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Uh, were we supposed to write "CD"?
I only wrote D. :c
I only wrote D. :c
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duh...Uh, were we supposed to write "CD"?
I only wrote D. :c
yep!!!And this is what happens when you panic after seeing a bulb question ._.
Silly me.
Isn't it supposed to be a balancing toy, though?must mention the type clockwise or anti clockwise.
it is anti clockwise by the way.
when we were all young we played with toys like these, funny we should sit for an exam with a question regarding one too.
Question 9: "Fail-short" current is given to be 0.9A and resistance of each lamp at this current has increased to 5% its original value. Original value: 8 ohm so the new value would be 8.4 ohm. After this it's just simple math.
V=RI
240=(8.4 x X)(0.9) is approximately 31.7. 40-31.7=8.3 bulbs. But since you cannot have an 8.3 bulb. the fuse will blow after the 9th one blows.
hopefully 15-20 max i hope .. maybe lost more idk but i should be B overallI lost around 8 marks in total(varient 2). What about you guys?
B? for 15 marks? No sir! Thresholds are supringly low for these papers. :xhopefully 15-20 max i hope .. maybe lost more idk but i should be B overall
so if i lose 15-25 marks on paper 3 , lets say 10 on paper 1 and like 8 on paper 6 does that mean its an A ?B? for 15 marks? No sir! Thresholds are supringly low for these papers. :x
I think the question asked for the maximum number of bulbs that can blow before the fuse blows, though? I'm not completely sure... if this is the case then it would be 8. Otherwise, you are correct.
ME TOO !Uh, were we supposed to write "CD"?
I only wrote D. :c
yea seems like but i don't think B .so
so if i lose 15-25 marks on paper 3 , lets say 10 on paper 1 and like 8 on paper 6 does that mean its an A ?
Yes that's what the question asked for. But with 8 bulbs, the current has not exceeded the fail-short current rating. Let's work backwards. 32 bulbs still working. So resistance is 8.4 x 32 = 268.8. So current in circuit is I = V/R which is 0.89A. Almost exceeded but not yet. When 9th bulb blows, it would definitely exceed.
ziad my 6eez its 8For the first one, it was 9 since you cant round it to 32 because they said minimum. and if its 32 it will blow so you should use 31. im sure
the answer is 8
let the total number of bulbs be x.
5% of original resistance is 8.4 ohms
voltage is same (source same) and the max current before the fuse melts is 0.9A
therefore current= voltage/total resistance
0.9 = 240/(x * 8.4)
the answer comes in 31.77 , round that off its 32. so the total number of bulbs is 32. difference 40-32= 8 bulbs
It's upwards, as the current induced in the coil , produces a magnetic field that OPPOSES the motion on the bar magnet.
therefore when it is entering the solenoid, it will produce north to repel
when it is leaving the solenoid, it will produce south to attract, overall all the forces exerted being upwards.
it's way more. you have to deduct background radiation which was around 14 counts per minute.
you find that from the graph (it is the point after which the counts do not decrease anymore)
52-14= 38
38/2= 19 counts per minute
find out how long it took to come to 19 counts per minute.
should be more than 3 days.
when it is low tide, meaning the sea water is at a lowe level, water moves FROM RIGHT to LEFT
I'm not sure about this, but I heard the red ray is supposed to refract more, so you draw one line refracting and a faint line for a partial reflection as well.
if you have drawn the reflection, you get one mark.
OR THEN NOT, remember to get that sequence correct in your drawing.
Y was 0, Z was 1.
hope that helped.
Now I know dumb ass, mr G told us todayziad my 6eez its 8
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