IGCSE/GCSE/O & A Level/IB/University Student Forum
Qualification => Subject Doubts => IGCSE/ GCSE => Sciences => Topic started by: moon on November 05, 2009, 05:41:20 pm
-
Plz I need help in these Questions. Q.4 a(ii), Q5 (b), Q9(b) ???
-
Q5bi ) half mass thus, 7500/2 = 3750kG
bii) power output second day = 10/100 * (3750/2 * 36) = 6750
fraction thus, 6750/54000 = 1/8
-
9bi - A
bii - attached
red-->blue-->green-->yellow
-
draw the scale diagram as shown
measure the length of the diagonal and calculate the force accordingly
measure the angle
shud be
-> 5500N
--> 30o
-
draw the scale diagram as shown
measure the length of the diagonal and calculate the force accordingly
measure the angle
shud be
-> 5500N
--> 30o
where is the road on the drawing???
-
9bi - A
bii - attached
where is the conventional current on the drawing?? here is the markscheme answer but I didn't understand it..
(ii) For answers A and B only in (i), not C or D:
Route to resistor: correct arrow on one downwards diode and
nothing wrong on this route B1
Route from resistor: correct arrow on one downwards diode and
nothing wrong on this route
-
check the diagram again. The road is marked in red.
and for the conventional current..i have made colored lines for ur better understanding.check again
-
but how can we draw the conventional current as specified in the mark scheme.. ???
-
but how can we draw the conventional current as specified in the mark scheme.. ???
the markscheme for this question is horrible..it doesnt specify anything...it just says arrow on one downwards diode, which i have shown, rest u just continue the path as shown in my diagram.
it says "nothing wrong on this route" which states clearly that they are not mentioning any correct answer for this. It depends on the examiner. I suppose my diagram is correct.
Did u understand the rest of the quetions?
-
can some1 help me in additional math....2002 November...paper 01....question 5.....i dont want de mark scheme answer...i want de process....
-
Q5bi ) half mass thus, 7500/2 = 3750kG
Why here y the mass was halved?? although the question mentions that the speed is halved ???
-
moon...if de speed is halved den de amount of air passes is also halved...hence de is also halved...
-
thank u A@di and vaan for ur help.hey, vaan u can send ur question in another topic to be seen. :)
-
can some1 help me in additional math....2002 November...paper 01....question 5.....i dont want de mark scheme answer...i want de process....
its a matrix one. i gotta draw 'em
having breakfast...will be back
-
thank u A@di and vaan for ur help.hey, vaan u can send ur question in another topic to be seen. :)
ur welcome
-
a) The number of each type of radiotors required by each type of house * b) the cost of each radiator * c) the number of each type of house *
a) the no. of each type of radiators required by each type of house can be shown as a 3X4 matrix
3 = types of house
4 = types of radiators
so the matrix shud be
A B C D
small (3 2 2 0)
med. (2 0 3 3)
large (0 1 6 3)
the second matrix can shud have 4 rows because the 1st one has 4 columns
2nd matrix i.e cost of each radiator can be represnted this way with 4 rows
(30)
(40)
(50)
(80)
first multiply the 1st and 2nd matrix
(3 2 2 0) * (30) = (270) Cost for 1 small
(2 0 3 3) (40) (450) Medium
(0 1 6 3) (50) (580) Large
(80)
once u multiply these matrices u get a matrix which represents the total cost for 1 of each type of house..but there is not 1 of each..there are many..so v need one more matrix which shows the number of houses for small, med. or large
as the matrix we just got has 1 column the new matrix to be multiplied with shud have 1 row which will represent the number of house.
that can be
(20 30 15)
so the final answer can be the matrix multiplication of
(270) * (20 30 15) = (27600)
(450)
(580)
i guess the markscheme missed the "0" ciz the answer says only 2760 whereas it shud be 27600
check it algebra way too, u shud get 27600
-
i have attached my answer wit it and also the ms ka line!!
chek whether the ans is correct!!
i guess ms is wrong!!
-
i have attached my answer wit it and also the ms ka line!!
chek whether the ans is correct!!
i guess ms is wrong!!
i cant find anything attached.
-
i cant find anything attached.
ops forogt!!
or i guess some error wid attachments!!
wait i'll modify my post!!
-
some proble wid internet!!
i cannot attcch the files
but i can say that they are simillar to aadi's ans!!
and the ms is also worng!!
hope they are attached now!!!
-
plz can u help in physics Q. 10 (b) Nov. 07 ???
-
u talking abt paper 3 rite??
-
attached
10bii) answere here
http://filesmelt.com/downloader/gates.jpg
-
y did u draw the NOT gate before the AND gate??? ???
-
to convert its low signal into high!!
and this makes the output high
which means that alarm will ring!!!
-
hey thanks aadi....
-
hey thanks aadi....
welcome
-
welcome
itna late??!
!?!!?!
!?!?!?!?!?@!