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Physics: Post your doubts here!

bamteck

bamteck

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emkay said:
1. line's are further apart in second order
2. Angle difference between the first order and the second order is greater, therefore less brightness is observed.
Click to expand...

Thanks dude ! Seems like you are good in physics ! :p
 
T

Trisha Graal Emile

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1. A metal ball is dropped into a tall cylinder of oil. The ball initially accelerates but soon reaches a terminal velocity.

a. By considering the forces in the metal ball bearing, explain why it first accelerates but then reaches terminal velocity.

b. Describe how you would show that the metal ball reaches terminal velocity.

1.
A metal ball is dropped into a tall cylinder of oil. The ball initially accelerates but soon reaches a terminal velocity.

a. By considering the forces in the metal ball bearing, explain why it first accelerates but then reaches terminal velocity.

b. Describe how you would show that the metal ball reaches terminal velocity.
 
queen of the legend

queen of the legend

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bamteck said:
http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s03_qp_2.pdf

Please someone help me for no. no. 3(a)(ii) and b(i) :(
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for question 3aii : its very simple if u read the text book you can find detailed explanation about this topic, anyhow .. take a ruler and put it on the line of the graph so that when line starts deviating from straight path its the point that should be marked L where hook's law will not apply because force is no longer proportional to extension . in the next question it asks about the spring constant and since we know that f=ke where k is a constant. to deduce value of k from graph we know that k=f/e. the gradient of the graph is e/f so that k = 1/gradient

bi: the circumference of the circle is pi x d ...the change in length of the wire occurs over the pulley ...we know the pulley turns an angle 6.5 so the amount of increase or extension is = 6.5/360 x 3pi..............6.5/360 will give us the amount of increase at the circumference ..............i hope u understood !
 
queen of the legend

queen of the legend

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mrnt3250 said:
Another one..
Oct/Nov 2010 - QP 22 - Question 1 - part (b).

I have no idea about doing questions related to horizontal components, the ones with tan, etc.
Could someone give me a complete explanation of how to deal with such questions?

QP: http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w10_qp_22.pdf
MS: http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w10_ms_22.pdf

Thanks.
Click to expand...
but the question is not about component of vectors ?! its about uncertainty isnt it ?
 
bamteck

bamteck

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queen of the legend said:
for question 3aii : its very simple if u read the text book you can find detailed explanation about this topic, anyhow .. take a ruler and put it on the line of the graph so that when line starts deviating from straight path its the point that should be marked L where hook's law will not apply because force is no longer proportional to extension . in the next question it asks about the spring constant and since we know that f=ke where k is a constant. to deduce value of k from graph we know that k=f/e. the gradient of the graph is e/f so that k = 1/gradient

bi: the circumference of the circle is pi x d ...the change in length of the wire occurs over the pulley ...we know the pulley turns an angle 6.5 so the amount of increase or extension is = 6.5/360 x 3pi..............6.5/360 will give us the amount of increase at the circumference ..............i hope u understood !
Click to expand...

Hey thank you soo much :)
 
Asheeta

Asheeta

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Hello, help needed for Nov 2010 paper 2 no.4 (b) the whole part
What is the region Es?? and how to find it?
please help and by the way can someone give a good definition for strain energy??
 
Asheeta

Asheeta

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bamteck said:
http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s06_qp_2.pdf

Help needed for no. 6 (a)(i) and (c) and 7(a) (b) (c) :(
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no. 6
so just stretch the dotted lines till fig 6.2
the upperpart is the same as in fig 6.1
in the 32.4 cm just draw a loop
that's it.

(c)so after the sketch you see that you have antinode on the top then a node a loop and a node
dist between antinode and node = (lambda/4)
dist between 2 nodes = (lambda/2)
(lambda/2)=32.4
therefore (lambda/4)=16.2cm
this means that the antinode will be 0.5 cm above the tube

hope you've understood!:)
 
bamteck

bamteck

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Asheeta said:
no. 6
so just stretch the dotted lines till fig 6.2
the upperpart is the same as in fig 6.1
in the 32.4 cm just draw a loop
that's it.

(c)so after the sketch you see that you have antinode on the top then a node a loop and a node
dist between antinode and node = (lambda/4)
dist between 2 nodes = (lambda/2)
(lambda/2)=32.4
therefore (lambda/4)=16.2cm
this means that the antinode will be 0/5 cm above the tube

hope you've understood!:)
Click to expand...

Hey thank you soo much :)

Btw, Strain energy is the energy stored in a body due to change of shape ! :)
 
A

AAAAAA

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bamteck said:
http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s06_qp_2.pdf

Help needed for no. 6 (a)(i) and (c) and 7(a) (b) (c) :(
Click to expand...

amplitude of wave A = 3 (maximum disp. from first graph)
amplitude of wave B=2

but phase difference between A and B is π radians.
It is said in the question that the waves are at a common point (point P)

therefore, from principle of superposition resultant amplitude = amplitude of A - amplitude of B

This is equal to 1

Intensity is proportional to amplitude squared

At A, intensity = I and amplitude=3
hence, I=k(3^2)
k=constant=I/9

so, if resultant amplitude is 1, intensity=k(1^2)

this is equal to I/9

this is for the NOV05 question..

Note: I have ignored factors of 10^-3 in amplitude because it would cancel out anyway.
 
bamteck

bamteck

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AAAAAA said:
amplitude of wave A = 3 (maximum disp. from first graph)
amplitude of wave B=2

but phase difference between A and B is π radians.
It is said in the question that the waves are at a common point (point P)

therefore, from principle of superposition resultant amplitude = amplitude of A - amplitude of B

This is equal to 1

Intensity is proportional to amplitude squared

At A, intensity = I and amplitude=3
hence, I=k(3^2)
k=constant=I/9

so, if resultant amplitude is 1, intensity=k(1^2)

this is equal to I/9

this is for the NOV05 question..

Note: I have ignored factors of 10^-3 in amplitude because it would cancel out anyway.
Click to expand...

Thank youuuuuuuuuuuuu :)
 
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