hey the acceleration of the block according is accoding to the refference frame and not the force as a result the block will be having an upward acceleration = g and thus the upward psuedoforce = mg

tn = n/ n^4 + n^2 + n = n/(n^2 + 1)^2 - n^2

=n/( n^2+1+n)(n^2+1-n)=(n^2+1 + n) - (n^2+1-n) =1/2( (1/n^2+1-n) -1/n^2+ 1+n) ) (0n separating)

                                             2 (n^2+1+n)(n^2 +1-n) 

Sn = 1/2[1/1 - 1/3 +1/3 - 1/7+1/7...................................................-1/n^2 +1+n]=1/2[1-1/n^2+n+1]=1/2[n^2+n+1-1]

                                                                                                                                                                             n^2 +n +1

if the block"s mass is m the the downward force acting on it id it:s weight i.e. mg. when the cable of the lift breaks then the block suffers free fall as a result according to yhe refference frame fixed with respect tolift an upward psuedoforce mg also acts on it as a result the deformation in the spring is 0.and no force is exerted by the spring on the block. so no S.H.M. is performed.

the compounds in the oxidation number of oxygen is

-2 are oxides

-1 are peroxides

-1/2 are superoxides

A/ square root 2 where A is the amplitude of the shm executed

i think that the strength of bonds depend on the attacking reagents and reaction conditions. for example a rxn involving breaking of a covalent bond in aromatic compound will require a huge amount of energy.

for covalent bonds breaking of the bonds depend on stability of the products formed.

while in general ionic bonds are termed stronger bonds due strong forces of attraction between the positive and the negative charge.

let 0/0 = k

0=kx0

since infinitely many values of k satisfy this eqn so o/o is called as an indeterminant form

try the same for infinity/infinity

definitely cosx

it must b 12C3 - 7C3 = 185

acc. to relative velocity,

Vm,o +Vo,g =Vm,g

Vm,o = -5i^-j^+4k^

this implies Vimage,g=-Vo,m=+5i^+j^-4k^

xy=k n y^2=x. solving them we get that they intersect at x=k raised to 2/3 n y = k raised to 1/3.

since the curves intersect at right angles this implies that the tangent drawn to them at these points to the curves are perpendicular to each other.supposing the equations to b y=mx+cand applying the condition of tangency differently for both the curves we get the equations of tangents respectively as x+k^1/3 y=2k^2/3 n x-2 k^1/3 y+k^2/3=0

applying the condition of tangency m1 x m2 =o, we get the desired result ie 8k^2=1 did u get it?

no not at all.though different subparticles acquire differnt velocities but the centre of mass of the particle/body continues to move on the same path with same initial velocity.

is ans 4.if it is do reply i will give u the complete solution

u are asking or telling

i think it is hyperbola

it is aimed above the target

no sufficient what?

tan(70-20)=tan50

tan70-tan20/1+tan20 tan70=tan50

tan20 tan70=cot70 tan70=1

tan70-tan20/2=tan50

tan70-tan20=2tan50

tan70=tan20+2tan50