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ah yes sorry my mistake sorry.well the equation that ik of is of De Broglie's wavelength and that's λ = h/p so
λ=h/mv
not E=h/mv
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ah yes sorry my mistake sorry.well the equation that ik of is of De Broglie's wavelength and that's λ = h/p so
and obviously the second graph would be the 1/x curve hence not touching the axis since the denominator would never = 0well the equation that ik of is of De Broglie's wavelength and that's λ = h/p
i havent studied anything abt E's proportion to p
lol no problem yh trueand obviously the second graph would be the 1/x curve hence not touching the axis since the denominator would never = 0
thank you very much for you help.
Idk if its in the syllabus but it has to do with transmission and receiving of waves(Waves)View attachment 63418
View attachment 63419
Is it in our syllabus?
If yes then where should I find it's notes or other resource?
anastasia grey113 (the solution producer)
They are asking for work done from mid point of AB to point P. Charges are same on both spheres and the distance as well ie 6um from both A and B to midpoint. At a point the total potential experienced by the e is equal to sum of potentials from both A and B charges to that point. Thus at point P the potential is equal to sum of the potential which is now 3um from A and 9um from B. The charges remain same. Now Energy= change in potential × charge on the e (the electric potential energy formula)
- (a)= 2.6*10^-17N
i need help in part (c) why did they multiply the the charge of point A BY 2? is it becuz they are also considering point B? but then why isn't the distance 12 micrometers instead of 6? since the separation is 12.
how did they calculate the second line of the markscheme?
actually u r not finding ratio in decibels what u r doing is working out ratio in watts.U got the answer 2.5 *10^4 ,now u put log in 2.5*10^4 and multiply with 10 u will get 44dbcan someone please explain how to do this question;
the markscheme did total attnetuation + signal to noise power ratio
but i thought i could do it this way:
what i did:
total attenuation: 84*0.19=16dB
-16=10log(signal output power/signal input power)
-16=10log(signal output power/9.7*10^3)
signal output= 2.5*10^-4 W
signal to noise ratio=10log(signal/noise)
28=10log(2.5*10^-4/noise power)
noise power= 3.9*10^-7W
hence ratio =
input power/noise outpout
9.7*10^-3/3.9*10^-7
=2.5*10^4 dB
however the ms says the answer is 44dB
Here.Can someone please draw a circuit diagram for this with a protective diode? The op-amp is inverting.
how would i draw the relay
Here.
The first diode is the rectifier diode. This will make sure your relay switches on only when there is a positive cycle of voltage.
Now some Electromagnetism comes into picture. A relay is an electromagnet, if there is a current in the coil it will act as a magnet, but if there is a large change in voltage there will be a change in magnetic flux linkage and an large emf will be induced. To avoid damaging the op amp, the other protective diode is used.
This is about your 2nd question first partsomeone please help
theyll be released on 17th of mayAnyone got f/m 2018 papers? TY
F/M means?theyll be released on 17th of may
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