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Physics: Post your doubts here!

Thought blocker

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Ahmed H. Al-Neel said:
broo wrong question i said 21 :p xD
Click to expand...
Okay so this problem is like parallel and series circuit. just imagine extension as resistors, and if you want overall k then find inverse of overall x. or otherwise the otherway if parallel then k would all add up for overall k. You do it yourself now and see if you got correct
 
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Asad Moosvi said:
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s12_qp_12.pdf

Question 38.

May someone explain how the resistances add up in the second circuit?
Click to expand...
sg-png.44814
 
Ahmed H. Al-Neel

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Thought blocker said:
Okay so this problem is like parallel and series circuit. just imagine extension as resistors, and if you want overall k then find inverse of overall x. or otherwise the otherway if parallel then k would all add up for overall k. You do it yourself now and see if you got correct
Click to expand...
thank youuu sooo much again! i got the answer :D
 
..sacrifice4Revenge..

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waves.jpg
displacement-distance graph
transverse wave.
which of these is correct:

A- the speed at P is max
B- displacement at Q is always zero
C- Energy at R is entirely kinetic
D- accelaration at S is max.

correct ans is D.
why isnt it B or A? why are they incorrect? and whats the logic behind D being correct
 
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mahmuda akter

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Thought blocker said:
10)
Consider barrel :
Use F = ma
That is mg - T = ma (T = tension): 120g - T = 120a
Make T the subject of formula so, later we can equate it in other equation or you can solve it by simultaneous equation method also.
But I personally go for Equate so, T = 120g - 120a

Consider stake :
Again use F = ma
so T = 80g + 80a

Equate it to get acceleration :
120g-120a=80g+80a
a=0.5g

Now we use V² - U² = 2as
so V² = 2 x 0.5 (10) x 9
V = 6m/s
Click to expand...
thank u very much
could u plz explain q-13 of m/j-12/12..
same force thing. i just dnt get them..:/
 
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sagar65265

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mahmuda akter said:
thank u very much
could u plz explain q-13 of m/j-12/12..
same force thing. i just dnt get them..:/
Click to expand...

The box accelerates to the right and the 2.0 kg mass accelerates downwards, since they are both connected by a string (and the string cannot stretch at will).
Suppose the mass moves down a little bit, the box will also move that same distance, that little bit, to the right.
So the acceleration of both the box and the mass is the same, but in different directions.
Alright, now let's write the equations of motion.

On the weight of mass 2.0 kg, we'll take the positive direction to be downwards and the negative direction to be upwards. Then, the following forces act on it:

i) Gravity - the weight of the 2.0 kg mass, mg = 2.0 * 9.81 = 19.62 Newtons, acts straight down. Therefore, it is (+19.62) Newtons.
ii)Tension - the force exerted by the string that tries to prevent downward motion is the tension force, and this pulls the 2.0 kg mass upwards. Let's write it as (-T) Newtons, since we do not know the magnitude but we know the direction.

The acceleration of this is "a" and the mass is 2.0 kilograms. So, using F(net) = ma, we can write

(+19.62 Newtons) + (-T Newtons) = 2.0 kilograms * a meters/second².
19.62 - T = 2a

This is our first equation. Now let's write the equations for the box. Supposing we take right direction to be positive and left direction to be negative, the forces are:
(only the forces in the horizontal direction - weight of the box and normal force on the box act vertically, so we don't need to write them now)

i) Tension from the rope, acting towards the right. This is the same tension that acts on the 2.0 kg mass, so it has the same magnitude as that force and we can write it as (+T) Newtons.
ii) Friction from the table, acting towards the left (friction acts towards the left since the velocity is towards the right, and friction always opposes the relative motion of the surfaces in contact, so it opposes the rightward motion of the box and acts to the left)- the magnitude of this force is given as 6.0 Newtons, and since it acts to the left (negative direction) we can write it as (-6.0) Newtons.

