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Physics: Post your doubts here!

immie.rose

immie.rose

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omaaaar said:
Hey, for part b)

This is the key for this question: For minima to occur: Path difference= 0.5Lambda, 1.5 Lambda. 2.5 Lambda and so on...

If you look at the figure, you'll see that it's a right angled triangle, use paythagors theorem. You'll get S2M= 128 cm

Path diff= 128 - 100= 28 cm


Your range is from 1khz to 4 kHz

Use v= lambda x frequency
Speed = 330 m/s

Range of lambda= 8.25 cm to 33cm

Now, start trying, remember path diff= nlambda

So 28= 0.5 x Lambda, 56 not in allowed range
28= 1.5 x lambda, 18.6 allowed range
28= 2.5 x lambda , 11.2 allowed range

So 2 minima occur
Click to expand...

Thanks a bunchhhhhh!! JazakAllah kher.
 
S

sagar65265

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_Ahmad said:
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s09_qp_2.pdf
Can you please help me with second varient Q7 b ii)
Click to expand...

You need to apply Kirchoff's Second Law for this question; if you put your finger at N, and move to M, then L, then K, and finally back to N, and sum up all the changes in potential energy of the current flowing through the wires then you should get zero (if you start at N, and return to N, then the potential has to be the same at the beginning and at the end, so the sum of changes has to be zero).

So, when you go from N to M, you can see that there is no resistance or impedance in the way (since the wires are assumed to have zero resistance). Therefore, no potential energy is lost or gained while going from N to M, so for this part we can write that the change in potential is = 0.

When you go then from M to L, the trick here is to notice that you are going from the positive terminal of the battery to the negative terminal of the battery. Since the positive terminal is considered to be at a higher potential than the negative terminal, you are going from a region of high potential to a region of low potential.
Therefore, you can write that the change in potential is negative.
But what is the value? It is simply the EMF of the battery. In other words, the potential decreases by the EMF of the battery E1, i.e. the change in potential is = -E1.

When you go from L to K, the same argument as N to M is made; the change in potential is = 0.

While going from L to N back again, you go from the negative terminal to the positive terminal, i.e. you go from a region of low potential to a region of higher potential. Therefore, the potential change is positive, and the magnitude is equal to the EMF of battery E2. In other words, the change in potential while crossing the battery E2 = +E2 Volts.

The last part of the journey from L to N involves going past the resistor; since you are going with the current, you are automatically going from a region of high potential to a region of lower potential, i.e. the potential difference decreases across the resistor (since current goes from a region of high potential to a region of low potential). The potential difference across the resistor, therefore, will be negative. The magnitude of this difference will be equal to I3 * r = RI3. Therefore, the change in potential is equal to -RI3.

Adding these up, we get -E1 + E2 - I3R = 0
Rearranging, we get E2 - E1 = I3R

Which is the answer.

Hope this helped!

Good Luck for all your exams!
 
A

_Ahmad

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sagar65265 said:
You need to apply Kirchoff's Second Law for this question; if you put your finger at N, and move to M, then L, then K, and finally back to N, and sum up all the changes in potential energy of the current flowing through the wires then you should get zero (if you start at N, and return to N, then the potential has to be the same at the beginning and at the end, so the sum of changes has to be zero).

So, when you go from N to M, you can see that there is no resistance or impedance in the way (since the wires are assumed to have zero resistance). Therefore, no potential energy is lost or gained while going from N to M, so for this part we can write that the change in potential is = 0.

When you go then from M to L, the trick here is to notice that you are going from the positive terminal of the battery to the negative terminal of the battery. Since the positive terminal is considered to be at a higher potential than the negative terminal, you are going from a region of high potential to a region of low potential.
Therefore, you can write that the change in potential is negative.
But what is the value? It is simply the EMF of the battery. In other words, the potential decreases by the EMF of the battery E1, i.e. the change in potential is = -E1.

When you go from L to K, the same argument as N to M is made; the change in potential is = 0.

While going from L to N back again, you go from the negative terminal to the positive terminal, i.e. you go from a region of low potential to a region of higher potential. Therefore, the potential change is positive, and the magnitude is equal to the EMF of battery E2. In other words, the change in potential while crossing the battery E2 = +E2 Volts.

