All scientists try to base their conclusions on

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).

When you push downward on a book at rest on a table, you feel an upward force. does this force depend on friction? defend your answer.

4r-134a enters the condenser of a residential heat pump at 800 kpa and 50°c at a rate of 0.022 kg/s and leaves at 750 kpa subcooled by 3°c. the refrigerant enters the compressor at 200 kpa superheated by 4°c determine (a) the isentropic efficiency of the compressor, (b) the rate of heat supplied to the heated room, and (c) the cop of the heat pump. also, determine (d) the cop and rate of heat supplied to the heated room if this heat pump operated on the ideal vapor-compression cycle between the pressure limits of 200 and 800 kpa. (0.757, 4.37 kw, 5.12, 6.18, 3.91 kw)