Physics: Post your doubts here!

[Snowysangel]

Finally I got all of the answers of the paper Paper was easy, here you go, hope this is correct... resistance is zero so pwer is zero obvious and in infinite resistance no current passes so zero rsistance i guess...

How come potential difference is highest when resistance is zero? Aren't potential difference and resistance directly proportional?

 

[Thought blocker]

How come potential difference is highest when resistance is zero? Aren't potential difference and resistance directly proportional?

sister there are few problem in what you said:
1 infinity is not a number
2 and you can say x=∞. correct way is x-->∞ which means it 'tends to infinity' so it never reaches infinity but it tends towards infinity.
3 in classical mechanics we use infinity to say very large compared to some thing. like we say S>>V means S is infinately bigger than V. like rest mass of proton compared to earth.

now we talk about equation. If R is infinite it means, by definition, that no current pass through it. a circuit which is not closed has infinite resistance. if that's the case then P=I²R so P---> 0*∞ which is not defined.

we say that no power is dissipated in volt meter because almost no current pass through it/has inifnite resistance. actually it has much more resistance then circuit but not truly infinite. like 3 Kilo Ohm voltmeter for circuit of combined resistance of 5 ohm.

Look, these equations are not used to define any of these quantities. like V=IR or P=w/t.
Basic concept:

-no power will be dissipated by the resistor if it's resistance is zero.
reason:

V=energy/charge. If R is zero then V is zero it means that it uses no energy for each columb of charge that passes through it. If it uses no energy it means it's power is zero.​

Why your argument is wrong(but equation is not wrong):

p=V²/R
R is zero power is infinity? no if R is zero then V is also zero. But zero / zero is not defined. so this equation is not defined for V or R zero or infinity. the actual equations says "power is directly proportional to square of V and inversely proportional to resistance for resistance and voltage are not equal to zero" there is a restriction to every equation. like sine of an angle can't be bigger than 1. this is restriction of sine inverse function.​

 

[DeViL gURl B)]

http://papers.xtremepapers.com/CIE/...nd AS Level/Physics (9702)/9702_w10_qp_22.pdf

NEED HELP IN THE FOLLOWING QUESTIONS... PLEASE HELP!
Question 1 (b) ---> though I've done this question hundred times .. But syllable confuses me like anything! -_-
Question 4(b) (ii)
Question 5 (c)
And ... Question 6 starting from (a) (ii) I don't get this question .. Please explain !
Help would be well appreciated .. Thank u

 

[zem]

Help neede in second part of b how do i find the area? Thankyou in advance

 

[Snowysangel]

Thanks but that doesn't answer my question...it actually makes it more confusing. In the potential divider question, the wire had some resistance too right? And to find the minimum potebtial difference between the two points, we took resistance of the potebtial divider as maximum and to find the maximum potential difference, we took the resistance to be zero....why? It was 5 c(II) in this paper http://papers.xtremepapers.com/CIE/...nd AS Level/Physics (9702)/9702_w11_qp_21.pdf

 

[Snowysangel]

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

NEED HELP IN THE FOLLOWING QUESTIONS... PLEASE HELP!
Question 1 (b) ---> though I've done this question hundred times .. But syllable confuses me like anything! -_-
Question 4(b) (ii)
Question 5 (c)
And ... Question 6 starting from (a) (ii) I don't get this question .. Please explain !
Help would be well appreciated .. Thank u

Ib...25sin35=T
4b...count the squares and multiply it by the area on 1 small square
5c....it is losing power because the amplitude decreases. Intensity =(amplitude)^2 so you can determine the ratio from the graph

 

[DeViL gURl B)]

T

Ib...25sin35=T
4b...count the squares and multiply it by the area on 1 small square
5c....it is losing power because the amplitude decreases. Intensity =(amplitude)^2 so you can determine the ratio from the graph

Thank u !

 

[DeViL gURl B)]

http://papers.xtremepapers.com/CIE/...nd AS Level/Physics (9702)/9702_w10_qp_22.pdf

Question 6 anyone? Please

 

[zem]

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

Question 6 anyone? Please

Which part?

 

[DeViL gURl B)]

Which part?

B (i)

 

[Princess Raven]

How do we calculate resultant intensity
After subtracting 2 amplitudes then what?

 

[Thought blocker]

B (i)

How u did a part ?

 

[DeViL gURl B)]

How u did a part ?

Well ... That was easy .. I can't do the second one :$

 

[Thought blocker]

Well ... That was easy .. I can't do the second one :$

meko ni ata a part, sikhado...

 

[MiniSacBall]

http://papers.xtremepapers.com/CIE/...and AS Level/Physics (9702)/9702_w04_qp_2.pdf

5) b) i) How is it possible to just multiply those values, i mean one is in m other is in mm. How could you simple do that. Anyways i don't know how do that question.
All i know is that at 1K it decreases by 0.012 mm, then the it should decrease by 350 * 0.012 mm = 4.2 mm. I don't know the rest of the part.
How multiplying mm by m will give you the change in length.
Explain please

 

[Suchal Riaz]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w04_qp_2.pdf

5) b) i) How is it possible to just multiply those values, i mean one is in m other is in mm. How could you simple do that. Anyways i don't know how do that question.
All i know is that at 1K it decreases by 0.012 mm, then the it should decrease by 350 * 0.012 mm = 4.2 mm. I don't know the rest of the part.
How multiplying mm by m will give you the change in length.
Explain please

where we multiply meter by mm?
we say 0.012 mm decrease per kelvin. so 0.012 mm/K * 350 K =4.2mm

 

[MiniSacBall]

where we multiply meter by mm?
we say 0.012 mm decrease per kelvin. so 0.012 mm/K * 350 K =4.2mm

Well, after this step why do we multiply by 0.62 m .

 

[robinhoodmustafa]

How do we calculate resultant intensity
After subtracting 2 amplitudes then what?

I even occurred such question.
btw I found this
http://www.physicsforums.com/showthread.php?t=408123

 

[robinhoodmustafa]

Help neede in second part of b how do i find the area? Thankyou in advance

You can calculate the area by Box Counting method.
Find the area of an individual box between the two curve. Highlight small size boxes and multiply the total with the area of 1 box.

or another method is to assume the curve to be straight and apply the triangle area formula but I am not sure in this whether they have given a range in MS.
edit:
I found a useful site which shows a similar problem and he has assumed the curve to triangle.
http://tap.iop.org/mechanics/materials/229/page_46534.html
scroll down . and read the heading "Energy stored in stretched material"
because it said as long the object obeys hookes law than it is considered as a triangle. because u see the unloading part on question shows that it returns back to its orignal length

 

[_Ahmad]

http://papers.xtremepapers.com/CIE/...nd AS Level/Physics (9702)/9702_w13_qp_21.pdf

Can any please explain Q5 b