Physics: Post your doubts here!

[midha.ch]

For the first question, work done always equals some of all the energy changes of the system. There was a decrease in GP and increase in EP. Thus WD equals their sum.

For the second question, both points are at the same distance from nodes. Considering that this is a stationary wave, the particles will only move up and down in the same place. I don't think there is any question of different frequencies as none of the two points are nodes.As the points are at similar distances from the nodes, there amplitude should be the same.

About the second part, I solved it wrong myself . I think it is 180 degrees because its a standing wave and not a progressive one. X is at its highest possible point, and Y at its lowest. Therefore there is a phase difference of 180.

can you explain the answer for 5a more clearly?? I didn't understand it

 

[usama321]

can you explain the answer for 5a more clearly?? I didn't understand it

http://www.physicsclassroom.com/Class/waves/h4.gif
See this maybe you will understand it better.

Both points will move the same distance up and down, as they are the same distance from nodes.

 

[Hassan Ali Abid]

http://papers.xtremepapers.com/CIE/...nd AS Level/Physics (9702)/9702_w11_qp_41.pdf Q#8 (C)
http://papers.xtremepapers.com/CIE/...nd AS Level/Physics (9702)/9702_w11_ms_41.pdf

http://papers.xtremepapers.com/CIE/...and AS Level/Physics (9702)/9702_w07_qp_4.pdf Q#7 (C)
http://papers.xtremepapers.com/CIE/...and AS Level/Physics (9702)/9702_w07_ms_4.pdf

http://papers.xtremepapers.com/CIE/...and AS Level/Physics (9702)/9702_s05_qp_4.pdf Q#7 (C)
http://papers.xtremepapers.com/CIE/...and AS Level/Physics (9702)/9702_s05_ms_4.pdf

http://papers.xtremepapers.com/CIE/...and AS Level/Physics (9702)/9702_w04_qp_4.pdf Q#6 (C)
http://papers.xtremepapers.com/CIE/...and AS Level/Physics (9702)/9702_w04_ms_4.pdf

I need help in these parts. Can anyone plz explain me in detail how to solve these parts ?

 

[midha.ch]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w11_qp_41.pdf Q#8 (C)
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w11_ms_41.pdf

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w07_qp_4.pdf Q#7 (C)
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w07_ms_4.pdf

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s05_qp_4.pdf Q#7 (C)
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s05_ms_4.pdf

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w04_qp_4.pdf Q#6 (C)
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w04_ms_4.pdf

I need help in these parts. Can anyone plz explain me in detail how to solve these parts ?

for the 1st question, lets say initially we had N1 number of Phosphorus nuclei
after 30 days, due to radioactive decay the number of P nuclei has decreased to N2
it's given in question that P nuclei forms Sulfur when it undergoes radioactive decay
that means the decrase in the number of P nuclei after 30 days is actually the number of S nuclei formed in 30 days
the difference in number of P nuclei after 30 days = N1-N2 = number of S nuclei formed after 30 days
thus we can say the ratio is {N1/(N1-N2)}

2nd question
Uranium splits to produce equal number of both Ba and Kr nuclie
so initially Number of Kr nuclei = Number of Ba nuclei = N1
Let after time t, N of Ba/N of Kr =8

N of Ba = N1 x e^(-λ1 x t)

λ2 = ln2/T1/2 of Ba

λ1 = 6.41x10-4

N of Kr = N1 x e^(-λ2 x t)

λ2 = ln2/T1/2 of Kr

λ2 = 0.231

N of Ba/N of Kr = 8

N of Ba = 8xNKr

Now put the values on both sides and equate them

N1 gets cancelled as it’s there on both sides

Remains an equation of the form e^a = e^b

Take ln both sides

ln(e^y) = y

using this formula get the value of t by solving the equation

 

[midha.ch]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w11_qp_41.pdf Q#8 (C)
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w11_ms_41.pdf

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w07_qp_4.pdf Q#7 (C)
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w07_ms_4.pdf

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s05_qp_4.pdf Q#7 (C)
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s05_ms_4.pdf

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w04_qp_4.pdf Q#6 (C)
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w04_ms_4.pdf

I need help in these parts. Can anyone plz explain me in detail how to solve these parts ?

your 3rd question is similar to 1st question. Here instead of the ratio, you are asked to find the time taken to reach that ratio
If you have understood my answer to first question it's easy to find the result for third question. But if you still have confuson then let me know.

I'll skip numerical calculations for your 4th question
Consider Initial activity is A1
after time t, Activity reduces to (A1 x 1/10)
put the values in the equation of activity
you have already found decay constant in previous parts
now just put the values and solve it.

If you are still unsure or confused about any of the answers, then let me know I'll be glad to help

 

[papajohn]

Question 7 part b...Please help usama321 and others.
http://papers.xtremepapers.com/CIE/...and AS Level/Physics (9702)/9702_w05_qp_2.pdf

 

[usama321]

Question 7 part b...Please help usama321 and others.
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w05_qp_2.pdf

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.

 

[Snackbox86]

Hey can i have help with questions 17,20,35,36 from 2013s p11
Thank again

 

[usama321]

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.

 

[1357911]

http://papers.xtremepapers.com/CIE/...nd AS Level/Physics (9702)/9702_w09_qp_41.pdf
http://papers.xtremepapers.com/CIE/...nd AS Level/Physics (9702)/9702_w09_ms_41.pdf
Q-8 cii

http://papers.xtremepapers.com/CIE/...and AS Level/Physics (9702)/9702_s09_qp_4.pdf
http://papers.xtremepapers.com/CIE/...and AS Level/Physics (9702)/9702_s09_ms_4.pdf
Q-9 b.. and Q-11 biii

Can anyone help me with these.

 

[Autumngirl]

Can someone please explain q25 from the paper ON2006 P1 to me? I don't take math so please give a detailed explanation.

 

[midha.ch]

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

 

[Autumngirl]

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

 

[ahmed abdulla]

For phy-paper5

our drawing does it have to be in pen/ pencil ??

 

[midha.ch]

For phy-paper5

our drawing does it have to be in pen/ pencil ??

Pencil
It's mentioned in the instructions.
If otherwise it'll be mentioned in the question I guess.

 

[sitooon]

how to find percentage uncertainty ?
http://papers.xtremepapers.com/CIE/...nd AS Level/Physics (9702)/9702_s10_qp_53.pdf
2e(ii)

 

[saadkhan97]

Re: Physics Help here! Stuck somewhere?? Ask here!

http://papers.xtremepapers.com/CIE/...nd AS Level/Physics (9702)/9702_s12_qp_11.pdf

QUESTION NUMBER 8

 

[midha.ch]

how to find percentage uncertainty ?
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s10_qp_53.pdf
2e(ii)

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

 

[♣♠ Magnanimous ♣♠]

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........

 

[midha.ch]

Well, Idk anythin abt uncertainty
and ur handwriting is nice but not better than mine........

Naa you can't say that!! Show a sample of your handwriting first
Before that I won't agree with you!
And uncertainty is another pointlessly important topic in A'levels
I'll get you easy-to-remember notes asap