ANY DOUBTS HERE!!!

[halosh92]

nov 2007 physics p2 cie
Q2dii
dont we have to subtract the forces? not add them?

[Meticulous]

No..you add them

f=force of boy, R=resultant, F=friction

therefore: f-F=R

f=F+R

got it?

[halosh92]

No..you add them

f=force of boy, R=resultant, F=friction

therefore: f-F=R

f=F+R

got it?

not rili , dont we have to find the resultant in this case..i mean friction is 23 and the force of boy is 75 ..??

[Meticulous]

Dont think of numbers

First:

resultant force= force by kid-friction

is that clear?

now add the resultant (mass x acceleration) to friction

[halosh92]

Dont think of numbers

First:

resultant force= force by kid-friction

is that clear?

now add the resultant (mass x acceleration) to friction

YEPP
THANXXX

[Meticulous]

Any time

[halosh92]

nov 2007 physics
Q4d

[Ghost Of Highbury]

Yes aadi ...what you've done is not ideally the correct way.

Resolve the forces vertically. There is no vertical motion...so no resultant vertical force

1650 sinx= 1800sin20

wudnt u add the weight of the barge too to the downward force??

1650sinx + 40,000 = 1800sin20??

[halosh92]

nov 2007 physics
Q4d

could someone solve this!!!!!!

[cooldude]

wudnt u add the weight of the barge too to the downward force??

1650sinx + 40,000 = 1800sin20??

nope, nid's rite, cuz she has considered the forces acting on the barge, while the weight is the force of the gravity acting on the barge, the weight is balanced by the buoyance force acting on the barge by the water (obviously the boat and the barge will be in water ), just think as though the boat and the barge was on the ground, the normal force wud balance the weight of the barge and in this case the buoyance force does that

[Ghost Of Highbury]

nope, nid's rite, cuz she has considered the forces acting on the barge, while the weight is the force of the gravity acting on the barge, the weight is balanced by the buoyance force acting on the barge by the water (obviously the boat and the barge will be in water ), just think as though the boat and the barge was on the ground, the normal force wud balance the weight of the barge and in this case the buoyance force does that

ok but many questions, like question 6...(same ex. 4A, q6) ...v had to take the weight into consideration!

is it something like, v dont consider the weight, wen two forces are acting on the object in the same direction.?? like here, two tugboats so no weight, but in q6...there are again 2 forces...push and pull...so while calculating the Normal contact force..y do we take the weight too??

[cooldude]

ok but many questions, like question 6...(same ex. 4A, q6) ...v had to take the weight into consideration!

is it something like, v dont consider the weight, wen two forces are acting on the object in the same direction.?? like here, two tugboats so no weight, but in q6...there are again 2 forces...push and pull...so while calculating the Normal contact force..y do we take the weight too??

yeah we gotta consider the weight in this, as the q is to find the normal force, thus we assume that the weight is not balanced by the component of the tension acting upwards, however in the q u asked previously we have to consider the weight and the buoyance force equal as the q has not been asked to consider the normal force.

[sweetie]

i need help in chem. ppr1( CIE)
May07
Q26, 28

[nid404]

i need help in chem. ppr1( CIE)
May07
Q26, 28

26) You really have to draw and check. The OH should be tertiary(since it doesn't react with MnO4)  Try making alcohols with the 5,6,7 and 8 carbons with a tertiary alcohol and a chiral carbon...

28) It has to be an aldehyde or a ketone to react with 2,4 DNP. so it's either B or C. Now to decolorise manganate ions, it has to form a carboxylic acid, which will happen only when an aldehyde is oxidised...so it's C.

[thecandydoll]

CIE AS CHEM PAPER 1.
Q5.
How do you get the answer explain?