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Physics - How'd it go?
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ah sure youll be grand! you can lose 60 marks and still get a1 sure.Q10ii: My kinetic energy worked out at 3.69x10^(-14) J
When I got to (iii) I was seriously considering just trying 10B!
RE: blerdii
It's really starting to look like i fecked up question 1...
but exact same answer for kinetic energy!*fingers crossed*
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Data-loggers are recommended, not required, in the syllabus. All state schools would have been given funding to get them, though not everyone uses them. You didn't really need to know much about data loggers for the question, just describe the motion given by the graph.MavisDavis wrote: »Also, I didn't know data-loggers were a required part of the course. I've always used ticker-tape timers and whatnot instead of them. Question 12 (a) was a bust for me - I've never seen a data-logger let alone used one!
positive and negative pions. the syllabus is clear that you are expected to know the specific composition of two baryons, but only says "Mesons composed of any quark and an anti-quark." Did a lot of people take issue with this question? I'll write a letter suggesting it is brought up in the marking scheme conference.s0ur_cherry wrote: »did anyone, for 10 (a) (iii) write more than one meson? i think it was just meant to be pion and anti-pion but it's not really clear, so in the exam i wrote pion +, pion -, kaon + and kaon - :S
there is more than one way to skin a cat!tracker-man wrote: »You were meant to solve q6 using ratios! I think! :pac:
that's the one in the bookthe ripple tank?D: i wrote all about a person swinging an object emitting a continous pitch, and when it came towards us it was higher etc.... 0 marks for me then
there is none!What is the unit of refractive index?0 -
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couple of questions here
how was the water cooled below room temperature - i said leave it in the fridge
i said sensitive thermomerer gives more accurate result
below 30 degrees = total internal reflection occurs, no emergent ray to measure angle
what was the conditions of equlibrium?
what device adjusts the radio tuning? please say diode
why does a note sound different wih different instruments?
tell me the direction of the charge P was to the right
hmm say your ok because there is no " wrong " way to cool water, i just had it in a beaker of ice 0 -
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was the variation in risistance around 300 ohms?
the temp was varied by applying heat ie a bunson burner
and the risistance was measured with an ohmmeter?
it was per unit kelvin so you had to divide it by 10 to get 2.5 ohms in my case per degree Celsius.. i think.. i forgot to do that part though
and i think you may have had to do something with the -273.15 to change for Kelvins but don't quote me on that...
most of the marks are for the graph and getting the 250-300 ohms ish ish variation.. shouldn't loose too much for forgetting the last bit0 -
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hmm say your ok because there is no " wrong " way to cool water, i just had it in a beaker of ice
Adding ice is the wrong way:P The way you find the mass of the ice is subtract the initial mass of the apparatus from the final mass i.e. the ice+water+calorimeter minus water+calorimeter = ice.
The way to find the mass of the steam is the same: steam+water+calorimeter minus h20+calorim = steam.
If you add ice to cool, then put steam in it, how do you know at the end where the extra mass came from? you can find out that say 100g of ice and steam were added, but how much of that 100g was ice and how much was steam?
You could probably argue that youd add the ice, wait for it to melt, then weigh the lot, THEN add the steam....but a fridge is easier!
q6 ya didnt need G or M....f = GM over D^2....for the guys to be weightless the gravitational force of the earth = gravitational force of the moon. so G(81)M / d^2 = GM / D^2....G and M cancel...end up with 9d = D...your given the dist between the centres...solve from there its jus algebra:)
The Q8 (i think?) one with the hairdryer....it said "the fan develops a fault"....how are we supposed to know whether the fault was a short, an open circuit, or the fan jamming? They all have different results but could all be read as "fault"0 -
it was per unit kelvin so you had to divide it by 10 to get 2.5 ohms in my case per degree Celsius.. i think.. i forgot to do that part though and i think you may have had to do something with the -273.15 to change for Kelvins but don't quote me on that...
most of the marks are for the graph and getting the 250-300 ohms ish ish variation.. shouldn't loose too much for forgetting the last bit
The variation in R was 300...the change in temperature was 10 degrees, given in the question. Thus the change in R per 1 degree was 30 ohms....
The Kelvin anad the Celcius are the same scale so theres no subtracting 273....they start at different places but a change in 10K is the exact same as a change in 1C....0 -
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