when his leg is stretched out before him, what is the torque exerted by the weight about his knee, 60 cm away from the weight? use g = 10 m/s2.

Answer/guglielmo marconi

We can not use the law of conservation of energy here. we will use the law of conservation of momenta. we will break into two components, the x component will represent east, and y component will represent north.solving this system will give us the velocity of the third mass.we can plug this backi into the first equation: with this angle we can now calculate the velocity:

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