ALL CIE PHYSICS DOUBTS HERE !!!

[Sue T]

A2 phys - th subject content says in section vi chp 26 quantum physics
(l) recall and solve problems using the realtion hf = E₁ - E₂
ny idea wth ths s bout? - all i no is :
hf = phi + 1/2mv²
is it related 2 tht or wat?

[6394]

plz explain the difference btwn "work dne by a system"
 and "work done on a system" ??
thnq

[Deadly_king]

plz explain the difference btwn "work dne by a system"
 and "work done on a system" ??
thnq

For the explanation, i'll take the system to be a gas.

1. Consider a gas which expands at constant pressure.



This will cause the final volume(V_f) to be larger than the initial volume.(V_i)

Hence this will be work done by gas = F x L = P x A x L = P ( V_f - V_i)

2. Consider a gas being compressed at constant pressure.

Work done on gas = P (V_i - V_f)

Work done on gas is just the opposite of work done by gas.

EXPANSION.

Work done by gas = P ( V_f - V_i) is positive while Work done on gas = P (V_i - V_f) is negative.

COMPRESSION.

Work done by gas = P ( V_f - V_i) is negative while Work done on gas = P (V_i - V_f) is positive.

Hope it helps

[aloha32]

Hi, I have 3 questions regarding ideal gases..please help!

1.Two vessels one having 3 times the volume of the other are connected by a narrow tube of negligible volume. Initially the whole system is filled with a gas at a temperature of 290 K. The smaller vessel is cooled to 250 k and the larger vessel is heated to 400 K. Find the final pressure of the system.

2. The molar mass of nitrogen is 28g. A sample of gas contains 6.02X10 ^23 molecules Calculate the number of moles of the gas , the mass of the gas and the volume of the gas at .110 Pa and 290 K

3. An oxygen cyclinder contains .50 kg of gas at a pressure of .50 PA and 7 degree centigrade. What mass of oxygen must be pumped into the cylinder to raise its pressure to 3 mega pascals at 27 degrees. If the molar mass of oxygen is 32g calculate the volume of the cylinder

I often get confused b/w the whole number of moles and number of molecules thing :S

Thanks!



To Monopoly : Are you sure it isn't 4.5 and 3.5 because then it makes sense..because the amplitude would vary between the voltages

[$H00t!N& $t@r]

hi i have a question..
O/N 2009 PAPER 1 VAR 11 QUESTION 9, 10, 13, 22

here is the link: http://www.freeexampapers.com/get_past_papers.php?l=Past_Papers%2FA+Level%2FPhysics%2FCIE%2F2009+Nov%2F9702_w09_qp_11.pdf

[The Golden Girl =D]

can you upload it Please ?

[elemis]

hi i have a question..
O/N 2009 PAPER 1 VAR 11 QUESTION 9, 10, 13, 22

here is the link: http://www.freeexampapers.com/get_past_papers.php?l=Past_Papers%2FA+Level%2FPhysics%2FCIE%2F2009+Nov%2F9702_w09_qp_11.pdf

Question 9

Use the formula : U₁ - U₂ = V₂ - V₁

That is, for an elastic collision the relative velocities of approach are equal to the relative velocities of separation.

NOTE : You are considering Velocity which is a vector quantity and who's sign changes with direction.

Question 10

Using impulse = force * time it

60 * 0.5 = 30 Ns   

Hence, it is given that the retarding force caused a 30 Ns change in momentum.

That is : 30(-v+3) = 30Ns     where v = final velocity

Hence, v = 2

Question 22

Stress/strain = Young's Modulus

Stress = 20 / [pi*(2.5*10^-4)²]   <---- let this be equation 1

Divide 1 by 2*10¹¹ = 5.1*10^-4

However, this is equal to strain which is basically change in length divided by original length. It is ratio so we multiply by 100 to give :

5.1*10^-2

[Sue T]

A2 ques:
ppr 4 mj 07 ques4
The output from the transformer is to be full-wave rectified. Fig. 4.1 shows part of the
rectifier circuit.
(attached - we had 2 put in th diodes on our own)
ii) The resistance of the resistor R is doubled. On Fig. 4.2, sketch the variation with
time t of the potential difference V across the resistor.
fig 4.2 is a graph also attached
my question is (ii)
the ans was:

(ii) sketch: same peak values
ripple reduced and reasonable shape
reduced ripple as in theres less 'curviness' or like it goes down lower?
and if its less 'curviness' then why?

[Sue T]

and 1 more pls:

Q)A cobalt-60 source having a half-life of 5.27 years is calibrated and found to have an
activity of 3.50 × 105 Bq. The uncertainty in the calibration is ±2%.
Calculate the length of time, in days, after the calibration has been made, for the stated
activity of 3.50 × 105 Bq to have a maximum possible error of 10%.

ANS)source must decay by 8%
A = A0 exp(–ln2 t / T½) or A/ A0 = 1 / (2t/T)
0.92 = exp(–ln2 × t / 5.27) or 0.92 = 1 / (2t/5.27)
t = 0.634 years
= 230 days

why must th source decay by 8%? how did they get tht?
my mock xam s 2moro - i need urgent help 

[$H00t!N& $t@r]

You dont need to do what requiem said.

A simpler method is to consider the velocities :

U₁ - U₂ = V₂ - V₁

Where U = initial velocity before collision
          V = final velocity after collision.

For an elastic collision the left hand side of the above equation should be equal to the right hand side.

Plug in the necessary values and remember velocity is a VECTOR.... you should get the right answer

I'm not getting the right answer  can you solve it for me 

[$H00t!N& $t@r]

Question 9

Use the formula : U₁ - U₂ = V₂ - V₁

That is, for an elastic collision the relative velocities of approach are equal to the relative velocities of separation.

NOTE : You are considering Velocity which is a vector quantity and who's sign changes with direction.

Question 10

Using impulse = force * time it

60 * 0.5 = 30 Ns   

Hence, it is given that the retarding force caused a 30 Ns change in momentum.

That is : 30(-v+3) = 30Ns     where v = final velocity

Hence, v = 2

Question 22

Stress/strain = Young's Modulus

Stress = 20 / [pi*(2.5*10^-4)²]   <---- let this be equation 1

Divide 1 by 2*10¹¹ = 5.1*10^-4

However, this is equal to strain which is basically change in length divided by original length. It is ratio so we multiply by 100 to give :

5.1*10^-2

Thanks for the answer.    However, question 9 is not quite clear... 

[donhassan]

hello everyone i appreciate everyones great effort in helping each other on this platform
i have set of doubts in may june 2003 cie paper question number 5  ,15,18,20,22

please please please try ur best to explain them in a very lucid way

[elemis]

Why is B wrong ?

Question 30

[astarmathsandphysics]

It says across the cell, not across the resistor

[elemis]

Question 5 (a)

I said the path difference must be a odd number of non integer half wavelengths

and

the vector sum of the amplitudes must be zero.

The mark scheme says :

either same amplitude / intensity at M
or ratio of amplitudes is 1.28 / ratio of intensities is 1.28^2

Can someone please explain the statements in bold ? How can same amplitudes lead to an intensity of zero ?