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Shouldn't it be 9000 Hz? The signal bandwidth is 18 kHz; the audio bandwidth is half of that, which means 9 kHz or 9000 Hz.
Yes. xD I answered part (ii) xDShouldn't it be 9000 Hz? The signal bandwidth is 18 kHz; the audio bandwidth is half of that, which means 9 kHz or 9000 Hz.
Audio (or message) bandwidth is the same as maximum frequency of the audio (or message) signal.
For this if you look at the samples taken you see that some peaks get lost (between 0.25ms and 0.50ms & between 1.00ms and 1.25ms) which we don't want. So there is a significant change in voltage measured and the sampling frequency(or sampling rate) is too low.9702/41/O/N/13
qtn 11 (a) & (c)
can someone plzz explain
We use A= -λN because we have Activity, Decay constant and N( the no of particles that are present).View attachment 61806
Can any1 pls xplain (b)?
y do we use A= -λN?
cuz N gives us undecayed nuclei nt amnt in atmosphere?
We use A= -λN because we have Activity, Decay constant and N( the no of particles that are present).
Using that formula we get N= 9.52 x 10^7 of Radon-222 atoms
So the ratio will be (2.52 ×10^25)/(9.52 x 10^7) giving 2.6 x 10^17. I believe this is the correct answer.
In 1m^3 of atmospheric air 9.52 x 10^7 atoms of Radon-222 are presentbt isn't N the no. of particles present in the sample n not in the air? Lyk the undecayed amnt?
I am an Engineering student. Got AAB in Phy, Math, and Chem respectively.Specifically looking for a candidate who appeared in A level exams and received an A and had Physics Chemistry and Biology just wanted few advices regarding how to approach the examintions and how to revise having 40 days in hand
(918 - 900) x 1000 = 18000yes but how?
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