ANY DOUBTS HERE!!!

[halosh92]

yup check the attachment

hey nid thankyou
but i got the "R" wrong could u plzz explain why its in such a direction
thx

[nid404]

hmm....well...it's reaction force...think abt mechanics now..lol...when u have a slope...and something is placed on it...visualize it's normal contact force...hope that helps

[halosh92]

hmm....well...it's reaction force...think abt mechanics now..lol...when u have a slope...and something is placed on it...visualize it's normal contact force...hope that helps

thx

[halosh92]

phyiscs cie AS
nov 2009 p2
Q3Cii
why do we use the vertical component of velocity when we calculate momentum here why not the horizontal component??
 
Q4cii
part 2

[nid404]

it's falling vertically down...so you use the vertical component


x when weight is F+3.8N =17.8-14.2=3.6cm=0.036m
x when weight is 3.8N=16.3-14.2=2.1cm=0.021m
2 epe= 1/2 kx² = 1/2 X 1.8 X 10^-2 X (change in x²)
                                    =1/2 X 1.8 X10^-2 X (0.036²-0.021²)
                                    =0.077 J

[Ghost Of Highbury]

A lamp is supported in equilibrium by 2 chains fixed to 2 points A and B at the same level.

i) The lengths of the chain are 0.3m and 0.4m and the distance between A and B is 0.5m. Given the tension in the longer chain is 36N, by resolving horizontally , find the tension in the shorter chain.

ii) By resolving vertically , find the mass of the lamp.

Please explain with A DIAGRAM!, thanks a lot!

[halosh92]

it's falling vertically down...so you use the vertical component


x when weight is F+3.8N =17.8-14.2=3.6cm=0.036m
x when weight is 3.8N=16.3-14.2=2.1cm=0.021m
2 epe= 1/2 kx² = 1/2 X 1.8 X 10^-2 X (change in x²)
                                    =1/2 X 1.8 X10^-2 X (0.036²-0.021²)
                                    =0.077 J

the answer am getting is 0.00000769
and for the extension why arent we taking only this extension:
x when weight is 3.8N=16.3-14.2=2.1cm=0.021m
as this is wat they only asked for...

[nid404]

Yup so here you go.

You the the angle between the two chains is 90. 0.3²+ 0.4²= 0.5²  yup?

now find angles a and b

tan a= 0.4/0.3=53.13
b= 180-90-53.13=36.87

now sina/36 = sinb/ T is the shorter chain

T=27N in the shorter chain

Now resolving forces vertically

36sin36.87+27sin53.13= mg
0.6X36+ 0.8X27=mg
mg=45
m=4.5 kg

[nid404]

the answer am getting is 0.00000769
and for the extension why arent we taking only this extension:
x when weight is 3.8N=16.3-14.2=2.1cm=0.021m
as this is wat they only asked for...

It says For the extension of the spring from a length of 16.3 cm to a length of 17.8 cm ...

[Ghost Of Highbury]

Yup so here you go.

You the the angle between the two chains is 90. 0.3²+ 0.4²= 0.5²  yup?

now find angles a and b

tan a= 0.4/0.3=53.13
b= 180-90-53.13=36.87

now sina/36 = sinb/ T is the shorter chain

T=27N in the shorter chain

Now resolving forces vertically

36sin36.87+27sin53.13= mg
0.6X36+ 0.8X27=mg
mg=45
m=4.5 kg

ok sry, i didnt put the q rite, it says by resolving horizontally show that the T in the shorter chain is 48N! not 27N

plus, the mass is 6kg :-/

[nid404]

ok sry, i didnt put the q rite, it says by resolving horizontally show that the T in the shorter chain is 48N! not 27N

plus, the mass is 6kg :-/

ok then...refer to the same diagram

a=53.13
b=36.87

now 36cos36.87=Tcos53.13
T=48 N

now resolve vertically
36sin36.87+48sin53.13=mg
  21.6    +38.4=mg
60=mg
m=6kg

[Ghost Of Highbury]

Okay, i tried that and i got it. But what puzzles me is the previous method u used, that is also a correct way to find the force rite?

F = 27N ?

By the way, thanks for the other answers..!

[nid404]

Okay, i tried that and i got it. But what puzzles me is the previous method u used, that is also a correct way to find the force rite?

F = 27N ?

By the way, thanks for the other answers..!

I am wondering...cause i used something similar in another question like this is the book...and got the right ans....I must check tho 

You're welcome 

[The SMA]

guys juz a short question on chemistry AS,
how do you arrange atoms/groups in an optical isomerism
for e.g (i) CH3COCOOH (ii) H3C-CBr(COOH)-CH2Br

if you could xplain in diagram it will be very much appreciated ^^

[nid404]

Optical isomer are possible only in compounds containing a chiral carbon

So it's not possible in the first one.

Here's how the second one would look