Physics: Post your doubts here!

[asma tareen]

I really need some guidance about the graphs the gradient of which graph gives what? The area under which graph gives what? and all such details. I'd be greatfull if someone could kindly help me out. asma tareen usama321 Thought blocker

For finding what the gradient gives us u should look at the values on y and x axis as the formula of gradient is rise/run meaning (value on y axis/value on x axis) e.g if the graph has force on y axis and area on x axis the formula of pressure can come in our mind as P=F/A so we can say gradient gives the pressure......u need to know all the formulae that can make graphs easy :|

 

[TheStallion-Reborn]

For finding what the gradient gives us u should look at the values on y and x axis as the formula of gradient is rise/run meaning (value on y axis/value on x axis) e.g if the graph has force on y axis and area on x axis the formula of pressure can come in our mind as P=F/A so we can say gradient gives the pressure......u need to know all the formulae that can make graphs easy :|

oh riight!! thankyouu! ^_^

 

[asma tareen]

oh riight!! thankyouu! ^_^

N.p.

 

[shoaib672002]

Thanks man.

love the signature

 

[Thought blocker]

love the signature

I know many of member said that Thanks

 

[♣♠ Magnanimous ♣♠]

y

love the signature

yours to too good.

 

[shoaib672002]

I know many of member said that Thanks

is that simile from my signature

 

[shoaib672002]

y

yours to too good.

Thanks

 

[Thought blocker]

is that simile from my signature

Nope.........!!! I thought you stole it..

 

[shoaib672002]

Nope.........!!! I thought you stole it..

Nope i found it on the internet , its just that we are alike

Cheers!!

 

[shoaib672002]

Nope.........!!! I thought you stole it..

Either way rock it is
dont stop now

 

[Thought blocker]

Nope i found it on the internet , its just that we are alike

Cheers!!

High_five

 

[Ali_00921]

Why does a capacitor store energy but not charge?
Thanks In Advance.

 

[Thought blocker]

Why does a capacitor store energy but not charge?
Thanks In Advance.

Well, a capacitor has a +'ve charged plate and a -'ve charged plate. If you add the TOTAL charge on a capacitor you get zero.

But the capacitor DOES store +'ve and -'ve charges. If you charge up a big capacitor and touch one of the plates, you'd ground it and get a nasty shock. So, the important thing is that each plate stores a charge.

Suppose I have a charged capacitor. I attach a bulb to one plate. Then I attach a wire from the second plate to the bulb, completing the circuit. The positive and negative charges on each plate now have a path to travel along so you get a current. And that current lights the bulb for a moment. So you can do work with this current, and therefore a capacitor is a device that stores energy. Energy is the ability to do work.

 

[Young Stunner]

http://papers.xtremepapers.com/CIE/...and AS Level/Physics (9702)/9702_s04_qp_1.pdf


Q25, 26, 33, 35

 

[sagar65265]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s04_qp_1.pdf


Q25, 26, 33, 35

Q25: 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 experience any 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.

Q26: There are two formulae we will need to use here; one relates Intensity of a wave to it's Amplitude, the other relates the Power of a wave to it's Intensity.

First, let's fix up an equation for the initial state/ situation (I = Intensity of Wave , A = Amplitude of Wave , k = arbitrary constant):

I = k * A^2

The other equation tells us (P = Power of Wave , S = Area component perpendicular to wave propagation direction):

P = I * S

Substituting the value of I from the first equation into the second,

P = k * A^2 * S

So all we have to do now is form two equations (one for each situation) and compare the two.
Initially,
P(i) = k * A^2 * S
Finally,
P(f) = k *4A^2 * 0.5S
Dividing the second equation by the first,
P(f) / P(i) = 2
Since P(i) = E,
P(f) = 2E =B

Q33: 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.

