explain why fluid resistance in water is greater than in air, and how streamlining reduces the effect of resistance in both

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).

Aplayground merry-go-round has a radius of 4.6 m and a moment of inertia of 200 kg-m2 and turns with negligible friction about a vertical axle through its center. a child applies a 26.0 n force tangentially to the edge of the merry-go-round for 15.0 seconds. if the merry-go-round is initially at rest, how much work did the child do on the merry-go-round?

In order to do work, the force vector must be question 1 options: in a different direction than the acceleration vector. in a different direction than the displacement vector. in the same direction as the displacement vector and the motion. in the same direction as the acceleration vector.