Figure 3 is a cubical box of edge length L containing n moles of an ideal gas with N molecules moving in all directions. A molecule of mass m and velocity v is about to collide with the shaded wall of area L2. Show that the pressure of the gas in the box is given by: p=nM(v2)avg3V where the terms have their usual meanings.

Using renewable energy sources. just did the test and got it right.

Ibelieve the answer above is c but im not sure

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