the tangent at the point 

to the circle x²+y²  = r² cuts the axes of coordinates in A and B. Proove that the area of the triangle OAB is r⁴/alpha.beta

An orbital does not abruptly end at some particular distance from the nucleas .The electron has probability of 95 to 99% within the contour.

a)The probability of finding the electron at x=y=0,z=a₀/2 in 1s orbital of hydrogen atom is 10^-8, the probability of finding it at x=y=a₀/2, z=0 is.............................????

i)10^-8                            ii)                          iii)      iv)cant be predicted.

b)If the distance of 99% contour from the nucleus along x-axis for d_x²-y² is minimum at

i)x=b/2,y=z=0             ii)y=b/2 ,x=z=0                      iii)y=z=b,x=0                        iv)x=b,y=z=0

What is the structure and geometry of BeF₂including any anomaly if found..(angle specifications are required)

the bond with lowest strength is..........??????

Q1.If the first and the nth terma of a G.P. are a and b ,respectively,and if P is the product of first n terms,prove P²=(ab)ⁿ

Q2.If S_pdenote the sum of the series 1+r^p +r^2p+........ad inf.,and s_p the sum of the series 1-r^p +r^2p-......ad inf.,prove that S_p+s_p =2S_2p

find the range of the function  defined by y=2x/(x²+1)

i have an idea , actually its d correct solution i m pretty sure abt it..but its a bit incomplete...

see..the first ball is released at a tym t(let) ..after t_{0 tym}  the second ball is relased .So wat v need 2 c is dat  in to tym how much dis did the first ball travel ...nd at this tym v knoe that the second ball is at zero displacement.

so..

x(distance travelled by the first ball in t₀ seconds)= -1/2gt₀^2..

if u carefully examine we will get all the all the distance in the form of a quadratic equation...y=kx^2..so the curve will be a parabola but i dont knoe exactly which points will the curve pass from..

hope it helped u ..rate if useful.

Let the vector B be xi^+yj^.

A vector =4i^+6j^

(A+B)vector= 10i^+9j^

therefore.. Bvector= (10-4)i^+(9-6)j^ =6i^+3j^

therefore,   mod B VECTOR =

tan= 3/6=1/2

therefore,,,,, =tan^-1 1/2.

rate if it was useful....