When you push a 1.90-kg book resting on a tabletop, you have to exert a force of 2.10 n to start the book sliding. once it is sliding, however, you can use a force of only 1.06 n to keep the book moving with constant speed. part a what is the coefficient of static friction between the book and the tabletop?

Trumpeter a holds a b-flat note on the trumpet for a long time. person c is running towards the trumpeter at a constant velocity. person b is running away from the trumpeter at the same rate. person d is standing still the whole time. which person hears a frequency that is lower than the b-flat?

Ahydroelectric plant takes energy from water and turns it into electrical energy. what are the transformations of energy in the water molecules that are used in the process of generating electricity this way?

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).