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Physics: Post your doubts here!

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1) rms of an ac is equal to the constant dc that produces same power or heating effect in a identical resistor. for a sinusoidal graph, rms voltage and rms current is (peakvoltage/under root of 2) and (peak current/under root 2) respectively
what if the wave is not sinusoidal how to find rms thn?
 
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make a power time graph. calculate the avg power. equate it to (I^2 * R) or (V^2/R). this current or voltage will be rms.
if the graph is rectified ac with linear change as it is in nov 2003 p4 q4, then (final+initial)/2
all of dis included in AS? dis is so new for me
edit: i just chkd the syllabus and its only in A2..... :sick:
 
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AoA wr wb

334sxoy.jpg


i dnt understand the last part
why is it not x-3=6.75
 
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thanku very much...but unfortunately i dont have tht book..:(....i just have a collection of 3 books pacific,cambridge one and bath physics.......
how much does one wave lags behind the other. one wave is equal to 2 pi radians. how much out of phase are two waves on a graph. use chadda's book cambridge endorsed one,
draw a graph , taking displacement on y axis and distance on x axis. draw a sin wave. the wave is 360 degree or 2 pi radians.
take points according on x- axis. take two points suppose A and B , a at start of the wave and b at the end of the wave, both being in same state of motion or at same displacement from mean position. they have a phase difference of 360 degree or 2 pi radians. just like this take any two points and read on the x axis how much they differ in radians or in degrees. this is how we calcular phase angle difference.
 
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Can someone please answer me question?

June09) 9) b): http://www.xtremepapers.com/papers/...and AS Level/Physics (9702)/9702_s09_qp_4.pdf

It says: A cobalt-60 source having a half-life of 5.27 years is calibrated and found to have an
activity of 3.50 × 105 Bq. The uncertainty in the calibration is ±2%.
Calculate the length of time, in days, after the calibration has been made, for the stated
activity of 3.50 × 105 Bq to have a maximum possible error of 10%.

I don't get what the Marking Scheme says: http://www.xtremepapers.com/papers/...and AS Level/Physics (9702)/9702_s09_ms_4.pdf
 

omg

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It's a bit confusing T^h/g... Is it T^(h/g) or (T^h)/g ???
If it's the former:
ln R = ln [T^(h/g)] = (h/g) ln T
On a graph of ln R against ln T the gradient is h/g.
If it's the latter:
ln R = ln [(T^h)/g] = h ln T - ln g
On a graph of ln R against ln T, the gradient is h and y-intercept is - ln g.
so in the first case we'll have no y-intercpt??
 
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Two solid substances P and Q have atoms of mass MP and MQ respectively. They have NP and NQ atoms per unit volume such that NP= NQ. It is found by experiment that the density of P is greater than that of Q.

Which of the following deductions from this experiment must be correct?

A MP > MQ

B NP > NQ

C MPNP > MQ NQ

D MP/NP = MQ/NQ
 
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its ok plz try solving the other 2 qs i posted

i dnt understand the last part why is it not x-3=6.75

It is called the "Potential Difference across the resistor ". So if 1 end of the resistor has a potential of 3 V and the potential difference is 6.75, then:
3-x= 6.75
x= 3 - 6.75

Concerning the potential divider question:

I'm not entirely sure of this answer but I understand that since the galvanometer is indicating a zero current, therefore the potential difference between the positive plate and B is zero ( same with the negative plate and A ). Why? Because if there was a potential difference there should be a current passing. This does mean that the potential difference across AB is equal to the potential difference across the cell being tested. Thus, you simply need to find the P.D. across AB.

By using ratio of length to resistance, we get the resistance of AB. Then again by using ratio of resistance to P.D. we get the P.D. across AB.

1 ---> 60
0.65 ---> ?

Then

60 + 20 ---> 12
39 ---> ?

Hope this helped. : )
 
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