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A2 Physics | Post your doubts here

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Assalamu alaikum,
What is impedence?? In Application Booklet, it's stated;
The ideal operational amplifier has infinite input imedence (i.e. no current enters or leaves either of the inputs )
Pls explain the above statement. Thanks
 
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Assalamu alaikum,
What is impedence?? In Application Booklet, it's stated;
The ideal operational amplifier has infinite input imedence (i.e. no current enters or leaves either of the inputs )
Pls explain the above statement. Thanks



impedence is a name given to resistance.

infinite means uncountable = verY high

and infinite impedence means very high resistance of the amplifier therefore no current can enter that point neither can leave !
so the ideal amplifier has infinite input impedence(resistance)
 
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impedence is a name given to resistance.

infinite means uncountable = verY high

and infinite impedence means very high resistance of the amplifier therefore no current can enter that point neither can leave !
so the ideal amplifier has infinite input impedence(resistance)
oh got it now! thanks!
 
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I am not sure if u guys are in a way of studying like this but i find these easy to learn in the following hence providing you all with a way too learn Mobile Circuit.... Hope it helps..

Meri - microfone
Ami Foran - a.f amplifie
Ati - ADC
pitti - Parralell to Digital:p
marti - modulator-ossilator
aur - amplifier :D
sunati - switch
aesay - aerial

Hope it help... Will keep on posting ........... ;)
lol its funny but interesting :p
 
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I find this question conceptual. Kindly please help:
Question no.1 part.c

Marking scheme says greater the radius, greater will be centripetal force. In the formula: F=mv^2/r centripetal force is INVERSELY proportional to r. So what's going on here? :eek:
 

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I find this question conceptual. Kindly please help:
Question no.1 part.c

Marking scheme says greater the radius, greater will be centripetal force. In the formula: F=mv^2/r centripetal force is INVERSELY proportional to r. So what's going on here? :eek:
we will not use this formula for centripetal force cox v changes with the radius. we will use mrw^2 cox mass is constant and w is also constant for every point
 
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we will not use this formula for centripetal force cox v changes with the radius. we will use mrw^2 cox mass is constant and w is also constant for every point
good question and answer! Just want to note out that;
Omega isn't constant. There's centripetal acceleration.
The word acceleration denotes that there's rate of change of velocity- i.e. there's rate of change of direction of the object moving in circular motion.
Pls do correct me if I'm wrong. :)
 
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good question and answer! Just want to note out that;
Omega isn't constant. There's centripetal acceleration.
The word acceleration denotes that there's rate of change of velocity- i.e. there's rate of change of direction of the object moving in circular motion.
Pls do correct me if I'm wrong. :)
Omega changes, but not necessarily with radius. :)
 
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So omega changes in its own rate, and F is directly proportional to the square of omega. Your answer is quite correct, increasing radius doesn't affect the value of omega. :)
Pls correct me if I'm wrong. :)
 
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good question and answer! Just want to note out that;
Omega isn't constant. There's centripetal acceleration.
The word acceleration denotes that there's rate of change of velocity- i.e. there's rate of change of direction of the object moving in circular motion.
Pls do correct me if I'm wrong. :)
omega is constant. it does not depend on radius. its a horizontal circle and for horizontal circles, omega is constant
 
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omega is constant. it does not depend on radius. its a horizontal circle and for horizontal circles, omega is constant
Omega does not depend on the radius, that's right. But there's still a centripetal on object right?? And we know, that F= ma, that when there's constant force, there's acceleration of the object. And when you recall the definition of centripetal acceleration, it's the rate of change of angular velocity. Angular velocity, that is omega , is changing at a constant rate. It's changing at a constant rate irrespective of the change of radius. Whether radius changes or not, The angular velocity , omega changes constantly. When omega is constant, there's no centripetal acceleration and with no centripetal acceleration, the object might just not move in a circular motion, but in translatory motion. :)
 
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Omega does not depend on the radius, that's right. But there's still a centripetal on object right?? And we know, that F= ma, that when there's constant force, there's acceleration of the object. And when you recall the definition of centripetal acceleration, it's the rate of change of angular velocity. Angular velocity, that is omega , is changing at a constant rate. It's changing at a constant rate irrespective of the change of radius. Whether radius changes or not, The angular velocity , omega changes constantly. When omega is constant, there's no centripetal acceleration and with no centripetal acceleration, the object might just not move in a circular motion, but in translatory motion. :)
umm ok. thnx for the info
 
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AOA,
http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s10_qp_41.pdf
Q3 (b) how do we know that acceleration is of constant magnitude
My guess:
because it is 'g' (acceleration due to gravity) ??? :D
hehe :D
The ball then moves
down plane LA and rises up plane RA to its original height.
The key word is right there, to it's original height. a= -(omega )^2 amplitude
In this defining equation of simple harmonic motion, omega is constant throughout the motion of the ball, right?? But accelerations are different at different points of the oscillation. Thus the only other factor, we can see from the equation that changes the acceleration of the ball is x(displacement from the mean position) . X changes and thereby acceleration changes. But in this case, x doesn't change. However they have damped the oscillations, the displacement from the mean position remains constant, that is, acceleration remains , a = -( omega ) ^2 amplitude.
Pls correct me if I'm wrong :)
 
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edit: defining equation of simple harmonic motion: a = - (omega ) ^2 x
a = -( omega )^2 amplitude is max acceleration of the oscillations! Apologies!
 
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hehe :D
The ball then moves
down plane LA and rises up plane RA to its original height.
The key word is right there, to it's original height. a= -(omega )^2 amplitude
In this defining equation of simple harmonic motion, omega is constant throughout the motion of the ball, right?? But accelerations are different at different points of the oscillation. Thus the only other factor, we can see from the equation that changes the acceleration of the ball is x(displacement from the mean position) . X changes and thereby acceleration changes. But in this case, x doesn't change. However they have damped the oscillations, the displacement from the mean position remains constant, that is, acceleration remains , a = -( omega ) ^2 amplitude.
Pls correct me if I'm wrong :)
x does change,
the ball covers displacement
 
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Take a look at this unofficial derivation. Not a fair work, but for understanding :D
Generally in simple harmonic motions, a = -( omega )^2 x
a = -constant.x
In this oscillation, x is also constant.
Hence a = -constant.constant.
Minus sign denotes the direction of acceleration so;
a = constant
Just to get the idea :)
 
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