Physics: Post your doubts here!

[♣♠ Magnanimous ♣♠]

what u think of your self

means.

 

[♣♠ Magnanimous ♣♠]

stupid khud to bohat smart samjtai ho

nahi to.

 

[mehria]

Re: Physics Help here! Stuck somewhere?? Ask here!

can some1 give me vector notes for physics for As level plzzz?????

http://www.s-cool.co.uk/a-level/phy...evise-it/vectors-and-scalars-whats-the-differ
http://www.s-cool.co.uk/a-level/phy...n/revise-it/resolving-vectors-into-components

 

[yousef]

Question 9 (b)

 

[mehria]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s03_qp_1.pdf
someone please help me with Q20 and 32

Q20:- As:- P=ρɡɦ
Where P= pressure
ρ= Density
ɦ= height
So:- ɦ= p/ρɡ
And 10% of p0 is multiplied with the equation

h=

p0 x p/ρɡ = p0/10ρɡ

Q32:- Whenever u are asked to find the Current in a circuit then always use terminal voltage instead of e.m.f of battery..as some of the voltage is used up by the internal resistor
i.e. I= V/R = 7.5/15= 0.5A

 

[sagar65265]

hey can u help me with this question
Q) a certain charge Q is dived into two parts q and Q-q,which are then separated by a certain distance.what must q be in terms of Q to maximize the electrostatic repulsion between the two charges?
Q)
two free particles (that is,free to move )with charges +q and +4q are at distance L apart . A third charge is placed so that the entire system is in equilibrium is unstable.

Q1) Suppose the particles are separated by a particular distance x, and their charges are as you have mentioned, one with a charge "q" and another with charge "Q-q".

Then, the coulomb force acting between them would be equal to:

F(Coulomb) = kq(Q-q)/x^2

Now, we want to know what values of q (in terms of Q) will result in the greatest force on each. Therefore, we want to find the maximum value of F(Coulomb) as q varies.

Therefore, we take the derivative of F(Coulomb) with respect to q and equate that to zero as follows:

d[F(Coulomb)]/dq = kQ/x^2 - 2kq/x^2

Setting this to zero, we get kQ/x^2 = 2kq/x^2
Cancelling out k and x^2, we get:
Q = 2q
q = Q/2

Substituting this value into the equation for F(Coulomb) we get
F(Coulomb) maximum value = kQ^2/(2x^2) Newtons

Q2) I can't really understand this question; is the final situation supposed to be stable or unstable? What is the magnitude of the charge to be placed between the +q and +4q charges?

Good Luck for all your exams!

 

[taffycandy]

than

Q1) Suppose the particles are separated by a particular distance x, and their charges are as you have mentioned, one with a charge "q" and another with charge "Q-q".

Then, the coulomb force acting between them would be equal to:

F(Coulomb) = kq(Q-q)/x^2

Now, we want to know what values of q (in terms of Q) will result in the greatest force on each. Therefore, we want to find the maximum value of F(Coulomb) as q varies.

Therefore, we take the derivative of F(Coulomb) with respect to q and equate that to zero as follows:

d[F(Coulomb)]/dq = kQ/x^2 - 2kq/x^2

Setting this to zero, we get kQ/x^2 = 2kq/x^2
Cancelling out k and x^2, we get:
Q = 2q
q = Q/2

Substituting this value into the equation for F(Coulomb) we get
F(Coulomb) maximum value = kQ^2/(2x^2) Newtons

Q2) I can't really understand this question; is the final situation supposed to be stable or unstable? What is the magnitude of the charge to be placed between the +q and +4q charges?

Good Luck for all your exams![/qu

 

[taffycandy]

thankyouuuuuuuu soooo muuucccchhhhh

 

[sagar65265]

thankyouuuuuuuu soooo muuucccchhhhh

You're welcome, but what about the second question (unless you've got it sorted out already)?

 

[unique111]

Somebody?

*the bottom of the ladder rests on rough ground where there is friction. The top of the ladder is at a height h above the ground and the foot .....

 

[unique111]

What is the concept behind potentiometer? What are the formulae?

 

[mahabaloch]

Q20:- As:- P=ρɡɦ
Where P= pressure
ρ= Density
ɦ= height
So:- ɦ= p/ρɡ
And 10% of p0 is multiplied with the equation

h=

p0 x p/ρɡ = p0/10ρɡ

Q32:- Whenever u are asked to find the Current in a circuit then always use terminal voltage instead of e.m.f of battery..as some of the voltage is used up by the internal resistor
i.e. I= V/R = 7.5/15= 0.5A

Thanks

 

[ShreeyaBeatz]

Hey
someone help me on this. We all know Power= Fv . The driving force and resistive forces are given. So isn't the F in power= Driving Force-Resistive force?
why in the markscheme is it given F= driving+resistive force.
I've attached the question for your easiness!

