Video

Could've been worse.

No problem at all. Hope you get good outcomes in august. Even if you don't, well done the perseverance. This trait will serve you well in the future.

number of photons per unit time as far as i'm aware. is it in this video?

@@MrNagaPhysics yes , you mentioned it in this video .

Hope it goes well tomorrow.

mrnagaphysics.app/main-resource-category/exam-question-packs. At the bottom of this list there are some EQ packs. Is that ok?

yh it should

agreed, sorry i made a mistake.

the 9 came out of nowhere

I got 6.92 x 10^-7

@@relja3206 ye that’s right

@@relja3206 no he’s not correct lmao

The equation should have 9/2 inside the root instead of a 9 alone. i got I got 6.92 x 10^-7 like relija3206 says below.

I agree, i will try my best in upcoming videos. Thank you for the feedback.

No problem. Glad it helped.

Paper 3A is hard to prep for but you need keep going with the practice. Keep doing the papers, practice ploting the graphs and analysing the results... Finish all the new spec paper 3. Try the questions on paper 3 and paper 6 from this spec. www.savemyexams.com/international-a-level/physics/edexcel/-/pages/past-papers/ Invest a lot of time into your optional unit, ensure you practice all the new spec and old spec questions on that. ( assumign you're AQA). You must be sick of doing past papers by now but i can't see what else you can do. The marginal gains count. Keep going only a week or two remaining.

Hi, thanks for your response I’m about to finish the new spec paper 3 and will have a look at the other. I’m doing turning points and your videos have helped alot and I’m going through the exam question packs on your website first before starting the papers.

@@osmansadat9650 with turning points after you've done the papers and if you have time you may want to look at some of the text books. Maybe CGP text book and Collins(this one seems to be free if you google it)

Hi, what should I do with the text book?

@@osmansadat9650 read some extra details. understand the history of it. turning points is a bit history based. think about how the examiner could ask questions based on what you read.

Yep, it was hard but not impossible. If you're doing it for a mock i would suggest not looking at it. Just work hard, learn and practice the content as best as you can.

not sure about that extra significant figure.

😂 thank you.

F = BILsin(theta) = 2.5T x 12A x 0.01m x sin(40) = 0.1928N is your calculator on degree mode?

@@MrNagaPhysics oh yessss sorry i didnt read the 'per cm' part and missed it, thank you!!

because the base is e. if the base had been 10 for example i would have used log base 10.

You're 100% right. hope you didn't spend ages wondering where the n came from. Thank you for pointing it out.

that factor of 2 is already within the equation now. Which is why it's a 4 at the front. so there are n gaps and n teeth. Total is 2n. Which i've already put into the equation so n is just 800. Do you see what i mean?

Sorry I don’t understand. But I see it was just a proof of the equation not what we need to do. So it just 800

@@osmansadat9650 n is defined as the number of teeth ( or gap it doesn't matter) in the derivation i had to use total number of gaps and teeth so i used 2n. Once the equation is dervied you can't switch the meaning of n from number of teeth to total number of teeth and gaps. to be honest you i don't think they will test derivation.

You're right. It will attract so yes we will need to do work to move it away. Thanks for pointing that out.

onyl when there is some kinda of change at constant volume and no particles entering or leaving. ( i.e heating a closed containers of fixed volume)

@@MrNagaPhysics Hello sir, your website is currently down right now

@@user-ll6rs2rf1z Please use www.mrnagaphysics.app

You don't need to have it to get the mark. I recommend including whatever is in the bracket in any future answers you give. This is incase the examboard decides to be more strict when coming up with mark schemes for upcoming years.

I'm drawing the dotted lines so they are perpendiclar to the rays. This means length of the rays are all the same except for that extra bit that i circle. This is the path difference between adjacent rays.

@@MrNagaPhysics yes but why would it be 90 degrees though? As in how one might prove that? Yes dropping a perpendicular would imply it being perpendicular but why would the perpendicular with a distance of lambda? Is it because we are approximating: because the D is large, the rays can be "said to be" approximately parallel?

Yes the rays are treated as parallel, the screen is flat and far away. i.stack.imgur.com/eQPHd.gif

But the point of a micrometer is that you don't need to round. You can measure more precisily with a micrometer so why round it to 11? But you're right, if you did round it up it would go to 11.

In the example or the questions at the end of the video? I checked both and they're both correct. Let me know why you think it's c.

You'r right it is. Thanks for pointing this out.

Glad it helped!

Another mistake on my part. Very sorry. The answer is 12.9 degrees. I rush it a bit when i made these videos. I will need to fix it eventaully.

@@MrNagaPhysics I really find these videos helpful as i have a few days for my exams and i thank you very much. You make sure to keep the key points and make it short with questions at the end. Which helps me understand the topic.

Well done. How about this vernier calliper reading? czcams.com/users/shorts4Q4MsoemTUY

you're right 🤦‍♂. it should be 0.036 = 0.5x0.8xv^2 + 0.5x20x0.025^2. Thanks for pointing it out.

In this case saying they are in phase, 360, 720 ... are all equivilant statements. They mean the same thing. So yes you can just say they are in phase. In progressive waves sometimes it's useful to say 360 , 720 and so on to see how many whole wavelengths seperate the two points. In stationary waves this is pointless because the whole loop is in phase. Does that help? Glad you found the video useful.

You're welcome.

Good luck!

Glad it helped 👍

Find the potential at 77pm using V = ( 6x1.6x10^-19)/4pi e r Because both Carbon nucelus will repl you will need to overcome that replusion from both so you will need to do double the work. So V x 2. Fianllly change in EPE = q x change in V and it's an electron you're moving so x1.6x10^-19. Does that help. The ideas is to add the two potentials to get the total potential and then move the electron from infinity to that point.

@@MrNagaPhysics Makes a lot more sense now, saving my physics a level you are sir, thanks a lot.

not all of the weight. Only a component of it is radial.

Yes but in this question we're looking for Ohms. Amps is already a base unit.

Glad it was helpful!

You might be right. I probably mixed up the digits in a rush. Thanks for correcting me.

Glad it was helpful!