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

Q35: Assuming that the ammeter and the voltmeter are both ideal, the resistance of the ammeter is Zero and the resistance of the voltmeter is Infinity. Following on from there, the potential difference across the ammeter is Zero and the current through the Voltmeter is Zero.

So these don not affect the circuit in any way except to give us a reading. When the variable resistance is dropped, applying Kirchoff's Second Law gives us:

i = V/(∑R) where ∑R is the equivalent resistance of the circuit, in this case the algebraic sum of the two resistance values.

Since the overall resistance of the circuit decreases, the current automatically decreases and the answers are narrowed down to C or D.

Furthermore, since we have

V = I * R

and the current across the fixed resistor increases while the resistance remains the same,

V(f) = I(f) * R

Dividing the second equation by the first, we get V(f) = [I(f) / I] * V.
Since [I(f) / I] > 1, V(f) > V and so the reading on the voltmeter also increases, telling us the answer is D.
Note: If the voltmeter was across the Variable Resistor, our calculation would have been a little longer, since the increase in Current and the decrease in Resistance might *possibly* make up for each other and result in Zero Change in the voltmeter reading.

Hope this helped!
Good luck for all your exams!

 

[zonaali]

Please help me with this one. Im really stuck here!

 

[zonaali]

Please help me with this one. Im really stuck here!

 

[zonaali]

please someone help me with this one...answer is A. Examinar has mentioned that Both masses are being accelerated, so the total mass being accelerated here is 3 kg not 2 kg

 

[sagar65265]

please someone help me with this one...answer is A. Examinar has mentioned that Both masses are being accelerated, so the total mass being accelerated here is 3 kg not 2 kg

The two most important things to note for this question are as follows:

i) Since the string is "mass-less", the tension in every part of the string is the same. In other words, the string pulls each block with the same force; it pulls the 2.0 kg block to the right with the same force as it tries to pull the 1.0 kg block upwards.

ii)Since the string is assumed to be in-extensible (in other words the length of the string cannot change, it cannot stretch at all) the acceleration magnitude of the two blocks are the same; if the 1.0 kg block moves a small distance downwards, the 2.0 kg block has to move the same distance in the horizontal direction. So, since both of them have to fall the same distance in the same time interval, they have the same velocity. From here, since any change in the speed of the 1.0 kg block during a particular time interval has to result in the SAME change in speed of the 2.0 kg block during the EXACT SAME time interval, the two accelerations at any point are the same.

Let's define the directions, first: for the 1.0 kg block the positive direction is downwards and the negative direction is upwards (it doesn't travel or accelerate sideways, so we don't need to worry about that direction). Also, for the 2.0 kg block, the positive direction is to the right and the negative direction is to the left.
We've chosen the axis so that the accelerations are all positive in sign, so that we don't have to juggle around any confusing negative values.

So, taking Newton's second law for the 1.0 kg block (Net Force = Mass * Acceleration):

(1.0 kg)(9.81 ms^-2) - T = (1.0 kg)a
9.81 - T = a

Alright, so that's the first equation. Considering the 2.0 kg block:

T = (2.0 kg)a
T = 2.0a

And there's the second one. Since there is no friction acting on the 2.0 kg block, that's it!

Now, since the two accelerations and the two tension forces are the same, we can substitute the second equation into the first as:

9.81 - 2.0a = 1.0a
9.81 = 3.0a

Therefore, a = 9.81/3 = 3.27 ms^-2

Since the acceleration here is constant and does not change with time, we can use the constant acceleration kinematics equations with:
u = 0 ms^-1
a = 3.27 ms^-2
s = 0.50 m (since both blocks travel this distance before the 1.0 kg block hits the ground and stops)
v = ?

v^2 = u^2 + 2 * a * s
v^2 = 0 + 2 * 3.27 * 0.5
v^2 = 3.27

Therefore, the final speed is √(3.27) = 1.8(08) ms^-1 = 1.8 ms^-1 = A.

Hope this helped!
Good Luck for all your exams!