Ahorizontal aluminum rod 4.8 cm in diameter projects 5.6 cm from a wall. a 1200 kg object is suspended from the end of the rod. the shear modulus of aluminum is 3.0 ? 1010 n/m2. neglect the rod's mass.(a) find the shear stress on the rod. (n/m2)(b) find the vertical deflection of the end of the rod. (m)

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).

Write the function getkthdigit(n, k) that takes a possibly-negative int n and a non-negative int k, and returns the kth digit of n, starting from 0, counting from the right

What distance does a car travel as its speed changes from 0 to 20 m/s in 17 s at constant acceleration?