physics mcq help needed! urgent!

[halosh92]

M/j 2006
22

calculate first the area for the triangle then calculate area of trapezium and then add

[[Spy]]

2003 june
Q5
Q8
someone plz explain
thx

Q5) my teacher even couldnt solve this, its something very weird :S
but here's my solution.....3% of 330 is 10, which is to two significant results, and therefore answer should be to two S.F meaning its C, but its a very stupid question and thats the only time it came :S

Q8) as you can see, the first stage involves a steady increase in velocity, meaning that the distance has to be increasing WITH AN INCREASING RATE. get it? since the velocity is the are in the D-T graph, then for each second there should be increasing area because velocity is increasing, and A shows this. if you look at the second part, its steady velocity meaning its not changing and therefore there is going to be a constant equal change in distance every second, and that's shown by a straight line. and for the last party it velocity is decreasing, BUT THAT DOES NOT MEAN DISTANCE IS INCREASING! you probably put A right? no the answer is C because in the third part it shows a increasing distance BUT WITH A DECREASING RATE until it doesnt change anymore (straight line)
hope its clear

[halosh92]

same paper Q15

[[Spy]]

same paper Q15

well, if you look at B, X Air resistance is correct because as an objects fall from a height, the total net force on it starts decreasing because air resistance starts from zero and then increases gradually which is shown in the graph for X, but in B it says Weight for force Y, weight does not change obiously no matter what unless your on an inclided plane ofcourse,

answer cant be C because we said C has to be resistance.

D cant be the answer because neither is X resitance and Y is weight which we said doesnt change here

as for A, then X is air resistance which is correct, and plus, force Y is net force i.e resultant force which like i said above, will start decreasing from a high value as resistance increases because as the object falls the resistance will keep increasing and therefore the total net force downwards decreases.

[halosh92]

well, if you look at B, X Air resistance is correct because as an objects fall from a height, the total net force on it starts decreasing because air resistance starts from zero and then increases gradually which is shown in the graph for X, but in B it says Weight for force Y, weight does not change obiously no matter what unless your on an inclided plane ofcourse,

answer cant be C because we said C has to be resistance.

D cant be the answer because neither is X resitance and Y is weight which we said doesnt change here

as for A, then X is air resistance which is correct, and plus, force Y is net force i.e resultant force which like i said above, will start decreasing from a high value as resistance increases because as the object falls the resistance will keep increasing and therefore the total net force downwards decreases.

thankyou
same paper Q36
i dont get the direction :S

and Q40

[[Spy]]

thankyou
same paper Q36
i dont get the direction :S

and Q40

Q36) it simple as that the electric field direction is always from positive to negative, and you can see that at the upper plate has a NEGATIVE voltage meaning the direction will be from Q to P, and so the answer is D

Q40) as we know a proton has the same charge as an electron (the value) thats why an atom is neutral, therefore here, we have to obtain a combination of up and down quarks so that achieve a value of "e" exactly, equal to the charge of the electron.

for A, (0*2/3)+(3*-1/3)= -1, it cant be the answer because we have to obtain a positive value since its a proton!

for B, (1*2/3)+(1*-1/3)= 1/3 - Not Possible because we need a value of exactly positive 1.

for C, (1*2/3)+(2*-1/3)= 0 - Not Possible because we need a value of exactly positive 1.

for D, (2*2/3)+(1*-1/3)= 3/3 = 1 and therefore D is the answer

[halosh92]

Q36) it simple as that the electric field direction is always from positive to negative, and you can see that at the upper plate has a NEGATIVE voltage meaning the direction will be from Q to P, and so the answer is D

Q40) as we know a proton has the same charge as an electron (the value) thats why an atom is neutral, therefore here, we have to obtain a combination of up and down quarks so that achieve a value of "e" exactly, equal to the charge of the electron.

for A, (0*2/3)+(3*-1/3)= -1, it cant be the answer because we have to obtain a positive value since its a proton!

for B, (1*2/3)+(1*-1/3)= 1/3 - Not Possible because we need a value of exactly positive 1.

for C, (1*2/3)+(2*-1/3)= 0 - Not Possible because we need a value of exactly positive 1.

for D, (2*2/3)+(1*-1/3)= 3/3 = 1 and therefore D is the answer

thx !! the second question is such simple common sense -.- 

[[Spy]]

thx !! the second question is such simple common sense -.- 

haha well there's always a common sense way and a scientific way

[halosh92]

haha well there's always a common sense way and a scientific way

hahah  too sad ive got neither...
2003 nov
Q12

2009 NOV variant 1
Q13

[ifraha18]

2003 nov q12...

elastic collison so velocity of approach is equal to velocity of separation..


ux+uy=vx+vy

[ifraha18]

nv 2007 q 2,11 and 40..??
plzz someone?

[halosh92]

2003 nov q12...

elastic collison so velocity of approach is equal to velocity of separation..


ux+uy=vx+vy

if we take this ----------> as +
and this <------------- as -
we dont get that answer :S

[ifraha18]

fr elastic colision simply follow v of aproach=v of separation...

[ifraha18]

nv 2007 q 2,11 and 40..??
plzz someone?

[da_rockstar]

well yea ur method is also rite in physics u can dervie everything from everything lol   
yea for the next part i dont know the question but if i an not wrong to find the total charge u have to add them
q of proton + q of electron
electron will be negative so
Q1+(-Q2)
so Q1-Q2
in a way ur subtracting them but i think u got my point



lol..yeah you are right dude.... we derive random stuff whenever we need it...lol.. got my biology practical tommorow....and this year i didn't do any practical...lol...hope all goes well.