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Physics: Post your doubts here!

DeViL gURl B)

DeViL gURl B)

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Thought blocker said:
5)
scan0002-jpg.23237

---------->Written form :
Frequency: 3000 oscillations in a minute so , 3000/60sec= 50
now Time period= 1/50= 0.02
and so for time base, as this is for one full wave, divide it by 2, 0.02/2= 0.01

7)The horizontal velocity will decrease to 0 as air resistance is acting against it so it will eventually decrease to 0 and the vertical velocity will increase until it reaches a constant value the question is actually related to terminal velocity theory

9)Use v^2 - u^2 = 2as
v has to be zero, since the train has to come to rest eventually.
s = x
The equation becomes: 2ax + u^2 = 0
make 's' distance the subject:
x = - (u^2)/2a
deceleration, i.e -a, is constant in the question.
take the constants out,
you get this relation: x ∝ u^2, i.e, s varies as the square of u.
So when u is increased by 20% (120/100 = 1.2), x will change by (1.20)^2 = 1.44x
Click to expand...
Dude, in question 5 .. I got till 0.02 seconds .. But then y did u divide by 2?
And then when you got 0.01 how's that supposed to make up with 10 in the answer given :$
 
Busy Buddy

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I'm really sorry if anyone of you get disturbed reading my message here. Good to see lot many friends working hard for paper 1 (ma sha Allah).
I have a wired doubt, its about physics p5. Yes, don't we lose the entire 15 marks in the first question if we write the independent variable wrong cause I did that mistake and wrote the whole concept wrong :(
(already both p2 & p4 went bad although p2 was easy)
 
DeViL gURl B)

DeViL gURl B)

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Thought blocker said:
5)
scan0002-jpg.23237

---------->Written form :
Frequency: 3000 oscillations in a minute so , 3000/60sec= 50
now Time period= 1/50= 0.02
and so for time base, as this is for one full wave, divide it by 2, 0.02/2= 0.01

7)The horizontal velocity will decrease to 0 as air resistance is acting against it so it will eventually decrease to 0 and the vertical velocity will increase until it reaches a constant value the question is actually related to terminal velocity theory

9)Use v^2 - u^2 = 2as
v has to be zero, since the train has to come to rest eventually.
s = x
The equation becomes: 2ax + u^2 = 0
make 's' distance the subject:
x = - (u^2)/2a
deceleration, i.e -a, is constant in the question.
take the constants out,
you get this relation: x ∝ u^2, i.e, s varies as the square of u.
So when u is increased by 20% (120/100 = 1.2), x will change by (1.20)^2 = 1.44x
Click to expand...
Dude where do u get 120 from in the last question u explained ? :$

Thought blocker
 
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DeViL gURl B)

DeViL gURl B)

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Busy Buddy said:
I'm really sorry if anyone of you get disturbed reading my message here. Good to see lot many friends working hard for paper 1 (ma sha Allah).
I have a wired doubt, its about physics p5. Yes, don't we lose the entire 15 marks in the first question if we write the independent variable wrong cause I did that mistake and wrote the whole concept wrong :(
(already both p2 & p4 went bad although p2 was easy)
Click to expand...
Well though I'm an AS student but .. Know one thing that u do get marks for the right concept as in I dunno what to I call it .. There's a name for that but.. U do get marks for usin the right concept sis :)
And nsha Allah we all will do good :)
 
Thought blocker

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Talha Irfan said:
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s03_qp_1.pdf

Q 5
Q 20
Q 30
Q 34
Click to expand...
5)
3% of 330 = 10 m/s
and speed of sound = 330 m/s
20)
P = ρgh
10%P = ρgh
h = (0.1 P)/ρg
Hence A
30)
Since the total time period = 8s, we use the mean current i.e. (100+20)/2 = 60 mA
Q = It
= 60 * 8
= 480 mC
34)
For the range we need to know Minimum V and Maximum V, That would be range itself, so:
Maximum Value of resistance would be 50 KΩ and minimum would be OΩ..
Total resistance would be 60 KΩ, hence :
Minimum V = 0/60 * 9 = 0 V and Max V = 50/60 * 9 = 7.5 V
So range would lie b/w Zero V to 7.5 V that is B :)
 
