Apendulum is 1.20 m long. if the pendulum is pulled until it makes a 25.0 degree angle to the vertical, what is the speed of the pendulum bob when it passes through its equilibrium position? hint: determine the vertical drop of the pendulum first.

Let l be the vertical distance of pendulum when it makes 25.0 degree angle to verticalcos(25.0) = l / 1.20 ml = (1.20m)(cos 25.0)so the height, h, of the pendulum above the equilibrium position is h = 1.20m - (1.20m)(cos 25.0)ep,i + ek, i = ep, f = ek, fwhen pendulum held at 25.0 degrees, initial kinetic energy is 0 jwhen pendulum passes through equilibrium point, final potential energy is 0jep, i = ek fmgh = 1/2 m v^2gh = 1/2 m v^2v = √(2gh)v = √[2 * 9.81 m/s² * ( 1.20m - (1.20m)(cos 25.0) )]v = 1.5 m/s

This is what i found on google “what happens to the bond angle when you add or remove an electron domain? the bond angle decreases as you add domains 5. can you force the atoms into new configurations by pushing atoms around? note: we can think of a bond or a lone pair of electrons as a “domain” of electrons.” i hope this you

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