IGCSE/GCSE/O & A Level/IB/University Student Forum
Qualification => Subject Doubts => IGCSE/ GCSE => Sciences => Topic started by: moon on October 31, 2009, 02:23:58 pm
-
please can somebody help me Nov. 98 Q3. a
-
please can somebody help me Nov. 98 Q3. a
I don't hve the paper....can u upload it
-
here is the question.
-
R is proportional to l
if u take 5ohm/p, length will be 0.6m for 3ohms
like if u take 10ohm/m.....length for 3 ohms would be 3/10...0.3 m
and if u take 20ohm/m....length for 3ohms would be 3/20.. 0.15m....I think would be suitable.
For least resistance in the resistor in itself....a smaller length would be preferred.
If anyone has another answer and a reason to justify it....go ahead...m not a master in physics :P.....but from whatever knowledge i have...I would chose this
-
how did u calculated the length?
-
how did u calculated the length?
Just cross multiply.
-
thanks a lot.
-
thanks a lot.
ya sure...np
-
but i think u shud choose a wire with more length ... cuz if u choose short length wires it cannot be accurate
as u did 0.6, 0.3 nd 0.15 .... so d 5ohm/m shud be used ...
isnt it ?? i might be wrong ??
-
yes, M.@.N.B.i.R. I guess u r rite!!! thanks 4 ur help.
-
do u have d mark scheme for this ppr ???
can u check .. nd lemme noe ??
-
yes, it should be W1 as a 60cm wire can be measured more accurately.
-
do u have d mark scheme for this ppr ???
can u check .. nd lemme noe ??
nd can u temme hw will u identify d length of d wire to be used ?
the length can be found out be simple cross multiplication
5ohms - 1m
3ohms - 3/5m = 0.6m = 60cm
-
woooohooo .. got it rite ...
yaa ... understood dat Thanks ...
-
the length can be found out be simple cross multiplication
5ohms - 1m
3ohms - 3/5m = 0.6m = 60cm
and what is the reason?
-
please can u help in Jun.1999 ppr 6 Q.3 (a) & (b ii) ???
-
can u put up d paper ?? or d question ??
-
please can u help in Jun.1999 ppr 6 Q.3 (a) & (b ii) ???
yes please...i don'y have da paper
-
ok, I will try.
-
here is the question ???
-
ok
1. a - Avg = (26.8 + 26.8 + 27 + 26.9 + 27 + 27) / 6 = 26.9 OC
- 0.1 OC
The largest temperature difference is very low ... Less than even 5% or even 1% of d avg temperature .. so we can take it as constant
b - 0 cm3 as the volume remained 14 aall d tym
3.1 x 2 = 6.2 cm3 This is because d volumes given are h/cm but we have a 2cm3 of tube so double.
13.9 x 2 = 27.8 cm3
Avg u can do it = 27.8/5 = 5.6 cm3
I guess this is d way ...
-
Avg = (26.8 + 26.8 + 27 + 26.9 + 27 + 27) / 6 = 26.9 OC
why did u divide it by 6???
Avg u can do it = 27.8/5 = 5.6 cm3
and here why did u divide it by 5???
-
c d table noe :
dere six values for room temperature :
for avg temperature .. u hav 6 observations ..
at 0hr , 1hr, 2hr, 3hr, 4hr and 5hr ...
nd divided by 5 ... cuz we r checking d rate of water lost in 5 hrs ... dats d experiment ... so its kept for 5 hrs ... ... so for 5 hrs .. 27.8 divided by 5 ...nd u hav 5 hrs of observation for d water loss so divided by 5
got it ?
read this again ... ive modified it
-
Why don't we take the initial time ??? ???
-
Why don't we take the initial time ??? ???
ok..its simple ratio
27.8cm3 of water lost in 5 hours
how much lost in 1 hour?
5 hours - 27.8
1 hours - 27.8/5 *1 = 5.56cm3/h
-
I've got it. Thank u both 4 helping me. :)
-
I've got it. Thank u both 4 helping me. :)
ur welcome..