The force in (d).i. is 75N

 

[mahabaloch]

thanks

 

[ShreeyaBeatz]

And yes the force in (d).i. is 75N.

 

[ShreeyaBeatz]

Please explain me: What happens when-
1.The width of each slit is increased but the separation remains constant.
2. seperation of the slits is increased

 

[sagar65265]

Somebody?

Well, for any stationary object, i.e. an object in equilibrium, the following two conditions have to be fulfilled:

i) No net force must act on that body (so that the centre of mass does not accelerate, or if the object is stationary, start moving);

ii)No net torque must act about ANY point chosen (for the angular rotation of a body to be zero; the center of mass can remain stationary and the object rotate about it, e.g. a ceiling fan).

This is an interesting point; IF and ONLY IF the object/ system at hand is stationary, you can take moments about ANY point, even if it is outside the system boundary, and still get the right answer (example coming up soon!).

So, to simplify matters here, there are three points about which the moment equations will be simplest:

i) Contact with floor;
ii)Contact with wall;
iii)Center of mass of ladder.

i) About an axis through this point into and out of the page, the weight tries to turn the body Anticlockwise, and the normal force from the wall tries to turn the ladder Clockwise.

An interesting thing about the torque of a force about an axis: To obtain that torque, join the point through which the axis passes to the point where the force is applied. Then, slide the force along it's direction (either in the direction it's pointing, or in the opposite direction) and do so until the line you've drawn, imaginary or real, is perpendicular to the force. The length of that line is the lever arm you use in the calculation.

So, in this case, the forces F and W(by the ground) have no torque. We can slide the Weight downwards, until the line joining that force and the point of contact is perpendicular to the force. Since the force is vertical, this happens when the line is horizontal. Therefore, the distance at the horizontal level is a so the torque is Wa.

For the force F(by the wall), we can slide the force towards the right or the left; sliding it to the right gives us a moment arm of h, since the force is horizontal, and so when the line joining the point of contact and the force vector is vertical, the length of that line is h, so the torque is Fh.

So, Fh = Wa

Well, not one of the options. So we repeat for the next point.

ii) About this point, the weight W rotates it Clockwise, Friction F rotates it Clockwise and F(by the Floor) W rotates it AntiClockwise.
So, for the point of contact with the wall, we move the Weight vector opposite it's pointing direction (we move it upwards) to give us a moment arm of a and a torque of +Wa, we slide the friction vector F to give us a moment arm of h and a torque of +Fh, and we slide the F(by the floor) W vector upwards to give us a moment arm of a+a = 2a and a torque of -2Wa.

Adding these us,

Wa + Fh + (-2Wa) = 0
And therefore,

Wa + Fh = 2Wa

Still, Cambridge examiners are devious guys. We could have rearranged that option from Wa + Fh = 2Wa to give us Fh = Wa, which we would have gotten first.
Oh well. It happens.

Either ways, going back to that earlier point, suppose we took moments about the corner in the wall.
Then we'd get:

Torque due to force F exerted by the wall = +Fh

Torque due to force W exerted by floor = -2Wa

Torque due to force W exerted by earth's gravity = +Wa

The sum would still have given us Fh = Wa.

Hope this helped!
Good Luck for all your exams!

 

[sagar65265]

Hey
someone help me on this. We all know Power= Fv . The driving force and resistive forces are given. So isn't the F in power= Driving Force-Resistive force?
why in the markscheme is it given F= driving+resistive force.
I've attached the question for your easiness!View attachment 38508

The force in (d).i. is 75N

Let's think of this question in terms of net force:

We know that the boy accelerates at a constant rate until his velocity reaches 5.6 ms^-1. In part (ii), his speed is 4.5 ms^-1, so he hasn't yet finished accelerating.
From the earlier part, we know that the acceleration of the boy is 1.12 ms^-2 (5.6/5.0 = 1.12) and so we know by Newton's Second Law that:

F(net) = ma

OR

(Net Force on Boy+Bicycle) = (Mass of Boy+Bicycle) * (Acceleration of Boy+Bicycle)

Since the acceleration is 1.12 ms^-2, we can write

(Net force on Boy+Bicycle) = 67 * 1.12 = 75.04 Newtons

So, the net force on the boy+bicycle must be 75.04 Newtons for as long as they are accelerating.

