Physics: Post your doubts here!

[Xylferion]

How do we calculate the time in part (d)(1) and (2)? Help please.

Well firstly. you need to look at all the data you already have.
From (b), you've got the initial acceleration which is 0.570 ms^-2.
You also have the constant velocity which is 2.00 ms^-1.
You have the deceleration too, which you calculated to be 0.306 ms^-2.

You already have the formula written down, but re-arrange it to make t the subject.
t = v-u/a

For 1) t = 2.00 - 0 ( it starts from rest ) / 0.570, this gives you t = 3.5 seconds.
For 2) t = 0-2.00 / (-)0.306 ( since its decelerating ), this gives you t = 6.5 seconds.

Hope that helped! ^_^

 

[MYLORD]

No problem ^_^. Can't help with 33 and 34 yet as I still have to go over my theory with those.

can u help me with something ?
r u taking A level or AS Level ?
if u r taking a level, then how r u preparing for paper ?

 

[manya]

can someone help me with this question

thankyou so much

 

[waleed302]

http://www.sheir.org/a-level-physics-43-june2010.pdf

Question 5, b2.

Can someone explain the method/concept, the mark scheme is useless.

Let me try:
C=Q/V, and they tell you for a capacity of 30 micro Farads the max voltage is 6V. So substitute these values: 30=Q/6. Q=180. So now we've got the charge on one of the capacitors. In parallel the combined capacitance is 20, and they're asking for the max voltage, so plug these values into the formula: C=Q/V... 20=180/V. So V=9. Hope that helps.

 

[Saad ur rehman]

Salamz.
I got doubts in Paper 1 of May june 2010 variant 1. Questions are 3, 9, 13, 16, 27, 33 and 39. I would be glad if someone helps

 

[Saad ur rehman]

physicist, thanks fo that blog i cleared a few doubts but i couldnt get the rest.

 

[Saad ur rehman]

btw, paper 11 and paper 12 seems to have the saem qu but the numbers are different. paper 12 is already explained there. try to compare the qu there. if its still not there, use the comment box at that blog

Bruh I did compare only 3 questions were not in the Paper 12. No worries i will find a solution.

 

[Medallion]

Is anyone having problems opening up the past papers. I was able to open the past papers for physics but now it just loads and never actually comes up and just says your server got disconnected...??? Why is this happening Plz help

 

[Saad ur rehman]

Is anyone having problems opening up the past papers. I was able to open the past papers for physics but now it just loads and never actually comes up and just says your server got disconnected...??? Why is this happening Plz help

Yes it is not working but there are other websites try www.examtestprep.com Hope this helps

 

[Mahnoorfatima]

http://maxpapers.com/syllabus-materials/physics-9702-a-level/attachment/9702_w08_qp_1/
Question 1,8,27,32 Please someone?

 

[Muskan Achhpilia]

20)
1*p/10 = rho * g * h
h = p/(10 * rho * g) = A

28) Taking O as theta and l as lambda
dsinO=nl
(1*10^(-6))sin35= 1l
l = 5.74*10^(-7) = 574nm = C

40)
YET NOT DONE IN OUR SYLLABUS. SORRY

Thank you very much!!

 

[The Sarcastic Retard]

http://maxpapers.com/syllabus-materials/physics-9702-a-level/attachment/9702_w08_qp_1/
Question 1,8,27,32 Please someone?

*NOTE "l" = lambda*
1) Examiner is asking for frequency indirecly, v = f*l
3*10^8/(600*10^9) = 5*10^14
8) Uniform increase in velocity means constant acceleration.
27) l/2 = 0.015m, l = 0.03m
v = fl
f = 3*10^8/0.03 = C
32)
The copper wires are in parrallel (||) so total resistance will be considerd for parallel. 1/R = 1/R1 + 1/R2 + ... + 1/Rn
6 copper wires. Hence 1/R = (1/10)*6 = 3/5 ohm.
Steel wire || to 6 copper wires = 1/100 + 3/5 = 61/100. R = 100/61 = 1.6 ohms.

 

[princess Anu]

help!

 

[princess Anu]

..

 

[princess Anu]

the ans to this should be B, no? :/

 

[The Sarcastic Retard]

..

Pi = mu
For elastic collision, Pi = Pf, No sticky collision as its perfectly inelastic and not in same direction after collision as it will be elastic collision (KEloss)
A: pf = mu
B: mu/3
C and D are not elastic collision. Ignore them. Answer is A

 

[princess Anu]

Pi = mu
For elastic collision, Pi = Pf, No sticky collision as its perfectly inelastic and not in same direction after collision as it will be elastic collision (KEloss)
A: pf = mu
B: mu/3
C and D are not elastic collision. Ignore them. Answer is A

how did you calculate? when do we have to subtract the speeds and when add?

 

[manutd96]

http://maxpapers.com/syllabus-materials/physics-9702-a-level/attachment/9702_s14_qp_42/

b)
A telescope gives a clear view of a distant object when the angular displacement between the
edges of the object is at least 9.7 × 10^−6 rad.
(i)
The Moon is approximately 3.8 × 10^5
km from Earth.
Estimate the minimum diameter of a circular crater on the Moon’s surface that can be
seen using the telescope.

someone explain this to me please! thanks! i dont get it why is the diameter r times theta here as i thought that is for arc length..?

 

[Mahnoorfatima]

how did you calculate? when do we have to subtract the speeds and when add?

In a perfectly elastic collisons, we always add the speeds if objects are travelling in OPPOSITE directions and subtract when they are in SAME direction. It's the opposite of the usual rule plus KE has to be conserved in elastic collisons. And for this question :
Intial momemtum= 2mu+(-mu) = mu
Final momentum= 2mu/3+ (-5mu/3)
thus Pi=Pf = mu=mu It's just the signs that confuse people. Hope you got it. =)

 

[Xylferion]

the ans to this should be B, no? :/

When you stretch a spring, we do work. If we then release the spring, some of the energy is recovered. Some is lost. When they say " net " work done, they're talking about the total work done.

The arrows going up signify that the spring is being loaded, and thus we call it the loading phase. The arrow that's going down is when the spring is released, it's now compressing. This is called the unloading phase. The area which is X and Y, is commonly known as " Hysteresis ".

Hysteresis is basically the energy lost during the stretching and releasing. This leaves us with Z, which is said to be the net work done.

My advice to you would be to think that whatever is outside that area of hysteresis, is the net work done. Hope that helps ^_^