A75 g, 30 cm long rod hangs vertically on a friction less, horizontal axel passing through the center. a 10 g ball of clay travelling horizontally at 2.5 m/s hits and stick to the very bottom tip of the rod. to what maximum angle, measured from vertical, does the rod rotates when the ball of clay is attached to the rod?

Angular momentum is conserved, just before the clay hits and just after; 
mv(L/2) = Iw 

I is the combined moment of inertia of the rod, (1/12)ML^2 , and the clay at the tip, m(L/2)^2 ; 
I = [(1/12)ML^2 + m(L/2)^2] 

Immediately after the collision the kinetic energy of rod + clay swings the rod up so the clay rises to a height "h" above its lowest point, giving it potential energy, mgh. From energy conservation in this phase of the problem; 
(1/2)Iw^2 = mgh 

Use the "w" found in the conservation of momentum above; and solve for "h" 
h = mv^2L^2/8gI 

Next, get the angle by noting it is related to "h" as; 
h = (L/2) - (L/2)Cos() 

So finally 
Cos() = 1- 2h/L = 1 - mv^2L/4gI 

m=mass of clay 
M=mass of rod 
L=length of rod 
v=velocity of clay

Cin carbonh in hydrogeno in oxygen

Answerthe force between the charges will be  f/9 (one-ninth of force f)explanationcoulomb's law relates the magnitudes  two point charges q₁ and q₂, the distance, r, between the charges, and the electrostatic force, f, between the two charges:     k represents coulomb's constant, k = 8.99×10⁹ n·m²/c².note that if we have r = 3r, then