A student must determine the angular displacement of a point on the edge of the desk from T equals zero to the instant of time that this comes to rest if the points initial velocity is changed to twoWD but it is angular acceleration is the same as shown. How can the graph be changed before the student can determine the angular displacement? Justify your selection

Since you have not included the experiment situation and the data, of course, i can not give you specific details about it. but i can you with this question.1) i will assume that the motion is projectile (parabolic) motion.2) yes, the horizontal distance depends on the horizontal velocity as, naming vx the horizontal velocity and t the time, it is: x = vx × t3) now, observe that the flight time depends on the vertical velocity as, naming voy the initial vertical velocity, t the time and y the vertical distance before reaching the ground: y = yo + voy×t - gt² / 2.the higher the initial vertical velocity (voy) the longer the object will be flying and so t will be higher in the equation for the horizontal distance.that explains the relationship between the horizontal distance and the horizontal velocity.

Umm, its not your fault, and its because some things in your brain.

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

Cheese biscuit chicken spicy but off cheese whiz china japan on north korea south korea kim john and kim kardashion they’re practically related to calculator needed a racer bum shoes