The acceleration of this box also has a magnitude "a" and the box has a mass of 8.0 kilograms, so we can write the equation of motion as

(+T Newtons) + (-6.0 Newtons) = 8.0a
T - 6 = 8a

This is our second equation. Now let's add this to the first equation to get

(19.62 - T) + (T - 6) = 2a + 8a
19.62 - T + T - 6 = 10a (Since T cancels out, we can write:)
19.62 - 6 = 10a
13.62 = 10a

a = 1.362 ms⁻² = 1.4 ms⁻² = A.

Hope this helped!
Good Luck for all your exams!
 
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AhsanAfzal said:
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s13_qp_13.pdf
Q25(why not A?), 30, 35 pls explain
sagar65265
Click to expand...

Q25)

For this question, you have to be very careful with the wording, of both the statement and the answer. The question says that the view of the calculator screen through the polaroid changes when the polaroid film is rotated. It does not tell us how the view changes, or whether the view changes by a little or a lot.

Carrying this point forward, it is impossible to say whether the radiation from the calculator screen (light, basically) has been polarized or not - polarized light basically vibrates in only one plane, back and forth along that plane. A wave can oscillate along the horizontal plane, but not in the vertical plane, or any other plane for that matter - even then, if it is viewed through the polaroid, there will be a time when the wave is not allowed to pass through the polaroid at all. On turning the polaroid just a teensy bit more, the wave will be partially allowed to pass through. So, even a wave that is polarized only in one dimension can still show changes when a polaroid filter is used to view it and the filter is rotated.

So, the wave may be polarized but it will still show a change when viewed through a polaroid filter if the filter is rotated. Thus A is not possible to confirm.

However, since the display changes, it is clear that the wave is polarized to some extent by the polaroid; some parts of the wave oscillations are blocked out and some are let through. This cannot happen with a longitudinal wave (a longitudinal wave cannot be polarized, as a rule), so B is ruled out - a longitudinal wave would show no difference whether a polaroid is used or not (and further, radiation from the electromagnetic spectrum comes in the form of transverse, not longitudinal waves).

Lastly, that D is either rubbish or something far beyond the syllabus, since the wavelength should only be concerned when the type of filter, the thickness of the filter, etc is known - it is not known in this case, and therefore you cannot say anything conclusive about D. Therefore, the best suited answer is C.

Q30)

We can either try out each of the options here and see which one is correct, or find out what the range of acceptable answers can be from the diagram and see which of the options fit - let's do the second for now.

What we see in the diagram is almost like a parabolic curve - the particle enters the field from some side and starts moving with an initial downward velocity. However, that downward velocity starts decreasing, until it is zero, and then the particle starts moving with an upwards velocity until it exits the field.

Note that since the downwards gradient starts decreasing in the first half of the journey, the acceleration is upwards. Further, the gradient starts tilting upwards more and more as the second half of the journey progresses, which means that the upwards velocity increases steadily and that the acceleration is constantly upwards. Therefore, the force on the object is also upwards. This means that either the charge is positive with the field upwards, or the charge is negative with the field downwards.

In both the situations above, the force on the charged particle is vertically upwards and the weight is downwards. The only option that contains either of these two situations is D, where the charge is negative (an electron) and the field is directed downwards. So our answer is D.

Note, it does not matter whether the particle comes from left to right or goes from the right to the left. The path will be the same.

Q35)

https://www.xtremepapers.com/commun...st-your-doubts-here.9860/page-513#post-815389
Let me know if the post above explains your doubt - if not, i'll post a more detailed reply as soon as I can.

Hope this helped!
Good Luck for all your exams!
 
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..sacrifice4Revenge.. said:
View attachment 44900
displacement-distance graph
transverse wave.
which of these is correct:

A- the speed at P is max
B- displacement at Q is always zero
C- Energy at R is entirely kinetic
D- accelaration at S is max.

correct ans is D.
why isnt it B or A? why are they incorrect? and whats the logic behind D being correct
Click to expand...
Its displacement graph, so A is rejected. And Q is not always zero, coz its oscillating. D coz Q and P has max amplitude....
 
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