Explain how to convert a galvanometer to an ammeter

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).

Unpolarized light whose intensity is 1.19 w/ is incident on a polarizer. (a) what is the intensity of the light leaving the polarizer? (b) if the analyzer is set at an angle of = 41.0∘ with respect to the polarizer, what is the intensity of the light that reaches the photocell?

Does work output exceed work input when a machine is used? a) yes, it does, as machines create energy. b) it depends on the type of machine used. c) no, it doesn’t, as machines are inefficient. d) no, it doesn’t, as machines cannot create energy.