-
We need your support!
We are currently struggling to cover the operational costs of Xtremepapers, as a result we might have to shut this website down. Please donate if we have helped you and help make a difference in other students' lives!
Click here to Donate Now (View Announcement)
physics help
- Thread starter mahad
- Start date
- Messages
- 726
- Reaction score
- 52
- Points
- 0
Due to the difference in angle opposite the resultant. Use Lami's theorem.
- Messages
- 2,884
- Reaction score
- 415
- Points
- 93
what is LAmis theorm????????????????
- Messages
- 726
- Reaction score
- 52
- Points
- 0
A/sina = B/sinb = C/sinc
where A,B, and C are the sides opposite the corresponding angles.
where A,B, and C are the sides opposite the corresponding angles.
- Messages
- 2,884
- Reaction score
- 415
- Points
- 93
ohhh i knew the theorm but not the name!!
haha i wont forget this now
u simple mean sine rule!!!
thanx !
haha i wont forget this now
u simple mean sine rule!!!
thanx !
- Messages
- 726
- Reaction score
- 52
- Points
- 0
That was just in mathematics. Lami applied it into the forces context and it became the Lami's theorem in Physics 
- Messages
- 2,884
- Reaction score
- 415
- Points
- 93
hahaaa didnt know what?
- Messages
- 2,884
- Reaction score
- 415
- Points
- 93
thata what even i didnt know abt it!
ok i came across a few more problems.
this is paper 4 physics may/june 09 question no. 3 part ii...
why do we subtract the masses and the powers...i just couldn't figure this out.
same paper question 6 part c..a full explanation required of this one..actually my concept of this root mean square current thing is a bit weak so if somebody can explain this and cast light on the topic, i would be very thankful.
and sadly SAME paper question 9..(a) ii. how do you apply A=lambdaN here. also it's B part.
NOV/OCT paper 4 physics 08.
question 7 part c. wasn't able to solve this one. =/
yeah that is about it. zohaib sir i would really appreciate if you explain this in detail.
many thanks in advance.
this is paper 4 physics may/june 09 question no. 3 part ii...
why do we subtract the masses and the powers...i just couldn't figure this out.
same paper question 6 part c..a full explanation required of this one..actually my concept of this root mean square current thing is a bit weak so if somebody can explain this and cast light on the topic, i would be very thankful.
and sadly SAME paper question 9..(a) ii. how do you apply A=lambdaN here. also it's B part.
NOV/OCT paper 4 physics 08.
question 7 part c. wasn't able to solve this one. =/
yeah that is about it. zohaib sir i would really appreciate if you explain this in detail.
many thanks in advance.
- Messages
- 2,884
- Reaction score
- 415
- Points
- 93
nay be thats y i didnt came across that word!!!
- Messages
- 380
- Reaction score
- 16
- Points
- 28
http://notezone.net/cambridgechem/chemi ... sis%5D.pdf can sum1 go in dix page 11 n do solve n gimme answers plzzzz i cant able to slove it n it might cum for exam help mee!!!!
http://notezone.net/cambridgechem/chemi ... sis%5D.pdf plz go dix link pg 11 n c answers below of it
First of all you need the mass of FA1. In the first table you have the mass of the empty test tube and the
mass of the test tube + FA1. All you do is subtract the values 12.80 - 10.50 giving you 2.30g of FA1.
In the second table you need the mass of FA1 after heating. Again, you just subtract the values 11.76 -
10.50 = 1.26g FA1 after heating.
1 d. (i) The mass of anhydrous XS04 present in the crystals is what is left over after heating = 1.26g
1 d. (ii) The mass of water driven from the crystals is the loss in mass that has occurred when the FA1 was
heated. Before heating there were 2.30g of FA1 and after heating there were 1.26g. So the mass of water
lost is 2.30 - 1.26 = 1.04g
1 e. To calculate the number of moles of water in the sample of FA1 you use the formula no. of moles =
mass/mass of 1 mole which gives you 1.04/18 = 0.058 mol
1 f. The number of moles in XSO4
7 moles of H20 gives 1 mole XSO4 (from the formula XSO4.7H20)
So 0.058 moles of water give 0.058/7 = 8.29 x 10 ^ -3 moles of XSO4
1 g. To calculate the relative formula mass of XSO4 you use the formula mass of 1 mole = mass/no. of
moles which gives you 1.26/8.29 x 10 ^ -3 = 152.7
1 h. To calculate the Ar of element X you first find the Mr of SO4 which is 32 + (4 x 16) = 96
You subtract this from the RFM from part g giving you 152.7 - 96 = 56.7
First of all you need the mass of FA1. In the first table you have the mass of the empty test tube and the
mass of the test tube + FA1. All you do is subtract the values 12.80 - 10.50 giving you 2.30g of FA1.
In the second table you need the mass of FA1 after heating. Again, you just subtract the values 11.76 -
10.50 = 1.26g FA1 after heating.
1 d. (i) The mass of anhydrous XS04 present in the crystals is what is left over after heating = 1.26g
1 d. (ii) The mass of water driven from the crystals is the loss in mass that has occurred when the FA1 was
heated. Before heating there were 2.30g of FA1 and after heating there were 1.26g. So the mass of water
lost is 2.30 - 1.26 = 1.04g
1 e. To calculate the number of moles of water in the sample of FA1 you use the formula no. of moles =
mass/mass of 1 mole which gives you 1.04/18 = 0.058 mol
1 f. The number of moles in XSO4
7 moles of H20 gives 1 mole XSO4 (from the formula XSO4.7H20)
So 0.058 moles of water give 0.058/7 = 8.29 x 10 ^ -3 moles of XSO4
1 g. To calculate the relative formula mass of XSO4 you use the formula mass of 1 mole = mass/no. of
moles which gives you 1.26/8.29 x 10 ^ -3 = 152.7
1 h. To calculate the Ar of element X you first find the Mr of SO4 which is 32 + (4 x 16) = 96
You subtract this from the RFM from part g giving you 152.7 - 96 = 56.7