Recent content by Samir_Khalilullah

sorry ... Should i post a new thread on this with the statement on the template?

here is my attempted solution. ## d^2 = z^2 + \frac {L^2} {3} ## ## C ## is coulomb constant since the point is symmetric, only the vertical component of the electric field remains. So, $$ E = 3 E_y =3 \frac {C Q cos \theta} {d^2} $$ $$ E= 3 \frac {C Q z} {d^3} $$ thus part (a) is done ( i...

I don't really know how to start here

Thanks, I think I'm getting somewhere. here's the graph i got. Though I'm a bit doubtful about the result of data number 4, g can't be that high. Do i have to draw a best fit line?? everything else seems to be almost in a straight line. Or do i just have to calculate the average?

ok i did what you told .i got , $$ g = \frac {\sqrt{2} \omega^2 R \left( h + \sqrt{h^2 + d^2} \right)} {2 d} $$ now what do i do? They have given me data of "d" and "T"(from which i can get ## \omega ##). i need to plot a linear equation in the graph and estimate its slope. pls help.

yeah i realized that. ## \tan 2 \theta = \frac d h ##. And i also know, ## \tan \theta = \frac {\omega^2 R} {g \sqrt{2}} ## . I can relate these two with $$ \tan 2 \theta = \frac {2 \tan \theta} {1 - {\tan^2 \theta}} $$ . But that would give me a complex equation. i need to get a linear equation...

(I'm new here so i cant really write with latex, sorry) So I balanced the forces at point P.(otherwise the height of water column would not remain same). we have centripetal acceleration (w^2)*x towards +ve x axis and gravity g at -ve y axis ( taking point P as centre). So the two forces...

also, after a bit of geometry, i did find the angle adjacent to h (at point P) to be 2 times theta. but i dont know how to proceed from there

my bad, it was supposed to be R/sqrt(2). well, that means, even if i change the angular velocity, the height of the water column would remain constant

I did (A) by balancing force on point p (in the figure). I found slope , tanθ = (w^2)x/g. I dont know what to do next. Pls help