AS Level Physics Circuits Question

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An 'accurately calibrated voltmeter' indicates that the device is correctly set to zero when no potential difference is present, ensuring precise measurements. If a voltmeter is not calibrated, it may introduce a systematic error, leading to consistently inaccurate readings. For example, a systematic error of +0.5 volts would misrepresent a 2-volt reading as 1.5 volts. This calibration is crucial for ensuring that calculations based on the voltmeter's readings are reliable. Understanding this concept is essential for solving circuits problems accurately.
QueenFisher
I have a circuits problem which I think i may be able to solve on my own, except that the voltmeter in question is described as an 'accurately calibrated voltmeter'. Can anyone explain to me what this means, and if it will affect my calculations?
Help greatly appreciated!
:cool:
 
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All it means is that the voltmeter is set to zero when there is no potenital difference across the terminals. So voltmeters that are not properly calibrated, may have a systematic error associated with them. ie. all the readings they provide are wrong by so many volts. eg. a systematic error of +0.5 volts would mean that a reading of 2 volts would actually be a reading of 1.5 volts.
 
Thread 'Variable mass system : water sprayed into a moving container'

Thread 'Variable mass system : water sprayed into a moving container'

Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
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