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means.what u think of your self
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means.what u think of your self
nahi to.stupid khud to bohat smart samjtai ho
http://www.s-cool.co.uk/a-level/phy...evise-it/vectors-and-scalars-whats-the-differRe: Physics Help here! Stuck somewhere?? Ask here!
can some1 give me vector notes for physics for As level plzzz?????
Q20:- As:- P=ρɡɦhttp://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
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![/qu
thankyouuuuuuuu soooo muuucccchhhhh
*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 .....
What is the concept behind potentiometer? What are the formulae?
ThanksQ20:- 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
Somebody?
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
Thanks! Your detailed method of explanation was very helpful.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!
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