In rayleigh scattering of light with frequency v ,intensity of scattered light is proportional to

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).

What are two different ways you could find the value of a? explain these methods.

Water is siphoned from a large tank and discharges into the atmosphere through a 50-mm diameter tube. the end of the tube is b = 2.6 m below the tank bottom which is a = 6.7 m deep, and viscous effects are negligible. determine the maximum height h over which the water can be siphoned without cavitation occurring. atmospheric pressure is 101.4 kpa, and the water vapor pressure is 1.79 kpa (absolute). report your answer in meters to two decimal places.