I understand that it is described as being a standing wave in the mark scheme, but how is this possible if the wavefronts are open to the elements ? Please explain? Thank you.
The answer on the mark scheme is pretty self explanatory... If you are talking about 21)d
i don't get it though it says something about it increasing in length or something but it doesn't.. the length is the same._. sorry, i'm really stupid something. i'd appreciate it if someone spelt it out to me :c
i don't get it though it says something about it increasing in length or something but it doesn't.. the length is the same._. sorry, i'm really stupid something. i'd appreciate it if someone spelt it out to me :c
So you have a long wire. If you put it in a zigzag shape this long wire will take up a lot less space as its all closely packed together. If you increase the length of this wire there still won't be that much of a difference to the space taken up. Using the equation R=pl/A if there is an increase in l, length, we know there would be a greater increase in r, resistance.
I understand that it is described as being a standing wave in the mark scheme, but how is this possible if the wavefronts are open to the elements ? Please explain? Thank you.
You times the wavelength by 2 because the question says the distance between two adjacent maxima. In a standing wave you know that two adjacent maxima would be the distance between two antinodes. Draw a quick sketch if you need and you will realise that this distance 0.05m is half a wavelength. Which means you'd need to times it by 2.
Also it is a standing wave... Sound waves can also be stationary.
You times the wavelength by 2 because the question says the distance between two adjacent maxima. In a standing wave you know that two adjacent maxima would be the distance between two antinodes. Draw a quick sketch if you need and you will realise that this distance 0.05m is half a wavelength. Which means you'd need to times it by 2.
Also it is a standing wave... Sound waves can also be stationary.
Does this answer your question?
Ahhhh okay the first bit makes sense, but the second part, I thought they only formed in tubes?
Can someone explain to me when working out the radiation flux why we use the height of the light source as the radius (to calculate area)? It doesn't come intuitively to me, I don't get it. Thanks!
Incase anyone wants an example Jan 09 Q19 a). You're given power:60 W (5% efficiency), height:2.5m, and made to assume area 4.pi.r^2
This was quite a mean question to give in unit 2 :/ basically in unit 5 you actually learn the equation you have to use: L/4πd2 (i just used this equation since im retaking and have already covered unit 5)
so the question is a bit misleading by telling you to use "r" instead of "d" however in the scenario given, d is equal to r, so it doesn't matter;
"Assume that a distance r from the source the energy is spread over a total area 4πr2 "
so you just have to be very careful reading the question and see that they specify that distance happens to be equal to r.
I was wondering for ages how someone doing unit 2 would know how to work that out, then i finally noticed the sneaky r in the question
Hey, guys, when you have a question that says "show that blah blah is approximately equal to blah blah" , for the next question do you use the approximate answer given in the question before or the value you calculated?
Hey, guys, when you have a question that says "show that blah blah is approximately equal to blah blah" , for the next question do you use the approximate answer given in the question before or the value you calculated?
You can use either. The mark scheme always credits you for either.
When asked to find the potential difference across a resistor in a potential divider circuit , can the potential divider formula be used to find the p.d.Normally 3 marks are given for this type of question. Or else we have to go the long way of finding the current in the circuit and then finding the p.d. across resistor.
Thanks! I assume you get ft marks if you use your value calculated and it was wrong?
But if its a 'show that' its unlikely your value will be wrong, unless you had no idea how to get the answer. If that's the case just use the value 'show that' value for the next question.
When asked to find the potential difference across a resistor in a potential divider circuit , can the potential divider formula be used to find the p.d.Normally 3 marks are given for this type of question. Or else we have to go the long way of finding the current in the circuit and then finding the p.d. across resistor.
I thought that the minimum vertical distance would be half a wavelength (as that corresponds to 180 degrees and would be the smallest unit of distance possible for this much out of phase) however the mark scheme gives the answer as a quarter of the wavelength, really confusing me, any help much appreciated.