Physics: Post your doubts here!

[selrey]

1. (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?

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)

 

[GarryTheGhost]

can 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

 

[GarryTheGhost]

someone please help

 

[GarryTheGhost]

how would i draw the relay

 

[nishant sapkota 123]

can 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

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 44db

 

[armeena]

Can someone please draw a circuit diagram for this with a protective diode? The op-amp is inverting.

 

[PixelsLevls]

Can someone please draw a circuit diagram for this with a protective diode? The op-amp is inverting.

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.

 

[PixelsLevls]

how would i draw the relay

Hi gary I've drawn a relay in my recent reply to armeena
Do exactly the same thing but don't forget the switch for the lamp

 

[armeena]

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.

thank you so much!

 

[PixelsLevls]

someone please help

This is about your 2nd question first part

(i)
We have a lot of information in the question itself.
We have been given a:

• Gold Nucleus
• Alpha Particle!

Lets list the Data for Gold Nucleus:

• Charge : 79e
• Convert to coloumbs: 79 * (1.6 * 10^-19)
• Answer: 1.264 * 10^-17

Lets list Data for Alpha Particle:

• This is a Helium Nucleus basically
• Helium has 2 Protons.
• So it has a charge of +2e
• Answer: 3.2 * 10^-19
• We also know the alpha particle's energy which is 7.7 * 10^-13

Okay Lets solve.

Electric potential at a point the in the work done per unit charge (very important)

So Ф = Joules/ Coloumb
So for the alpha particle, its electric potential due to the Gold nucleus is:

1. Ф = 7.7 * 10^-13/ 3.2 *10^-19
2. Answer: 2406250 J/C

Now lets use our original Electric potential formula which was derived from the Coloumb's force

Ф = Q/(4piEpsilon) r
Solving for r

r = Q/ (4piEpsilon)Ф
Q will be the charge on the gold nucleus
So r =4.72* 10^-14
So if the distance b/w the alpha particle and gold nucleus is (r) then the radius has to be smaller!!

 

[Asad.Babur]

I have ATP tomorrow with almost zero preparation. I'm about to finish the book in a hour or so. I'll look at 4 years of past papers and do some practice.

Suggest something urgently please.

AND btw, can I leave the nuclear/atomic energy chapters? I don't think they come in ATP/P42

 

[ba-lock-ey]

Anyone got f/m 2018 papers? TY

 

[anastasia grey113]

Anyone got f/m 2018 papers? TY

theyll be released on 17th of may

 

[Asad.Babur]

theyll be released on 17th of may

F/M means?

 

[armeena]

F/M means?

february/march

 

[Asad.Babur]

february/march

Yeh konsa naya season hai bay?

 

[Dtrickster]

Can someone please show me how to draw a voltmeter in parallel to 3 resistors in parallel

 

[PixelsLevls]

I have ATP tomorrow with almost zero preparation. I'm about to finish the book in a hour or so. I'll look at 4 years of past papers and do some practice.

Suggest something urgently please.

AND btw, can I leave the nuclear/atomic energy chapters? I don't think they come in ATP/P42

No bad idea, they are actually present as one whole question. It's fairly simple
I suggest you go through it.

 

[Asad.Babur]

ATP WALAY BATAEIN KESA GAYA!

 

[Zaki ali asghar]

How was 9702_42 everyone? Quite good for me!!