Author Topic: ANY DOUBTS HERE!!!  (Read 34520 times)

Offline halosh92

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Re: ANY DOUBTS HERE!!!
« Reply #240 on: May 31, 2010, 03:01:55 pm »
nov 2007 physics p2 cie
Q2dii
dont we have to subtract the forces? not add them?
everyday we wake up is a miracle, then how do we say miracles dont happen?????

Offline Meticulous

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Re: ANY DOUBTS HERE!!!
« Reply #241 on: May 31, 2010, 03:19:12 pm »
No..you add them

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

therefore: f-F=R

f=F+R

got it?

Offline halosh92

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Re: ANY DOUBTS HERE!!!
« Reply #242 on: May 31, 2010, 03:38:29 pm »
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 ..??
everyday we wake up is a miracle, then how do we say miracles dont happen?????

Offline Meticulous

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Re: ANY DOUBTS HERE!!!
« Reply #243 on: May 31, 2010, 03:54:21 pm »
Dont think of numbers

First:

resultant force= force by kid-friction

is that clear?

now add the resultant (mass x acceleration) to friction

Offline halosh92

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Re: ANY DOUBTS HERE!!!
« Reply #244 on: May 31, 2010, 03:58:54 pm »
Dont think of numbers

First:

resultant force= force by kid-friction

is that clear?

now add the resultant (mass x acceleration) to friction

YEPP :D
THANXXX
everyday we wake up is a miracle, then how do we say miracles dont happen?????

Offline Meticulous

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Re: ANY DOUBTS HERE!!!
« Reply #245 on: May 31, 2010, 04:02:09 pm »
Any time :)


Offline halosh92

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Re: ANY DOUBTS HERE!!!
« Reply #246 on: May 31, 2010, 05:26:01 pm »
nov 2007 physics
Q4d
everyday we wake up is a miracle, then how do we say miracles dont happen?????

Offline Ghost Of Highbury

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Re: ANY DOUBTS HERE!!!
« Reply #247 on: May 31, 2010, 07:44:22 pm »
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??
divine intervention!

Offline halosh92

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Re: ANY DOUBTS HERE!!!
« Reply #248 on: June 01, 2010, 08:41:36 am »
nov 2007 physics
Q4d

could someone solve this!!!!!! ??? ??? ??? ???
everyday we wake up is a miracle, then how do we say miracles dont happen?????

Offline cooldude

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Re: ANY DOUBTS HERE!!!
« Reply #249 on: June 01, 2010, 08:55:47 am »
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 :P ), 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

Offline Ghost Of Highbury

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Re: ANY DOUBTS HERE!!!
« Reply #250 on: June 01, 2010, 10:06:26 am »
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 :P ), 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??



« Last Edit: June 01, 2010, 10:12:47 am by A@di »
divine intervention!

Offline cooldude

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Re: ANY DOUBTS HERE!!!
« Reply #251 on: June 01, 2010, 05:55:31 pm »
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.

Offline sweetie

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Re: ANY DOUBTS HERE!!!
« Reply #252 on: June 02, 2010, 10:43:06 pm »
i need help in chem. ppr1( CIE)
May07
Q26, 28

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Re: ANY DOUBTS HERE!!!
« Reply #253 on: June 03, 2010, 05:57:09 am »
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.

Offline thecandydoll

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Re: ANY DOUBTS HERE!!!
« Reply #254 on: June 04, 2010, 07:03:34 am »
CIE AS CHEM PAPER 1.
Q5.
How do you get the answer explain?