Physics: Post your doubts here!

[Saad Mughal]

Ok so A simple question..
Why would Power be zero when resistance would be O or also infinity?
Why would power be zero in both the cases? :/

There are two formulas for power that involve resistance,
P = I^2*R (for series circuits, since current is constant). If resistance is 0 here, the power output is 0.
P = V^2/R (for parallel circuits, since voltage is constant). If resistance is infinite here, the power output is 0.

 

[Snackbox86]

hey, Can I have an an explanation to question 7 all P23 2010
The answer in the MS said (ii) F=E.g
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w10_qp_23.pdf
here is the ms
http://papers.xtremepapers.com/CIE/...nd AS Level/Physics (9702)/9702_w10_ms_23.pdf

thank you

 

[sagar65265]

hey, Can I have an an explanation to question 7 all P23 2010
The answer in the MS said (ii) F=E.g
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w10_qp_23.pdf
here is the ms
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w10_ms_23.pdf

thank you

7a(ii):
1.
(I'm pretty sure the E.g thing is a typing error, since there's no "g" variable in the question - the answer should be E.q or qE).
Since the moving particle is positively charged and the upper plate is negatively charged (not to forget, the lower plate is positively charged), the moving particle will be attracted upwards (by the negative plate) and repelled in the same direction (by the lower plate).
This force does not depend on the position of the particle or it's velocity, and it is constant all throughout.
The magnitude of the force exerted on a particle with charge q when the strength of the electric field at it's location is E, is given by:

F = q * E

So, since the electric field strength is constant and directed upwards (between the plates) all along the length of the plates, this is the force exerted by the plates on the particle. Therefore, F = qE.

2.
Since the force exerted on the particle is always in the vertical plane, the horizontal component of the particle's velocity will remain constant throughout it's journey, because no force component (neither the force of gravity or the electric force have a horizontal component) will influence it's velocity.
So, since the horizontal displacement of the particle as it travels from one side of the plates to the other is equal to L and this is related to the velocity by

v = s / t = L / t

we can write

t = L / v.

7b(i)

The law of conservation of momentum says that "the momentum - or quantity of motion - of a particle or the center of mass of a system of particles will not change unless a net external force acts on that particle or system of particles".
In other words, "the momentum of a closed system will not change unless a net external force acts on that system of particles".
It can be extended to the situation where two or more objects, taken as components of a complete system, interact by collision or action-at-a-distance forces; if no force exerted by a body from outside the system exists, then the momentum of the system (of all the components of the system) summed up can never change.

So, if no external forces act on a system, momentum before interaction = momentum after interaction.

7b(ii)

This question requires the use of the Impulse theorem, which can be written in symbols as

Δ(mv) = F(net) * t

And so, we can write

Δ(mv) = qE * L/v = qLE/v

7b(iii)

The charged particle is not an isolated system; since a net force exerted by a body other than itself acts on the particle, it cannot be considered a close system and as such it's momentum will change, so the conservation of momentum doesn't apply here.
However, if you take all the objects involved to be part of the system (i.e. if you say that the particle as well as the plates form the system) all the forces involved in the diagram are between the objects of the system, and no external force acts on it. Therefore, it's momentum is conserved.

(I'll try the first part after some time, the path thing).

Hope this helped!
Good Luck for all your exams!

 

[Faithix MFSH]

Guys can someone help me here?
This question is from W13 physics 42.
Regarding the ADC 4bit question, how do I continue the table for the next five samples? How do I convert the voltage values into binary digits or bits?

 

[sma786]

Guys can someone help me here?
This question is from W13 physics 42.
Regarding the ADC 4bit question, how do I continue the table for the next five samples? How do I convert the voltage values into binary digits or bits?

How did you attach the picture?

 

[EddieX]

Guys can someone help me here?
This question is from W13 physics 42.
Regarding the ADC 4bit question, how do I continue the table for the next five samples? How do I convert the voltage values into binary digits or bits?

you basically just divide continuously by 2 ...i.e. 0.25 ms is 10mv...10/2 give you 5,0...then 5/2 give you 2,1...and 2/2..gives you 1,0...so it would be 1010
ill take a picture later on i dont think its clear in this explanation

 

[EddieX]

you basically just divide continuously by 2 ...i.e. 0.25 ms is 10mv...10/2 give you 5,0...then 5/2 give you 2,1...and 2/2..gives you 1,0...so it would be 1010
ill take a picture later on i dont think its clear in this explanation

 

[sma786]

There are two formulas for power that involve resistance,
P = I^2*R (for series circuits, since current is constant). If resistance is 0 here, the power output is 0.
P = V^2/R (for parallel circuits, since voltage is constant). If resistance is infinite here, the power output is 0.

