For question 12 since it’s asking for flux linkage would you not also have to divide 1.04x10^-3 by the number of turns (200) to leave with just phi so the answer should be B if I’m not mistaken
Current can only flow one way through LED. Shown by the bar on top of the triangle. If “only two are lit” which is stated, in order for P or R to be lit. Q Also has to be lit. That’s not 2 LEDs thas 3. Also pretty sure current takes most direct path.
@danielbaker583because think of current as “wanting to get back to the negative terminal”. It won’t go through R because that is not the path of least resistance. Each LED will have some internal resistance
I don’t understand why you don’t convert to metres on question 21 for the 0.85 and 2.1, as converting to metres gives a different answer to leaving it.
@@yamatosennin9249 Same here lol. I thought it was a ratio, didn't think they were giving us the actual values of the density and and speed of ultrasound
what's with the random Irish accent at 6:20 😂
At this point I just want to join the army.
yea paper 1 got me questioning if this is all worth it lmao
@@lilboom7565 have u done the pure math, dude im fuked.... totally fuked up :-/
@@lilboom7565 💀fax
What did u man ending up getting for physics
@@Anyting9OnTopGrrr B. By the way, it turns out that Sunak heard me, just a bit late.
for question 6, why does the value get reduced by 10 percent each time?
Really appreciate the video!
One question, how come you subtract the e.m.f in question 18?
its because the currents are flowing in opposite directions so the one with more emf will overcome the other
For question 12 since it’s asking for flux linkage would you not also have to divide 1.04x10^-3 by the number of turns (200) to leave with just phi so the answer should be B if I’m not mistaken
hi will you do the 2022 and 2023 papers?
7:42 , why not 100hz?
Because the time period of the wave is 5ms, f=1/T means the frequency is 200Hz
wait what's the logic behind 3:00 is it taking the most direct path or something
Current can only flow one way through LED. Shown by the bar on top of the triangle. If “only two are lit” which is stated, in order for P or R to be lit. Q Also has to be lit. That’s not 2 LEDs thas 3. Also pretty sure current takes most direct path.
@@angejo5591 oh that makes sense ty
@@angejo5591 why cant it go through R?
@danielbaker583because think of current as “wanting to get back to the negative terminal”. It won’t go through R because that is not the path of least resistance. Each LED will have some internal resistance
I don’t understand why you don’t convert to metres on question 21 for the 0.85 and 2.1, as converting to metres gives a different answer to leaving it.
It doesn't actually give a different answer, its 0.85 per cm so 0.85 x 10(2) and 2.1 cm is 2.1 x 10(-2)
i made same mistake, the constant was in per cm already
@@s0ph_r0se1 thanks so much
for question 19 part b , didnt they ask for absolute uncertainty? How do we give the uncertainty as absolute uncertainty?
(percentage uncertainty)/100 and then multiply that to your value for B
Why, when finding the ultrasound intensity, was the impedance z1 1?
Just realised tissue F was a thing
@@yamatosennin9249 Same here lol.
I thought it was a ratio, didn't think they were giving us the actual values of the density and and speed of ultrasound
6:56, 3 x 60 gives us 60?!????
Nah, I messed up by writing 3x again
happy days
thank you!
thanks for uploading