What is a solution? A)shown in picture B)shown in picture C)shown in picture D)shown in picture

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).

Aboat radioed a distress call to a coast guard station. at the time of the call, a vector a from the station to the boat had a magnitude of 45.0 km and was directed 15.0° east of north. a vector from the station to the point where the boat was later found is = 30.0 km, 15.0° north of east. what are the components of the vector from the point where the distress call was made to point where the boat was found? in other words, what are the components of vector c = b - a?

Aracecar driver has to hold on tightly when going around a banked curve. approximately what is the centripetal force on a 2220.0 kg car going around a circle with a diameter of 190.0 meters at 25.0 m/s?