Physics: Post your doubts here!

[anastasia grey113]

Holmes
it says it samples at a frequency of 44.1kHz so that means it takes 44.1 samples in 1 second and each sample is of 16 bts.
So for 1 second we have =16 x 44100 = 705600 bits
And for 5 mins 40 s = 340 seconds, we hav = 705600 x 340 = 2.4 x 10^9 bits

 

[Holmes]

Holmes
it says it samples at a frequency of 44.1kHz so that means it takes 44.1 samples in 1 second and each sample is of 16 bts.
So for 1 second we have =16 x 44100 = 705600 bits
And for 5 mins 40 s = 340 seconds, we hav = 705600 x 340 = 2.4 x 10^9 bits

thx

 

[anastasia grey113]

Q6 part c)
I viewed the mark scheme but i still don't get it

 

[Gouki_Destruction]

View attachment 63417
Q6 part c)
I viewed the mark scheme but i still don't get it

Im still an Olevels student , but my knowledge suggests its bcz photons are said to b massless , or as small as 1X10^-18 , while an electron weighs approx. 1/1840 of a proton . Considering this theory , and equating it by the formula mv-mu , the change in momentum is found . Naturally , if the masses are different, the momentum change has to different too , thus taht is my suggestion , probably wrong i guess

 

[Hamnah Zahoor]

36)
Applying KVL around the circuit gives
2E = IR + Ir1 + Ir2

Since the voltmeter reading zero this means there is no voltage difference between its contacts points which gives
Vrise=Vdrop
E = Ir1

Substituting this into the first equation gives
2Ir1 = IR + Ir1+ Ir2
R = r1 - r2 (B)

37)
Applying KVL around the circuit gives
2E - E = 3IR
E = 3IR

The PD between P and Q is due to the battery and resistor between them
PD = E - IR
= E - 1/3 E (substituiting first equation)
= 2/3 E (C)

I would recommend practicing how to use KCL and KVL using YouTube videos.

Thankyou.

 

[anastasia grey113]

Im still an Olevels student , but my knowledge suggests its bcz photons are said to b massless , or as small as 1X10^-18 , while an electron weighs approx. 1/1840 of a proton . Considering this theory , and equating it by the formula mv-mu , the change in momentum is found . Naturally , if the masses are different, the momentum change has to different too , thus taht is my suggestion , probably wrong i guess

lol idk i have this feeling that u can explain it to me
here's wut the mark scheme says
(c) momentum is a vector quantity B1
either must consider momentum in two directions or direction changes so cannot just consider magnitude B1 [2]
u got any idea wut this means?

 

[Holmes]

lol idk i have this feeling that u can explain it to me
here's wut the mark scheme says
(c) momentum is a vector quantity B1
either must consider momentum in two directions or direction changes so cannot just consider magnitude B1 [2]
u got any idea wut this means?

that's what I was just about to post

 

[Holmes]

Is it in our syllabus?
If yes then where should I find it's notes or other resource?
anastasia grey113 (the solution producer)

 

[GarryTheGhost]

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)

i think anything to do with mobile phones isn't in the syllabus.

 

[GarryTheGhost]

GarryTheGhost thats how v draw em
View attachment 63402
View attachment 63403
THE 2ND ONE IS NOT THAT ACCURATE BUT JUST TO GIVE U AN IDEA

GarryTheGhost thats how v draw em
View attachment 63402
View attachment 63403
THE 2ND ONE IS NOT THAT ACCURATE BUT JUST TO GIVE U AN IDEA

how would we draw the graph of the last question?

 

[Ebrahim12]

s09/11
How do we solve this? I tried this
Torque by the cord = Torque by F on the spindle
(900)(0.4) = F(1.2)
But that gives C, answer is B

 

[Gouki_Destruction]

lol idk i have this feeling that u can explain it to me
here's wut the mark scheme says
(c) momentum is a vector quantity B1
either must consider momentum in two directions or direction changes so cannot just consider magnitude B1 [2]
u got any idea wut this means?

Yeah i got it . Wat they basically required was the concept of vectors . Uk a vector quantity is magnitude + direction , and momentum is a vector quantity as we equate it by MV , mass X velocity , so if either quantity is affected by the environment or an external force , the resultant must also change , right ? So the examiner wanted details bout vector

 

[Gouki_Destruction]

lol idk i have this feeling that u can explain it to me
here's wut the mark scheme says
(c) momentum is a vector quantity B1
either must consider momentum in two directions or direction changes so cannot just consider magnitude B1 [2]
u got any idea wut this means?

Yeah i got it . Wat they basically required was the concept of vectors . Uk a vector quantity is magnitude + direction , and momentum is a vector quantity as we equate it by MV , mass X velocity , so if either quantity is affected by the environment or an external force , the resultant must also change , right ? So the examiner wanted details bout vectors

 

[Gouki_Destruction]

In terms of the Q , the photon was the external force acting on the electron , thus it shudve changed direction , thus they both wud nt have same change of momentum

 

[GarryTheGhost]

View attachment 63420
s09/11
How do we solve this? I tried this
Torque by the cord = Torque by F on the spindle
(900)(0.4) = F(1.2)
But that gives C, answer is B

Its hard to visualize this question so dont worry if you didn't get this question correct.
Look at the spindle and forces you can see that the direction it must spin for the weight to lift up is clockwise. Now look at the 900N focus on the pulley for a second and imagine when the weight is getting lifted the pulley string must move in an anticlockwise motion.
Anticlockwise motion =clockwise motion
0.2(900)=F(0.6)+F(0.6)
180=F(0.6+0.6)
180=F(1.2)
150=F
answer is 150N

 

[anastasia grey113]

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)

yh i agree with GarryTheGhost this is no longer part of the syllabus.

 

[GarryTheGhost]

can someone please explain why the charges are same?
E=V/d
but the graph goes negative as it approaches sphere B so shouldn't they both be opposite signs?

 

[amina1300]

can someone please explain why the charges are same?
E=V/d
but the graph goes negative as it approaches sphere B so shouldn't they both be opposite signs?

The electric field strength becomes zero at the center because the Electric field lines cancels out as they are of the same magnitude but only in opposite directions. This isnt possible for opposite charges

 

[anastasia grey113]

can someone please explain why the charges are same?
E=V/d
but the graph goes negative as it approaches sphere B so shouldn't they both be opposite signs?

Yes they should be the same because if they had been different, there would be no zero E at some point.
The diagram below is how the direction of electric field is for both the particles IF THEY HAD BEEN OF THE SAME CHARGE.
Also it becomes negative because at first, the electric field is in on direction but after the field of the second particle takes over, the electric field's direction becomes opposite.

 

[GarryTheGhost]

for part c it says the A=56 where did they get that from?
binding energy for La was (7.85Mev)