Thnku!

 

[A star]

hows the prep for p4 comming along A2s:/ nuclear and quantum physics samjh araha hae ksi ko to maera paper kar ae plsss :3

 

[Thought blocker]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s12_qp_21.pdf
q3cfull.. q5bii,iii,iv please ZaqZainab

 

[Saad Mughal]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s12_qp_21.pdf
q3cfull.. q5bii,iii,iv please ZaqZainab

Question 3, part (c):
(i) The extension is proportional to the load (straight linear line on graph) (mass here, hence load), therefore yes, the spring obeys Hookes law.

(ii) F = kx.
Use any point on the line, for example (0.20, 27.5).
Since mass is on the x-axis and length on the y-axis, you need to change these to fit into the formula.
F = (0.20)(9.8) = 1.96 N.
x = extended length - original length = 27.5 - 20 = 7.5 * 10^-2 m.
Put these in the formula and you get,
k = 1.96/(7.5 * 10^-2) = 26.1 ~ 26 N/m.

(iii) Strain Energy = 1/2 * F * x
Force = (0.40)(9.8) = 3.92 N.
x = 35 - 20 = 15 * 10^-2 m.
Strain Energy = 0.5 * 3.92 * 15*10^-2 = 0.294 J.

 

[Thought blocker]

Question 3, part (c):
(i) The extension is proportional to the load (straight linear line on graph) (mass here, hence load), therefore yes, the spring obeys Hookes law.

(ii) F = kx.
Use any point on the line, for example (0.20, 27.5).
Since mass is on the x-axis and length on the y-axis, you need to change these to fit into the formula.
F = (0.20)(9.8) = 1.96 N.
x = extended length - original length = 27.5 - 20 = 7.5 * 10^-2 m.
Put these in the formula and you get,
k = 1.96/(7.5 * 10^-2) = 26.1 ~ 26 N/m.

(iii) Strain Energy = 1/2 * F * x
Force = (0.40)(9.8) = 3.92 N.
x = 35 - 20 = 15 * 10^-2 m.
Strain Energy = 0.5 * 3.92 * 15*10^-2 = 0.294 J.

Wb q5 ?

 

[Saad Mughal]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s12_qp_21.pdf
q3cfull.. q5bii,iii,iv please ZaqZainab

Question 5:
(ii) The resistance of the wire to the point J is (0.9)*(4) = 3.6 ohms.
Potential difference = IR = (0.286)(3.6) = 1.03 V.

(iii) The value of E has to be the same as the potential difference across the wire XJ. To ensure that current is 0 (reverse polarities). Hence, 1.03 V.

(iv) Since there is no current passing through cell B, hence the internal resistance can be ignored (since potential difference will be 0 because 0 current).

 

[Thought blocker]

Question 5:
(ii) The resistance of the wire to the point J is (0.9)*(4) = 3.6 ohms.
Potential difference = IR = (0.286)(3.6) = 1.03 V.

(iii) The value of E has to be the same as the potential difference across the wire XJ. To ensure that current is 0 (reverse polarities). Hence, 1.03 V.

(iv) Since there is no current passing through cell B, hence the internal resistance can be ignored (since potential difference will be 0 because 0 current).

ty

 

[Browny]

Can anyone please give me some notes on the potentiometer and it various functions as measuring a EMF of a battery, comparing EMFs of 2 batteries?

 

[Saad Mughal]

Can anyone please give me some notes on the potentiometer and it various functions as measuring a EMF of a battery, comparing EMFs of 2 batteries?

David Sang, Physics Coursebook.
Page 189. Probably the best concise details (according to the syllabus) you'll find on the topic.

 

[AnujaK]

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[saqib.tariq]

Hi everyone
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[Thought blocker]

Question 5 bii) iii) iv)

 

[AnujaK]

Thank you very much I appreciate your feedback

gr8 work