-56]

i really dont understand 3 and 5.. please could you elaborate
and fir 4, the mark scheme says horizontal.. doesnt that mean 90 degrees? so why is it 36?

thankx a lot

For q5 (c):
To find the resultant intensity, you have to find the resultant amplitude first, since they are 180 degrees out of phase, find the difference between the amplitudes of 2 peaks (or troughs) which is (-) 1, so I(resultant)/I=1(sqr)/3(sqr)  simplified gives the resultant intensity of 1/9.
(you could also use the intensity 4/9 * I but with amplitude of 2 squared)
Hope that was helpful.

[$!$RatJumper$!$]

ok thankx that makes sense now and jus a query, when finding resultant amplitudes, dont we ADD the amplitudes. like in this question, i think they added them in order to get their resultant amplitude as 0.0001

[TJ-56]

ok thankx that makes sense now and jus a query, when finding resultant amplitudes, dont we ADD the amplitudes. like in this question, i think they added them in order to get their resultant amplitude as 0.0001

That is correct, we add them.
Take for example the 1st peak on wave A, wave A's amp is 3, and wave B's amp is -2 at that time, so 3 + (-2) is 1 x 10^-4

[$!$RatJumper$!$]

ok so is it a rule that we should ALWAYS ADD amplitudes to find the resultant?

[TJ-56]

ok so is it a rule that we should ALWAYS ADD amplitudes to find the resultant?

I think so yeah, theres another question regarding this topic on w02 qp2 q5 b iii , so, positive