However, we are also told that at this speed, a force of 23 Newtons acts against the boy's speed; there is a negative force of 23 Newtons on the system, which tries to decelerate it. Therefore:

Force applied by the boy - 23 Newtons = 75.04

Therefore, to continue accelerating himself and the bicycle at a constant rate of 1.12 ms^-2 when a resistive force of 23 Newtons acts on them, the boy has to apply 98.04 Newtons of force (more accurately, though, the ground has to apply this force).

So, If we apply P = Fv here, we get
P = 98.04 * 4.5 = 441.18 Watts = 440 W.

The key point here is that in P = Fv * cos(θ),

P = the power of a particular force (which is the work is done per unit time) in Watts;
F = the magnitude of that force in Newtons;
v = the rate at which the position vector of the object being accelerated by that force and
θ = the angle between the Force and Velocity vectors.

Here, since the angle between the two is zero, cos(θ) is equal to 1 and P = Fv.

Hope this helped!
Good Luck for all your exams!

 

[daredevil]

Heyy here;s a question from M/J/13 paper 42 ...

Q. state and explain the effect on the transmitted analogue waveform of increasing, for the ADC and the DAC, both the sampling frequency and the number of bits in each sample.


the answer in the marking scheme is:
increasing number of bits reduces step height
increasing sampling frequency reduces step depth / width
reproduction of signal is more exact


I get the sampling frequency part but not the part about increasing no. of bits... can anyone explain please??

 

[unique111]

Well, for any stationary object, i.e. an object in equilibrium, the following two conditions have to be fulfilled:

i) No net force must act on that body (so that the centre of mass does not accelerate, or if the object is stationary, start moving);

ii)No net torque must act about ANY point chosen (for the angular rotation of a body to be zero; the center of mass can remain stationary and the object rotate about it, e.g. a ceiling fan).

This is an interesting point; IF and ONLY IF the object/ system at hand is stationary, you can take moments about ANY point, even if it is outside the system boundary, and still get the right answer (example coming up soon!).

So, to simplify matters here, there are three points about which the moment equations will be simplest:

i) Contact with floor;
ii)Contact with wall;
iii)Center of mass of ladder.

i) About an axis through this point into and out of the page, the weight tries to turn the body Anticlockwise, and the normal force from the wall tries to turn the ladder Clockwise.

An interesting thing about the torque of a force about an axis: To obtain that torque, join the point through which the axis passes to the point where the force is applied. Then, slide the force along it's direction (either in the direction it's pointing, or in the opposite direction) and do so until the line you've drawn, imaginary or real, is perpendicular to the force. The length of that line is the lever arm you use in the calculation.

So, in this case, the forces F and W(by the ground) have no torque. We can slide the Weight downwards, until the line joining that force and the point of contact is perpendicular to the force. Since the force is vertical, this happens when the line is horizontal. Therefore, the distance at the horizontal level is a so the torque is Wa.

For the force F(by the wall), we can slide the force towards the right or the left; sliding it to the right gives us a moment arm of h, since the force is horizontal, and so when the line joining the point of contact and the force vector is vertical, the length of that line is h, so the torque is Fh.

So, Fh = Wa

Well, not one of the options. So we repeat for the next point.

ii) About this point, the weight W rotates it Clockwise, Friction F rotates it Clockwise and F(by the Floor) W rotates it AntiClockwise.
So, for the point of contact with the wall, we move the Weight vector opposite it's pointing direction (we move it upwards) to give us a moment arm of a and a torque of +Wa, we slide the friction vector F to give us a moment arm of h and a torque of +Fh, and we slide the F(by the floor) W vector upwards to give us a moment arm of a+a = 2a and a torque of -2Wa.

Adding these us,

Wa + Fh + (-2Wa) = 0
And therefore,

Wa + Fh = 2Wa

Still, Cambridge examiners are devious guys. We could have rearranged that option from Wa + Fh = 2Wa to give us Fh = Wa, which we would have gotten first.
Oh well. It happens.

Either ways, going back to that earlier point, suppose we took moments about the corner in the wall.
Then we'd get:

Torque due to force F exerted by the wall = +Fh

Torque due to force W exerted by floor = -2Wa

Torque due to force W exerted by earth's gravity = +Wa

The sum would still have given us Fh = Wa.

Hope this helped!
Good Luck for all your exams!

Thanks! Your detailed method of explanation was very helpful.
Could you help me with my other query too? It's posted above.