The distances (y), in miles, of two cars from their starting points at certain times (x), in hours, are shown by the equations below: car a y = 60x + 10 car b y = 40x + 70 after how many hours will the two cars be at the same distance from their starting point and what will that distance be?

Part f - example: finding two forces (part i) two dimensional dynamics often involves solving for two unknown quantities in two separate equations describing the total force. the block in (figure 1) has a mass m=10kg and is being pulled by a force f on a table with coefficient of static friction îľs=0.3. four forces act on it: the applied force f (directed î¸=30â above the horizontal). the force of gravity fg=mg (directly down, where g=9.8m/s2). the normal force n (directly up). the force of static friction fs (directly left, opposing any potential motion). if we want to find the size of the force necessary to just barely overcome static friction (in which case fs=îľsn), we use the condition that the sum of the forces in both directions must be 0. using some basic trigonometry, we can write this condition out for the forces in both the horizontal and vertical directions, respectively, as: fcosî¸â’îľsn=0 fsinî¸+nâ’mg=0 in order to find the magnitude of force f, we have to solve a system of two equations with both f and the normal force n unknown. use the methods we have learned to find an expression for f in terms of m, g, î¸, and îľs (no n).

Within a pendulum, as potential energy decreases, energy increases. a. heat b. kinetic c. frictional d. gravitational

Astone weighing 1.5 kilograms is resting on a rock at a height of 20 meters above the ground. the stone rolls down 10 meters and comes to rest on a patch of moss. the gravitational potential energy of the stone on the moss is joules.