Acar travels at a speed of 25 m/s on a flat stretch of road. the driver must maintain pressure on the accelerator to keep the car moving at this speed. what is the net work done on the car over a distance of 250 m?

When an object moves in uniform circular motion, the direction of its acceleration is a) in the same direction as its velocity vector. b) in the opposite direction of its velocity vector c) is directed toward the center of its circular path. d) is directed away from the center of its circular path. e) depends on the speed of the object.

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).

The slope of a velocity time graph over any interval of time gives the during that interval