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Physics, 22.06.2019 00:30
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).
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Physics, 22.06.2019 07:50
Calculate the ratio of h+ ions to oh– ions at a ph = 6. find the concentration of h+ ions to oh– ions listed in table b of your student guide. then divide the h+ concentration by the oh– concentration. record this calculated ratio in table a of your student guide. compare your approximated and calculated ratios of h+ ions to oh– ions at a ph = 6. are they the same? why or why not? record your explanation in table a. what is the concentration of h+ ions at a ph = 6? mol/l what is the concentration of oh– ions at a ph = 6? mol/l what is the ratio of h+ ions to oh– ions at a ph = 6? : 1
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Physics, 22.06.2019 17:20
In a system with only a single force acting upon a body, what is the relationship between the change in kinetic energy and the work done by the force? answers: work is equal to the change in kinetic energy.work depends on the square of the change in potential energy.work is equal to the negative of the change in kinetic energy.work is equal to the square of the change in kinetic energy
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