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Physics: Post your doubts here!

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For b(i), they have already stated that the wire is of uniform resistance. Thus, as the distance between two points on the wire increases, according to pL/A, the resistance and as a result, the P.D across those two points will also increase.

For the second part, you have to understand how the voltmeter actually works. It shows how much of a voltage drop there is between two points. If there is no voltage difference between two points, then it would give us a reading of zero.

Now we already know that the voltage at point B is 2.7, and 1.8 has already been dessipated across CB. The voltmeter would only give us a reading of zero if the voltage at point M is also 2.7, resulting in no potential "difference" between the two points. As a result, there would be a pd of 2.7 across M and Q

This part is quite easy, as we already know that voltage is proportional to length. just use ratios here.

Now the last part. Warming the thermistor would decrease its resistance, resulting in a less PD across it as the voltage remains constant. Thus there would be a higher PD between CB, and there would be less voltage at point B. Thus we will have to increase the distance between PM or decrease it between MQ, so that the PD on both sides of the voltmeter is same.
 
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Hey can i have help with questions 17,20,35,36 from 2013s p11
Thank again
 

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Hey can i have help with questions 17,20,35,36 from 2013s p11
Thank again
17: The trick is to remember that while taking GP, we take the height of the object from the bottom of the object. Thus at the start the energy of the ball would be
m * 9.81 * .72 = .75

m = .75/9.81*.72

Now, when the ball is at height of 37 cm all of its kinetic energy would have conerted to GP. Therefore
m * 9.81 * .37 = kinetic energy
kinetic energy = 0.385 = B

20: Answer is B. The particles do require some heat energy to turn to vapors, however evaporation can occur at any temperature. As the particles with the highest energy leave the surface, there is less energy left in the liquid and its temperature lowers.

35: For this question, you have to understand that voltmeter measures the difference in voltage between two points. In the first wire, voltage of 1.5V has already been spent at the first resistor(same resistors so half voltage spent there). So at the upper terminal of the voltmeter, we now know that the voltage left is 3 - 1.5 = 1.5 V

In the lower wire, the total resistance of all the resistors here is 2.25 kohm. Now, at the point till the terminal of the voltmeter, there is a total resistance of 1.5 Kohm. Thus the pd across these three resistors would 1.5/2.25 * 3 = 2V. The voltage left would be 3-2 = 1 V.
Now the difference between the two points is 1.5 - 1 = .5 V

36: I am not sure about it, so i checked the ms. I think it is D because using the voltage formula, we can check that the pd across the voltmeter is 1.99 V. Thus as it would measure the voltage difference across its two terminals, it would give us a reading of 2V.
 
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Can someone please explain q25 from the paper ON2006 P1 to me? I don't take math so please give a detailed explanation.
 
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Can someone please explain q25 from the paper ON2006 P1 to me? I don't take math so please give a detailed explanation.
well there is a formula to find the answer and I never really got the hang of the formula so I'll explain you another simple way (The one I follow :))
The graph is similar to a sine graph! So if you consider it sine graph, then X is at 90 degrees or π/2 rad
And this shows Y is at 630 degrees or 7π/2 rad
Difference of position between these two points is 3π
Thus answer is C
 
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Thank
well there is a formula to find the answer and I never really got the hang of the formula so I'll explain you another simple way (The one I follow :))
The graph is similar to a sine graph! So if you consider it sine graph, then X is at 90 degrees or π/2 rad
And this shows Y is at 630 degrees or 7π/2 rad
Difference of position between these two points is 3π
Thus answer is C
Thank you, your simple and concise explanation was exactly what I needed :)
 
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if you write the intial equation in terms of l, you'll get
l = (T^2 x g)/4pi
Skip the constants when finding uncertainty
Using the method in pic, we can say
(del(l))/l = (2del(l))/l + (delg)/g

find del(l)
then (del(l))/l x 100 = % uncertainty
Well, Idk anythin abt uncertainty :(
and ur handwriting is nice :) but not better than mine........ :p
 
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Well, Idk anythin abt uncertainty :(
and ur handwriting is nice :) but not better than mine........ :p
Naa you can't say that!! Show a sample of your handwriting first :p
Before that I won't agree with you! :cool::p:D
And uncertainty is another pointlessly important topic in A'levels :mad:
I'll get you easy-to-remember notes asap :)
 
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if you write the intial equation in terms of l, you'll get
l = (T^2 x g)/4pi
Skip the constants when finding uncertainty
Using the method in pic, we can say
(del(l))/l = (2del(l))/l + (delg)/g

find del(l)
then (del(l))/l x 100 = % uncertainty
For the second rule , when its a^m/b^n >> you meant to write MINUS right :LOL:
 
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For the second rule , when its a^m/b^n >> you meant to write MINUS right :LOL:
it doesn't matter because originally our del(a) and del(b) both has a ± sign
considering that, evn if you write a minus sign, the ± cancels it out.
So, overall all we do is, take the sum and when writing the final answer simply put a ± before the uncertainty value :p
 
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