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Thought blocker

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ZaqZainab said:
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s04_qp_1.pdf
Q9,25 and 33
Please :(
Thought blocker
Click to expand...
9) First find the time, using S = Ut + 0.5t^(2) ; S = 1.25 , U = zero, t = ?
That is 1.25= 0.5 x 9.81 x t^(2), t = 0.5 seconds.
Now for the Velocity, we know it equals Displacement / time taken ----> V = 10 / 0.5 = 20 m/s so in this type of question either of 2 quantities are given, we just need to find 3rd and input them all in second eqn to get the final result we are asked.

25)The image of a wave has been given, and so has the direction it is traveling in.

Note that since this is a transverse wave and the disturbances are traveling from left to right, the particles at any point are vibrating in a direction perpendicular to that propagation.
In other words, if you focused on any one segment of the rope and that one segment alone (let's say by painting it a different color from that of the rope), then you would find it does not experienceany sideways displacement; it only moves up, then down, then up, then down, etc.

At the same time, the waveform shifts towards the right, since that is the direction in which it is traveling.
So just imagine the waveform displayed in the question shift to the right; imagine each part of the waveform progress, and you would then see that P is moving downwards; since the trough just before P has to travel to the right, P has to "fall" into that trough, and if it "falls" into that trough, it is moving downwards.

Of course, this eliminates all the options in the question, leaving behind only A, the correct answer, but it is worth discussing the motion of Q:

At the instant of the displayed image, Q is a segment at the crest of the waveform; it is debatable that after some time, Q has to fall towards the equilibrium position, but taking a closer look at the wave motion (and seeing that we can estimate the motion of any one segment as Simple Harmonic Motion, SHM) it turns out that Q would be stationary.

It's just like the motion of a pendulum; at any extreme, the rate of change of it's position - it's velocity - is zero, but due to the forces acting on the pendulum (in THIS case, the force is the tension in the string) this changes slowly to make it change position. So that's most likely why the velocity of Q is zero at the instant shown.



33)Seriously, this one is a doozy :) the wording is not very nice, to be honest, and it makes the question rather confusing.
What it actually asks for (as far as I can tell) are the variables required to find the different between OPEN circuit voltage and CLOSED circuit voltage. In other words, the difference between the potential difference measured when the circuit is OPEN and when it is fully connected with an external resistor.
(I thought it spoke about the normal decrease in voltage with running duration, which can be expected as a battery runs down. My mistake!)

A Google Images search for "internal external resistance" provides some very clear images, and i'm using the attached one for reference.
img635-jpg.38354

When the circuit is open, no current flows through the battery, and the initial reading the equivalent to the EMF of the battery; no energy is lost per second to heat in the internal resistance, so it is equal to the EMF.
Let's write this down as V.

Once the circuit is connected as above, a current indeed does flow through all components, and a potential difference is maintained across each too. However, since the battery consists of an internal resistance (this internal resistance is NEVER separate from the battery; it can be considered as such, but the battery consists both of the cells and the internal resistance), there is a drop in potential across this internal resistance before the current even leaves the battery. Thus, the potential difference across the terminals of the battery drop.

Applying Kirchoff's Second Law while moving from B to A, what we get is (V is still the EMF, r(internal) is the internal resistance):

(Potential at A) + V - i * r(internal) = (Potential at B)

All this says is that the potential at A is one thing, it changes by so and so amount, as a result of which is becomes another value, the Potential at B.

So, since the Potential Difference is the (Potential at B) - (Potential at A), we have:

PD = V - i * r(internal)

Thus, the drop in potential is

V - (V- i * r(internal)) = i * r(internal)

The change is negative, but the decrease is positive, so the signs change there. Therefore, all you need to find the drop in potential are the final current, and the internal resistance of the battery = C
Hope you get it, sagar explained me this, now I can explain to you ;)
 
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