h2 physics

absol SG

above in those pictures, i need help on 2b(ii) [picture1],

2b(iii), 3b [picture2],

1a(ii) and the whole of (b) [picture3]

thanks in advance and i think im going to fail physics at this rate.lol.

absol SG

above in those pictures, i need help on 2b(ii) [picture1],

2b(iii), 3b [picture2],

1a(ii) and the whole of (b) [picture3]

uh you can tell me if there are better ways to upload the pics such that it is easier to view if you guys cant see it properly...

thanks in advance and i think im going to fail physics at this rate.lol.

eagle SG

hey

haven't had time recently to look through the questions...

I try to see if got time tonight...

Meanwhile, paging for AudioBoxing and Scruuphysics for help

lavastar SG

for picture 3, the gravity qn,

1bi) Note that the satellite is launched from west to east, same as the direction of Earth rotation. Hence this will give it some initial velocity in its direction of travel. ( picture u urself rotating in a circle and fly off in the same direction as its direction of rotation, easier isnt it? if u wan to understand by formulas, notice that earth has a certain angular velocity as it is rotating, and the object is at a certain radius of rotation. hence v=rw) Therfore less energy needed to be supplied to satellite to reach its destination.

1bii) this question involves the use of keplar law. T^2 = (4pi^2/GM)r^3

u used the wrong value of T though, T is the time for the earth to undergo one full revolution ( day to night to day again). that should be 24 X 60 X 60s.

note that after u find r, the r repesent the distance from the centre of the earth to the satellite. u need to minus the radius of the earth to determine its height above

and btw, dun worry abt this paper. its very difficult. NJC 2008 h2 prelim right? i tried until i gave up last time lol but i still did well for my physics

absol SG

still needing help (: tyty. esp for the moments qns...i...totally cannot do.zzzz.....

eagle SG

still needing help (: tyty. esp for the moments qns...i...totally cannot do.zzzz.....

I explain the moments question for you

C is the c.g., so it's used for the downwards force for anti clockwise moments.

( 5 - 2.4 ) * cos 20 degrees =======> Note that answer use sin 70 degrees, but you get the same value as cos 20
This is for the perpendicular distance between the line of action of the force to pivot.

Force is weight, equals 204 * 9.81 (W=mg)

So now for clockwise moments, Fy * 2.4 should be self explanatory