The last part of the journey from L to N involves going past the resistor; since you are going with the current, you are automatically going from a region of high potential to a region of lower potential, i.e. the potential difference decreases across the resistor (since current goes from a region of high potential to a region of low potential). The potential difference across the resistor, therefore, will be negative. The magnitude of this difference will be equal to I3 * r = RI3. Therefore, the change in potential is equal to -RI3.

Adding these up, we get -E1 + E2 - I3R = 0
Rearranging, we get E2 - E1 = I3R

Which is the answer.

Hope this helped!

Good Luck for all your exams!
Click to expand...

What if we go from N to K to L to M and then N

PLEASE HELP!!!!
 
MiniSacBall

MiniSacBall

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waleed302 said:
But isn't the resistance of C decreasing with voltage? Why would the PD through it be greater than in R? Also, isn't the current through R constant? Why would there be a 'curve' for R?
Click to expand...

Dude just jumping onto parts of questions will confuse you do the whole question,
Well first do b(i)
On Fig. 6.1, draw a line to show the variation with potential difference V of the
current I in resistor R.

Once you have drawn this curve, you will find that, for example take the I as 2 A. now take the P.D (Volts) of C and R from the graph! Example: At 2A, P.D for C is 28 while for R is 20.
So now recall the formula of Power = IV, Currently, we are only looking at I and V we don't need resistance, if you want to count resistance then count for each resistor its resistance at specific current and volts.
As P = IV ,
For C => 2 x 28 = 56
For R => 2 x 20 = 40
As we know Power is the rate of energy lost (dissipated)
Thus, C has higher rate of Energy dissipation, than R.

If you want to take resistance into account then, take a specific current, take readings of volts for both resistors, then calculate their resistance from V= IR.
Then, calculate the power using any power formula, there you go. You will find same results. Instead of guessing answers on the internet, just print and do it.
Hope that solves your problem! :)
 
MiniSacBall

MiniSacBall

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MiniSacBall said:
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s10_ms_23.pdf

Q1 a ii)
How to do it i mean apart from adding 0.02, why do we add 0.01, is it because of the +-1 digit.
Click to expand...

Can Any one help me in this one. :unsure:

And,
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s10_qp_23.pdf
Q3 a)
How to calculate the area, by counting the squares right, i have got like 3600+ and the answer is only 39m
The heck!!
 
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MissBellum

MissBellum

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Can someone explain how to solve Oscillation questions and some Alternating current questions that use radians. Im not sure how to work it on my calculator, i always get confused on these questions :confused: My working is always 100% correct but the answer is never right :oops: Could someone show me a few examples so i dont make this mistake in my exam tomorrow :sick:
 
W

waleed302

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MiniSacBall said:
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s10_ms_23.pdf

Q1 a ii)
How to do it i mean apart from adding 0.02, why do we add 0.01, is it because of the +-1 digit.
Click to expand...
Yes, after you add 0.02, you have to take into consideration the other information they give you which is +-1 digit. Since it's asking for the maximum, you'd obviously use +1 digit. I know it's kind of confusing but I see no other explanation for this!
 
B

boxfire1995

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sagar65265 said:
I'm not able to understand your question; hopefully this guess might help, if not i'm sorry, could you explain again?

Uranium-235, according to the given equation, can decay into Barium-141 and Krypton-92; each nucleus in the sample used will undergo this reaction, and since the equation is balanced for conservation of charge and mass, each nucleus of Uranium, when hit by a neutron, will produce 1 nucleus of Barium, another of Krypton and 3 neutrons, along with some energy.

So, when 1.2 grams of Uranium-235 decay, we need to find out how many nuclei of Barium-141 are produced. Since we have a one-one relation (i.e. one nucleus of
U-235 produces one nucleus of Ba-141) we can say that the number of Barium nuclei produced is equal to the number of Uranium nuclei that take part in the nuclear reaction. In this case, all of them react, so the number of atoms (and correspondingly, the number of nuclei) of U-235 is given by

Avogadro's Number * No. of moles

The number of moles of nuclei in the Uranium sample is given by

(Mass of Sample)/(Mass of 1 mole of element)

Substituting the values, we get

6.02 * 10^23 * 1.2/235 = 3.07 * 10^21 nuclei.

Hope this helped!

Good Luck for all your exams!
Click to expand...
Thank you! I was over looking the molar ratio part. JazakAllah! May Allah help you like you helped me. :)
 
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