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question 4c)i) i got a different answer of 0.143239448cm/sec i got part ii) correct, are the answers wrong or me XD

hopefully it attracts peoples attention hehe well ive never seen the above type question be4 and i only have simple harmonic motion left, so i missed something what topic does this go under?

i)express sin4t + sqrt3(cos4t) in the form Rsin(4t +a) where a is in radians ii)hence solve sin4t + sqrt3(cos4t)=0 part ii) i could do without the first proof, but i would still like to know how to do it with the above proof. mainly becasue the way i did it was easy divide both sides by cos4t...

i went 2pi(integral)between 2 and 0, x^2(2-x) =2pi[2x^3/3-x^4/4]between 2 and 0 simplify i got 8pi/3

O.o i got 8pi/3 what did i do wrong

ahh thanks

okay q. calculate hoe far an astronaut would need to be above the earth in order for his weight to be 0.01 his weight on the earths surface i did it like this, Force or weight on the surface is F1. the force at the distance from the surface where the astronauts weight is 0.01 that on the...

when was this in our syllabus lol? in the bold part did u mean 360000 if u put that in then u get the right answer of 62.2ms^-2 the last part i knew how to do, this new answer will change the answer for c obviously. thanxs for ur help its the correct way

my teacher says this is a band 5 at its lowest (10/15 across the board)

gah i dnt get this A rocket has a mass of 30000kg, including 25000kg of fuel. It develops a 360000N thrust. calculate: a)its acceleration at lift-off b)the theoretical maximum acceleration of the rocket c)the g-force by an astronaught under maximum conditions i just wanna see what everone else...

woops i didnt say thnx. Thankyou

ahh damn just a simple error. thats what i did but somewhere i made a mistake expanding hte bracket

i got a ridiculus answer of 123.6 could some1 plz explain it. thnx part (iii) this is the diagram, dnt worry bout the lil bit of scribble questions here. also dnt worry bout the little scribble on this too, i was trying to find how to do the last part on (ii) and realised it was sum and...

k lol i got another 2 questions but ill post them up later after i have another crack at it. they are from the same paper

this was in my schools trial paper last year for 2u. its q10a. how come we got the same max volume of 4 but different times? i differentiated volume then let it equal 0. there are 2 answers pi/6 and 7pi/6 pi/6 gives v=0, while the other one gives v=4 so ones obviously max and ones min. wats taht...

y'=2x - 1 y'' = 2 y'-y''=2x-1-2 2x-3 = (2y - x)/x multiply both sides by 1 (x/x)=1 so we can multiply the 1st side by (x/x) and the second side by 1. (its the same thing (x(2x-3))/x=(2y-x)/x (2x^2-3x)=(2y-x)/x (2x^2-2x-x)/x (2(x^2-x)-x)/x y=x^2-x sub it in =2y-x/x

if u want u could whack some numbers into ur calc and plot pts

how do u do part (iii) the way i did it i got 7pi/6 for time and for max volume i got 4m^3 also is part (ii) emptying because the rate of change when t=o isa negative value? thnx in advance